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RRB JE 2019 Quantitative Aptitude Quiz

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RRB JE 2019 Quantitative Aptitude Quiz

Introduction

What is Quantitative Aptitude? Quantitative aptitude is one of the top competitive subject which will students really want to learn and to be expert. Quantitative Aptitude Tests evaluate numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance exams for different fields.
The article RRB JE 2019 Quantitative Aptitude Quiz provides Quantitative aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. Railway Recruitment Board (RRB) has released RRB JE 2019 Official Notification to recruit eligible candidates for the post of it Railway JE Recruitment 2018-19 will be held to recruit candidates for the posts of Junior Engineer(JE), Junior Engineer(Information Technology), Depot Material Superintendent(DMS) and Chemical & Metallurgical Assistant(CMA). Quantitative Aptitude plays major role to qualify examination. The article RRB JE 2019 Quantitative Aptitude Quiz will assist the students understand the type of questions expected from the topic Quantitative Aptitude .
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Quiz

RRB JE 2019 Quantitative Aptitude Quiz - Set 1

1. If 7x - 5y = 13 and 2x - y = 5, then (x - y) is

Answer - D
Explanation -
7x - 5y = 13 - - - - - - - equation 1
2x - y = 5 - - - - - - - - - - - - - equation 2
Subtract equation 2 from equation 1
5x - 4y = 8 - - - - - - - - - - - - - equation 3
Subtract equation 2 from equation 3
3x - 3y = 3
x - y = 1
hence option D is the correct answer
2. If the amount received at the end of 2nd and 3rd year at Compound Interest on a certain Principal is Rs 24200, and Rs 26620 respectively, what is the rate of interest?

Answer - A
Explanation -
Let the rate of interest = r
Principal amount = p
Amount = [latex]P({1 + \frac{r}{100} }^{t})[/latex]
Amount at the end of two year = 24200 = [latex]P({1 + \frac{r}{100} }^{2})[/latex] equation 1
Amount at the end of 3 year = 26620 = [latex]P({1 + \frac{r}{100} }^{3})[/latex] equation 2
Divide equation 2 by equation 1
[latex](\frac{26620}{24200}) = 1 + \frac{r}{100}[/latex]
[latex]\frac{r}{100} = \frac{26620}{24200} - 1 = \frac{2420}{24200}[/latex]
[latex]\frac{r}{100} = \frac{1}{10}[/latex]
r = 10%
so required rate of interest = 10%
3. If x : y be the ratio of two whole numbers and z be their HCF, then the LCM of those two numbers is

Answer - D
Explanation -
First no. x Second no. = L.C.M. x H.C.F.
So, L.C.M. = xyz
4. An engineering student has to secure 25% marks to pass. He gets 47 and fails by 43 marks. What are the maximum marks of the examination?

Answer - D
Explanation -
Let the maximum marks be max.
According to the question,
Passing marks = Marks of student + 43
25% of max = 47 + 43
([latex]\frac{1}{4}[/latex]) × max = 90
max = 360 marks
5. The sum of the ages of husband and wife at present is 56. Ten years ago the product of their ages was 320. What is the age of the husband and the wife?

Answer - C
Explanation -
Let the present age of husband = x
And present age of wife = y
x + y = 56
x = 56 - y - - - - - - equation1
(x - 10)(y - 10) = 320
xy - 10x - 10y + 100 = 320
xy - 10(x + y) = 220
xy - 10(56) = 220
xy = 780 - - - - - - - equation 2
Put value of x from equation 1 into equation 2
(56 - y)y = 780
[latex]{y}^{2}[/latex] - 56y + 780 = 0
[latex]{y}^{2}[/latex] - 30y - 26y + 780 = 0
y = 30 or 26
so we get x = 26 or 30
hence the required ages are 30 and 26
6. A shopkeeper, sold dried apricots at the rate Rs 1210 a kg and bears a loss of 12%. Now if he decides to sell it at Rs 1331 per kg, what will be the result?

Answer - Option D
Explanation -
Old selling price = 1210 and loss percentage = 12%
Loss percentage = [latex]\frac{CP - SP}{CP}* 100[/latex]
This means (100 - 12)% of CP = 1210
88% CP = 1210 * [latex]\frac{100}{88}[/latex]
CP = 1210
CP = 1375
New selling price = 1331
Loss = CP-SP = 1375 - 1331 = 44
Loss percentage = [latex]44 * \frac{100}{1375}[/latex]
loss percentage = 3.2 %
7. Tom is chasing Jerry. In the same interval of time Tom jumps 8 times while Jerry jumps 6 times. But the distance covered by Tom in 7 jumps in equal to the distance covered by Jerry in 5 jumps. The ratio of speed of Tom and Jerry is

Answer - Option D
Explanation -
Given;
[latex]\Rightarrow \frac{Tom}{Jerry} = \frac{5}{7}[/latex]
Let jerry's 1 leap = 7 meter and Tom's 1 leap = 5 meter
then, ratio of speed of Tom and Jerry = [latex] \frac{8 * 5}{6 * 7} = \frac{20}{21}[/latex]
8. Ruchir walks at 20 km/hr and Rukma cycles at 25 km/hr towards each other. What was the distance between them when they started if they meet after 48 minutes?:

Answer - Option C
Explanation -
Let the distance between them at the beginning was x km
Relative speed of Rukma = 25 + 20
= 45 km/hr
Time-taken = 48 min
1 min = 1/60 hr
i.e, 48 min = [latex]\frac{48}{60}[/latex]
= 0.8 hr
We know that, distance = time-taken × speed
= 0.8 × 45
= 36 km
9. A painter can paint a fence in 24 hours. After 6 hours he takes a break. What fraction of the fence is yet to be painted?

Answer - Option C
Explanation -
Painter completed the total work in 24 hours
Work completed in one hour = [latex]\frac{1}{24}[/latex]
Work completed in six hour = 6 [latex]\frac{1}{24} = \frac{1}{4}[/latex]
remaining work = 1 - [latex]\frac{1}{4} = \frac{3}{4}[/latex]
fraction of work yet to be completed = [latex]\frac{3}{4}[/latex] = 0.75
10. [latex]\frac{sin A}{1 + cos A}[/latex] is equal to

Answer - Option A
Explanation -
[latex]\frac{sin A}{1 + cos A}[/latex]
multiply numerator and denominator by 1-cosA
we get [latex]\frac{sin A(1 - cos A)}{1 + {cos}^{2} A} = \frac{sin A(1 - cos A)}{ {sin}^{2} A} = cosec A - cot A[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 2

1. Find the difference between the unit digits of a and b a= [latex](437)^{253} [/latex].[latex](134)^{337} [/latex] b= [latex](437)^{253} [/latex][latex](134)^{337} [/latex].[latex](132)^{653} [/latex][latex](559)^{381} [/latex]

Answer - D
Explanation - the unit digits repeats themselves after every power multiple of 4
Now , for a unit digit of a = 7*4 = 28 = unit digit is 8
Now , for b unit digit of b = 2*9 = 18 = unit digit is 8
Therefore difference between unit digits of a and b = 8 - 8 = 0
2. Unit digit in [[latex](253)^{98} [/latex]-[latex](106)^{100} [/latex]+[latex](708)^{403} [/latex]-[latex](16)^{4} [/latex]+259]

Answer - A
Explanation - Unit digit of the expression
= [[latex](3)^{98} [/latex]-[latex](6)^{100} [/latex]+[latex](8)^{403} [/latex]-[latex](6)^{4} [/latex]+9]
= [[latex](3^{(4(24))} [/latex]-[latex](6)^{100} [/latex]+[latex](8)^{(4(100))} [/latex]-[latex](6)^{4} [/latex]+9]
= 1 * 9 - 6 + 6 * 2 - 6 + 9
= 9 - 6 + 2 - 6 + 9
= 8
3. Unit digit in [latex]264^{(5!(6!))}[/latex]* [latex]264^{(103(1024))}[/latex]= ?

Answer - A
Explanation -
[latex](5!)^{(6!)}[/latex] is an even number and [latex](103)^{(1024)}[/latex] is an odd number.
We know that,
4 even always gives 6 as last digit and 4 odd always gives 4 as last digit.
= [latex](4)^{(5!(6!))}[/latex]* [latex](4)^{(103(1024))}[/latex]
= 4even * 4odd
= 6 * 4
= 4
4. Unit digit in [latex](17)^{(256)}[/latex] is:

Answer - D
Explanation -
Unit digit of a number is the remainder when that number is divided by 10.
Rem = [latex]\frac{{17}^{256}}{10}[/latex]
= [latex]\frac{{17}^{(256)}}{10} = \frac{{({7}^{2})}^{123}}{10}[/latex]
= [latex]\frac{{49}^{(123)}}{10} = \frac{{-1}^{123}}{10}[/latex] = -1
= 10 -1
= 9(unit digit)
5. Unit digit in [latex](6474)^{(1863)}[/latex]*[latex](605)^{(427)}[/latex]*[latex](281)^{(491)}[/latex] is:

Answer - A
Explanation -
[latex](6474)^{(1863)}[/latex]*[latex](605)^{(427)}[/latex]*[latex](281)^{(491)}[/latex]
= [latex](6474)^{(1863/4)}[/latex]*[latex](605)^{(427)}[/latex]*[latex](281)^{(491)}[/latex]
= [latex](6474)^{(3)}[/latex]* 5* 1[(1863/4)=3]
= 4 * 5 * 1
= 0
6. Find unit digit in [latex](39)^{18(17)}[/latex] is:

Answer - A
Explanation -
([latex](18)^{(17)}[/latex]/4) = (18*18*18*...17 times)/4
= So power of 39 is completely divisible by 4.If power even and there is 9 at unit place of the base number, 1 will come at unit place of the resultant.
= [latex](39)^{(4)}[/latex] = 1
Hence the unit digit = 1
7. Find unit digit in [latex](459)^{(213)}[/latex]*[latex](545)^{(315)}[/latex]*[latex](723)^{(395)}[/latex]*[latex](634)^{(833)}[/latex] is:

Answer - D
Explanation -
[latex]\frac{213}{4}[/latex] = 1 (Remainder) = [latex]{9}^{1}[/latex] = 9 (unit digit)
[latex]\frac{315}{4}[/latex] = 3 (Remainder) = [latex]{5}^{3}[/latex] =5 (unit digit)
[latex]\frac{395}{4}[/latex] = 3 (Remainder) = [latex]{3}^{3}[/latex] = 7 (unit digit)
[latex]\frac{833}{4}[/latex] = 1 (Remainder) = [latex]{4}^{1}[/latex] = 4 (unit digit)
Requred
9 * 5 * 7 * 4 = 0 (unit digit)
8. Find unit digit in [latex]{{27}^{37}}^{35}[/latex] by 4?

Answer - C
Explanation -
7 has a cyclicity of 4 for unit digit
So [latex]{7}^{1} \rightarrow[/latex] 4
[latex]{7}^{2} \rightarrow[/latex] 9
[latex]{7}^{3} \rightarrow[/latex] 3
[latex]{7}^{4} \rightarrow[/latex] 1
ans so on
So seeing the divisibility of the power with 4
[latex]{{27}^{37}}^{35}[/latex] = [latex]\frac{(37*37*37....35 times)}{4}[/latex]
Hence in given question when we divide [latex](27)^{37(35)} {7}^{1}[/latex] = 7 as unit digit
9. Find unit digit in [latex](264)^{(102)}[/latex][latex](264)^{(103)}[/latex] is:

Answer - A
Explanation -
[latex]{(264)}^{102} + {(264)}^{103}[/latex] If the unit number is 4 in the base of the exponential number, (1)When the power is even, the unit's digit is 4.(2) If the power is odd, the unit's digit is 6.That is[latex]{(264)}^{102}[/latex][latex]{(264)}^{103}[/latex] = 6 + 4 = 10 That is, the unit number will be zero.
10. Find unit digit in [latex]{(2467)}^{153}[/latex]*[latex]{(341)}^{72}[/latex] is:

Answer - A
Explanation -
= Unit digit of [latex]{(2467)}^{153} \times {(341)}^{72}[/latex]
= Unit digit of [latex]{(7)}^{153} \times {(1)}^{72}[/latex]
= Unit digit of [latex]{{(7)^{4}}}^{38} \times {(7)}^{1} \times {(1)}^{(72)}[/latex]
= Unit digit of [latex]{(1)}^{(153)} \times 7 \times {(1)}^{72}[/latex]
So, final unit digit = 7 * 1 = 7

RRB JE 2019 Quantitative Aptitude Quiz - Set 3

1. x + [latex]\frac{1}{x}[/latex], then the value of [latex]x^{7}[/latex] + [latex]\frac {1}{x^ {5}}[/latex] is

Answer - Option B
Explanation -
x + [latex]\frac{1}{x}[/latex] = 2
[latex]x^{2}[/latex] + 1 = 2x
[latex]x^{2}[/latex] - 2x + 1 = 0
[latex](x - 1)^{2}[/latex] = 0
x - 1 =0
x = 1
therefore, [latex]x^{7}[/latex] + [latex]\frac{1}{x^{5}}[/latex] = 1 +1=2
2. If [latex](2a - 3)^{2}[/latex] + [latex](3b + 4){2}[/latex] + [latex](6c + 1){2}[/latex] = 0 then the value of [latex]\frac{a^{3} + b^{3} + c^{3} - 3abc}{a^{2} + b^{2} + c^{2}}[/latex] is:

Answer - Option D
Explanation -
[latex](2a - 3)^{2}[/latex] + [latex](3b + 4){2}[/latex] + [latex](6c + 1){2}[/latex]
therefore, 2a - 3 = a = [latex]\frac {3}{ 2}[/latex],
3b + 4 = 0 = b [latex]\rightarrow \frac {-4}{3}[/latex]
6c + 1 = 0 [latex]\rightarrow[/latex] c = -[latex]\frac {1}{6}[/latex]
(a + b + c) = [latex]\frac {9 - 8- 1}{6}[/latex] = 0
therefore, [latex]{a^{3} + b^{3} + c^{3} - 3abc}[/latex] = 0
[latex]\frac{a^{3} + b^{3} + c^{3} - 3abc}{a^{2} + b^{2} + c^{2}}[/latex] + 3 = 0 + 3 = 3
3. What is the value of positive square root of 14 + [latex]\sqrt {6}{5}[/latex]

Answer - Option A
Explanation -
= 14 + [latex]\sqrt {6}{5}[/latex]
= [latex]3^{2}[/latex] + [latex]\sqrt{5}^{2}[/latex] + 2 * 3* [latex]\sqrt{5}[/latex]
= (3 + [latex]\sqrt{5}^{2}[/latex])
4. A sum of money placed at compound interest doubles itself in 5 years. It will amount to eight times of itself in:

Answer - Option A
Explanation -
A = P[latex](1 + \frac{R}{100})^{T}[/latex]
2 = 1[latex](1 + \frac{R}{100})^{5}[/latex]
Cubing both sides
[latex]2^{3}[/latex] = [latex](1 + \frac{R}{100})^{15}[/latex]
5. How many factors of 106 is/are prime number?

Answer - Option A
Explanation -
Factor of 106 = 2 × 53
Both 53 and 2 are prime
Hence there are 2 prime factor of 106
6. The population of a town is 9000. It the number of females increases by 5% and the males by 7.5%, what will be the total population after increase. The number of females currently is 3000.

Answer - Option A
Explanation -
In the village
Females = 3000
Males + 9000 - 3000 = 6000
After respective increases,
Population of Village
= 3000 * [latex]\frac {105}{100}+ {6000 * 107.5}{100}[/latex]
= 3150 + 6450 = 9600
7. Rohit buys an odd bat for Rs 1230 and spends Rs 180 on its repairs. If he sells it for Rs 1128, then what is his loss percentage?

Answer - Option C
Explanation -
Total cost price of bat =1230 + 180 = 1410
Selling price = 1128
Loss= 1410 - 1128 = 282
Loss percentage = [latex]\frac{282}{1410}[/latex] * 100 = 20%
8. [latex]\frac{4}{5}[/latex] part of a tank is filled with oil. After taking out 42 litres of oil the tank is 3/4 part full. What is the capacity (in litres) of the tank?

Answer - Option C
Explanation -
Let the capacity of tank be 20x
Then initial quantity of oil in the tank= [latex]\frac{4}{5}[/latex] * 20X = 16X
Final quantity = [latex]\frac{3}{4}[/latex] * 20X = 15X
16x - 15x = 42
X = 42
So capacity of tank=20x = 20 × 42 = 840
9. To do a certain work, B would take time thrice as long as A and C together and C twice as long as A and B together. The three men together complete the work in 10 days. The time taken by A to complete the work separately is

Answer - Option B
Explanation -
If B does the work in 3x days, (A + C) will do the same work in x days.
If C does that work in 2y days. (A + B) will do it in y days.
therefore, [latex]\frac{1}{x}\frac{1}{3x}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{4}{3x}[/latex] = [latex]\frac{1}{10}[/latex]
3x = 40
x = [latex]\frac{40}{3}[/latex]
Again, [latex]\frac{1}{y}\frac{1}{2y}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{3}{2y}[/latex] = [latex]\frac{1}{10}[/latex] = y = 15
therefore(A + B+ C)'S = 1 Day's work = [latex]\frac{1}{10}[/latex]
[latex]\frac{1}{A} + \frac{1}{40} + \frac{1}{30}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{1}{A}[/latex] = [latex]\frac{1}{10} - \frac{1}{40} - \frac{1}{30}[/latex]
[latex]\frac{12 - 3- 4}{120}[/latex] = [latex]\frac{5}{120}[/latex] = [latex]\frac{1}{24}[/latex]
therefore, A alone will complete the work in 24 days.
10. If sin(A - B) = Sin A Cos B - Cos A Sin B, then Sin [latex]15^{0}[/latex]will be

Answer - Option D
Explanation -
Let, A= [latex]45^{0}[/latex]
B= [latex]30^{0}[/latex]Sin (A-B)
=sin A. cos B-cos A. sin B Sin [latex]45^{0} - 30^{0}[/latex]
= sin[latex]15^{0}[/latex]
= [latex]\frac{1}{\sqrt{2}} * \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} * \frac{1}{2} [/latex]
[latex]\frac{\sqrt{3}}{2\sqrt{2}} * \frac {1}{2\sqrt{2}}[/latex] = [latex]\frac{\sqrt{3} - 1}{2\sqrt{2}}[/latex]

RRB JE 2019 Quantitative Aptitude Quiz -Set 4

1. What is the HCF (highest common factor) of 57 and 513?

Answer - Option B
Explanation -
Factors of 57 = 3 × 19
Factors of 513 = 3 × 3 × 3 × 19
Since the common factor between the two is 3 × 19 = 57
Thus, HCF of 57 and 513 = 57
2. The two numbers are 63 and 77, HCF is 7. Find the LCM.

Answer - Option B
Explanation -
Relation b/w H.C.F, L.C.M & two no is given by
H.C.F. × L.C.M. = a × b
The numbers a = 63 and b = 77
HCF = 7
So, L.C.M. = [latex]\frac{63 * 77}{7}[/latex] = 693
3. What is the HCF (highest common factor) of 133 and 112?

Answer - Option B
Explanation -
Factor of 133 = 7× 19
Factor of 112 = 7× 16
Hence the HCF of 133 and 112 = 7
4. HCF and LCM of two numbers are 11 and 825 respectively. If one number is 275, find the other number.

Answer - Option C
Explanation -
Let the second number be n.
Formula : First Number x Second Number = LCM X HCF
275 x n = 825 x 11
n = 33
5. The sum of two numbers is 36 and their H. C. F and L. C. M are 3 and 105 respectively. The sum of the reciprocals of the two numbers will be:

Answer - Option D
Explanation -
HCF = 3
Numbers = 3x and 3y where x and y are co-primes.
therefore 3x + 3y = 36
x + y =12
= 3xyz = 105 and, LCM
Dividing (i) by (ii)
[latex]\frac{x + y}{xy}[/latex] =[latex]\frac{12}{35}[/latex]
[latex]\frac{1}{x}[/latex][latex]\frac{1}{y}[/latex] = [latex]\frac{12}{35}[/latex]
[latex]\frac{1}{3x}[/latex] + [latex]\frac{1}{3y}[/latex] = [latex]\frac{4}{35}[/latex]
6. A farmer has 945 cows and 2475 sheep. He farms then into flocks, keeping cows and sheep separate and having the same number of animals in each flock. If these flocks are as large as possible, then the maximum number of animals in each flock and total number of flocks required for the purpose are respectively

Answer - Option C
Explanation -
First of all we find the HCF of 945 and 2475
945 = 3 * 3 * 3 * 5 * 7
2475 = 3 * 3 * 5 * 5 * 11
HCF = 45
Maximum number of animals in each flock = 45
Required total number of flockS= [latex]\frac{945}{45}[/latex] + [latex]\frac{2475}{45}[/latex]
21 + 55 = 76
7. Find the least number which must be subtracted from 2423 so that the resultant number when divided by 15, 25 and 40 leaves remainder of 7 in each case.

Answer - Option C
Explanation -
The number which divided by 15, 25 and 40 leaves remainder 7 in each case
= n * LCM (15, 25, 40) + 7.
(where n is a positive integer)
= n *600 + 7
the closest such number is 2407.
Hence, the least number to be subtracted
= 2423 - 2407
= 16.
8. There are three classes A, B and C with 72, 96 and 144 students respectively. Find the minimum number of small groups in which they can be divided such that each group has students from one class only and each small group has equal number of students.

Answer - Option D
Explanation -
The minimum groups will be formed only when number of students in each group is largest which will be the HCF
HCF of 72, 96 and 144 = 24
Therefore, number of students in each group is 24.
Hence, the number of groups required is
= [latex]\frac{72}{24} + \frac{96}{24} + \frac{144}{24}[/latex]
= 3 + 4 + 6
= 13
9. How many numbers between 400 and 800 are divisible by each 4, 5 and 6?

Answer - Option A
Explanation -
LCM of 4, 5, 6 = 60
So 420, 480, 540, 600, 660, 720, 780
i.e, 7 no is divided between 400 - 800
10. The ratio of two numbers is 3 : 4 and their H.C.F. is 4; then their L.C.M. is:

Answer - Option D
Explanation -
Ratio of two numbers is 3 : 4
let the numbers be 3x and 4x.
HCF of (3x, 4x) = x = 4 (given)
Hence numbers are 12 and 16
The LCM of 12 and 16 = 48
Hence Option D is correct

RRB JE 2019 Quantitative Aptitude Quiz - Set 5

1. The sum of the ages of husband and wife at present is 100. Ten years ago the ratio of their ages was 9:7. What is the age of the husband?

Answer - Option B
Explanation -
Let 10 years ago, the age of husband and wife be 9x and 7x respectively.
Given: (9x + 10) + (7y + 10) = 100
[latex]\rightarrow[/latex] 16x = 100 - 20
[latex]\rightarrow[/latex] x = 5
Thus, present age of husband = 9x + 10
= 9 × 5 + 10
= 55 years
2. The ratio of present ages of Rangana and Sayed is 7:5. After 11 years the ratio of their ages will be 4:3. What is Rangana's present age?

Answer - Option B
Explanation -
The ratio of present ages of Rangana and Sayed is 7:5. Let their present ages are 7x and 5x respectively. After 11 years, the ratio of their ages = 4:3, i.e.
[latex]\frac{7x + 11}{5x + 11}[/latex] = [latex]\frac{4}{3}[/latex]
21x + 33 = 20x + 44
x = 11
Thus, the present age of Rangana = 7x = 77 years
3. The ratio of present ages of Ratnabali and Shaukat is 8:5. After 22 years the ratio of their ages will be 10:9. At present, what is Ratnabali’s age ?

Answer - Option D
Explanation -
Let the present age of Ratnabali= 8x
And present age of Shaukat= 5x
As per question
[latex]\frac{8x + 22}{5x + 22}[/latex] = [latex]\frac{10}{9}[/latex]
72x + 198 = 50x + 220
22x = 22
x = 1
hence present age of Ratnabali = 8× 1= 8 years
4. The ratio of present ages of Ranjini and Shahid is 5:4. After 13 years the ratio of their ages will be 6:5. What is Ranjini's present age?

Answer - Option B
Explanation -
Let present age of Rajini = 5x
And present age of Shahid = 4x
After 13 years,
Age of Rajini = 5x + 13
Age of Shahid = 4x + 13
Ratio of their age after 13 years is 6 : 5
[latex]\frac{(5x +13)}{(4x+13)}[/latex] = 6/5 [latex]\frac{6}{5}[/latex]
25x + 65 = 24x + 78
x = 13
Rajini’s present age = 5x = 5 × 13 = 65 years
5. The ratio of present ages of Rambha and Sarvesh is 8:5. After 7 years the ratio of their ages will be 5:4. What is Rambha's present age?

Answer - Option C
Explanation -
Let present age of Rambha and Sarvesh is 8x and 5x
After 7 years
[latex]\frac{8x + 7}{5x + 7} = \frac {5}{4}[/latex]
32x + 28 = 25x +35
7x = 7
X = 1
So Rambha present age will be 8x = 8 years
6. The sum of the ages of husband and wife at present is 56. Ten years ago the product of their ages was 320. What is the age of the husband and the wife?

Answer - Option C
Explanation -
Let the present age of husband = x
And present age of wife = y
x + y = 56
x = 56 - y - - - - - - equation1
(x - 10)(y - 10) = 320
xy - 10x - 10y + 100 = 320
xy - 10(x + y) = 220
xy - 10(56) = 220
xy = 780 - - - - - - - equation 2
Put value of x from equation 1 into equation 2
(56 - y)y = 780
[latex]{y}^{2}[/latex] - 56y + 780 = 0
[latex]{y}^{2}[/latex] - 30y - 26y + 780 = 0
y = 30 or 26
so we get x = 26 or 30
hence the required ages are 30 and 26
7. The ratio of present ages of Ramita and Satyajit is 9 : 7. After 9 years the ratio of their ages will be 5 : 4. What is Ramita's present age?

Answer - Option A
Explanation -
Let the present age of Ramita and Satyajit be 9x and 7x respectively.
According to the question, after 9 years the ratio of their ages will be 5:4, i.e.
[latex]\frac{9x + 9}{7x + 9} = \frac {5}{4}[/latex]
4(9x + 9) = 5(7x + 9)
36x + 36 = 35x + 45
x = 9
Thus, present age of Ramita = 9x = 9 × 9
= 81 years
8. A got married 8 years ago. A’s present age 1[latex]\frac{1}{4}[/latex]is times his age at the time of marriage. A’s son’s age is [latex]\frac{1}{10}[/latex] times his present age. His son’s age in years, is

Answer - Option C
Explanation -
Let A’s age eight years ago X years be.
X + 8 = [latex]\frac{5X}{4}[/latex]
therefore, 4x + 32 + 5x
x = 32 years
therefore, Son's present age = [latex]\frac{1}{10}[/latex](x + 8)
= [latex]\frac{1}{10}[/latex]* 40 = 4
9. Twice the age of A is 2 years more than that of B. If the product of ages of A and B is 180 years, what will be the age of A after 2 year?

Answer - Option C
Explanation -
2A = B + 2
A * B = 180
A(2A - 2) = 180
[latex]A^{2}[/latex] - a- 90 = 0
A = 10 years
After 2 years A’s age = 10 + 2 = 12 years
10. 2 years ago, the ratio of ages of Rocky and Jacky was 6:7. 2 years hence the ratio will become 7:8, What is the present age of Rocky?

Answer - Option D
Explanation -
Let 2 years ago, ages of Rocky and Jacky were R and J respectively. A/Q,
R : J = 6 : 7
[latex]\frac{6x + 4}{7x + 4}[/latex] = [latex]\frac{7}{8}[/latex]
48x + 32 = 49x + 28
x = 4
resent age of Rocky = R + 2 = 6X + 2 = 26

RRB JE 2019 Quantitative Aptitude Quiz - Set 6

1. A is 4 years younger than B while C is 4 years younger than D but 1/5th times as old as A. If D is 8 years old, how many times as old is B as C?

Answer - Option A
Explanation -
A = (B - 4),C = (D - 4)& C = [latex]\frac{1}{5}[/latex]A
C = 8 - 4 = 4
A = 5C = 5 * 4 = 20
B = A + 4 = 20 + 4 + 24
[latex]\frac{B}{C}[/latex] = [latex]\frac{24}{4}[/latex] = 6
B = 6C
2. 6 years ago, age of a mother was 3 times that of her daughter. If sum of their present ages is 60 years, what are the present age of mother and daughter respectively?

Answer - Option C
Explanation -
Mother’s age 6 years ago = M
Daughter’s age 6 years ago = D
[latex]\frac{M}{D}[/latex] = 60
x = [latex]\frac{48}{4} = 12[/latex]
Present age of mother = 3x + 6 = 42 years
Present age of daughter = x + 6 + 18 years
3. The ration of ages of the father and mother was 11:10 when their son was born. The ratio of ages of the father and mother will be 19:18 when the son will be twice his present age. What is the ratio of present ages of father and mother?

Answer - Option A
Explanation -
Let the age of father =11x and that of mother be 10x when there son is born
When son is twice his age, that is after 2 year the ratio of father and mother age=19:18
So, [latex]\frac{11x + 2}{10x + 2} = \frac{19}{18}[/latex]
198x - 190x = 2
X = [latex]\frac{1}{4}[/latex]
So, father age = [latex]\frac{11}{4}[/latex] and mother age = [latex]\frac{10}{4}[/latex]
So father present age = [latex]\frac{11}{4}[/latex] + 1 = [latex]\frac{15}{4}[/latex]
And mother present age = [latex]\frac{10}{4}[/latex]+ 1 = [latex]\frac{14}{4}[/latex]
So required ratio = 15 : 14
4. If 6 years are subtracted from the present age of Gagan and the remainder is divided by 18, then present age of his grandson Anup is obtained. If Anup is two years younger to Madan whose age is 5 years, then what is Gagan’s present age?

Answer - Option B
Explanation -
Madan’s age =5 years. Then Anup’s age = 5 - 2 = 3 years. Let x be the present age of
Gagan. According to question,
[latex]\frac{x-6}{18}[/latex] =3
x = 54 + 6 = 60 years
5. A person was asked to state his age in years. His reply was “take my age 3 years hence, multiply it by 3 and then subtract 3 times my age 3 years ago and you will know how old I am.” What was the age of the person?

Answer - Option A
Explanation -
Let his present age = x years
According to question,
(x + 3) × 3- 3 × (x - 3) = x
x=18 years
6. If 10 years are subtracted from the present age of A and the remainder is divided by 6, then the present age of his grandson B is obtained. If B is 2 years younger to C whose age is 7 years, then what is A’s present age?

Answer - Option A
Explanation -
[latex]\frac{A-10}{6}[/latex] = B, B = C - 2, C = 7 years
B = 7 - 2 = 5 years
[latex]\frac{A-10}{6}[/latex] =5
A = 40 years
7. The average age of a husband-wife and their son was 42 years. The son got married and exactly after 1 year a child was born to them. When the child became 5 years old, the average age of the family became 36 years. What was the age of bride at the time of marriage?

Answer - Option C
Explanation -
[latex]\frac{H+W+S}{3}[/latex] = 42
H+W+S = 126
Let age of son’s bride at the time of marriage =B year. After 1 year a child was born to them and there are 5 family members now. After 5 years of his birth i.e,after 6 years of son’s marriage average age of the family
[latex]\frac{(H + 6) + (W + 6) + (S + 6) + (B + 6) + 5}{5}[/latex] = 36
(H + W + S + B) = 180 - 29 = 151
126 + B = 151
B = 151 - 126 = 25 years
8. If the ages of A and C are added to twice the age of B, the total becomes 59. If the ages of B and C are added to thrice the age of A, the total becomes 68 and if the age of A is added to thrice the age of B and thrice and age of C, the total becomes 108. What is the age of A?

Answer - Option C
Explanation -
A + C + 2B = 59, 3A + B + C=68 and A + 3B + 3C=108
B + C = 68 - 3A and A + 3(B+C) = 108
A + 3(68 - 3A) = 108
8A = 204 - 108
A=[latex]\frac{96}{8}[/latex] = 12 years
9. Mr X has three sons namely P, Q and R. P is the eldest son of Mr. X while R is the youngest one. The present ages of all three of them are square numbers. The sum of their ages after 5 years is 44. What is the age of P after three years?

Answer - Option C
Explanation -
Let the present age of P, Q and R are x, y and z (square numbers) respectively.
(x + 5) + (y + 5) + (z + 5) = 44
x + y + z = 44 – 15 = 29
Perfect square numbers are: 1, 4, 9, 16, 25 etc. The only possible values of x, y and z are 16, 9 and 4 respectively. Present age of P is 16 years (oldest). After 3 year his age would be 19 years.
10. Three years ago, the average age of Ramesh's family having 5 members was 17 years. Ramesh becomes father but the average age of his family is same today. What is the present age of baby?

Answer - Option B
Explanation -
Three years ago total age of family having 5 members =5 × 17 = 85 years At present there are 6 members is the family but average is same. Therefore,
[latex]\frac{85 + 3 + 3 + 3 + 3 + 3 + Baby}{6}[/latex] = 17
Baby = 102 - 100 = 2 years

RRB JE 2019 Quantitative Aptitude Quiz - Set 7

1. In a class of 55 students there are 34 girls. The average weight of these girls is 51 Kg and average weight of the full class is 55.2 kgs. What is the average weight of the boys of the class?

Answer - Option A
Explanation -
Average weight of the boys = [latex]\frac{sum of weights of full class - sum of weights of girls}{number of boys}[/latex]
= [latex]\frac{55.2 \times 55 - 34 \times 51}{55 - 34}[/latex]
= [latex]\frac{3036 - 1734}{21}[/latex]
= 62 kg
2. The average revenues of 11 consecutive years of a company is Rs 69 lakhs. If the average of first 6 years is Rs 64 lakhs and that of last 6 years is Rs 76 lakhs, what is the revenue for the 6th year?

Answer - Option D
Explanation -
Revenue of the sixth year = Revenue of the first 6 years + Revenue of the last 6 years – Revenue of total 11 years
= 6 × 64 + 6 × 76 – 11 × 69
= Rs 81 lakhs
3. The mean of marks secured by 55 students in division-A of class X is 55, 45 students of division-B is 51 and that of 40 students of division-C is 49. What is the mean of marks of the students of three divisions of Class X?

Answer - Option C
Explanation -
Mean of all the three divisions = [latex]\frac{sum of total marks secured by all the students}{total number of students}[/latex]
= [latex]\frac{55 \times 55 + 45 \times 51 + 40 \times 49}{55 + 45 + 40}[/latex]
= [latex]\frac{7280}{140}[/latex]
= 52
4. The mean of marks secured by 55 students in division A of class X is 58, 45 students of division B is 54 and that of 75 students of division C is 52. Find the mean of marks of the students of three divisions of class X.

Answer - Option B
Explanation -
Total marks= mean× number of students
Total marks in three division of class X= 55 × 58 + 45 × 54 + 75 × 52
3190 + 2430 + 3900= 9520
Total number of students in class X = 55 + 45 + 75= 175
Mean marks of students of three division of class x = [latex]\frac{9520}{175}[/latex]
= 54.4
5. The mean of marks secured by 65 students in division A of class X is 54, 30 students of division B is 50 and that of 55 students of division C is 48. Find the mean of marks of the students of three divisions of Class X.

Answer - Option C
Explanation -
Total marks secured by class A = 65 × 54 = 3510
Total marks secured by class B = 30 × 50 = 1500
Total marks secured by class C = 55 × 48 = 2640
Average of class X = [latex]\frac{Total marks secured by class A + Total marks secured by class B Total marks secured by class C}{Total number of students in class X}[/latex]
Average of class X = [latex]\frac{3510 + 1500 + 2640}{65 + 30 + 55}[/latex]
= [latex]\frac{7650}{150}[/latex]
= 51
6. Three years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family remains the same today. The age of the baby today is

Answer - Option A
Explanation -
Present age of 5 members = (5 * 17 + 3 * 5) = 100 Years
Present age of 5 members and a baby = 17*6 = 102 years
Age of baby = 102 - 100 = 2 years.
7. A man engaged a servant on the condition that he would pay him `90 and a turban after service of one year. He served only for nine months and received the turban and an amount of `65. The price of turban is

Answer - Option C
Explanation -
Let monthly salary = x
So,
12x = 90 + turban
9x = 65 + turban
then 3x = 25
12x = 100
= 90 + turban = 100
turban = 10
8. The costs of daily ticket of local train is Rs. 50 and Monthly Pass costs is Rs. 1125. If I buy the Monthly Pass and travel for 30 days in a month then I save?

Answer - Option B
Explanation -
Costs of daily ticket = Rs 50
Monthly pass costs = 1125
Cost of monthly travelling = 50 × 30 = 1500
Amount saved by travelling through pass = 1500 - 1125 = 375
Required percentage =[latex]\frac{375}{1500}[/latex] * 100 = 25%
9. The average of 19 consecutive even integers is 50. The highest of these integers is

Answer - Option B
Explanation -
Given: The average of 19 consecutive even integers is 50.
Then, their sum = 19 × 50
= 950
Let the first digit be a. Common difference between these digits, d = 2
Sum of terms of AP = [latex]\frac{n}{2}[/latex][2a + (n – 1)d]
[latex]\frac{19}{2}[/latex][2a + (19 – 1)2] = 950
2a + 36 = 100
a = 32
Also, if l is the last term, then
Sum = [latex]\frac{n}{2}[/latex][a + l]
950 = [latex]\frac{19}{2}[/latex][32 + l]
l = 68
10. The average revenues of 13 consecutive years of a company is Rs 82 lakhs. If the average of first 7 years is Rs 77 lakhs and that of last 7 years is Rs 89 lakhs, find the revenue for the [latex]{7}{th}[/latex] year.

Answer - Option C
Explanation -
Revenue of [latex]{7}{th}[/latex] year = sum of revenues of first 7 years and last 7 years – sum of revenues of 13 years
= 7 × 77 + 7 × 89 – 13 × 82
= 1162 - 1066
= Rs 96 lakhs

RRB JE 2019 Quantitative Aptitude Quiz - Set 8

1. Three persons walk from place A to place B. their speed are in the ratio 4 : 3 :5. The ratio f the times taken by them to reach B will be :

Answer - Option C
Explanation -
Time taken is inversely proportional to relevant speeds. = [latex]\frac{1}{4}[/latex] : [latex]\frac{1}{3}[/latex] : [latex]\frac{1}{5}[/latex]
i.e, Required ratio = [latex]\frac{1}{4} \times 60 : \frac{1}{3} \times 60 : \frac{1}{5} \times [/latex] 60
= 15 : 20 : 12
2. The ratio of boys and girls in a college is 5 : 3. If 50 boys leave the college and 50 girls join the college, the ratio becomes 9 : 7. Initial number of boys in the college was how much?

Answer - Option C
Explanation -
The ratio of boys and girls in a college is 5 : 3
If X be the number of total students in college, then number of boys = [latex]\frac{5x}{8}[/latex]
Number of girls = X - [latex]\frac{5x}{8}[/latex] = [latex]\frac{3x}{8}[/latex]
If 50 boys leave the college and 50 girls join the college,
Now number of boys = [latex]\frac{5x}{8}[/latex] - 50 = [latex]\frac{5x - 400}{8}[/latex]
Number of girls = [latex]\frac{3x}{8}[/latex] + 50 = [latex]\frac{3x + 400}{8}[/latex]
Given ratio between boys and girls = 9:7
[latex]\frac{5x - 400}{8}[/latex] : [latex]\frac{3x + 400}{8}[/latex] = 9 : 7
7 * [latex]\frac{5x - 400}{8}[/latex] = 9[latex]\frac{3x + 400}{8}[/latex]
35x - 2800 = 27x + 3600
8x = 6400
x = 800
Therefore number of boys in college = [latex]\frac{5x }{8}[/latex] = [latex]\frac{5x * 800}{8}[/latex] = 500
3. A, B and C are batsmen. The ratio of the runs scored by them in a certain match are given below : A : B = 5 : 3 and B : C = 4 : 5. In all they scored 564 runs. The number of runs scored by B is :

Answer - Option D
Explanation -
A + B + C = 564
A : B = 5 : 3 = 20 : 12 and B : C = 4 : 5 = 12 : 15
A : B : C = 20 : 12 : 15
Proportional sum of A, B and C = 20 + 12 + 15 = 47
Hence Run scored by B
= 12 * [latex]\frac {564}{47}[/latex] = 12 * 12 = 144
4. A, B, C enter into a partnership with shares in the ratio [latex]\frac {7}{2}[/latex] : [latex]\frac {4}{3}[/latex] : [latex]\frac {6}{5}[/latex] After 4 months, A increases his share by 50%. If the total profit at the end of the year was 43,200, then the B’s share in the profit is

Answer - Option D
Explanation -
A, B, C center into a partnership with shares in the ratio [latex]\frac {7}{2}[/latex] : [latex]\frac {4}{3}[/latex] : [latex]\frac {6}{5}[/latex]
After 4 months, A increases his share by 50%
Hence, A share at the end of year
= [latex]\frac {7x}{2}[/latex] * 4 + [latex]\frac {21x}{4}[/latex] * 8
14x + 42x = 56x
B' share = [latex]\frac {4x}{3}[/latex] * 12 = 16m
C, share = [latex]\frac {6x}{5}[/latex]* 12 = [latex]\frac {72x}{5}[/latex]
Now B profit = [latex]\frac {16x}{56x + 16x + \frac {72}{x}}[/latex] * 43200
= [latex]\frac {16x * 5}{432}[/latex] * 43200
= 8000 Rs.
5. In a particular type of alloy the ratio of lead to tin is 5 : 6. The amount of lead should be added to 12.1 kg of this material to make the ratio 1 : 1 is

Answer - Option A
Explanation -
In a particular type of alloy the ratio of lead to tin is 5 : 6
Total amount of material = 12.1 Kg
Amount of lead = 5 × [latex]\frac {12.1}{11}[/latex] = 5 × 1.1 = 5.5 Kg
Amount of tin = 6 × [latex]\frac {12.1}{11}[/latex] = 6 × 1.1 = 6.6 Kg
Difference between tin and lead = 6.6 – 5.5 = 1.1 kg
Therefore 1.1 kg of lead should be added in material to make the ratio 1 : 1.
6. If A : B = 1 : 2, B : C = 3 : 4 C : D = 6 : 9 and D : E = 12 : 16 then A : B : C : D : E is equal to

Answer - Option D
Explanation -
A : B = 1 : 2 = 3: 6 = 9 : 18
B : C = 3 : 4 = 6: 8 = 18 : 24
C : D = 6 : 6 = 24: 36
D : E = 12 : 16 = 36 : 48
therefore, A : B : C : D : E = 9 : 18 : 24: 36 : 48
= 3 : 6 : 8: 12 : 16

RRB JE 2019 Quantitative Aptitude Quiz - Set 9

1. What number should be added to each of 6, 14, 18 and 38 so that the resulting numbers make a proportion?

Answer - Option B
Explanation -
From the given alternatives
[latex]\frac {6 + 2}{14 + 2}[/latex] = [latex]\frac {18 + 2}{38 + 2}[/latex]
[latex]\frac {8}{16}[/latex] = [latex]\frac {20}{40}[/latex]
[latex]\frac {1}{2}[/latex] = [latex]\frac {1}{2}[/latex]
2. If 5.5 of a = 0.65 of b, then a : b is equal to:

Answer - Option C
Explanation -
5.5 of a = 0.65 of b
a : b = 0.65 : 5.5
a : b = 13 : 110
Hence Option C is correct
3. A invests Rs. 64000 in a business: After few months B Joined him with Rs. 48000. At the end of year, the total profit was divided between them in the ratio 2 :1. After how many months did B join?

Answer - Option B
Explanation -
B entered after x months.
Ratio of profits
= 6400 * 12 : 48000(12 - X)
= 16 : (12 - X)
therefore [latex]\frac {16}{12 - x}[/latex] = [latex]\frac {2}{1}[/latex] = 24 - 2x = 16
2x = 8
x = 4 months
4. if [latex]\frac {a}{b}[/latex][latex]\frac {2}{3}[/latex] and [latex]\frac {b}{c}[/latex] = [latex]\frac {4} {5}[/latex], then (a +b) : (b + c) =?

Answer - Option D
Explanation -
a : b = 2 : 3
b : c = 4 : 5
a : b : c = 2×4 : 3×4 : 3×5
= 8 : 12 : 15
[latex]\frac {a + B}{b + C}[/latex] = [latex]\frac {8 + 12}{12 + 15}[/latex] = [latex]\frac {20}{27}[/latex]
Hence Option D is correct
5. A certain amount of money is divided among x, y and z. If x receives 25% more than y and y receives 25% less than z, they x : y : z is equal to

Answer - Option B
Explanation -
x : y = 125 : 100 = 5 : 4
y : z = 75 : 100 = 3:4
x : y : z = 15 : 12 : 16
6. A sum of 53 is divided among A, B and C in such a way that A gets Rs 7 more than what B gets and B gets Rs 8 more than what C gets. The ratio of their shares is

Answer - Option B
Explanation -
Short Trick:-
C get C
B = C + 8
A = C + 8 + 7 = C + 15
So, C + C + 8 + C + 15 = 53
3C + 23 = 53
3C = 30
C gets 10, B gets 18 and A gets 25
Ratio : A : B : C =25 : 18 : 10
7. If a:b = 3:8 then find the value of (5a-3b):(2a+b).

Answer - Option C
Explanation -
[latex]\frac {5a - 3b}{2a + b}[/latex] = [latex]\frac {5\frac{a}{b}-3}{2 \frac {3}{8}+ 1}[/latex]
= [latex]\frac {5\frac{3}{8}-3}{2 \frac {3}{8}+ 1}[/latex]
= [latex]\frac {\frac{15}{8}-3}{\frac {6}{8}+ 1}[/latex]
= [latex]\frac{\frac{15 - 24}{8}}{\frac{6 + 8}{8}}[/latex]
= -[latex]\frac{9}{14}[/latex]
8. If x : y = 2 : 5 then (5x + 3y) : (5x - 3y) is equal to _____.

Answer - Option D
Explanation -
[latex]\frac{5x + 3y}{5x - 3y}[/latex] = [latex]\frac{5\frac{x}{y}+ 3}{5\frac{x}{y} - 3}[/latex]
= [latex]\frac{5\frac{2}{5}+ 3}{5\frac{2}{5} - 3}[/latex]
= [latex]\frac{5}{-1}[/latex]
= -5
9. A person bought some rice and wheat for Rs. 380. The ratio of weight of rice and wheat is 4 : 3 and the price of equal amount of rice and wheat is in the ratio 5 : 6. The rice was bought of worth

Answer - Option C
Explanation -
A person bought some rice and wheat for Rs. 380
Ratio of weight of rice and wheat = 4 : 3
Ratio of price of equal amount of rice and wheat = 5 : 6
= (4 * 5) : (3 * 6)
= 20 : 18
= 10 : 9
worth of rice = [latex]\frac {380}{19}[/latex] * 10 = 200 Rs
10. If 5A = 12B = 13C; FIND A : B: C

Answer - Option C
Explanation -
It is given that, 5A = 12B = 13C
Now, using hit and trial, we can find the understated values of the variables
A = 12 × 13 = 156
B = 13 × 5 = 65
C = 5 × 12 = 60
A : B : C = 156 : 65 : 60

RRB JE 2019 Quantitative Aptitude Quiz - Set 10

1. Ticket for an adult is Rs 1600 and a child is Rs 1200. 1 child goes free with two adults. If a group has 23 adults and 10 children what is the discount the group gets?

Answer - Option B
Explanation -
Number of kids who went for free = 23/2 = 11 kids
But because there are 10 kids only, so all 10 kids will go for free
Cost of ticket of 1 kid is Rs 1200 and that of an adult = Rs 1600
Thus, total cost = 1600 × 23 + 1200 × 10 = Rs 48800
And cost of ticket only for adults = Rs
Discount = [latex]\frac {12000 * 100} {48800}[/latex]
= 24.59%
2. A student multiplied a number by 5/8 instead of 8/5. What is the percentage error in the calculation?

Answer - Option A
Explanation -
Given that, the student multiplied a number by 5/8 instead of 8/5. Then,
Error = [latex]\frac {correct - incorrect} {correct} * 100[/latex]
= [latex]\frac {\frac{8}{5} - \frac{5}{8}} {\frac{8}{5}} * 100[/latex]
= [latex]\frac {64 - 25} {64} * 100[/latex]
= 60.94%
3. An engineering student has to secure 36% marks to pass. He gets 53 and fails by 37 marks. Find the maximum marks.

Answer - Option B
Explanation -
According to the question,
Passing marks = 53+37 = 90
36% of maximum marks = 90
100% = [latex]\frac {90 \times 100} {36}[/latex] = 250
Maximum marks = 250
4. The price of an article is cut by 3%. To restore to its original value, the new price must be increased by

Answer - Option C
Explanation -
Let initial price of the article was Rs 100, then
New price = Rs 97
To restore original price, price should be increased by [latex]\frac {3} {27} * 100[/latex] = 3.09%
5. A number is divided into two parts in such a way that 80% of [latex] {1}^ {st}[/latex] part is 3 more than 60% of 2nd part and 80% of [latex] {2}^ {nd}[/latex] part is 6 more than 90% of the [latex] {1}^ {st}[/latex] part. Then the number is

Answer - Option C
Explanation -
First part = Rs. x and second part = Rs. y
i.e, [latex]\frac {x * 80} {100}[/latex] = [latex]\frac {y * 60} {100}[/latex] + 3
[latex]\frac {4x} {5}[/latex] =[latex]\frac {3y} {5}[/latex] + 3
4x - 3y = 15 .........(i)
[latex]\frac {4y} {5}[/latex] =[latex]\frac {9x} {10}[/latex] + 6
8y - 9x = 60
By equation (i) × 8 + (ii) × 3,
32x – 24y = 120
24y – 27x = 180
5x = 300 x = 60
From equation (i)
4 × 60 – 3y = 15
3y = 240 - 15 = 225
i.e, x + y = 60 + 75 = 135
6. If population of women in a village is 90% of population of men, what is the population of men as a percentage of population of women?

Answer - Option D
Explanation -
Required % = [latex]\frac {100 \times 100} {90}[/latex] =111 (Approx)
7.In measuring the sides of a rectangle, there is an excess of 5% on one side and 2% deficit on the other. Then the error percent in the area is

Answer - Option
Explanation -
Required percentage error = [latex]\frac {x + y + xy} {100}[/latex]
[Negative sign is used for decrease]
= 5 - 2 -[latex]\frac {5 \times 2} {100}[/latex]% = 2.9%
Required percentage error
8. For an examination it is required to get 36 % of maximum marks to pass. A student got 113 marks and failed by 85 marks. The maximum marks for th examination are :

Answer - Option B
Explanation -
If maximum marks be x, then,
[latex]\frac {36 * x} {100}[/latex] = 113 + 85 = 198
x = [latex]\frac {198 * 100} {36}[/latex] = 550
9. Due to an increase of 50% in the price of eggs, 4 eggs less are available for 24. The present rate of eggs per dozen is:

Answer - Option C
Explanation -
Ratio of price increase: 2 : 3
Ratio of decrease in quantity: 3 : 2
Difference of 1 is equal to 4
So, quantity after decrease = 8 eggs
8 eggs in 24, So Rs 3 per egg
Dozen egg = 12*3 = 36 Rs
10.If A’s income is 20% more than that of B, by how much percent is B’s income less than that of A?

Answer - Option C
Explanation -
Required percentage = [latex]100 * \frac {x} {100 + x}[/latex]
[latex]100 * \frac {20} {100 + 20}[/latex]
[latex] \frac {50} {3}[/latex] = [latex]16 \frac {2} {3}[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 11

1. The price of a certain television set is discounted by 10% and the reduced price is then discounted by 10%. The equivalent successive discount is ______.

Answer - Option B
Explanation -
Let X be the Market price of a television.
First discount= 10%
= x - [latex]\frac {10X} {100}[/latex] = [latex]\frac {90X} {100}[/latex] = [latex]\frac {9X} {10}[/latex]
Hence Price after first discount
Now one more discount 0f 10%
= [latex]\frac {9X} {10}[/latex] - [latex]\frac {9X} {10}[/latex] * [latex]\frac {10} {100}[/latex]
Mow price after second discount
= [latex]\frac {9X} {10}[/latex] - [latex]\frac {9X} {100}[/latex] = [latex]\frac {81x} {100}[/latex]
= X - [latex]\frac {81x} {100}[/latex] = [latex]\frac {19x} {100}[/latex]
Hence total discount = [latex]\frac {19X} {100}[/latex] * [latex]\frac {100} {X}[/latex] = 19
Therefore total discount percentage
2. A shopkeeper gains 56 on a toy after allowing 23% discount on its marked price. If his gain is 10%, then the marked price of

Answer - Option B
Explanation -
Let ‘X’ be the market price of a toy.
Discount = 23%
Hence discount amount=[latex]\frac {23X} {100}[/latex]
Hence selling price = X - [latex]\frac {23X} {100}[/latex] = [latex]\frac {77X} {100}[/latex]
Gain = 10%
Gain amount = 56
Let C. P. = Y
Hence gain = [latex]\frac {10Y} {100}[/latex] = [latex]\frac {Y} {10}[/latex]
But given: [latex]\frac {Y} {10}[/latex] = 56 [latex]\rightarrow[/latex] Y = 560
As we know that S. P. = C. P. + Profit
[latex]\frac {77X} {100}[/latex] = 560 + 56
77X = 61600
X = [latex]\frac {61600} {77}[/latex] = [latex]\frac {5600} {7}[/latex] = 800
Hence Market price of toy = 800.
3. A fruit seller makes a profit of 20% by selling mangoes at a certain price. If he charges 1 more for each mango, he can make a profit of 40%. Find the selling price of a mango in the first case.

Answer - Option A
Explanation -
Let there are ‘n’ mangos and ‘x’ be the selling price of one mango.
Hence S. P. of all mangos = nx
Profit = 20%
Hence C. P. of ‘n’ mangos = [latex]\frac {100nx} {120}[/latex] = [latex]\frac {5nx} {6}[/latex]
If S.P. of each mango is increase by 1, profit = 40%
S.P. of all mangos = n(x + 1)
= [latex]\frac {5nx} {6}+ \frac{40(\frac{5nx}{6})}{100} = \frac{5nx}{6} + 2 \frac{\frac{5nx}{6}}{5}[/latex]
n(x + 1) = [latex]\frac {5nx} {6} + \frac {2nx} {6}[/latex]
n(x + 1) = [latex]\frac {7nx} {6}[/latex]
6(x + 1) = 7n
x = 6 Rs
Hence price of each mango = 6
4. A man purchased a bed sheet for Rs 450 and sold it at a gain of 10% calculated on the selling price. The selling price of the bed sheet was

Answer - Option D
Explanation -
[latex]\frac {Gain} {sp} * 100[/latex] = 10
[latex]\frac {Gain} {sp} = \frac {1} {10}[/latex]
So, CP = 10 - 1 = 9 units
SP = 10 units
The value of 9 unit is 450 Rs
So, Value of 10 units = 500 Rs
5. Aniruddha sold a bicycle at a gain of 8%. Had it been sold for 75 more, the gain would have been 14%. The cost price of the bicycle was

Answer - Option B
Explanation -
Let CP = 100 unit
Old SP = 108 unit (8% profit)
New SP = 114 unit (14% profit)
According to question,
New SP = Old SP + 75
114 unit = 108 unit + 75 Rs
6 unit = 75 Rs
1 unit = [latex]\frac {75} {6}[/latex] Rs
100 unit = [latex]\frac {75 * 100} {6}[/latex] Rs
CP = 100 unit = 1250
6. The single discount equivalent to two successive discounts of 20% and 5% is

Answer - Option A
Explanation -
Single equivalent discount = 20 + 5 - [latex]\frac {20 * 5} {100}[/latex] = 24%
7. When a shopkeeper gives 10% discount on the list price of a toy, his gain is 20%. If he had given a discount of 20%, his percentage of gain would have been

Answer - Option A
Explanation -
Let the MP of toy be Rs 100 Rs
SP after 10% discount = 90
120% CP = 90
CP = 75 Rs
SP after 20 % discount = 80 Rs
Gain% =[latex]\frac {5} {75}[/latex] * 100 = [latex]\frac {20} {3}[/latex]%
8. A radio dealer sold a radio at a loss of 2.5%. Had he sold it for Rs. 100 more, he would have gained [latex] 7 \frac {1} {2}[/latex]% In order to [latex] 12 \frac {1} {2}[/latex]% gain he should sell it for

Answer - Option B
Explanation -
Let CP = x, then
107.5% of x - 97.5% of x = 100
10% of x = 100
100% x = 1000
Desired SP = 112.5 % of CP = 1125
9. The cost price of 24 apples is the same as the selling price of 18 apples. The percentage of gain is:

Answer - Option D
Explanation -
Let the CP of 1 apple = 1
i.e, CP of 18 apples = 18
SP of 18 apples = 24
[latex] 100 * \frac {6} {18}[/latex] = [latex] \frac {100} {3}[/latex] = 33.3%
10. A shopkeeper sells his goods at 15% discount. The marked price of an article whose selling price is 629 is:

Answer - Option A
Explanation -
Let the marked price be x
i.e, [latex]\frac{x * 85}{100}[/latex] = 629
x = [latex]\frac{629 * 100}{85}[/latex] = 740

RRB JE 2019 Quantitative Aptitude Quiz - Set 12

1. Which of the following successive discount series is the best of all for a customer?

Answer - Option A
Explanation -
(i) Single equivalent discount for 30% and 20% = (30 + 20 - [latex]\frac{30 * 20}{100}[/latex])% = 44%
Single equivalent discount for 44% and 10% = (44 + 10 - [latex]\frac{44 * 10}{100}[/latex])% = 49.6%
(ii) Single equivalent discount for 25% and 20% = (25 + 20 - [latex]\frac{25 * 20}{100}[/latex])% = 40%
Single equivalent discount for 40% and 15% = (44 + 15 - [latex]\frac{44 * 15}{100}[/latex])% = 49%
(iii) Single equivalent discount for 30% and 10% = (30 + 10 - [latex]\frac{30 * 10}{100}[/latex])% = 37%
Single equivalent discount for 37% and 15% = (37 + 15 - [latex]\frac{37 * 15}{100}[/latex])% = 46.45%
(iv) Single equivalent discount for 25% and 15% = (25 + 15 - [latex]\frac{25 * 15}{100}[/latex])% = 36.25%
Single equivalent discount for 36.25% and 10% = (36.25 + 10 - [latex]\frac{36.25 * 10}{100}[/latex])% = 42.63%
2. The selling price of 12 articles is equal to the cost price of 15 articles. The gain per cent is

Answer - Option C
Explanation -
Let cost price of each article = Rs. 1
Cost price of 12 articles = Rs. 12
Selling price of 12 articles = Rs. 15
Profit =15-12=3
Required percentage of profit = [latex]\frac{3 * 100}{12}[/latex]% = 25%
3. A vendor sells lemons at the rate of 5 for 14, gaining thereby 40%. For how much did he buy a dozen lemons ?

Answer - Option C
Explanation -
C.P. of 5 lemons
= [latex]\frac{100}{140}[/latex] = Rs. 10
C.P. of 12 lemons
[latex]\frac{10 * 12}{5}[/latex] =24
4. The percentage of profit, when an article is sold of 78, is twice than when it is sold for 69. The cost price of the articls is :

Answer - Option D
Explanation - Let the C.P. of article be Rs. x.
[latex](\frac{78 - x}{x}) * 100[/latex]
= 2 ([latex]\frac{69 - x}{x}[/latex] * 100)
78 - x = 2 * 69 - 2x
2x - x = 138 - 78
x = 60
5. Mohan sold his watch at 10% loss. If he had sold it for Rs. 45 more, he would have made 5% profit. The selling price of the watch was

Answer - Option D
Explanation -
Difference of 15% = 45 Rs
1% = 3Rs
100% = 300 Rs
So, CP = 300 Rs
SP after 10% loss = 300 - 30 = Rs 270
6. A trader purchases a watch and a wall clock for 390. He sells them making a profit of 10% on the watch and 15% on the wall clock. He earns a profit of 51.50. The difference between the original prices of the wall clock and the watch is equal to

Answer - Option C
Explanation -
If the C.P. of wrist watch be ? x, then
C.O. of wall clock = (390 - x)
[latex]\frac{x * 10}{100} + \frac{(390 - x) * 15}{100}[/latex] = 51.50
10x + 5850 - 15x = 5150
5x = 5850 - 5150
5x = 700
x = [latex]\frac{700}{5}[/latex] = 140
C.P. of wall clock = 390 – 140 = 250
Required difference = 250 – 140 = 110
7. A profit of 10% is made after giving a discount of 5% in a T.V. If the marked price of the TV is Rs. 2640.00, the cost price of the TV was:

Answer - Option A
Explanation -
MP = Rs 2640
D = 5%
SP = 2640×95/100 = Rs 2508
Profit = 10%
CP = 2508 ×100/110 = Rs 2280
Hence Option A is correct
8. A dishonest fruit vendor sells his goods at cost price but the uses a weight of 900 gm for the kg. Weight. His gain per cent is :

Answer - Option B
Explanation -
Let goods cost the vendor 1 Rs / kk
i.e, He sells for Rs. 1 which cost him = 0.9
i.e, Gain% = [latex]\frac{1 - 0.9}{0.9} * 100 = \frac{100}{9} = 11 \frac{1}{9}[/latex]
9. By selling an article for 960 a man incurs a loss of 4%; what was the cost price?

Answer - Option A
Explanation -
C.P.= [latex]\frac{(100 × 960)}{(100 - 4)}[/latex] = Rs.1000
10. Kamala bought a bicycle for 1,650. She had to sell it at a loss of 8%. She sold it for

Answer - Option B
Explanation -
Kamala sold her Bicycle = [latex]\frac{1650 * 92}{100}[/latex] = 1518

RRB JE 2019 Quantitative Aptitude Quiz - Set 13

1. Two trains, of same length, are running in parallel tracks in opposite directions with speed 65 km/hour and 85 km/hour respectively. They cross each other in 6 seconds. The length of each train is

Answer - Option C
Explanation -
Trains are running in opposite direction hence their relative speed will be the sum of their respective speeds.
Relative speed = 65 + 85 = 150 Km/hour
Both trains are of same length (say X), if they crosses each other they travels 2X distance
Time = 6 sec = [latex]\frac{6}{3600}[/latex] hour = [latex]\frac{1}{600}[/latex] hour
Distance = speed × time
2X = 150 × [latex]\frac{1}{600}[/latex] = 0.25 km = 250m
X = 125 meter
2. A car goes 20 metres in a second. Find its speed in Km/hr.

Answer - Option B
Explanation -
Since, A car goes 20 metres in a second. Therefore,
Speed of the car = 20 m/sec
= 20 * [latex]\frac{18}{5}[/latex] = 72 km/hr
3. Two trains 100 metres and 95 metres long respectively pass each other in 27 seconds when they run Cin the same direction and in 9 seconds when they run in opposite directions. Speeds of the two trains are

Answer - Option B
Explanation -
Let speed of first train = [latex]{s}_{1}[/latex]
Let speed of second train = [latex]{s}_{2}[/latex]
[latex]{s}_{1}[/latex] - [latex]{s}_{2}[/latex] = [latex]\frac{195}{27}[/latex] = [latex]\frac{65}{9} \rightarrow {65}{9} * {18}{5}[/latex] = 26km/hr [latex]\rightarrow[/latex] Eq (1)
similerly [latex]\rightarrow[/latex]
[latex]{s}_{1}[/latex] + [latex]{s}_{2}[/latex] = [latex]\frac{195}{9} * \frac{18}{5} [/latex] = 78 km/hr [latex]\rightarrow[/latex] Eq (2)
Add Eq (1) and Eq (2)[latex]\rightarrow[/latex]
[latex]{s}_{1}[/latex] + [latex]{s}_{2}[/latex] = 78
[latex]{s}_{1}[/latex] - [latex]{s}_{2}[/latex] = 26
Sothat, [latex]{s}_{1}[/latex] = 52 km/hr
[latex]{s}_{2}[/latex] = 26 km/hr
4. A train travelling with uniform speed crosses two bridges of lengths 300m and 240m in 21 seconds and 18 seconds respectively. The speed of the train is:

Answer - Option A
Explanation - Let the length of the train be x metre.
[latex]\frac {x+ 300}{21}[/latex] = [latex]\frac {x + 240}{18}[/latex]
i.e, Speed of train = [latex]\frac {x+ 300}{7}[/latex] = [latex]\frac {x + 240}{6}[/latex]
7x + 1680 = 6x + 1800
7x – 6x = 1800 – 1680
x = 120
[latex]\frac {x+ 300}{21}[/latex] = [latex]\frac {420}{21}[/latex] = 20 m/sec
i.e, Speed of train = [latex](\frac {20 * 18}{5})[/latex] kmph
5. A train of length 500 feet crosses a platform of length 700 feet in 10 seconds. The speed of the train is

Answer D
Explanation
Speed of the train = [latex]\frac {(500+700)}{10}[/latex] = 120 ft/sec
6. A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60 km/h and rest at 45 km/h. Find the total distance travelled by him.

Answer C
Explanation
Let the total distance covered by a man = x km
Since, he covers the first half of the journey at 60 km/h. therefore,
Time taken to cover the x/2 distance at the speed of 60 km/hr = [latex]\frac {\frac {x}{2}}{60}[/latex]
Time taken to cover rest x/2 distance at the speed of 45 km/hr = [latex]\frac {\frac {x}{2}}{45}[/latex]
But, he travels for 5 hours 15 minutes. Therefore
= [latex]\frac {x} {120}[/latex] + [latex]\frac {x}{90}[/latex] = 5 hr 15 min
= [latex]\frac {x} {120} [/latex] + [latex]\frac {x}{90}[/latex] = 5 [latex]\frac {1}{4}[/latex]
= [latex]\frac {3x + 4x} {360}[/latex] = [latex]\frac {21}{4}[/latex]
[latex]\frac {7x}{360}[/latex] = [latex]\frac {21}{4}[/latex]
X = [latex]\frac {3 * 360}{4}[/latex] = 270 km
7. A cyclist, after cycling a distance of 70 km on the second day, finds that the ratio of distances covered by him on the first two days is 4 : 5. If he travels a distance of 42 km. on the third day, then the ratio of distances travelled on the third day and the first day is:

Answer C
Explanation
Let the common ratio be x
Distance covered in first day = 4x
Distance covered in second day = 5x
According to the question
5x = 70 km
So, X = 14 km
Distance covered in first day = 4x = 4×14 = 56 km
Distance covered in third day = 42 km
Required Ratio = Distance covered in third day/ Distance covered in first day
I.e, [latex]\frac {42}{56}[/latex] = 3 : 4
Hence Option C is correct
8. A train is running at a uniform speed of 60 km/hr. If the length of the train is 73 m, then the time taken by the train in crossing 77m Ling Bridge is

Answer A
Explanation
Speed of the train = 60 km/hr = 60 × 5/18 = 50/3 m/sec
Length of train = 73m
Length of bridge= 77
Time taken =150 ÷ 50/3 = 9 sec
9. A man walks ‘a’ km in ‘b’ hours. The time taken to walk 200 metres is how much?

Answer B
Explanation
Speed of man = [latex]\frac {a}{b}[/latex] Km/hr
Hence time taken to walk 200 meter ([latex]\frac {1}{5}[/latex] Km) = distance/speed hrs.
[latex]\frac {1}{5}[/latex] * [latex]\frac {a}{b}[/latex] = [latex]\frac {b}{a}[/latex] hours
10. A missile travels at 1260 km/h. How many metres does it travel in one second?

Answer D
Explanation
1 km/hr = [latex]\frac {5}{18}[/latex] m/s
1260 km/hr = 1260 × = [latex]\frac {5} {18}[/latex]
= 350 m/s
Thus, in one second the missile travels 350 meters

RRB JE 2019 Quantitative Aptitude Quiz - Set 14

1. A missile travels at 1422 km/h. How many meters does it travel in one second?

Answer A
Explanation
Speed of missile = 1422 * [latex]\frac {5}{18}[/latex] = 395 m/s
Speed in meter per second = 1422
Distance travelled in one second = 395 x 1 = 395 meters.
Note:
To convert speed in km/h to m/s we multiply the speed in km/h by [latex]\frac {5}{18}[/latex]
2. To cover a distance of 333 km in 2 hours by a car, what should be the average speed of the car (in meter/second)?

Answer B
Explanation
Speed of the car in km/hr = [latex]\frac {333}{2}[/latex]
= 166.5 km/hr
We know that, in 1km/hr = [latex]\frac {5}{18}[/latex] m/sec
Therefore, 166.5 km/hr = 166.5 × ([latex]\frac {5}{18}[/latex])
= 46.25 m/sec
3. Two trains start from stations A and B and travel towards each other at the same time at speeds of 50 kmph and 60 kmph respectively. At the time of their meeting, the second train has travelled 120 km more than the first. The distance between A and B is ?

Answer C
Explanation
Let the distance travelled by first train be x.
Then, the distance travelled by second train =x+120
Time taken by first train = [latex]\frac {x} {50}[/latex]
Time taken by second train= [latex]\frac {x + 120} {60}[/latex]
[latex]\frac {A}{Q}[/latex],
Distance travelled by first train=600
Distance travelled by second train = x + 120 = 720
The distance between A and B = 600 + 720 = 1320
4. 7 hrs after a goods train passed a station, another train travelling at a speed of 54 km/hr following that goods train passed through that station. If after passing the station the train overtakes the goods train in 11 hours. What is the speed of the goods train?

Answer C
Explanation
Distance travelled by the goods train in (11 + 7) = 18 hrs. is same as that travelled by the other train in 11 hrs. Let speed of the goods train be x, then
18x = 54 × 11
[latex]\rightarrow[/latex] x = 33 km/hr
5. Two cars travel from from city A to city B at a speed of 42 and 60 km/hr respectively. If one car takes 2 hours lesser time than the other car for the journey, then the distance between City A and City B is :

Answer B
Explanation
Let the distance between the city A and city B be D km
As per question
[latex]\frac {D}{42}[/latex] - [latex]\frac {D}{60}[/latex] = 2
[latex]\rightarrow[/latex]18D= 2 × 42 × 60
D= 280km
6. Rakesh goes on a trip on his motorcycle and rides for 368 kms. If he rides for 5 hours at a speed of 49 km/hr, at what speed he travels for the remaining 3 hours of the journey?

Answer B
Explanation
Let speed with which he travelled for 3 hr be x, then
Distance travelled with speed 49 km/hr for 5 hr + Distance travelled in 3 hr = Total distance travelled
[latex]\rightarrow[/latex] 49 × 5 + 3 × x = 368
[latex]\rightarrow[/latex] 3x = 368 - 245
[latex]\rightarrow[/latex] 3x = 123
[latex]\rightarrow[/latex] x = 41 km/hr
7. Two boat are travelling with speed of 36 km/hr and 54 km/hr respectively towards each other. What is the distance (in metres) between the two boats one second before they collide?

Answer C
Explanation
Speeds of the boats are 36km/h and 54 km/h respectively.
Then, their relative speed = 36 + 54
= 90 km/h
[latex]\rightarrow[/latex] 1 km/h = 5/18 m/s
[latex]\rightarrow[/latex] 90 km/h = 90 × 5/18
= 25 m/s
So, 1 sec before, boats were 25 metres apart.
8. Durga walks 5 km from her home to school in 60 minutes, then bicycles back to home along the same route at 15 km per hour. Her sister mriti makes the same round trip, but does so at half of Durga's average speed. How much time does Smriti spend on her round trip?

Answer C
Explanation -
Durga’s average speed = [latex](\frac{2xy}{x + y})[/latex]kmph
[latex]\frac{2 * 5 * 15}{5 + 15}[/latex]kmph
[latex]\frac{150}{20}[/latex]kmph = [latex]\frac{15}{2}[/latex]kmph
Distance of school = 5 km.
Smriti’s speed = [latex]\frac{15}{4}[/latex]kmph
[latex]\rightarrow[/latex] Required time = [latex]2(\frac{\frac{5}{15}}{4})[/latex] hours
[latex]\frac{2 * 5 * 4}{15}[/latex] = [latex]\frac{8}{3}[/latex] hours
[latex]\frac{8 * 60}{3}[/latex]minutes
= 160 minutes
9. In a 200 m sprint race, Amit can beat Rajiv by 20 m and can beat Satyendra by 30 m. If Satyendra and Rajiv participate in a 100m race, how much head start can Rajiv give to Satyendra?

Answer - Option D
Explanation -
When Rajiv runs 180 m Satyendra can run 170 m in the same duration
Therefore when Rajiv will run 100 m Satyendra can run [latex]\frac{(170*100)}{180}[/latex] = 94.44 m
Required head start = 100-94.44 = 5.56 m
10. In what time will a 100 m long train running with a speed of 50 km/hr cross a pillar

Answer - Option C
Explanation -
We know that, during cross a pillar, a train will cover the distance as same as its length.
Speed of the train = 50 km/hr = [latex]\frac{5}{18} * 50 = \frac{250}{18}[/latex]m/sec
The distance covered by the train = length of the train = 100 m
Therefore, required time = [latex]\frac{100}{\frac{250}{18}} = \frac{100 * 18}{250}[/latex] = 7.2 m/sec
Note:
To convert speed from km/h to m/s we multiply speed in km/h by [latex]\frac{5}{18}[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 15

1. A man can row 30 km downstream and return to the starting point in 8 hours. If the speed of the boat in still water is four times the speed of the current, then the of the speed current is

Answer - Option B
Explanation -
If the speed of stream = x kmph,
then speed of boat in still water = 4x kmph
Rate downstream = 4x + x = 5x kmph
Rate upstream = 4x - x = 3x kmph
i.e, [latex]\frac{30}{3x} + \frac{30}{5x}[/latex] = 8
[latex]\frac{10}{3x} + \frac{6}{x}[/latex] = 8
[latex]\frac{16}{x} [/latex] = 8
x =[latex]\frac{16}{8} [/latex] = 2 kmph
2. A boat travels 24 km upstream in 6 hours and 20 km downstream in 4 hours Then the speed of boat in still water and the speed of water current are respectively

Answer - Option B
Explanation -
Let the speed of boat in still water = X kmph
Speed of water = Y kmph
Formula: speed = distance / time
Relative speed of boat in upstream = X – Y = 24/6 = 4 kmph …… (1)
Relative speed of boat in downstream = X+ Y = 20/4 = 5 kmph …… (2)
After solving these two equations :
X = 4.5 kmph and Y = 0.5 kmph
3. If a boat goes 100 km down stream in 10 hours and 75 km upstream in 15 hours, then the speed of the stream is

Answer - Option B
Explanation -
Rate downstream = [latex]\frac{100}{10}[/latex] = 10 kmph
Rate upstream = [latex]\frac{75}{15}[/latex] = 5 kmph
Speed of current = [latex]\frac{(speed in downstream - speed in upstream)}{2}[/latex]
= [latex]\frac{1}{2}(10 - 5)[/latex]kmph
= [latex]\frac{1}{2} 5[/latex] = 2.5 kmph
4. A man rows 40 km upstream in 8 hours and a distance of 36 km downstream in 6 hour Then speed of stream is

Answer - Option A
Explanation -
Let the speed of boat in still water = X km/hr
Speed of water = Y km/hr
Formula: speed = [latex]\frac{distance}{time}[/latex]
Relative speed of boat in upstream = X – Y = 40/8 = 5 km/hr …… (1)
Relative speed of boat in downstream = X+ Y = 36/6 = 6 km/hr …… (2)
After solving these two equations :
x = 5.5 km/hr and Y = 0.5 km/hr
Hence speed of stream = 0.5 km/ hr
5. The speed of a boat in still water is 10km/hr. It covers upstream a distance of 45 km in 6 hours. The speed (in km/hr) of the stream?

Answer - Option A
Explanation -
Let the speed of the stream be x km/hr
[latex]\frac{45}{6}[/latex] = 10 -x
60 - 6x = 45
x = [latex]\frac{15}{6}[/latex] = 2.5 km/hr
6. Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:

Answer - Option B
Explanation -
Down-stream rate = [latex]\frac{9}{2}[/latex] = 4.5 kmph
Upstream rate = [latex]\frac{9}{6}[/latex] = 1.5 kmph
The speed of the boat = (4.5 – 1.5) kmph = 3 kmph
7. Rajesh rows in still water with a speed of 4.5 kmph to go to a certain place and comes back. Find his average speed for the whole journey, if the river is flowing with a speed of 1.5 kmph?

Answer - Option B
Explanation -
Let the distance in one direction = k kms
Speed in still water = 4.5 kmph
Speed of river = 1.5 kmph
Hence, speed in upstream = 4.5 - 1.5 = 3 kmph
Speed in downstream = 4.5 + 1.5 = 6 kmph
Time taken by Rajesh to row upwards = [latex]\frac{k}{3} hrs[/latex]
Time taken by Rajesh to row downwards = [latex]\frac{k}{6} hrs[/latex]
Now, required Average speed =Total distance/Total speed
= [latex]\frac{2k}{\frac{k}{3}} + \frac{k}{6} = 2k * \frac{18}{6k} + 3k[/latex] = 4 kmph
Therefore, the average speed of the whole journey = 4kmph.
8. A boat travels 24 km upstream in 6 h and 20 km downstream in 4 h. then, the speed of a boat in still water and the speed of current are respectively.

Answer - Option B
Explanation -
Distance travelled by boat in upstream = 24 km
Time taken = 6 h
Speed of the boat in upstream = [latex]\frac{24}{6}[/latex] = 4 km/h
And distance travelled by boat in downstream = 20 km
Time taken = 4h
Speed of the boat in downstream = [latex]\frac{20}{4}[/latex] km/h = 5 km/h
Now, speed of the boat in still water = [latex]\frac{1}{2}[/latex] [ speed of the boat in upstream + speed of the boat in downstream]
= [latex]\frac{1}{2}[/latex] [4 + 5] = [latex]\frac{1}{2}[/latex] × 9 = 4.5 km/h
And speed of the current = [latex]\frac{1}{2}[/latex] [speed of the boat in downstream – speed of the boat in upstream]
= [latex]\frac{1}{2}[/latex][5 – 4] = [latex]\frac{1}{2}[/latex] × 1 = 0.5 km/h
9. P, Q and R are the three towns on the bank of a river which flows uniformly. Q is equidistant from P and R. I row from P to Q and back in 10 hours and I can row from P to R in 4 hours. Compare the speed of my boat in still water with that of the river.

Answer - Option B
Explanation -
Since time taken to cover distance from P to R= 4h
so, time taken to cover distance from P to Q= 2h
and time taken to cover distance from Q to P= 10-2= 8h
so direction of flow of current is from P to Q to R
Let the speed of the boat be v1 and the speed of the current be v2 and d be the distance between the cities.
now , = [latex]\frac{d}{{v}_{1} + {v}_{2}}[/latex] = 2 and [latex]\frac{d}{{v}_{1} - {v}_{2}}[/latex] = 8
= [latex]\frac{{v}_{1} + {v}_{2}}{{v}_{1} - {v}_{2}}[/latex]
[latex]{v}_{1} : {v}_{2}[/latex] = 5 :3
10. A man can cross a downstream river by steamer in 40 minutes and same by boat in 1 hour. If the time of crossing the river in upstream direction by steamer is 50% more than downstream time by the steamer and the time required by boat to cross the same river by boat in upstream is 50% more than the time required in downstream by boat. What is the time taken for the man to cross the river downstream by steamer and then return to same place by boat half the way and by steamer the rest of the way?

Answer - Option B
Explanation -
Downstream (steamer) = 40min
Downstream (boat) = 60 min.
Upstream (steamer) = 60 min.
Upstream (boat) = 90min
Required time = 40 + 30 + 45 =115 min.

RRB JE 2019 Quantitative Aptitude Quiz - Set 16

1. Mihir can do a piece of work in 30 hours. If he is joined by Jigya who is 50% more efficient, in what time will they together finish the work?

Answer - Option B
Explanation -
Short Trick:-
Jigya is 50% more efficient , So their ratio is Mihir : Jigya = 2 : 3
Total work =2 * 30
Time taken by both = 60/ 5 = 12 hours
2. Abhinav is 2 times as good a workman as Balwant and therefore is able to finish a job in 36 days less than Balwant. Working together, they can do it in

Answer - Option D
Explanation -
Let Balwant finishes his job in 36days, then Abhinav finishes the job in (x – 36) days
According to the question,
[latex]\frac{2}{x} = \frac{1}{x - 36}[/latex]
[latex]\rightarrow[/latex] 2x – 72 = x
[latex]\rightarrow[/latex] x = 72 days and (x – 36) = 36 days, i.e. Abhinav will take 36 days and Balwant will take 72 days to finish the job.
[latex]\rightarrow[/latex] if they work together, they will take
= [latex]\frac {1}{\frac {1}{36} + \frac {1}{72}}[/latex]
= [latex]\frac{1}{\frac{2 + 1}{72}}[/latex]
= 24 days
3. Pankaj has done 1/2 of a job in 12 days, Sainath completes the rest of the job in 6 days. In how many days can they together do the job?

Answer - Option C
Explanation -
Let the total work be 24 units, then
Efficiency of Pankaj = [latex]\frac{\frac{24}{2}}{12}[/latex]
= 1 unit/day
And efficiency of Sainath = [latex]\frac{12}{6}[/latex]
= 2 units/day
If they work together their efficiency will be 3 units/day
Thus, number of days to finish the work if they work together = [latex]\frac{24}{3}[/latex]
= 8 days
4. Amar is 2 times as good a workman as Badal and therefore is able to finish a job in 15 days less than Badal. In how many days can they together complete the job?

Answer - Option C
Explanation -
Given that, Amar is 2 times as good a workman as Badal and therefore is able to finish a job in 15 days less than Badal.
[latex]\rightarrow[/latex] Efficiency of Amar = [latex]\frac{1}{15}[/latex] units
And that of Badal = [latex]\frac{1}{30}[/latex] units
[latex]\rightarrow[/latex] number of days required to finish the job = [latex]\frac{1}{\frac{\frac{1}{15}}{\frac{1}{30}}}[/latex]
= 10 days
5. Manoj can do a piece of work in 42 hours. If he is joined by Jayashree who is 100% mor efficient, in what time they will finish the work together?

Answer - Option C
Explanation -
Manoj can do a piece of work in 42 hour
Working efficiency of manoj = [latex]\frac{1}{42}[/latex]
Jayashree is 100% more efficient than manoj so Jayashree will take 21 hour
Efficiency of Jayashree = [latex]\frac{1}{21}[/latex]
Combined efficiency of manoj and Jayashree = [latex]\frac{1}{21} + \frac{1}{42} = \frac{3}{42} = \frac{1}{14}[/latex]
Hence time taken by both = 14 hour
6. Painter 'A' can paint a house in 10 days and 'B' can do it in 20 days. With help of 'C', they did the job in 5 days only. Then, 'C' alone can do the job in:

Answer - Option A
Explanation -
A can paint house in 10 days
So working efficiency of A = [latex]\frac{1}{10}[/latex]
B can complete the work in 20 days
So working efficiency of B = [latex]\frac{1}{20}[/latex]
A, B and C they complete the work in 5 days
So combine efficiency of A + B + c = [latex]\frac{1}{5}[/latex]
Efficiency of C = [latex]\frac{1}{5} - \frac{1}{20} - \frac{1}{10}[/latex]
[latex]\frac{4 - 1 - 2}{20}[/latex] = [latex]\frac{1}{20}[/latex]
Efficiency of C = [latex]\frac{1}{20}[/latex]
Hence C will complete the work in 20 days
7. Mayur can complete a work in 27 hours. If he is joined by Jayantika who is 100% more efficient, in what time will they together finish the work?

Answer - Option C
Explanation -
let total work be 27 units
Number of hours needed by Mayur to finish the work = 27
Then, efficiency of Mayur = [latex]\frac{27}{27}[/latex] = 1 unit/h
Given: Jayantika is 100% more efficient than Mayur, so
Efficiency of Jayantika = 2 units/h
Now, number of hours needed to finish the work if they work together = [latex]\frac{27}{1 + 2}[/latex]
= [latex]\frac{27}{3}[/latex] = 9 hours
8. A does 60 % of a work in 15 days. He then calls B, and they together finish the remaining work in 5 days. How long B alone would take to do the whole work ?

Answer - Option A
Explanation -
A complete 60% work in 15 days
A will complete 100% work in = [latex]\frac{15}{60} * 100[/latex] = 25 days
A and B together complete the 40% work in 5 days
They will complete 100% work in= [latex]\frac{15}{40} * 100[/latex]
One day work of A and B together= [latex]\frac{1}{12.5} [/latex]
One day work of A = [latex]\frac{1}{25} [/latex]
One day work of B = [latex]\frac{1}{12.5} -\frac{1}{25}[/latex] = [latex]\frac{1}{25} [/latex]
Hence B complete the whole work in 25 days
9. A, B and C can complete a work in 10, 12 and 15 days respectively.All of them started working together but A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work was?

Answer - Option D
Explanation -
A’s 1 day work = [latex] \frac{1}{10}[/latex] ; B’s 1 day work = [latex]6 \frac{1}{12}[/latex]; A’s 1 day work = [latex]6 \frac{1}{15}[/latex]
Suppose number of days required to complete the whole work was x days.
Therefore,
[latex]\frac{(x-5)}{10} + \frac{(x-3)}{12} + \frac{(x)}{15}[/latex] = 1
[latex]\frac{(6x-30+5x-15+4x)}{60}[/latex] = 1
15x = 105
x = 7
10. A stock of food grain lasts for 21 days for village A. The same stock would last for 28 days for village B. If the food grains given to one village are used for both the villages together, the stock would last for

Answer - Option B
Explanation -
Consumption of A for 1 day = [latex]\frac{1}{21}[/latex]
Consumption of B for 1 day= [latex]\frac{1}{28}[/latex]
Consumption of A+B=[latex]\frac{1}{21} + \frac{1}{28}[/latex] = [latex]\frac{1}{12}[/latex]
The stock would last for 12 days

RRB JE 2019 Quantitative Aptitude Quiz - Set 17

1. A, B and C are three taps connected to a tank. A and B together can fill the tank in 6 hours, B and C together can fill it in 10 hours and A and C together can fill it in [latex]7 \frac{1}{2}[/latex]hours. In how much time will C alone fill the tank?

Answer - Option D
Explanation -
A and B together can fill the tank in 6 hours
Hence part of the tank can be filled by A and B together in 1 hour = [latex]\frac{1}{6}[/latex]
[latex]\rightarrow[/latex] A+B = [latex]\frac{1}{6}[/latex] ….(1)
B and C together can fill it in 10 hours
Hence part of the tank can be filled by B and C together in 1 hour = [latex]\frac{1}{10}[/latex]
[latex]\rightarrow[/latex] B+C = 1/10 …(2)A and C together can fill it in [latex]7 \frac{1}{2}[/latex] hours
Hence part of the tank can be filled by A and C together in 1 hour = [latex] \frac{2}{15}[/latex]
[latex]\rightarrow[/latex] A+C = [latex] \frac{2}{15}[/latex] …(3)
Adding eqn. (2) and (3)
2C + (A+B)=[latex] \frac{1}{10}[/latex] + [latex] \frac{2}{15}[/latex] = 3 + [latex] \frac{4}{30}
[/latex] = [latex] \frac{7}{30}[/latex] But A + B = [latex] \frac{1}{6}[/latex]
Hence: 2C + [latex] (\frac{1}{6})[/latex] = [latex] \frac{7}{30}[/latex]
2C = [latex] \frac{7}{30} - \frac{1}{6} = 7 - \frac{5}{30} = \frac{2}{30} = \frac{1}{15}[/latex]
C = [latex]\frac{1}{30}[/latex]
Therefore C can fill 1/30 part of tank along in 1 hour or C can fill full tank in 30 hours.
2. A, B and C together earn 150 per day while A and C together earn 94 and B and C together earn 76. The daily earning of ‘C’ is

Answer - Option B
Explanation -
Let X, Y and Z be the earning per day of A, B and C respectively.
Earning of A, B and C together per day = (X + Y + Z = 150) ....(1)
Earning of A and C together per day = (X + Z = 94).....(2)
Earning of B and C together per day = (Y + Z = 76) ....(3)
Eqn. (2) + Eqn. (3) – Eqn. (1)
(X + Z) + (Y + Z) - (X + Y + Z) = 94 + 76 – 150
Z = 170 – 150 = Rs 20
Hence the daily earning of ‘C’ is 20.
3. Akash is 3 times as good a workman as Baldev and therefore is able to finish a job in 40 days . Working together, they can finish it in

Answer - Option A
Explanation -
Let working capacity of baldev be x unit per day
then working capcity of akash = 3xunit per day
total work=40 * 3x = 120x unit
number of days taken by both working together = [latex]\frac{120x}{(x+3x)}[/latex] = 30
4. Working 8 hours a day, Anu can copy a book in 18 days. How many hours a day should she work so as to finish the work in 12 days?

Answer - Option A
Explanation -
18 days are required to complete work working 8 hours a day
12 days are required to complete work in [latex]\frac{(8 × 18)}{12}[/latex] = 12 hours
5. B is 50% more efficient than A and A can alone do a work in 33 days. Then find thenumber of days, in which A and B, working together can finish the job.

Answer - Option B
Explanation -
Let B can do a work in x days.
Then, A can do a work in [latex]x + \frac{50}{100} * x = x + \frac{x}{2} = \frac{3x}{2}[/latex] days
Therefore,
[latex]\frac{3x}{2}[/latex]
x = [latex]\frac{66}{3}[/latex] = 22 days
A’s 1 day work = [latex]\frac{1}{33}[/latex]
B’s 1 day work = [latex]\frac{1}{22}[/latex]
(A +B)’s 1 day work together = [latex]\frac{1}{33} + \frac{1}{22} = \frac{5}{66}[/latex]
Thus, A+B can do a work together in [latex]\frac{66}{5}[/latex] = 13.2
6. X alone can complete a piece of work in 40 days. He worked for 8 days and left. Y alone completed the remaining work in 16 days. How long would X and Y, together take to complete the work?

Answer - Option A
Explanation -
Part of the work done by X in 8 days = [latex] \frac{8}{40} = \frac{1}{5}[/latex]
Remaining work = [latex]1 - \frac{1}{5} = \frac{4}{5}[/latex]
This part of work is done by Y in 16 days.
Time taken by Y in doing 1 day = [latex] \frac{16 * 5}{4} = 20[/latex]days
Work done by X and Y in 1 day = [latex] \frac{1}{40} + \frac{1}{20} = \frac{1 + 2}{40} = \frac{3}{40}[/latex]
Hence, both together will complete the work in [latex] \frac{40}{3} = 13 \frac{1}{3}[/latex]days
7. Total weekly emoluments of the workers of a factory are ? 1534. Average weekly emolument of a worker is ? 118. The number of workers in the factory is:

Answer - Option C
Explanation -
Number of workers in the factory = [latex]\frac{1534}{118}[/latex] = 13
8. 5 men and 2 women working together can do four times as much work per hour as a man and a woman together. The work done by a man and a woman should be in the ratio :

Answer - Option B
Explanation -
Given that: work of 5 men and 2 women = 4(work of a man and a woman)
[latex]\rightarrow[/latex] work of (5 - 4) men = work of (4 - 2 ) women
[latex]\rightarrow[/latex] work of one man = work of 2 women
[latex]\rightarrow[/latex] Work of one man : work of two women = 1:1
[latex]\rightarrow[/latex] Work of one man : work of one woman = 2:1
9. A boy and a girl together fill a cistern with water. The boy pours 4 litres of water every 3 mintures and the girl pours 3 litres every 4 minutes. How much time will it take to fill 100 litres of water in the cistern?

Answer - Option C
Explanation -
Water filled by 1 boys in 1 min = [latex]\frac{4}{3}[/latex] litres
and water filled 1 girl 1 min = [latex]\frac{3}{4}[/latex] litres
In 1 min .water filled by both = [latex]\frac{25}{12}[/latex] litres
Time taken to fill 100 litres = 100 x [latex]100 * \frac{12}{25}[/latex]= 48 min
10. A can do a piece of work in 20 days which B can do in 12 days. B worked at it for 9 days. A can finish the remaining work in

Answer - Option A
Explanation -
A can do a piece of work in 20 days
Hence A’s work of one day = [latex]\frac{1}{20}[/latex]
B can do the same piece of work in 12 days
Hence B’s work of one day = [latex]\frac{1}{12}[/latex]
B worked at it for 9 days, therefore total work done by B = 9 × [latex]\frac{1}{12}[/latex] = [latex]\frac{3}{4}[/latex]
Remaining work = 1 –[latex]\frac{3}{4}[/latex] = [latex]\frac{1}{4}[/latex]
Now A alone do the remaining work, therefore time taken by A to do this ¼ piece of work = 20 × [latex]\frac{1}{4}[/latex] = 5 days

RRB JE 2019 Quantitative Aptitude Quiz - Set 18

1. A trader had 9 Quintals of wheat. He sold a part of it at 10% profit and the rest at 20% profit, so that he made a total profit of 14 %. How much wheat did he sell at 20% profit?

Answer - Option B
Explanation -
Given that,
Total quantity of wheat = 9 quintals
Profit on part 1 = 10% and that on part 2 = 20%

Thus, quantity of part 2 = [latex]\frac{2}{5}[/latex] × 900
= 360 kg
2. The ratio of syrup and water in a mixture is 3 : 1, then the percentage of syrup in this mixture is

Answer - Option D
Explanation -
The ratio of syrup and water in a mixture is 3 : 1
Let the syrup in mixture = 3x and water in the mixture = x
Then, the required percentage of the syrup in mixture = [latex] \frac{3x}{3x + x} * 100[/latex]%
= [latex] \frac{3x}{4x} * 100[/latex]%
= 75%
3. A milkman makes 20% profit by selling milk mixed with water at Rs. 9 per litre. If the cost price of 1 litre pure milk is Rs. 10, then the ratio of milk and water in the said mixture is

Answer - Option A
4. Two containers have acid and water mixed respectively in the ratio 3 : 1 and 5 : 3. To get a new mixture with ratio of acid to water as 2 : 1, the two types have to be mixed in the ratio

Answer - Option A
5. A container contains two liquids A and B in the ratio 7 : 5. When 9 litres of mixture is drawn off and the container is filled with 9 litre of pure B, the ratio of A and B becomes 7: 9. How many litres of liquid A was in the container initially?

Answer - Option A
Explanation -
Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left = [latex]7X - \frac{7}{12} * 9[/latex] = [latex]7X - \frac{21}{4}[/latex] litres.
Quantity of B in mixture left = [latex]5X - \frac{5}{12} * 9[/latex] = [latex]5X - \frac{15}{4}[/latex]litres.
[latex]\frac{7x - \frac{21}{4}}{5x - {15}{4} + 9}[/latex] = [latex]\frac{7}{9}[/latex]
[latex]\frac{28x - 21}{20x + 21}[/latex] = [latex]\frac{7}{9}[/latex]
x = 3
So, the can contained 21 litres of A.
6. In two alloys A and B. the ratio of zinc to tin is 5 : 2 and 3 : 4 respectively. Seven kg of the alloy A and 21 kg of the alloy B are mixed together to form a new alloy. What will be the ratio of zinc and tin in the new alloy?

Answer - Option D
Explanation -
In alloy A ration of Zink and Tin = 5 : 2
Hence amount of Zink in 7 Kg of alloy A = 7 × [latex]\frac{5}{7}[/latex] = 5 Kg
Amount of Tin in 7 Kg of alloy A = 7 - 5 = 2 Kg
In alloy B ration of Zink and Tin = 3 : 4
Hence amount of Zink in 21 Kg of alloy B = 21 × [latex]\frac{3}{7}[/latex] = 9 Kg
Amount of Tin in 21 Kg of alloy B = 21 - 9 = 12 Kg
Seven kg of the allay A and 21 kg of the allay B are mixed together to form a new alloy
Hence in the mixture of Alloy A and B amount of Zink = 5 + 9 = 14Kg
Amount of Tin in the mixture of Alloy A and B = 2+12 = 14 Kg
Therefore ratio of Zink and Tin in the mixture of Alloy A and B = 14 : 14 = 1 : 1
7. The ratio in which a man must mix rice at Rs. 10.20 per kg and Rs. 14.40 per kg so as to make a mixture worth Rs. 12.60 per kg. is

Answer - Option D
Explanation -

Required ratio = 1.8 : 2.4
= 3 : 4
Hence Option D is correct
8. One type of liquid contains 20 % water and the second type of liquid contains 35 % of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. The water in the new mixture in the glass is

Answer - Option D
Explanation -
In 10 liters of first type of liquid;
Water = [latex]\frac{1}{5}[/latex] × 10 = 2 liters
In 4 liters of second type of liquid;
Water = 4 × [latex]\frac{35}{100}[/latex] = [latex]\frac{7}{5}[/latex] liters
Total amount of water =2 + [latex]\frac{7}{5}[/latex] = [latex]\frac{17}{5}[/latex] liters
Required percentage = [latex]\frac{\frac{17}{5}}{14}[/latex] × 100
=[latex] \frac{170}{7}[/latex] = [latex]24 \frac{2}{7}[/latex]%
Hence Option D is correct
9. A container contains 60 kg of milk. From this container 6 kg of milk was taken out and replaced by water. This process was repeated further two times. The amount of milk left in the container is

Answer - Option C
Explanation -
Amount of milk left after 3 operations= 60 × [latex]{\frac{1 - 6}{60}}^{3}[/latex]
=60 × [latex]\frac{54}{60} * \frac{54}{60} * \frac{54}{60}[/latex]
= 60 × [latex]\frac{9}{10} * \frac{9}{10} * \frac{9}{10}[/latex]
= 43.74 kg
10. In a mixture of 60 litres, the ratio of milk and water is 2 : 1 How much more water must be added to make its ratio 1 : 2?

Answer - Option D
Explanation -
In original mixture,
Milk = 40 litres
Water = 20 litres
If x litres of water is mixed,
[latex]\frac{40}{20 + x} = \frac{1}{2}[/latex]
20 + x = 40 * 2 = 80
x = 80 -20 = 60 liters

RRB JE 2019 Quantitative Aptitude Quiz - Set 19

1. One type of liquid contains 25% of milk, the other contains 50% of milk. A container is filled with 6 parts of the first liquid and 4 parts of the second liquid. The percentage of milk in the mixture is:

Answer - Option C
Explanation -
Using Weighted Average:
weighted average = [latex]\frac{(n{1}_{}{A}_{1} + {n}_{2}{A}_{2})} {({n}_{1} + {n}_{2})} [/latex]
Weighted Average = [latex]\frac{(6*25+4*50)}{(6+4)}[/latex]
= [latex]\frac{(150+200)}{(10)}[/latex]
= 35%
2. Ram invests Rs.9000 at the rate of 5%.How much more should he invest at the rate of 8% so that he can earn at total profit of 6%:

Answer - Option A
Explanation -
Therefore 2 = 9000
Then 1 = 4500
So he invest Rs.4500 at the rate of 6%
3. Zinc is 17 times heavy as water and silver is 8 times heavy as water.In what ratio should these be mixed to get an alloy 15 times as heavy as water?

Answer - Option B
4. Shyam buys a pen at 4% discount and a copy at 12% discount.She overall gets a discount of 10%.If the marked price of both are Rs.200/item then find the total selling price of copies, if 8 items in total are purchased?

Answer - Option D
Explanation -
8 items are purchased means no of Pens = 2
and no of copy = 6
SP of copies = [latex]\frac {6* 200* 88}{100}[/latex] = 1056
5. An alloy contains only zinc and copper. One such alloy weighing 25 gm contains zinc and copper in the ratio of 2 : 3 by weight. If 20 gm of zinc is added then find what amount of copper has to be removed from the alloy such that the final alloy has zinc and copper in the ratio of 4 : 1 by weight?

Answer - Option C
Explanation -
In [latex]{1}^{st}[/latex] alloy zinc= [latex]\frac{2}{5}[/latex] ×25=10gm
Copper= [latex]\frac{3}{5}[/latex] × 25=15gm
Let copper be removed by x gm
Then ATQ
[latex]\frac{10 + 20}{15 - x} = \frac{4}{1}[/latex]
X=[latex]\frac{30}{4}[/latex]gm = 7.5gm
6. A 24 liters cylinder contains a mixture of oxygen and nitrogen, the volume of oxygen being 36% of total volume. A few liters of the mixture is released and an equal amount of nitrogen is added. Then the same amount of the mixture as before is released and replaced by nitrogen for the second time .As a result the oxygen content becomes 4% of the total volume. How many liters of mixture is released each time?

Answer - Option C
Explanation -
[latex]\frac{Remaining oxygen}{original oxygen} = 1 - \frac{Vol. of each time released}{total vol of vessel}[/latex]
[latex]\frac{4 %}{36 %} = (1 - {\frac{x}{24}}^{2})[/latex]
X= 16 liter
Therefore amount of mixture released each time =16 liter
7. A man purchased two chairs in Rs.1200; he sells the first chair at [latex]\frac{4}{5}[/latex] of its cost price and the second chair at [latex]\frac{5}{4}[/latex] of its cost price. If during the whole transaction he earns a profit of Rs.120.Find the cost price of cheaper chair?

Answer - Option B
Explanation -
[latex]{1}^{st}[/latex] chair = [latex]\frac{4}{5}[/latex] * 100 = 20%
[latex]{2}^{nd}[/latex] chair = [latex]\frac{5}{4}[/latex] * 100 = +25%
Profit% = [latex]{1}^{st}[/latex] chair = [latex]\frac{120}{12oo}[/latex] * 100 = 10%
3 = Rs. 1200
1 = Rs. 400 = cost price of [latex]{1}^{st}[/latex] chair
2 = Rs. 800 = cost price of [latex]{2}^{nd}[/latex] chair
8. A goldsmith has two qualities of gold, one of 22 carats and another of 16 carats purity. In what proportion should he mix both to make a jewelry of 18 carat purity?

Answer - Option A
Explanation -
Let the ratio of 22 carat gold be a.
[latex]\rightarrow[/latex] Ratio of 16 carat gold = 1 – a
Given, 22 carat gold and 16 carat gold are mixed to obtain 18 carat gold.
[latex]\rightarrow[/latex] 22 × a + 16 × (1 – a) = 18
[latex]\rightarrow[/latex] 22a + 16 – 16a = 18
[latex]\rightarrow[/latex] 6a = 2
[latex]\rightarrow[/latex] a = [latex]\frac{1}{3}[/latex]
Ratio of 16 carat gold = 1 – a = [latex]\frac{2}{3}[/latex]
Ratio of 22 carat gold and 16 carat gold in the mixture = [latex]\frac{\frac{1}{3}}{\frac{2}{3}}[/latex] = 1 : 2
9. The volume of a mixture of X and Y in the ratio 16:25 is 155L more than the volume of mixture of Y and Z in the ratio 5:17 and the quantity of Z in mixture is 102L. If two mixture are mixed find the new ratio between Z and Y.

Answer - Option C
Explanation -
Let the quantity of Y and Z mixture = a
[latex]\frac{17a}{22}[/latex] = 102
a = 132 L
Quantity of Y = 132-102 = 30 L
Volume of mixture = 132 + 155= 287 L
Quantity of
Y = 25 * [latex]\frac{287}{41}[/latex] = 175 L
X = 287 - 175= 112 L
Total Y = 175 + 30 = 205 L
Total Z = 102 L
Ratio = [latex]\frac{Z}{Y} = \frac{102}{205L}[/latex]
10. Three containers whose volumes are in the ratio of 2:3:4 are full of mixture of spirit and water. In the first container the ratio of spirit and water is 4:1, in [latex]{2}^{nd}[/latex]container the ratio is 11:4 and in the [latex]{3}^{rd}[/latex] container ratio is 7:3. All the three mixtures are mixed in a big container. The ratio of spirit and water in the resultant mixture is?

Answer - Option C
Explanation -
Let the volume of the containers be 200 L, 300 L and 400 L respectively
So spirit in [latex]{1}^{ST}[/latex] container= [latex]\frac{4}{5}[/latex] * 200 = 160 L
Water =40 L
Spirit in [latex]{2}^{nd}[/latex] container= [latex]\frac{11}{15}[/latex] * 300 = 220 L
Water =80 L
Spirit in 3rd container= [latex]\frac{7}{10}[/latex] * 400=280 L
Water =120 L
Spirit in new mixture= 660 L
ater in new mixture= 240 L
So, ratio= 660:240= 11:4

RRB JE 2019 Quantitative Aptitude Quiz - Set 20

1. There is 40% increase in an amount in 10 years at simple interest. What will be the compound interest of Rs. 30000 after 3 years at the same rate?

Answer - Option A
Explanation -
There is 40% increase in an amount in 10 years, it mean yearly increase will be [latex]\frac{40}{10}[/latex]% = 4 %
Hence yearly rate of interest = 4 %
Principal amount=30000
Time =3 years
Compound interest= p[latex]{(1 + \frac {r}{100})}^{t}[/latex] - 1
30000 [latex]{(1 + \frac {4}{100})}^{3}[/latex] - 30000
33745.92 - 30000 = 3745.92
So required compound interest=3745.92
2. A sum fetched a total simple interest of Rs. 8100 at the rate of 6% per year in 9 years. What is the sum?

Answer - Option
Explanation -
Let the sum be P
We know that SI = [latex]\frac{P * R * T}{100}[/latex]
Here SI =8100
r-6%
t=9 years
hence 8100 * 100 = P * 6 * 9
P=15000
Hence required sum is 15000
3. John invested a sum of money at an annual simple interest rate of 10%. At the end of four years, the amount invested plus interest earned was Rs 770. The amount invested was how much?

Answer - Option C
Explanation -
P+ SI = Rs 770
(770 – P) = [latex]\frac{P * 10 * 4}{100}[/latex]
P = Rs 550
4. A sum becomes 2,916 in 2 years at 8% per annum compound interest. The simple interest at 9% per annum for 3 years on the same amount will be

Answer - Option B
Explanation -
A = P[latex]{(1 + \frac{R}{100})}^{T}[/latex]
2916 = x [latex]{(1 + \frac{8}{100})}^{2}[/latex]
2916 = x [latex]{(\frac{27}{25})}^{2}[/latex]
x = x [latex]\frac{2916 * 25 * 25}{27 * 27}[/latex]
= ?2500
i.e, S.I = [latex]\frac{P * R * T}{100}[/latex] = ?675
5. A sum was lent at simple interest at a certain rate for 2 years. Had it been lent at 3% more rate, it would have fetched Rs. 300 more. The original sum of money was:

Answer - Option A
Explanation -
For 2 years we got Rs 300 more
So for 1 year we must get Rs 150 extra
Assume we deposited Rs 100 in bank. If bank gives us 3% extra rate it gives us Rs 3 extra
To get Rs 150 extra we need to invest 150/3 = 50 times of Rs 100 that is Rs 5000
Hence Option A is correct
6. Simple interest on a certain amount is [latex]\frac{9}{16}[/latex] of the principle. If the numbers representing the rate of interest in present and time in years be equal, then time, for which the principle is lent out, is:

Answer - Option D
Explanation -
Let sum = x, then, S.I. = [latex]\frac{9}{16}[/latex]x
Let rate = R% and time = R years.
i.e, [latex]\frac{x * R * R}{100} = \frac{9x}{16}[/latex]
[latex] {R}^{2} = \frac{900}{16} = R = \frac{30}{4}[/latex]
Hence, time = 7 [latex]\frac{1}{2}[/latex] years
7. A certain sum of money lent out at simple interest amounts to ?1380 in 3 years and ?1500 in 5 years. Find the rate per cent, per annum.

Answer - Option D
Explanation -
S.I. for 2 years = 1500 – 1380 = 120
S.I. for 1 year = 120/2 = 60
S.I. for 3 years = 60*3 = 180
Amount after 3 years = 1380
Principal = 1380 - 180 = 1200
And interest per year is already known to be 60.
S.I. = PRT/100
60 = [latex]\frac{1200*R*1}{100}[/latex]
R = [latex]\frac{6000}{1200}[/latex]
R = 5%
8. A person deposited Rs. 500 for 4 years and Rs. 600 for 3 years at the same rate of simple interest in bank. Altogether he received Rs. 190 as interest. The rate of simple interest per annum was

Answer - Option B
Explanation -
90 = [latex]\frac{500 * 4 * r}{100} + \frac{600 * 3 * r}{100}[/latex]
=>20r + 18r = 190
=>38r = 190
=>r = [latex]\frac{190}{38}[/latex] = 5%
Hence Option B is correct
9. If x, y, z are three sums of money such that y is the simple Interest on x and z is the simple interest on y for the same time and at the same rate of interest, then the relationship between x,y & z will be?

Answer - Option D
Explanation -
SI = [latex]\frac{P * T * R}{100}[/latex]
Y = [latex]\frac{x * T * R}{100}[/latex]
Z = [latex]\frac{y * T * R}{100}[/latex]
[latex]\frac{y}{z} = \frac{x}{y}[/latex]
[latex]{y}^{2}[/latex] = 2x
10. The difference between the simple interests received from two different banks on ? 500 for 2 years is ? 2.50. The difference between their per annum rates of interest is:

Answer - Option B
Explanation -
[latex]\frac{500 * 2 * {R}_{1}}{100} - \frac{500 * 2 * {R}_{2}}{100}[/latex] = 2.5
[latex]10({R}_{1} - {R}_{2})[/latex] = 2.5
[latex]{R}_{1} - {R}_{2} = \frac{2.5}{10}[/latex] = 0.25

RRB JE 2019 Quantitative Aptitude Quiz - Set 21

1. A sum of money was invested at a certain rate of simple interest for 2 years. Had it been invested at 1 % higher rate, it would have fetched 24 more interest. The sum of money is _____.

Answer - Option A
Explanation -
At 1% additional rate of interest, the interest is Rs 24.
[latex]\frac{P * 1 * 2}{100} = 24[/latex]
P = [latex]\frac{2400}{2}[/latex]
= Rs. 1200
2. In how many years will a sum of money double itself at 12% per annum at simple interest?

Answer - Option
Explanation -
In case of simple interest, money is going to double, it means the interest will be equal to principal.
Present rate of interest = 12%
t= time period
p= [latex]\frac{P * 12 * t}{100} = 24[/latex]
t=[latex]\frac{100}{12} = 24[/latex]= 8 years 4 months
3. If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is 31, the sum is

Answer - Option C
Explanation -
Given: rate of interest ‘r’ = 10% per annum
Time period ‘T’ = 3 year
Let Principle amount = X
Simple interest = [latex]\frac{X * T * r}{100} = \frac{30X}{100} = 0.3X[/latex]
Compound interest = [latex]X ({ 1 + \frac{ r}{100}})^{T} - X = X ({ 1 + \frac{10}{100}})^{3} - X[/latex]
[latex]{\frac {110}{100}}^{3} - X = 1.331X - X = 0.331X[/latex]
Given : difference between compound interest and simple interest = ?31
0.331X – 0.30X = 31
0.031X = 31
X = [latex]\frac{31}{0.031}[/latex] = 1000
Hence sum = 1000
4. In how many years will the simple interest on a sum of money be equal to the principal at the rate of [latex]16 \frac{2}{3}[/latex]% per annum ?

Answer - Option C
Explanation -
100 = 100 [latex]\frac{50}{3} * \frac{x}{100} [/latex]
x = 6 years
5. Deepak lent Rs. 8800 to Jaichand for 13 years and Rs 5500 to Kapil for 12 years on simple interest at the same rate of interest and received Rs 14430 in all from both of them as interest. The rate of interest per annum is __________.

Answer - Option D
Explanation -
Let rate of interest be = R
SI = [latex]\frac{PTR}{100}[/latex]
Simple interest for Deepak = [latex]\frac{(8800 × 13 × R)}{100}[/latex]
Simple interest for Kapil = [latex]\frac{(5500 × 12 × R)}{100}[/latex]
14430 =[latex]\frac{(8800 × 13 × R)}{100} + \frac{(5500 × 12 × R)}{100}[/latex]
14430 = 1144R + 660R
14430 = 1804 R
R = [latex]\frac{14430}{1804}[/latex] = 7.99 ~ 8%
6. Deepinder lent Rs 8200 to Jairaj for 16 years and Rs 4900 to Karna for 15 years on simple interest at the same rate of interest and received Rs 19446.5 in all from both of them as interest. The rate of interest per annum is:

Answer - Option C
Explanation -
Let the rate of interest be r
We know that Simple interest = [latex]\frac{Prt}{100}[/latex]
Total simple interest = 19446.5
= 19446.5 = [latex]\frac{8200 * 16 * r}{100} = \frac{4900 * 15 * r}{100}[/latex]
1944650 = 131200r + 73500r
204700r = 1944650
R = 9.5
7. If a sum of money amounts to ? 12,900 and ? 14,250 at the end of 4th year and 5th year respectively at a certain rate of simple interest, then the rate of interest is

Answer - Option C
Explanation -
SI Rate = [latex]\frac{(14250 - 12900) * 100}{12900 * 5 - 14250 * 4}[/latex]% = 18%
8. P borrowed a sum of money form Q at simple interest the rate of interest is 10% p.a. for the first 2 years and 12% p.a. for the next 3 years and 15% p.a. thereafter. If paid 5332 as interest after 7 years, then find the sum (in)

Answer - Option B
Explanation -
Let the sum be p
[latex]\frac {p * 2 * 10}{100} + \frac {p * 3 * 12}{100} + \frac {p * 2 * 15}{100}[/latex]= 5332
[latex]\frac {20p}{100} + \frac{36p}{100} + \frac {30p}{100}[/latex]= 5332
[latex]\frac {86p}{100}[/latex] = 5332
P = 5332 × [latex]\frac {100}{86p}[/latex] = Rs. 6200
9. Ram earned Rs 4800 as simple interest on a certain sum of money in four years at 8% rate per annum. What would be the compound interest Ram will receive if he invest thrice the sum and at the same rate of interest in two years?

Answer - Option D
Explanation -
Principal= [latex]\frac{4800 * 100}{8 * 4}[/latex] = RS 15000
thrice the sum is invested ,
so Principal becomes = 15000*3 = 45000
Compound Interest = 45000[latex](\frac{1 + 8{100}) 2 - 1[/latex]
Compound Interest= 45000 [latex](\frac{729}{625}) - 1[/latex]
= 45000* [latex]\frac{104}{625}[/latex]
= Rs 7488
10. A man borrowed some money from a private organization at 8% simple interest per annum. He lent 75% of his money to another person at 10% compound interest per annum. If the difference between the CI on 3/4 th amount and the SI on whole amount, that was borrowed is Rs 561.5 in 4 years. Then Find the total amount Borrowed?

RRB JE 2019 Quantitative Aptitude Quiz - Set 22

1. On what sum of money will the difference between the simple interest and the compound interest for 2 years at 8% per annum be equal to 8?

Answer - Option C
Explanation -
Difference between CI and SI for two years = [latex]P{(\frac{R}{100})}^{2}[/latex]P(R/100)2
Difference = [latex]\frac{P * 8* 8}{10000}[/latex]
8 = [latex]\frac{P * 8* 8}{10000}[/latex]
P = [latex]\frac{10000}{8}[/latex] = Rs. 1250
2. The principal that amounts to Rs.4913 in 3 years at [latex]6 \frac{1}{4}[/latex] % per annum compound interest compounded annually, is :

Answer - Option D
Explanation -
Principle = Rs. [latex]6 \frac{4913}{{(1 + \frac {25}{4 * 100})}^{3}}[/latex] = Rs. [latex](4913 * \frac{16}{17} * \frac{16}{17}* \frac{16}{17})[/latex] = Rs. 4096
3. If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is ? 31, the sum is

Answer - Option C
Explanation -
Given: rate of interest ‘r’ = 10% per annum
Time period ‘T’ = 3 year
Let Principle amount = X
Simple interest = [latex]\frac{X * T * r}{100} = \frac{30X}{100} = 0.3X[/latex]
Compound interest =
[latex]X{1 + \frac{r}{100}}^{T} - X = X {1 + \frac{10}{100}}^{3} - X[/latex]
[latex]X{\frac{110}{100}}^{T} - X = 1.331X -X = 0.331X[/latex]
Given : difference between compound interest and simple interest = ?31
[latex]\rightarrow[/latex] 0.331X – 0.30X = 31
[latex]\rightarrow[/latex] 0.031X = 31
[latex]\rightarrow[/latex] X = [latex]\frac{31}{0.031}[/latex] = 1000
Hence sum = 1000
4. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs.1. the sum (in Rs.) is:

Answer - Option A
Explanation -
Let the sum be Rs.x. Then,
C.I. = [latex]x{(1 + \frac{4}{100})}^{2} -x = (\frac{676}{625}x - x) = \frac{51}{625}x[/latex]
S.I. = [latex](\frac{x * 4 * 2}{100}) = \frac{2x}{25}[/latex]
i.e, [latex](\frac{51x }{100}) - \frac{2x}{25}[/latex] = 1
or x = 625
5. At some rate per annum, the compound interest for 2 years on Rs. 1500 is ? 449.40 The rate of interest per annum is :

Answer - Option C
Explanation -
A = P [latex]x{(1 + \frac{R}{100})}^{T} [/latex]
= 1500 + 449.40
= 1500 [latex]x{(1 + \frac{R}{100})}^{2} [/latex]
= [latex]\frac{1949.4}{15000} = {(1 + \frac{R}{100})}^{2}[/latex]
= [latex]{\frac{57}{50}}^{2} = {(1 + \frac{R}{100})}^{2}[/latex]
= 1 + [latex] \frac{R}{100} = \frac{57}{50} [/latex]
[latex] \frac{R}{100} = \frac{57}{50} [/latex] R = 14
6. The compound interest on 6,400 for 2 years at[latex]7 \frac{1}{2}[/latex] per annum is (in Rs)

Answer - Option B
Explanation -
C.I.= 6400 [latex]{(\frac{1 + 15}{200})}^{2} - 1[/latex] = 6400 × 0.155625 = Rs.996
7. A sum of money becomes 1.331 times in 3 years as compound interest. The rate of interest is

Answer - Option C
Explanation -
If principal = Rs 1000 ; amount = Rs 1331
[latex]\rightarrow[/latex] A = [latex]P{(1 + \frac{r}{100})}^{T} [/latex]
[latex]\rightarrow[/latex] [latex]\frac{1331}{1000} = {(1 + \frac{r}{100})}^{3}[/latex]
[latex]\rightarrow[/latex] [latex]{\frac{1331}{1000}}^{3} = {(1 + \frac{r}{100})}^{3}[/latex]
[latex]\rightarrow[/latex] 1 + [latex]\frac{r}{100} = \frac{1}{10}[/latex]
[latex]\rightarrow[/latex] [latex]\frac{r}{100} = \frac{1}{10}[/latex]
[latex]\rightarrow[/latex] r = [latex]\frac{1}{10} * 100[/latex] = 10%
Hence Option C is correct
8. Simple interest on a certain sum of money for 3 years at 18 % per annum is half the compound interest on Rs. 9000 for 2 years at 10 % per annum. The sum placed on simple interest is :

Answer - Option C
Explanation -
Compound interest= [latex]P{(1 + \frac{r}{100})}^{2} - P [/latex]
Compound interest at 9000 for two year= [latex]9000 {(1 + \frac{10}{100})}^{2} - 9000 [/latex]
CI = 9000 * 1.1 * 1.1 - 9000
CI= 10890 - 9000 = 1890
Simple interest = [latex]\frac{CI}{2} = \frac{1890}{2} = 945[/latex]
SI= [latex]\frac {P * r * t}{100}[/latex]
945 = [latex]\frac {P * 18 * 3}{100}[/latex]
P = [latex]\frac {945 * 100}{18 * 3} = 1750[/latex]
Sum placed on simple interest= 1750
9. If the amount received at the end of [latex]{2}^{nd}[/latex] and [latex]{3}^{rd}[/latex] year at compound interest on a certain Principal is Rs 30250, and Rs 33275 respectively, what is the rate of interest?

Answer - Option A
Explanation -
Let the rate of interest be r% and sum be p, then
We have,
P + CI of 3 years = Rs 30250 ----------(i)
P + CI of 2 years = Rs 33275 ----------(ii)
Subtracting (ii) from (i), we get
CI of [latex]{3}^{rd}[/latex] year = 33275 – 30250
= Rs 3025
Thus, the CI calculated in the [latex]{3}^{rd}[/latex] year which is Rs.3025 is basically the amount of interest on
the amount generated after 2 years, which is Rs 30250
? Rate of interest = [latex]\frac{3025 * 100}{30250}[/latex]
= 10%
10. There is 40% increase in an amount in 8 years at simple interest. What will be the compound interest of Rs. 20000 after 3 years at the same rate?

Answer - Option A
Explanation -
As there is 40% increase in an amount in 8 years,
So yearly increase in amount = [latex]\frac{40}{8}[/latex] = 5% = rate of interest
Compound interest = [latex]P{(1 + \frac{r}{100})}^{t} - P [/latex]
Compound interest = [latex]20000 {(1 + \frac{5}{100})}^{3} - 20000 [/latex]
= 20000 * [latex]\frac{21}{20} * \frac{21}{20} * \frac{21}{20} - 20000[/latex]
= 23152.5 - 20000
Compound interest = 3152.5

RRB JE 2019 Quantitative Aptitude Quiz - Set 23

1. A man borrowed some money from a private organization at 8% simple interest per annum. He lent 75% of his money to another person at 10% compound interest per annum. If the difference between the CI on 3/4 th amount and the SI on whole amount, that was borrowed is Rs 561.5 in 4 years. Then Find the total amount Borrowed?

Answer - Option C
Explanation -
CI in 1 year = 10%
CI in 2 years = 21%
For 4 years @ 10%= 21+21+[latex]\frac{21*21}{100}[/latex] = 46.41%
SI in 1 year = 8%
SI in 4 years = 32%
Now,
46.41% of [latex]\frac{3}{4P}[/latex] – 32% of P = 561.5
= 46.41% of 3P – 32 *4P% = 561.5*4
= 139.23%P – 128P% = 2246
= 11.23P% = 2246
P = 2246 * 100 * [latex]\frac{100}{1123}[/latex] = Rs 20000
2. The compound interest on Rs.2400 at 10% per annum, for a certain period of time is Rs.504. Find the time in years.

Answer - Option B
Explanation -
CI = P([latex]{(1 + \frac{R}{100})}^{t} -1[/latex])
Where, P is principal, t is time in years and r is rate% p.a.
Given, P = Rs. 2400, rate = 10% and CI = Rs. 504
504 = 2400([latex]{(1 + \frac{10}{100})}^{t} -1[/latex])
[latex]\rightarrow[/latex] 0.21 =[latex]{(1.1)}^{t}[/latex] – 1
[latex]\rightarrow[/latex] 1.21 = [latex]{(1.1)}^{t}[/latex]
[latex]\rightarrow[/latex] [latex]{(1.1)}^{2} = {(1.1)}^{t}[/latex]
[latex]\rightarrow[/latex] t = 2 years
3. if the difference between compound interest at 8 %p.a. and simple interest at 13/2 if the difference between compound interest at 8p.a. and simple interest at 13/2 p.a. on a certain sum of money for 2 years is 1820, then find the sum (in rs.)?p.a. on a certain sum of money for 2 years is 1820, then find the sum (in rs.)?

Answer - Option A
Explanation -
Let, P = 100
C.L = 100 [latex]{(1.08)}^{2}[/latex] – 100 = 16.64
S.L = 2 × [latex]\frac{13}{200}[/latex] × 100 = 13
When the difference is 3.64, the sum is 100
When the difference is 1820 the sum is
[latex]\frac{100}{3.64}[/latex] × 1820 = 50000
Alternative method.
P([latex]{(1 + \frac{8}{100})}^{2}[/latex] – 1) - [latex]\frac{P * 2 * 13}{100 * 2}[/latex]= 1820
P = 50, 000
4. 2000 was lent at compound interest, interest being compounded annually for 3 years. The respective rates of interest for the first, the second and the third years are 10% pa 20% pa and 30% pa had the sum of 2000 been lent at 20% pa simple interest for 3 years, how much more/less interest would have been realized?

Answer - Option C
Explanation -
The value that the sum of 2000 would have amounted to at the end of 3 years
= 2000 (1 + [latex]\frac{10}{100}[/latex]) (1 +[latex]\frac{20}{100}[/latex] ) (1 + [latex]\frac{30}{100}[/latex] ) = 2000 (1.716)
The value that sum would have amounted to the end of 3 years had it been lent at 20% pa simple interest =2000
(1 + [latex]\frac{3(20)}{100}[/latex]) = 2000 (1.6)
The difference of the interests in the two cases = The difference of the amounts in the two cases = 2000(1.716 – 1.6) = 2000 (0.116) = 232
The sum amounts to a higher value when lent at compound interest than when lent at simple interest
Had the sum been lent at 20% pa simple interest for 3 years, the interest realized would have been 232 less.
5. Mr. Girish deposits an amount of Rs 96,000 to obtain compound interest at the rate of 10 p.c.p.a. for 3 years. What total amount will Mr. Girish get at the end of 3 years?

Answer - Option D
Explanation -
C.I. = 96000 × [latex]\frac{33.10}{100}[/latex] = 31776
So amount = 96000 + 31776
= 127776
6. What would be the compound interest accrued on an amount of Rs. 14,800 @13.5 p.c.p.a.at the end of two years? (Rounded off to two digits after decimal)

Answer - Option D
Explanation -
CI = 14800(1+[latex]{(\frac{13.5}{100})}^{2}[/latex] -1)
= 14800[1.288225 - 1] = 14800 × 0.288225
= Rs 4265.7
7. What would be the compound interest obtained on an amount of ? 1,210 at the rate of 6 p.c.p.a. after a year?

Answer - Option D
Explanation -
C.I. for one year
= [latex]\frac{PTR}{100}[/latex]
= [latex]\frac{1200 * 6 * 1}{100}[/latex]
= Rs. 72.6
8. The compound interest accrued in two years on a principal of ?15,800 is ?7716.72. What is the rate of interest pcpa?

Answer - Option A
Explanation -
Amount = P[latex]({1 + \frac{r}{100}}^{2}[/latex]
15800 + 7716.72 = 15800 [latex]{(1 + \frac{r}{100}})^{2}[/latex]
or, 23516.72 = 15800 [latex]{(1 + \frac{r}{100}})^{2}[/latex]
[latex]\frac{23516.72}{15800} = {(1 + \frac{r}{100}})^{2}[/latex]
[latex]1.4884 = {(1 + \frac{r}{100}})^{2}[/latex]
Taking square root of both the sides,
[latex]1.22 = 1 + \frac{r}{100}[/latex]
122 = 100 + r
or, r = 22%
9. The simple interest accrued on a sum of certain principle is Rs 7,200 in six years at the rate of 12 p.c.p.a. What would be the compound interest accrued on that principle at the rate of 5 p.c.p.a. in 2 years?

Answer - Option B
Explanation -
Required difference = 456-272 = 184
Let the principal be Rs x . Then,
[latex]\rightarrow \frac{x * 12 * 6}{100} = Rs. 7200 \rightarrow x = Rs 10000[/latex]
Required CI = [latex] 10000 (1 + {(\frac{5}{100})}^{2} - 1)[/latex] = Rs. 1025
10. What will be compound interest on a sum of Rs. 1550 at the rate of 10 p.c.p.a. after 3 years?

Answer - Option B
Explanation -
A = 1550 [latex]{(1 + \frac{10}{100})}^{3}[/latex]
= 1550 [latex]{(\frac{11}{10})}^{3}[/latex]
= 1550 * [latex]\frac{10}{100} * \frac{10}{100} * \frac{10}{100}[/latex]
= [latex]\frac{1550 * 1331}{1000} = 2065.05[/latex]
Interest = 2063.5 – 1550 = 513.05 Rs.

RRB JE 2019 Quantitative Aptitude Quiz - Set 24

1. If [latex]{cos}^{2}\alpha + {cos}^{2}\beta = 2[/latex] then the value of [latex]{tan}^{3}\alpha + {sin}^{3}\beta[/latex] is

Answer - Option B
Explanation -
[latex]{cos}^{2}\alpha + {cos}^{2}\beta = 2[/latex]
The maximum value of cosx cannot be more than 1.
[latex]{cos}^{2}\alpha = 1 = {cos}^{2}\beta[/latex]
[latex]\alpha = n\pi; \beta = m\pi [/latex]{n = 0, 1, 2,......; m = 0. 1,2,.....}
[latex]{tan}^{3}\alpha + {sin}^{3}\beta = {tan}^{3}n \pi + {sin}^{5}m \pi = 0 + 0 = 0[/latex]
2. If A= tan 11º tan 29º, B = 2 cot 61º cot 79º, then:

Answer - Option C
Explanation -
As we know that; tan (90 - x) = cotx
A = If A = tan 11º tan 29º,
B = 2 cot 61° cot 79º = 2tan(90 - 61º)tan(90 - 79º)= 2 tan 11º tan 29º
[latex]\Rightarrow [/latex] B = 2A
3. The value of tan 10° tan 15° tan 75° tan 80° is

Answer - Option B
Explanation -
We know that tan (90 - x) = cotx
Hence tan10° = tan (90 - 80) = cot 80°
& tan15 = tan (90 - 15) = cot 75°
Therefore
tan 10° tan 15° tan 75° tan 80°
= cot80° cot75° tan 75° tan 80° = 1
as cotx = [latex]\frac{1}{tanx }[/latex]
[latex]\Rightarrow [/latex] cotx.tanx= 1
4. If [latex]{sec}^{2}\theta + {tan}^{2}\theta= 7, then the value of \theta when 0° \leq \theta \leq 90°[/latex]

Answer - Option A
Explanation -
[latex]{sec}^{2}\theta + {tan}^{2}\theta = 7[/latex]
[latex]1 + {tan}^{2}\theta + {tan}^{2}\theta = 7[/latex]
[latex]2{tan}^{2}\theta = 7 - 1 = 6[/latex]
[latex]{tan}^{2}\theta= \frac {6}{3} = 3[/latex]
[latex]tan\theta = \sqrt {3} [/latex]
[latex]\theta = 60°[/latex]
5. If [latex]{tan}^{2}\alpha = 1 + 2{tan}^{2}\beta (\alpha, \beta[/latex] are positive acute angles), then [latex]\sqrt{2}cos \alpha - cos\beta [/latex] is equal to

Answer - Option A
Explanation -
As [latex]{sec}^{2}x = 1 + 2{tan}^{2}x [/latex]
[latex]{tan}^{2}\alpha = 1 + 2{tan}^{2}\beta[/latex]
[latex]\Rightarrow {sec}^{2}\alpha - 1 = 1 + 2({sec}^{2}\beta - 1) [/latex]
[latex]\Rightarrow {sec}^{2}\alpha - 1 = 1 + 2{sec}^{2}\beta - 2 [/latex]
[latex]\Rightarrow {sec}^{2}\alpha = 2{sec}^{2}\beta [/latex]
[latex]\frac {1}{{cos}^{2}\alpha} = \frac {2}{{cos}^{2}\beta}[/latex]
[latex]\Rightarrow {cos}^{2}\beta = 2{cos}^{2}\alpha [/latex]
[latex]\Rightarrow 2{cos}^{2}\alpha - {cos}^{2}\beta = 0 [/latex]
[latex]\Rightarrow (\sqrt{2}cos\alpha - cos\beta)(\sqrt{2}cos\alpha + cos\beta) = 0 [/latex]
[latex]\Rightarrow (\sqrt{2}cos\alpha - cos\beta) = 0 [/latex]
6. If [latex]2({cos}^{2}\theta - {sin}^{2}\theta) = 1 (\theta[/latex] s a positive acute angle), then cot? is equal to

Answer - Option D
Explanation -
[latex]2({cos}^{2}\theta - {sin}^{2}\theta) = 1 [/latex]
[latex]\Rightarrow{cos}^{2}\theta - (1 - {cos}^{2}\theta )= \frac {1}{2} [/latex]
[latex]\Rightarrow 2{cos}^{2}\theta = 1 + \frac {1}{2} = \frac {3}{2}[/latex]
[latex]\Rightarrow 2{cos}^{2}\theta = \frac {3}{4}[/latex]
[latex]\Rightarrow 2{sec}^{2}\theta = \frac {4}{3}[/latex]
[latex]\Rightarrow 1 + {tan}^{2}\theta = \frac {4}{3}[/latex]
[latex]\Rightarrow tan\theta = \sqrt {3} = cot\theta = \sqrt {3}[/latex]
7. he numerical value of [latex]\frac{5}{{sec}^{2}\theta} + \frac{2}{1 + {cot}^{2}\theta } + 3 {sin}^{2}\theta[/latex] is:

Answer - Option A
Explanation -
[latex]\frac{5}{{sec}^{2}\theta} + \frac{2}{1 + {cot}^{2}\theta } + 3 {sin}^{2}\theta = 5 {cos}^{2}\theta + 2 {sin}^{2}\theta + 3 {sin}^{2}\theta = 5[/latex]
8. If [latex] tan\theta = \frac {4}{3}[/latex], then the value of [latex]\frac{3sin\theta + 2cos\theta}{3sin\theta - 2cos\theta}[/latex] is

Answer - Option C
Explanation -
[latex] tan\theta = \frac {4}{3}[/latex](Given)
i.e, [latex]\frac{3sin\theta + 2cos\theta}{3sin\theta - 2cos\theta} = \frac{3tan\theta + 2}{3tan\theta - 2}[/latex]
[latex]\frac{3 * \frac{4}{3} + 2}{3 * \frac{4}{3} - 2} = \frac{4 + 2}{4 - 2} = 3[/latex]
9. The numerical value of [latex]\frac{9}{{cosec}^{2}\theta} + 4{cos}^{2}\theta + \frac{5}{1 + {tan}^{2}\theta}[/latex] will be?

Answer - Option B
Explanation -
[latex]\frac{9}{{cosec}^{2}\theta} + 4{cos}^{2}\theta + \frac{5}{1 + {tan}^{2}\theta}[/latex]
= [latex]9{sin}^{2}\theta + 4{cos}^{2}\theta + \frac{5}{{sec}^{2}\theta}[/latex]
= [latex]9{sin}^{2}\theta + 4{cos}^{2}\theta + 5{cos}^{2}\theta [/latex]
= [latex]9{sin}^{2}\theta + 9{cos}^{2}\theta [/latex]
= [latex]9({sin}^{2}\theta + {cos}^{2}\theta) [/latex]
= 9 * 1 = 9
10. If [latex]\frac{2sin\theta - cos\theta}{cos\theta + sin\theta } = 1[/latex] then value of cot ? is :

Answer - Option A
Explanation -
[latex]\frac{2sin\theta - cos\theta}{cos\theta - sin\theta} = 1 \Rightarrow 2sin\theta - cos\theta = cos\theta + sin\theta[/latex]
[latex]sin\theta = 2cos\theta \Rightarrow cot\theta = \frac{1}{2} (\frac{cos\theta}{sin\theta} = cot\theta)[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 25

1. The distance between two pillars of length 16 metres and 9 metres is x metres. If two angles of elevation of their respective top from the bottom of the other are complementary to each other, then the value of x (in metres) is

Answer - Option C
Explanation -
If the angles are complementary then the distance between two pillars of length 'a' and 'b' is vab.
In this question,
x = [latex]\sqrt{(16*9)} = \sqrt{144}[/latex] = 12
2. The length of the shadow of a vertical tower on level ground increases by 10 meters when the altitude of the sun changes from 45 ° to 30 °. Then the height of the tower is

Answer - Option C
Explanation -

The ratio of sides in these triangles are shown in red colour.
[latex] (\sqrt{3} - 1)[/latex] unit = 10m
1 unit = [latex]\frac{10} {(\sqrt{3} - 1)}[/latex] m
h = 1 unit = [latex]\frac{10 (\sqrt{3} - 1)} {2}[/latex] m
h = [latex]5 (\sqrt{3} + 1)[/latex] m
3. A man standing at a point C is watching the top of a tower, which makes an angle of elevation of 30 °. The man walks some distance towards the tower and then his angle of elevation of the top of the tower is 60 °. If the height of the tower is 30 m, then the distance he moves is

Answer - Option D
Explanation -

B 60º D 30º C
AB = Tower = 30 m
CD = X m
ACB = 30º
ADB = 60º
From ABD
tan 60º = [latex]\frac{AB}{BD}[/latex]
[latex]\sqrt {3} = \frac{30}{BD}[/latex]
BD [latex] = \frac{30}{\sqrt {3}} = 10\sqrt {3}[/latex]m
From ? ABC
tan 300 = [latex]\frac{AB}{BD}[/latex]
[latex] = \frac{1}{\sqrt {3}} = \frac{30}{10\sqrt {3} + x}[/latex]
[latex]x = 30\sqrt {3} - 10\sqrt {3} [/latex]
[latex]x = 20\sqrt {3}[/latex] m
4. If a pole of 12 m height casts a shadow of [latex]4 \sqrt{3} [/latex] m long on the ground, then the sun’s angle of elevation at that instant is

Answer - Option B
Explanation -

Tan C = [latex] = \frac{AB}{BC} = \frac{12}{4\sqrt {3}} = \frac {3}{\sqrt {3}} = \sqrt {3}[/latex]
or, C = 60º
5. From two points on the ground and lying on a straight line through the foot of a pillar, the two angles of elevation of the top of the pillar are complementary to each other. If the distances of the two points from the foot of the pillar are 12 meters and 27 meters and the two points lie on the same side of the pillar, then the height (in meters) of the pillar is

Answer - Option B
Explanation -

Let, ACB = [latex]\theta[/latex]
i.e, ACB = 90 - [latex]\theta[/latex]
BC = 12 meter
BD = 27 meter
AB = Pillar = h meter
From [latex]\triangle[/latex]ABC,
[latex]tan \theta = \frac{AB}{BC} = \frac{h}{12}[/latex] .....(i)
From ABC,
[latex]tan (90 - \theta) = \frac{AB}{BD} \Rightarrow cot \theta \frac{h}{27}[/latex] .....(ii)
i.e, [latex]tan \theta .cot \theta = = \frac{h}{12} * \frac{h}{27}[/latex]
[latex]{h}^{2}[/latex] = 12 * 27
h = [latex]\sqrt{12 * 27} [/latex]
= [latex]\sqrt{2 * 2 * 3 * 3 * 3 * 3} [/latex]
= 2 * 3 * 3
6. The shadow of a tower standing on a level plane is found to be 40m longer when the sun's altitude is 45 °, than when it is 60 °. The height of the tower is

Answer - Option C
Explanation -

Let, ACB = 60º; BC = Xmeters
CD = 40 m, AB = Tower = h metre
From [latex]\triangle[/latex] ABC
[latex]tan 60º = \frac{AB}{BC} \Rightarrow \sqrt {3} \frac{h}{x}[/latex]
[latex] h = \sqrt {3}x[/latex]
From [latex]\triangle[/latex] ABD
[latex]tan 45º = \frac{AB}{BD} [/latex]
[latex] \Rightarrow h = x + 40 = \frac{h}{\sqrt {3}} + 40[/latex]
[latex] \Rightarrow h = \frac{h}{\sqrt {3}} = 40[/latex]
[latex] \Rightarrow \frac{40\sqrt {3} }{\sqrt {3} - 1} + 40[/latex]
[latex] = \frac{40\sqrt {3} (\sqrt {3} + 1)}{(\sqrt {3} - 1)(\sqrt {3} + 1)} + 40[/latex]
[latex] = \frac{40\sqrt {3} (\sqrt {3} - 1)}{3 - 1} + 40[/latex]
= [latex]20 (3 + \sqrt{3}) [/latex] m
7. A ladder Leaning against a house has angle of elevation 60º and the foot of the ladder is 6.5 metres from the house. The length of the ladder is

Answer - Option B
Explanation -

AC = Ladder, AB = House
In [latex]\triangle[/latex] ABC
cos 60º = [latex]\frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{6.5}{AC} [/latex]
[latex] \Rightarrow [/latex] = 2 * 6.5
= 13 m
8. The angle of elevation of the sun when the length of the shadow of a pole is equal to its height is :

Answer - Option B
Explanation -

AB = BC
[latex] tan \theta = \frac{AB}{BC}[/latex]
[latex] \Rightarrow tan \theta =[/latex] tan 45º
[latex] \Rightarrow \theta =[/latex] 45º
9. The cliff of a mountain is 180 m high and the angles of depression of two ships on the either side of cliff are 30º and 60º respectively. What is the distance between the two ships? ([latex]\sqrt{3}[/latex] = 1.732)

Answer - Option C
Explanation -

AC = Cliff = 180 m
In [latex]\triangle[/latex] ABC
tan 60º = [latex]\frac{AD}{BD}[/latex]
[latex]\Rightarrow \sqrt {3} = \frac{180}{BD}[/latex]
[latex]BD = \frac{180}{\sqrt {3}} = 60\sqrt {3}[/latex] m
From [latex]\triangle[/latex] ACD,
tan 30º = [latex]\frac{AD}{CD}[/latex]
[latex]\frac{1}{\sqrt {3}} = \frac{180}{CD}[/latex]
CD= 180 [latex]\sqrt{3} [/latex]metre
[latex]\triangle[/latex] BC = BD + DC
= 60 [latex]\sqrt{3} [/latex] + 180 [latex]\sqrt{3} [/latex]
= 240 [latex]\sqrt{3} [/latex]metre
=(240 x 1.732) metre
= 415.69 metre
10. The angles of elevation of the top of a tree 220 meters high from two points lie on the same plane are 30 ° and 45 °. What is the distance (in metres) between the two points?

Answer - Option D

RRB JE 2019 Quantitative Aptitude Quiz - Set 26

1. The distance between the tops of two building 38 metres and 58 metres high is 52 metres. What will be the distance (in metres) between two buildings?

Answer - Option D
Explanation -

[latex]{52}^{2} = {x}^{2} + {h}^{2}[/latex]
2704 = [latex]{x}^{2} + {20}^{2} = {x}^{2} + 400[/latex]
[latex]{x}^{2}[/latex] = 2704 - 400 = 2304
x = 48 meter
2. The angle of elevation of an aeroplane from a point on the ground is 60º. After flying for 30º seconds, the angle of elevation changes to. If the aeroplane is flying at a height of 4500m, then what is the speed (in m/s) of aeroplane?

Answer - Option B
Explanation -

AB=ED= height of aeroplane= 4500m
In triangle ABC
Tan60° = [latex]\frac{AB}{BC}[/latex]
[latex]\sqrt{3} = \frac{4500}{BC}[/latex]
BC=1500 [latex]\sqrt{3} [/latex] m
In triangle CED
In triangle CED
Tan30 ° = [latex]\frac{ED}{DC}[/latex]
[latex]\frac{1}{\sqrt{3}} = \frac{4500}{BC}[/latex]
DC=4500 [latex]\sqrt{3} [/latex] m
Distance covered in 30 second = BC = 4500 [latex]\sqrt{3} m-1500 \sqrt{3} = 3000 \sqrt{3} [/latex] m
Speed of aeroplane= [latex]\frac {3000 \sqrt{3}}{30} [/latex] m
3. A kite is flying in the sky. The length of string between a point on the ground and kite is 420 m. The angle of elevation of string with the ground is 30º Assuming that there is no slack in the string, then what is the height (in metres) of the kite?

Answer - Option A
Explanation -

Here length of string= AC=420m
In triangle ABC , AB= height of kite
So sin30 ° = [latex]\frac{AB}{AC}[/latex]
[latex]\frac{1}{2} = \frac{AB}{420}[/latex]
AB=210 m
So height of kite is 210m
4. If the angle of elevation of sun changes from 45º to 60º, the length of the shadow of a pillar decreases by 20 meters. The height of the pillar is:

Answer - Option D
Explanation -
Let h be the height of the pillar. Then in [latex]\triangle[/latex]ABC
Tan 45° = [latex]\frac{h}{BD + 20}[/latex]
h = BD + 20
In [latex]\triangle[/latex]ABd,
Tan 60° = [latex]\frac{h}{BD} = \sqrt{3}[/latex]
h = BD [latex]\sqrt{3} = (h -20) \sqrt{3}[/latex]
h = [latex]h \sqrt{3} - 20 \sqrt{3}[/latex]
h = [latex]\frac{20 \sqrt{3}}{\sqrt{3} - 1} = 10\sqrt{3}(\sqrt{3} + 1) [/latex]
5. From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Answer - Option C
Explanation -

[latex]\frac{AB}{AP} = Tan 30° = \frac{1}{\sqrt{3}}[/latex]
AP = AB [latex]\sqrt{3}[/latex]
AP = 100 [latex]\sqrt{3}[/latex] = 173 m
6. The angle of elevation of the sun, when the length of the shadow of a tree [latex]\sqrt{3} [/latex] times the height of the tree, is:

Answer - Option A
Explanation -
Let AB be the tree and AC be its shadow.

Let [latex]\triangle[/latex] ACB = [latex]\theta, \frac{AC}{AB} = \sqrt{3} \rightarrow cot \theta = \sqrt{3}[/latex]
[latex]\theta = 30º[/latex]
7. A tree is cut partially and made to fall on ground. The tree however does not fall completely and is still attached to its cut part. The tree top touches the ground at a point 10m from foot of the tree making an angle of 30 °. What is the length of the tree?

Answer - Option A
Explanation -
AC = [latex]\frac{10}{COS 30 °} = 20\sqrt{3} m[/latex]
AB = 10 tan 30 ° = [latex]10 \sqrt{3} m[/latex]
Length of the tree = AC + BC = [latex]\frac{30}{\sqrt{3}} = 10\sqrt{3} m[/latex]
8. At a point on a horizontal line through the base of a monument the angle of elevation of the top of the monument is found to be such that its tangent is [latex]\frac{1}{5}[/latex]. On walking 138 meters towards the monument the secant of the angle of elevation is found to be [latex]\frac{\sqrt{193}}{12}[/latex]. The height of the monument (in meter) is:

Answer - Option A
Explanation -

tan ? D = [latex]\frac{AB}{BD} = \frac{1}{5}[/latex] Now,
sec ? C = [latex]\frac{AB}{BD} = \frac{\sqrt{193}}{12}[/latex]
AB = [latex]\sqrt{{AC}^{2} - {BC}^{2}} = \sqrt{193 - 144} = 7[/latex] Perpendicular AB in both cases must be same, therefore
[latex]\frac{AB}{BD} = \frac {1}{5} * \frac{7}{7} = \frac {7}{35}[/latex]
BD - BC = 35r - 12r = 23r = 138
AB = 7r = 7 * [latex]\frac {138}{23} = 42 meter[/latex]
9. The angle of elevation of the top of a tower from a point A on the ground is 30 °. On moving a distance of 20 meters towards the foot of the tower to a point B, the angle of elevation increases to 60 °. The height of the tower is:

Answer - Option C
Explanation -

tan 30 °= [latex]\frac{1}{\sqrt{3}} = \frac{h}{20 + x} \Rightarrow x = h \sqrt{3} - 20[/latex] Now,
tan 60 ° = [latex]\sqrt{6} = \frac{h}{x}\Rightarrow x = \frac {h}{\sqrt{3}}[/latex]
[latex]h \sqrt{3} - 20 = \frac {h}{\sqrt{3}}[/latex]
3h - h = 20 [latex] \sqrt{3}[/latex]
h = 10 [latex] \sqrt{3}[/latex]
10. The shadow of a tower is [latex]\sqrt{3} [/latex] times its height. Then the angle of elevation of the top of the tower is

Answer - Option B
Explanation -

[latex]tan \theta = \frac {h}{h \sqrt{3}} = \frac {1}{ \sqrt{3}} = tan30°[/latex]
[latex]\theta = 30°[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 27

1. A circle is touching the side BC of [latex]\triangle[/latex]ABC at P and is also touching AB and AC produced at Q and R respectively. If AQ = 6 cm, then perimeter of [latex]\triangle[/latex] ABC is

Answer - Option C
Explanation -
Given: A circle is touching the side BC of [latex]\triangle[/latex] ABC at P and is also touching AB and AC produced at Q and R respectively. If AQ = 6 cm

By the property of triangle
AQ = AR = 6 cm; BQ = BP and CR= CP
AQ +AR = AB + BQ + AC + CR
Hence 12 = AB + AC + BP + CP = AB + AC + BC
Hence AB + AC + BC = 12 cm
Perimeter of triangle = 12 cm
2. [latex]\triangle[/latex] ABC is similar to [latex]\triangle[/latex] PQR. Length of AB is 18 cm and length of the corresponding side PQ is 12 cm. If area of [latex]\triangle[/latex] ABC is 324 sq cm, what is the area of [latex]\triangle[/latex] PQR?

Answer - Option B
Explanation -
Since, [latex]\triangle[/latex]ABC is similar to [latex]\triangle[/latex]PQR. And, AB = 18 cm and PQ = 12 cm
[latex]\frac{ar (\triangle ABC)}{ar (\triangle PQR)} = \frac{{AB}^{2}}{{PQ}^{2}}[/latex]
[latex]\frac{324}{ar (\triangle PQR)} = \frac{{18}^{2}}{{12}^{2}}[/latex]
[latex]\frac{324}{ar (\triangle PQR)} = \frac{324}{144}[/latex]
ar (\triangle PQR) = 144[latex]{cm}^{2}[/latex]
3. E is the mid-point of the median AD of [latex]\triangle[/latex] ABC. BE is joined and produced to meet AC at F. F divides AC in the ratio :

Answer - Option B
Explanation -
Since, E is the mid-point of the median AD of [latex]\triangle[/latex]ABC. BE is joined and produced to meet AC at F. here, BD = DC and AE = ED

In [latex]\triangle[/latex] ADG
EF [latex]\parallel[/latex] DG
According to the theorem,
[latex]\frac{AF}{FG} = \frac{1}{1}[/latex]
FG = AF
FG = GC = AF
In [latex]\triangle[/latex] BCF
DG [latex]\parallel[/latex] BF and D is mid point of BC so G is also the mid point of FC.
FG = GC = AF
So [latex]\frac{AF}{FC} = \frac{1}{2}[/latex]
4. For a triangle, base is [latex]6 \sqrt{3}[/latex]cm and two base angles are 30 ° and 60. Then height of the triangle is

Answer - Option B
Explanation -

sin 30º = [latex]\frac{AC}{BC}[/latex]
[latex]\Rightarrow \frac{AC}{6 \sqrt{3}} \Rightarrow AC = 3 \sqrt{3}[/latex]
sin 60º = [latex]\frac{ACD}{AC}[/latex]
[latex]\Rightarrow \frac {\sqrt{3}} {2} \frac{AD}{3 \sqrt{3}} [/latex]
[latex]\Rightarrow AD = \frac{3\sqrt{3} * \sqrt{3}}{2} = 4.5 cm[/latex]
5. A straight line parallel to the base BC of the triangle ABC intersects AB and AC at the points D and E respectively. If the area of ?ABE be 36 sq. cm then the area of ?ACD is

Answer - Option B
Explanation -

[latex]\triangle[/latex] DBC and [latex]\triangle[/latex] EBC lie on the same base and between same parallel lines.
[latex]\triangle[/latex] DBC = [latex]\triangle[/latex] BEC
[latex]\triangle[/latex] ABC – [latex]\triangle[/latex] ADC = [latex]\triangle[/latex] ABC – [latex]\triangle[/latex] ABE
[latex]\triangle[/latex] ADC = [latex]\triangle[/latex]ABE = 36 sq cm
6. The perpendiculars, drawn from the vertices to the opposite sides of a triangle, meet at the point whose name is

Answer - Option D
Explanation -

O = Ortho - centre
7. The ratio of length of each equal side and the third side of an isosceles triangle is 3 : 4. If the area of the triangle is [latex]18 \sqrt{5}[/latex] square unit, the third side is

Answer - Option D
Explanation -
Sides = 3x, 3x and 4x
Semi perimeter = [latex]\frac{3x + 3x + 4x}{2} = 5x[/latex]
area of triangle = [latex]\sqrt{5x(5x - 3x)(5x - 3x)(5x - 4x)}[/latex]
[latex]\sqrt{5x * 2x * 2x * x}[/latex]
[latex]20{x}^{4} = 18 \sqrt{5} * 18 \sqrt{5}[/latex]
[latex]20{x}^{4} = 81[/latex]
[latex]{x}^{2} = 9[/latex]
[latex]x = \sqrt {5} = 3[/latex]
Third side = 4x = 4 * 3 = 12 units
8. The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ?PSR is

Answer - Option A
Explanation -
Side PR= QR= RS
Hence in triangle PRS ; PR = RS ; so ?P = ?S= x
PRS = 180 PRQ = 180 - 60 – 120
Now in triangle PRS = PRS + ? P + ?S= 180
120 + x + x = 180
x = 30° = ?PSR
9. ABC is an isosceles triangle With AB = AC. The side BA is produced to D such that AB = AD If ?ABC = 30 °, then BCD is equal to

Answer - Option B
Explanation -
AB = AC = AD
Angle ABC = Angle ACB = 30°
Angle BAC = 180 – 60 = 120°
Angle DAC = 180 – 120 = 60°
Angle ADC + Angle ACD = 120
Angle ACD = 120/2 = 60°
Angle BCD = Angle ACB + Angle ACD = 30 + 60 = 90°
10. The in-radius of an equilateral triangle is of length 3 cm. Then the length of each of its medians is

Answer - Option D
Explanation -

Since, in equilateral triangle in-centre, centroid, circumference, and ortho-centre lie on the same point.
Therefore, centroid divides the median in the ratio 2:1
i.e, Length of each median = (2 + 1) × in-radius
= 3 × 3 = 9 cm
Thus, length of each of its medians is 9 cm.
hence, option D is correct.

RRB JE 2019 Quantitative Aptitude Quiz - Set 28

1. The diagonal of a square is [latex]10 \sqrt{2}[/latex]cm. Find its perimeter?

Answer - Option D
Explanation -
Diagonal of a square = a [latex]\sqrt{2}[/latex] = 10 [latex]\sqrt{2}[/latex] cm
a = 10 cm
Perimeter of the square = 4a
= 4 × 10
= 40 cm
2. The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, then the distance between the longer sides is

Answer - Option D
Explanation -

We have,
Area of parallelogram = base × height
27 * 12 = 9 cm
x = [latex]\frac{27 * 12}{36} = 9 cm[/latex]
Hence, option D is correct.
3. The ratio of the areas of the circum circle and the in circle of a square is?

Answer - Option A
Explanation -

Let ABCD be a square with sides ‘2a’.
AB = BC = CD = DA = 2a
AC be the diagonal of square, hence
[latex]AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{4{a}^{2} + 4{a}^{2}} = 2 \sqrt{2}a[/latex]
AO = OC = [latex]\sqrt{2} a[/latex]
AO is the radius of circum Circles of square and OP is the radius of in Circles of same square.
OP = [latex] \frac{1}{2}[/latex](AB) = a
Hence area of circum Circles = [latex]\pi (2{a}^{2}) = 2\pi {a}^{2}[/latex]
Area of in Circles = [latex]\pi ({a}^{2}) = \pi {a}^{2}[/latex]
Hence required ratio (ratio of the areas of the circum Circles and the in Circles of a square) = [latex]2 \pi {a}^{2} : \pi {a}^{2} = 2 : 1[/latex]
4. A quadrilateral is inscribed in a circle. If the opposite angles of the quadrilateral are equal and length of its adjacent sides are 6 cm and 8 cm, what is the area of the circle?

Answer - Option B
Explanation -
Length of diagonal of the quadrilateral = [latex]\sqrt{{8}^{2} + {6}^{2}}[/latex]
= 10 cm
i.e, Radius of the circle = [latex]\frac{10}{2}[/latex]
= 5 cm
i.e, Area of the circle =[latex]\pi {r}^{2} [/latex]
= 25[latex]\pi[/latex] sq. cm
5. A parallelogram ABCD has sides AB = 24 cm and AD = 16 cm. The distance between the sides AB and DC is 10 cm. Find the distance between the sides AD and BC.

Answer - Option C
Explanation -

Area of the parallelogram = Base × Height
= 24 * 10 = 240 sq. cm
If the required distance becm, then
240 = 16 * x
x = [latex]\frac{240}{16} = 15[/latex] cm.
6. ABCD is a rhombus. A straight line through C cuts AD produced at P and AB produced at Q. I DP = [latex]\frac{1}{2}[/latex] AB, then the ratio of the lengths of BQ and AB is

Answer - Option A
Explanation -

Angle P is common in triangle DPC and APQ. Apart from that DC || AB hence other two angles are also same. triangle DPC and APQ are similar.
Let AB = BC = CD = DA = 2 and PD = 1 to satisfy the question condition.
Now using similarity
[latex]\frac{PD}{PA} = \frac{DC}{AQ}[/latex]
[latex]\frac{1}{3} = \frac{2}{AQ}[/latex]
AQ = 6
BQ = AQ - AB = 6 - 2 = 4
BQ : AB = 4 : 2 = 2 : 1
7. A quadrilateral ABCD circumscribes a circle and AB = 6 cm. CD = 5 cm and AD = 7 cm. The length of side BC is

Answer - Option A
Explanation -
Let ABCD is a quadrilateral in which a circle is inscribed
Given AB = 6cm; CD = 5cm

By the property of circle, sum of opposite sides are equal
AB + CD = BC + AD
6 + 5 = BC + 7
BC = 11 - 7 = 4cm
8. The area of a rhombus is 216 cm and the length of its one diagonal is 24 cm. The perimeter (in cm) of the rhombus is

Answer - Option B
Explanation -
Let ABCD be a rhombus, all sides of rhombus equal and its diagonal cuts each other at right angle. Let AC = 24cm

Area of rhombus = 1/2 (product of diagonals) = [latex]\frac{1}{2}[/latex] (AC × BD)
216 = [latex]\frac{1}{2}[/latex] (24 × BD)
24 × BD = 512
BD = 18cm
Now in right angled Triangles AOB
[latex]{AB}^{2} = {AO}^{2} + {BO}^{2}; AO[/latex]
= [latex]\frac{1}{2}[/latex](AC)
= 12cm; BO [latex]\frac{1}{2}[/latex] (BD)
= 9cm
[latex]{AB}^{2} = {12}^{2} + {9}^{2}[/latex]
= 144 + 81 = 225
AB = 15cm
Hence perimeter of rhombus = 4 × side
= 4 × 15 = 60cm
9. Two adjacent sides of a parallelogram are of lengths 15 cm and 18 cm. If the distance between two smaller sides is 12 cm, then the distance between two bigger sides is

Answer - Option B
Explanation -
Adjacent sides of parallelogram = 15 cm and 18 cm
Distance between shorter sides = 12 cm
Area of parallelogram = b × h
= 15 × 12 cm2 = 180 [latex]{cm}^{2}[/latex]

Again, area of parallelogram = b × h
180 = 18 × h
h = [latex]\frac{180}{18}[/latex]
h = 10 cm
Therefore, the distance between its longer side = 10 cm.
10. Q is a point in the interior of a rectangle ABCD. If QA = 3 cm, QB = 4 cm and QC = 5 cm, then the length of QD in centimetre is

Answer - Option A
Explanation -

Hint: Use pythagorous theore in all 4 small rectangles to derive this formula.
[latex]{QD}^{2} + {QB}^{2}= {QA}^{2} + {QC}^{2}[/latex]
[latex]{QD}^{2} + {16}^{2}= 9 + 15[/latex]
[latex]{QD}^{2} = 34 - 16 = 18[/latex]
[latex]{QD}^{2} = \sqrt{18} = 3 \sqrt{2}[/latex]cm

RRB JE 2019 Quantitative Aptitude Quiz - Set 29

1. What is the area of the sector whose central angle is 90 ° and radius of the circle is 14 cm?

Answer - Option C
Explanation -
Area of circle = [latex]\pi {r}^{2}[/latex]
As the central angle is 90 ° which is one fourth of 360 °. It means the sector is quadrant and area of
quadrant will be one fourth of the area of complete cirle
Hence required area = [latex]\frac{1}{4} * \pi {r}^{2} = \frac{1}{4} * \frac{22}{7} * 14 * 14 = 154 sq cm[/latex]
2. A gear 12 cm in diameter is turning a gear 18 cm in diameter. When the smaller gear has 42 revolutions, how many has the larger one made?

Answer - Option A
Explanation -
Smaller gear perimeter = 12 [latex]\pi[/latex]
Larger gear perimeter = 18 [latex]\pi[/latex]
No of revolutions of larger = [latex]\frac{44 * 12 * \pi}{18 \pi}[/latex]
3. Two circles touch each other externally at point A and PQ is a direct common tangent which touches the circles at P and Q respectively. Then ?PAQ =

Answer - Option B
Explanation -
Two circles with centres [latex]{O}_{1}[/latex] and [latex]{O}_{2}[/latex] touch each other externally at point A and PQ is a direct common tangent which touches the circles at P and Q respectively.
Let ?PAQ = [latex]\alphaº [/latex]
and [latex]?{O}_{1}AP = ?{O}_{1}PA = {\Thetaº}_{1}[/latex]
and [latex]?{O}_{2}AQ = ?{O}_{2}QA = {\Thetaº}_{2}[/latex]
then, in [latex]\triangle APQ -[/latex]
?APQ + ?AQP + ?PAQ = 180º
(90º - [latex]{\Thetaº}_{1}) + (90º - {\Thetaº}_{2}[/latex]) + [latex]\alphaº = 180[/latex]
[latex]\alphaº = {\Thetaº}_{1} + {\Thetaº}_{2}[/latex] ...(i)
The straight line [latex]{O}_{1}A?{O}_{2}[/latex] joining the centres of the circles.
[latex]{\Thetaº}_{1} + {\Thetaº}_{2} + \alphaº = 180º[/latex]
[latex]\alphaº + \alphaº = 180º[/latex]
?PAQ = [latex]\alphaº[/latex]
= [latex]\frac{1}{2} * 180º = 90º [/latex]
Hence, option D is correct.
4. A 7 m wide road runs outside around a circular park, whose circumference is 176m.The area of the road is: [use [latex]\pi \frac{22}{7}[/latex]]

Answer - Option A
Explanation -
If the radius of the circular park be metre, then
[latex]2 \pi r = 176[/latex]
[latex]2 * \frac{22}{7} * r = 176[/latex]
[latex]\frac{176 * 7}{2 * 22} = 28[/latex] metre
Radius of the park with road = 28 + 7 = 35 metre
Area of the road = [latex]\frac{22}{7} ({35}^{2} - {28}^{2}) {m}^{2}[/latex]
= [latex]\frac{22}{7} * 63 * 7 = 1386 {m}^{2}[/latex]
5. Two circles [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex] touch each other internally at P. Two lines PCA and PDB meet the circles [latex]{C}_{1}[/latex] in C, D and [latex]{C}_{2}[/latex] in A, B respectively. If [latex]\triangle[/latex] BDC = 120º, then the value of [latex]\triangle[/latex] ABP is equal to

Answer - Option A
Explanation -

[latex]\triangle PC{O}_{1} \simeq \triangle PA{O}_{2} and \triangle PD{O}_{1} \simeq \triangle PB{O}_{2} [/latex]
So [latex]/frac {PC}{PA} = \frac{r}{R}[/latex]
[latex]\frac{r}{R} = \frac{PD}{PB}[/latex]
So [latex]\frac{PC}{PA} = \frac{PD}{PB}[/latex]
Now

In [latex]\triangle[/latex] PCD and [latex]\triangle[/latex] PAB ;
P is common and PC/PA = PD/PB So [latex]\triangle[/latex] PCD = [latex]\triangle[/latex] PAB So CD II AB.
So, ABP + BDC =180 or,
ABP + 120 = 180 or, ABP = 60
6. Find the radius of the circle if the length of its arc is 33 cm and the corresponding central angle is 90 °?

Answer - Option A
Explanation -
We know that,
Angle = [latex]\frac{arc}{radius}[/latex]
Here, angle = 90 ° = [latex]\frac{\pi}{2}[/latex]
And arc = 33 cm
i.e, Radius = [latex]\frac{33}{\frac{\pi}{2}}[/latex]
= 21 cm
7. In circular measure, the value of the angle 11 °15’ is

Answer - Option A
Explanation -
[latex]11 + \frac{15}{60} = \frac{45}{4}[/latex]degrees
[latex]\frac{45}{4} \times \frac{\pi}{180} = \frac{\pi}{15} [/latex] radians
8. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then ?APB is

Answer - Option D
Explanation -

OA = OB = r
OP = 2r
sin APO = [latex]\frac {OA}{OP} = \frac {r}{2r} = \frac {1}{2}[/latex]
?APO = 30º
i.e, ?APB = 60º
9. In a circle of radius 21 cm, an arc subtends an angle of 72 ° at the centre. The length of the arc is

Answer - Option B
Explanation -
[latex]\theta = 72º = 72 \times \frac {\pi}{180} radians[/latex]
[latex]= \frac {2\pi}{5}[/latex] radians i.e, [latex]\theta = \frac {s}{r}[/latex]
s = [latex]= \theta r = \frac {2 \pi}{5} \times 21[/latex]
[latex]= \frac {2}{5} \times \frac {22}{7} \times 21[/latex]
[latex]= \frac {132}{5} = 26.4 cm[/latex]
10. If the four equal circles of radius 3 cm touch each other externally, then the area of the region bounded by the four circles is

Answer - Option B
Explanation -

Area of squard abed = 2r [latex]\times[/latex]
= 4 [latex]{r}^{2}[/latex]
= 4 [latex]\times[/latex] 9 = 36 sq.cm
Area of shaded part = Area of square abed - Area of the circle
[latex]{(2r)}^{2} - \pi {r}^{2}[/latex]
= [latex](4 - \pi) {r}^{2}[/latex]
= [latex]9(4 - \pi)[/latex]sq cm (r = 3 cm, given)

RRB JE 2019 Quantitative Aptitude Quiz - Set 30

1. The cross section of a canal is in the shape of an isosceles trapezium which is 2 m wide at the bottom and 3 m wide at the top. If the depth of the canal is 1 m and it is 100 m long, what is the maximum capacity of this canal?

Answer - Option B
Explanation -
Area of isosceles trapezium = [latex]\frac{1}{2} (a + b) \times h[/latex]
Here a = 2, b = 3 and h = 1
Area trapezium = [latex]\frac{1}{2} (2 + 3) \times 1 = 2.5 sqm[/latex]
Required volume = area × length of canal
[latex]\Rightarrow 2.5 \times 100 = 250 [/latex]cubic meter
2. Volume of a cylinder is 770 cubic cm. If circumference of its base is 22 cm, what will be the curved surface area of the cylinder? (Take p = 22/7)

Answer - Option A
Explanation -
Let the radius of base = r and height of cylinder = h
Circumference of base = 2p r = 22cm, here r is the radius of base
[latex]2 \times \frac{22}{7} \times r = 22[/latex]
r = 3.5 cm
volume of cylinder = [latex]\pi \frac{r}{2} h[/latex]
[latex]\Rightarrow 770 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h[/latex]
[latex]\Rightarrow[/latex] h = 20cm
curved surface area of cylinder = 2prh
CSA = [latex]2 \times \frac{22}{7} \times \frac{7}{2} \times 20 = 440 sqcm[/latex]
3. The total surface area of a hemisphere is 1848 sq cm, what is its radius? (Take n = 22/7)

Answer - Option D
Explanation -
Total surface area of a hemisphere = [latex]3 \pi {r}^{2}[/latex]
[latex]\Rightarrow \times \frac{22}{7} \times {r}^{2}[/latex]
[latex]\Rightarrow[/latex] r = 14 cms
4. A pentagonal prism has 15 edges. How many vertices does it have?

Answer - Option B
Explanation -
In geometry, the pentagonal prism is a prism with a pentagonal base. It is a type of heptahedron.

According to Euler's formula:
F + V - E = 2
F = Number of faces (7)
E = Edges (15 given)
V = Vertices
F + V - E = 2
7 + V - 15 = 2
V = 2 + 15 - 7 = 10
Thus, B is correct.
5. Sum of lengths of all edges of a cube is 84 cm, find its volume?

Answer - Option B
Explanation -
Given: sum of all edges of the cube, 12 a = 84 cm
[latex]\Rightarrow[/latex] a = 7 cm
Then, its volume = [latex]{a}^{3}[/latex]
= [latex]{7}^{3}[/latex]
= 343 cubic cm
6. The total surface area of a hemisphere is 41.58 sq cm. Find its curved surface area.

Answer - Option A
Explanation -
Total surface area = 3 × p × [latex]{r}^{2}[/latex]
Curved surface area = 2 × p × [latex]{r}^{2} = 41.58 × (\frac{2}{3}) = 27.72 {cm}^{2}[/latex]
7. A cube and a sphere have equal surface areas. The ratio of their volumes will be _____.

Answer - Option B
Explanation -
Let the length of the side of the cube to be 'l' units & radius of the sphere to be 'r' units.
Then,
Surface area of the cube = 6 [latex]{I}^{2}[/latex]
Surface area of sphere = 4p [latex]{R}^{2}[/latex]
[latex]\Rightarrow \frac{{I}^{2}}{{R}^{2}} = \frac{4 \pi}{6} [/latex]
i.e, [latex]\frac{Volume of the cube}{Volume of the sphere}[/latex]
= [latex]\frac{{I}^{3}}{\frac{4}{3}} \pi {r}^{3}[/latex]
= [latex]\frac{3}{4 \pi} \times \frac{8 \pi \times \sqrt {\pi}}{6 \times \sqrt {\pi}}[/latex]
= [latex]\sqrt {\pi} : \sqrt {6}[/latex]
8. If the radii of the circular ends of a truncated conical bucket which is 45 cm high be 28 cm and 7 cm, then the capacity of the bucket in cubic centimeter is

Answer - Option A
Explanation -
Volume of bucket = [latex]\frac{1}{3} \pi h ({{r}_{1}}^{2} + {{r}_{2}}^{2} + {r}_{1} {r}_{2})[/latex]
= [latex]\frac{1}{3} \times \frac{22}{7} \times 45 ({28}^{2} + {7}^{2} + 28 \times 7)[/latex]
= [latex]\frac{1}{3} \times \frac{22}{7} \times 45 \times 1029[/latex]
= 4810 cu. cm
9. A circus tent is cylindrical up to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical part is 63 m then the total area of the canvas required to make the tent is how much?(Take p = 22/7)

Answer - Option A
Explanation -
Total area of the canvas = 2prh + prl
= p r (2h + l)
= [latex]\frac{22}{7} \times \frac{105}{2} (2 \times 3 + 63)[/latex]
= [latex]\frac{22}{7} \times \frac{105}{2} \times 69[/latex]
= [latex]11 \times 15 \times 69 = 11385 sq.m[/latex]
10. If the ratio of the diameters of two right circular cones of equal height be 3 : 4, then the ratio of their volumes will be

Answer - Option B
Explanation -
Ratio in the diameter of two right circular cone = 3 : 4
Hence ratio in radius is same as ratio in diameter = 3 : 4
Hence radius of first cone = [latex]\frac{3X}{7}[/latex]
Radius of second cone = [latex]\frac{4X}{7}[/latex]
They have equal height h (say)
Volume of cone = [latex]\frac{1}{3} \pi {r}^{2} h[/latex]
Now ratio in their volume
= [latex]\frac{1}{3} \pi {(\frac {3X}{7})}^{2} h : \frac{1}{3} \pi {(\frac {4X}{7})}^{2} h[/latex]
= [latex]{3}^{2} : {4}^{2} \Rightarrow 9 : 16[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 31

1. Let O be the in-centre of a triangle ABC and D be a point on the side BC of [latex]\triangle[/latex] ABC, such that OD ? BC. If ?BOD = 15 °, then ?ABC =

Answer Option C
Explanation :

BO is the internal bisector of ?B. ?ODB = 90°; ?BOD = 15°, ?OBD = 180° - 90° - 15° = 75°
[latex]\triangle[/latex] ABC = 2 * 75° = 150°
2. D is any point on side AC of ABC. IF P, Q, X, Y are the midpoints of AB, BC, AD and DC respectively, then the ratio of PX and QY is

Answer Option B
Explanation :

PX [latex]\parallel[/latex] BD and PX = [latex]\frac{1}{2}[/latex] BD
QY [latex]\parallel[/latex] BD and QY = [latex]\frac{1}{2}[/latex] BD
i.e, PX : QY = 1:1
3. The sides of a triangle are in the ratio [latex]\frac{1}{3}[/latex] : [latex]\frac{1}{4}[/latex] : [latex]\frac{1}{5}[/latex] and its perimeter is 94 cm. The length of the smallest side of the triangle is:

Answer Option B
Explanation :
[latex]\frac{1}{3}[/latex] : [latex]\frac{1}{4}[/latex] : [latex]\frac{1}{5}[/latex] = [latex]\frac{1}{3} \times 60[/latex] : [latex]\frac{1}{4} \times 60[/latex] : [latex]\frac{1}{5} \times 60[/latex]
20 : 15: 12
i.e, 20x + 15 + 12x = 94
47x = 94
= [latex]\frac{94}{47} = 2[/latex]
i.e, The smallest side = 12x
= 12 × 2 = 24 cm
4. If in a [latex]\triangle[/latex]ABC, the medians CD and BE intersect each other at O, then the ratio of the areas of ?ODE and ?ABC is

Answer Option C
5. If the length of the side of an equilateral triangle is 8 cm, what is its area?

Answer Option C
Explanation :
Given the side of the equilateral triangle, a = 8 cm
[latex]\Rightarrow [/latex] Area = [latex](\frac{\sqrt{3}}{4}) {a}^{2}[/latex]
[latex]\Rightarrow [/latex] Area = [latex](\frac{\sqrt{3}}{4}) {8}^{2} = 16 \sqrt{3} sq cm[/latex]
6. ABC is an isosceles right angled triangle with ?B = 90 °. On the sides AC and AB, two equilateral triangles ACD and ABE have been constructed. The ratio of areas of ?ABE and ?ACD is

Answer Option C
Explanation :

AB = x units
BC = x units
AC = [latex]\sqrt{2}[/latex] x units
[latex]\frac{\Delta ABE}{\Delta ACD}[/latex] = [latex]\frac{\sqrt{3} \frac{{x}^{2}}{4}}{\sqrt{3} \frac{({\sqrt{3}2)}^{2}}{4}}[/latex] = [latex]\frac {1}{2}[/latex]
7. If a, b and c are the sides of a triangle and [latex]{a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca[/latex] then the triangle is

Answer Option C
Explanation :
[latex]{a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca[/latex]
[latex]2{a}^{2} + 2{b}^{2} + 2{c}^{2} = 2ab + 2bc + 2ca[/latex]
[latex]{a}^{2} - 2ab + {b}^{2} + {b}^{2} - 2bc + {c}^{2} + {c}^{2} - 2ca + {a}^{2} = 0[/latex]
[latex]{a -b}^{2} + {b -c}^{2} + {c -a}^{2} = 0[/latex]
a - b = 0; a = b
b - c = 0; b = c
c - a = 0; c = a
i.e, a = b= c
8. ABCD is a trapezium in which AB || DC and AB = 2 CD. The diagonals AC and BD meet at O. the ratio of areas of triangles AOB and COD is

Answer Option C
Explanation :
AB || DC (AC is a transversal)
Angle DCA = Angle CAB(alternate interior angles)
In Triangle AOB and Triangle COD
Angle OAB= Angle OCD ( Proved)
Angle AOB= Angle COD (Vertically opposite angles)
By AA of similarity, Triangle AOB ~ Triangle COD
Now using Area theorem of similar triangle
[latex]\frac{Area(AOB)}{Area(COD)} = \frac{{AB}^{2}}{{CD}^{2}} = \frac{{(2CD)}^{2}}{{CD}^{2}}[/latex] (AB 2CD)
[latex]\rightarrow \frac{Area(AOB)}{Area(COD)} = \frac{4}{1} = \frac{4}{1}[/latex]
Hence option C is correct
9. AD is the Median of a triangle ABC and O is the centroid such that AO = 10 cm. Length of OD (in cm) is

Answer Option C
Explanation :
AD is median of the triangle.
OA : OD = 2 : 1
So, OD = [latex]\frac{10}{2}[/latex] = 5cm
10. The points D and E are taken on the sides AB and AC of [latex]\triangle ABC [/latex] such that AD = [latex]\frac{1}{3}[/latex] AB, AE = [latex]\frac{1}{3}[/latex] AC. If the length of BC is 15 cm, then the length of DE is:

Answer Option D
Explanation :
Let ABC be a Triangles, D and E are taken on the sides AB and AC
Given: AD = [latex]\frac{1}{3}[/latex] AB, AE = [latex]\frac{1}{3}[/latex] AC
BC = 15cm

In [latex]\triangle[/latex] ABC and [latex]\triangle[/latex] ADE
[latex]\triangle[/latex] BAC= [latex]\triangle[/latex] DAE
[latex]\frac{AD}{AB} = \frac{1}{3} ; \frac{AE}{AC} = \frac{1}{3}[/latex]
Hence these two triangles are similar.
Hence;
[latex]\frac{AD}{AB} \frac{AE}{AC} = \frac{DE}{BC} = \frac{1}{3}[/latex]
Hence DE = [latex] \frac{1}{3}[/latex] (BC) = [latex] \frac{15}{3}[/latex] = 5

RRB JE 2019 Quantitative Aptitude Quiz - Set 32

1. The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 9 cm. Determine the corresponding side of the second triangle.

Answer - Option B
Explanation -
If the required side be x cm, then
[latex]\frac {30}{20} =\frac {9}{x}[/latex]
3 = 9 * 2
x = [latex]\frac {9 * 2}{3}[/latex] = 6 cm
2.The area of similar triangles ABC and DEF are 20 [latex]{cm}^{2}[/latex] and 45 [latex]{cm}^{2}[/latex] respectively. If AB = 5 cm, then DE is equal to :

Answer - Option B
Explanation -
[latex]\frac {\triangle ABC}{\triangle DEF} =\frac {{AB}^{2}}{{DE}^{2}}[/latex]
[latex]\frac {20}{45} =\frac {25}{{DE}^{2}}[/latex]
[latex]{DE}^{2} = \frac {45 * 25}{20} = \frac {225}{4}[/latex]
i.e, DE = [latex] \frac {15}{2} = 7.5 cm[/latex]
3. If area of an equilateral triangle is a and height is b then the value of [latex] \frac {{b}^{2}}{a} [/latex] will be how much?

Answer - Option C
Explanation -

AD = b
BD = DC = x
tan 60 = [latex] \frac {AD}{BD}[/latex]
[latex] \sqrt {3} = \frac {b}{x}[/latex]
x = [latex] \frac {b}{\sqrt {3}}[/latex]
i.e, BC = 2x = [latex] \frac {2b}{\sqrt {3}}[/latex]
i.e, Area of the triangle
[latex] \frac {1}{2} * BC * AD[/latex]
a = [latex] \frac {1}{2} * \frac {2b}{\sqrt {3}} * b[/latex]
[latex]\frac {{b}^{2}}{a} = \sqrt {3}[/latex]
4. If the orthocenter and the centroid of a triangle are the same, then the triangle is:

Answer - Option C
Explanation -

Here, AB = BC = CA
So, the concerned triangle is equilateral triangle.
5. What is the area of a largest triangle that can be inscribed in a circle of radius r?

Answer - Option C
Explanation -
The only possible triangle with maximum area is equilateral triangle. Since all the centers in an equilateral triangle coincide.
So, O is circumcenter as well as centroid. Now, AP is a medium and O is centroid. Therefore,
[latex]\frac{AO}{OP} = \frac{2}{1}[/latex]
[latex]\frac{r}{OP} = \frac{2}{1}[/latex]
[latex]OP = \frac{r}{2}[/latex]
BC = 2BP = 2 [latex]\sqrt{{BO}^{2} - {OP}^{2}}[/latex]
= 2 [latex]\sqrt {{r}^{2} - {(\frac{r}{2})}^{2}} = \sqrt {3}r[/latex]
Area of [latex]\triangle[/latex] ABC
[latex] = \frac{ \sqrt{3}}{4} {BC}^{2} = \frac{ \sqrt{3}}{4} {(\sqrt{3 r})}^{2} = \frac{3 \sqrt{3}}{4} {r}^{2}[/latex]
6. [latex]\triangle[/latex] ABC is a right triangle. [latex]\angle ACB = 90°, CD \perp AB, BC = a, AC = B[/latex] and CD = P, then:

Answer - Option B

RRB JE 2019 Quantitative Aptitude Quiz - Set 33

1. The length of aside of a rhombus is 10 m and one of its diagonal is 12 m. The length of the other diagonal is

Answer - Option C
Explanation -
Let one diagonal = x ; and second diagonal = y
Side of rhombus= 10
[latex]{x}^{2} + {y}^{2} = {4a}^{2} [/latex]
[latex]{12}^{2} + {y}^{2} = 4{10}^{2} [/latex]
[latex]{y}^{2} [/latex] = 256
y = 16 m
2. The length and breadth of a rectangle are increased by 20% and 25% respectively. The increase in the area of the resulting rectangle will be:

Answer - Option B
Explanation -
Effect on area of rectangle = (20 + 25 + [latex]\frac{20 * 25}{100} [/latex] )% = 50%
3. AB is a chord to a circle and PAT is the tangent to the circle at A. If BAT = 75o and BAC = 45o, C being a point on the circle, then ABC is equal to

Answer - Option C
Explanation -

If a line touches a circle and from the point of contact a chord is drawn, the angles which this chord makes with given line are equal respectively to the angles formed in the corresponding alternate segments.
i.e, ACB = BAT = 75°
ABC = 180° - 45° - 75° = 60°
4. If a chord of a circle of radius 5 cm is a tangent to a circle of radius 3 cm, both the circles being concentric, then the length of the chord is

Answer - Option C
Explanation -
Let [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex] are the circles of radius 5 cm and 4cm respectively with centre O.

PQ be a chord of circle [latex]{C}_{1}[/latex], PQ is tangent at the point R of circle [latex]{C}_{2}[/latex].
As we know that tangent makes an angle of 90° with the radius of respective circle.
Hence PQ makes 90° with OR.
Therefore Triangles ORQ is a right angle Triangles.
Hence by Pythagoras theorem:
[latex]{RQ}^{2} = {OP}^{2} - {OR}^{2} [/latex]
= 25 – 9 = 16
RQ = 4cm
Hence PQ = 2RQ = 2 × 4 = 8cm
5. The length of two chords AB and AC of a circles are 8 cm and 6 cm and ?BAC = 90°, then the radius of circle is

Answer - Option D
Explanation -
A circle with centre O has two chords AB = 8cm and Ac = 6cm

Given: = 900
Therefore Triangles ABC is right angle Triangles.
By Pythagoras theorem
[latex]{BC}^{2} = {AB}^{2} - {AC}^{2} [/latex]
[latex]{BC}^{2}[/latex] = 64 + 36
[latex]{BC}^{2}[/latex] =100
BC = 10 cm
BC is the diameter of circle, hence radius of circle = [latex]\frac{diameter}{2}[/latex]
Radius = [latex]\frac{10}{2}[/latex] = 5cm
6. The radius of two concentric circles are 9 cm and 15 cm. If the chord of the greater circle be a tangent to the smaller circle, then the length of that chord is

Answer - Option A
Explanation -

Let [latex]{C}_{1} and {C}_{2}[/latex]are the circles of radius 15 cm and 9 cm respectively with centre O.
PQ be a chord of circle [latex]{C}_{1}[/latex]. PQ is tangent at the point R of circle [latex]{C}_{2}[/latex].
We know that tangent makes an angle of 90° with the radius of respective circle. Hence, PQ makes 90° with OR.
Therefore triangle ORQ is a right angle triangle.
Hence, by Pythagoras theorem:
[latex]{RQ}^{2} = {OQ}^{2} - {OR}^{2} [/latex]
= 225 – 81 = 144
RQ = 12cm
Hence, PQ = 2 RQ = 2 × 12 = 24 cm
7. A chord 12 cm long is drawn in a circle of diameter 20 cm. The distance of the chord from the centre is

Answer - Option A
Explanation -
OA = radius= 10 cm
CE= RD = 6 cm
i.e, OE = [latex]\sqrt{{10}^{2} - {6}^{2}}[/latex]
= [latex]\sqrt{100 - 36}[/latex]
= [latex]\sqrt{64}[/latex] = 8 cm
8. In what ratio is the segment joining (-1,-12) and (3,4) divided by the x-axis?

Answer - Option C
Explanation -

Now, Using section formula
[latex]\frac{3m - n} {m + n} = x[/latex]
[latex]\frac{4m - 12n} {m + n} = 0[/latex]
4m - 12n =0
4m = 12 n
[latex]\frac{m} {n} = \frac{3} {1}[/latex]
Therefore line is divided in the ratio 3 : 1
9. The graph of the equation intersects the x-axis at the point

Answer - Option B
Explanation -
When a straight line cuts x-axis, the coordinates of point of intersection = (x, 0),
i.e. y = 0
Putting y = 0 in 4x – 5y = 20
4x = 20 x = 5
Point of intersection = (5, 0)
Note: Putting y = 0 in 4x – 5y = 20, point of intersection on x-axis = (5, 0)
Putting x = 0 in 4x – 5y = 20, point of intersection on y-axis = (0, -4)
Look at the graph of the equation:

10. Chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PD = 5 cm, then the length of PB is

Answer - Option B
Explanation -

AB = 6 cm; CD = 3cm
PD = 5 cm; PB= ?
PA * PB = PC * PD
(PB - 6)PB = 2 * 5
[latex]{PB}^{2}[/latex] - 6PB = 2 * 5
PB = [latex]\frac {6 ± \sqrt {36 + 40}}{2}[/latex]
= [latex]\frac {6 ± \sqrt {76}}{2}[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 34

1. In the adjoining figure, sides AB and AC of a [latex]\triangle[/latex]ABC are produced to P and Q respectively. The bisectors of PBC and QCB intersect at O. then BOC is equal to:

Answer - Option A
Explanation -
1 = 90° - [latex]\frac{1}{2} ?3[/latex]
2 = 90° - [latex]\frac{1}{2} ?4[/latex]
Now in BOC
1 + 2 + BOC = 180°
BOC = 180° - (2 + 1)
= 180° - [90° - [latex]\frac{1}{2} 4 + 90° - \frac{1}{2} 3 [/latex]]
Or, BOC = [latex]\frac{1}{2}(180° - A)[/latex]
So, A + 3 = 180°
BOC = 90° - [latex]\frac{1}{2} A[/latex]
2. Two circle of radii 5 cm and 3 cm intersects at two points and the distance between their centres is 4 cm. find the length of common line segment.

Answer - Option B
Explanation -
Given [latex]{C}_{1}{C}_{2}[/latex] = 4 cm
[latex]{C}_{1}P[/latex] = 3 cm (Radius)
= 5 cm (Radius)
Figure according to the question.

Common line segment PQ = 1 + 3 = 4cm
Difficult level : Esay
3, Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. find the distance between their centres.

Answer - Option C
Explanation -

Let O and [latex]{O}^{1}[/latex] be the centre of the circle of radii 10 cm and 8 cm and PQ = 12c m.
PL = [latex]\frac{1}{2} PQ = 6 cm[/latex]
In right triangle OLP, we have
[latex]{OP}^{2} = {OL}^{2} + {OP}^{1}{LP}^{2}[/latex]
OL = [latex]\sqrt{{10}^{2} - {6}^{2}} = \sqrt{64} = 8 cm[/latex] and in right triangle , we have
[latex]{O}^{1}{L}^{2} + {LP}^{2} = {O}^{1}{P}^{2}[/latex]
[latex]{O}^{1}L = \sqrt{{8}^{2} - {6}^{2}} = \sqrt{28} = 5 .29 cm[/latex]
[latex]O{O}^{1} = OL + LQ = (8 + 5.29) = 13.29 cm[/latex]
4. In the adjoining figure BAC = 60° and BC = a, AC = b and AB = c, then

Answer - Option B
Explanation -
cos A = [latex]\frac{{b}^{2} + {c}^{2} - {a}^{2}}{2bc} [/latex]
[latex]\frac{1}{2} = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2bc}[/latex]
[latex]{a}^{2} = {b}^{2} + {c}^{2} - bc[/latex]
5. An equilateral triangle of side 6 cm has its corners cut off to form a regular hexagon. Area (in [latex]{cm}^{2}[/latex]) of this regular hexagon will be

Answer - Option A
Explanation -
Side of the regular hexagon = [latex]\frac{(1*6)}{3}[/latex] = 2
Area of the hexagon = [latex]\frac{(3\sqrt{3} * {a}^{2})}{2}[/latex]

[latex]\frac{3 \sqrt{3}}{2} * 2* 2 = 6 \sqrt{3}[/latex]
6, The length of a chord of a circle is equal to the radius of the circle. The angle which this chord subtends in the major segment of the circle is equal to

Answer - Option A
Explanation -

AO = OB = AB
i.e, AOB = 600
i.e, ACB = 300
7. In [latex]\triangle[/latex] PQR, right-angled at Q, PR= 3 [latex]\sqrt{5}[/latex]cm. If the difference between the lengths of the perpendicular sides is 3cm, then find the value of sinP + sinR?

Answer - Option B
Explanation -

Let lengths of perpendicular sides be ‘x’ and ‘x+3’. So using Pythagoras Theorem,
we get [latex]{x}^{2} + {x + 3}^{2} = {(3\sqrt{5})}^{2}[/latex]. So, ‘x’ = 3cm. So,
sin P + sin R = [latex]\frac{QR}{PR} + \frac{PQ}{PR} +\frac{3}{3 \sqrt{5}} + \frac{6}{3 \sqrt{5}} + \frac{3}{ \sqrt{5}}[/latex]
8. At the centers two circles, two Arcs of equal length subtend angles of 60° and 75° respectively, the ratio of the radii of the two circles is –

Answer - Option B
Explanation -
Radius of the first circle is and second circle is
[latex]{q}_{1} = 60 and {q}_{2} = 75[/latex](given)
[latex]{arc}_{1} = 2\pi{r}_{1} * \frac{{Q}_{1}}{360} = 2\pi{r}_{1} * \frac {60}{360}[/latex]
[latex]{arc}_{2} = 2\pi{r}_{2} * \frac{{Q}_{2}}{360} = 2\pi{r}_{2} * \frac {75}{360}[/latex]
[latex]{arc}_{1} = {arc}_{2}[/latex]
[latex] 2\pi{r}_{2} * \frac {60}{360}\frac{{Q}_{2}}{360} = 2\pi{r}_{2} * \frac {75}{360}[/latex]
[latex] \frac {{r}_{1}}{{r}_{2}} = \frac {5}{4}[/latex]
= 5 : 4
9. The radius of two concentric circles are 9 cm and 15 cm. If the chord of the greater circle be a tangent to the smaller circle, then the length of that chord is

Answer - Option A
Explanation -

BO = OC = 15 cm. OD = 9 cm. BD = [latex] \sqrt{{15}^{2} - {9}^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12[/latex] cm.
BC = 2 × 12 = 24 cm.
10. In the adjoining figure a parallelogram ABCD is shown. AB = 24 cm and AO = BO = 13 cm. Find BC :

Answer - Option B
Explanation -
AO = BO = 13 cm

[latex]\rightarrow[/latex] AC = BD = 26 CM
Now , since the diagonals are equal, it means the given figure is actually a rectangle.
[latex]{BC}^{2} = {AC}^{2} - {AB}^{2}[/latex]
[latex]{BC}^{2} = {26}^{2} - {24}^{2}[/latex]
[latex]{BC}^{2}[/latex] = 100
[latex]\rightarrow[/latex] BC = 10 cm

RRB JE 2019 Quantitative Aptitude Quiz - Set 35

1. Chord PQ is the perpendicular bisector of radius OA of circle with center O (A is a point on the edge of the circle). If the length of Arc PAQ =[latex]\frac {2\pi}{3}[/latex]. what is the length of chord PQ?

Answer - Option B
Explanation -

PQ is perpendicular bisector of OA.
OP = OQ = PA = AQ
OPAQ is a rhombus.
2 PAQ = Reflex POQ
(The angle subtended at the centre by an arc is twice to that at the circumference)
2 PAQ = 360° - POQ
3 PAQ = 360°
(? PAQ = ? POQ)
? PAQ = 120° = ? POQ = [latex]\frac {\pi 2}{3}[/latex]
Again redius (r)= [latex]\frac {t}{0} = \frac {\frac {\pi 2}{3}}{\frac {\pi 2}{3}} = 1[/latex]
From ? OPB
OP = 1 unit
? POB = 60°
? sin 60° = [latex]\frac {PB}{OP}[/latex]
? PB = [latex]\frac {\sqrt{3}}{2}[/latex]
? PQ = 2 × = [latex]\sqrt{3}[/latex] unit
2. What can be the maximum number of common tangent which can be drawn to two non-intersecting circles?

Answer - Option B
Explanation -

4 tangents.
3. There is a circular garden of radius 21 metres. A path of width 3.5 metres is constructed just outside the garden. What is the area (in [latex]{metres}^{2}[/latex]) of the path?

Answer - Option A
Explanation -

Area of path AB = p[latex]{R}^{2}[/latex] - p[latex]{r}^{2}[/latex]
= p([latex]{24.5}^{2} - {21}^{2}[/latex])
= p(600.25-441)
= [latex]\frac{22}{7}[/latex]*159.25 = 500.5[latex]{metres}^{2}[/latex]
4. Two circles touch each other at point X. A common tangent touch them at two distinct points Y and Z. If another tangent passing through X cut YZ at A and XA= 16 cm, then what is the value (in cm) of YZ?

Answer - Option D
Explanation -

We know that XA = YA = ZA
Required
YZ = AY + AZ = 2 × XA = 2 × 16 = 32
5. The sum of radii of the two circles is 91 cm and the difference between their area is 2002 [latex]{cm}^{2}[/latex]. What is the radius (in cm) of the larger circle?

Answer - Option D
Explanation -
[latex]{R}_{1} + {R}_{2}[/latex]= 91 ………(i)
[latex]\pi{{R}_{1}}^{2} - \pi{{R}_{2}}^{2}[/latex] = 2002
[latex]{{R}_{1}}^{2} - {{R}_{2}}^{2}[/latex] = 2002 * [latex]\frac {7}{22}[/latex]
[latex]({R}_{1} + {R}_{2}) ({R}_{1} - {R}_{2})[/latex]=91×7
[latex]({R}_{1} - {R}_{2})[/latex] = 7 …(ii)
Solve eqn. (i) & (ii)
[latex]{R}_{1}[/latex]= 49 (Larger Circle)
[latex]{R}_{2}[/latex]= 42 (Smaller Circle) .
6. Two circle of same radius 5 cm intersect each other at A and B. If AB = 8 cm, then distance between their centers is:

Answer - Option A
Explanation -
AB = 8 cm, OA = AP = 5 cm. OP and AB are perpendicular bisector of each other.
OP = 2 * OC = [latex]\sqrt{{OA}^{2} - {AC}^{2}} = \sqrt{25 - 16} = 6 cm[/latex]
7. In the given figure, ?ONY = 50° and ?OMY = 15° Then the value of ?MON is

Answer - Option D
Explanation -
Since OM = OY = ON = radius of circle.
?OMY = ?OMY = 15°
?ONY = ?OYN = 50°
?MON = 180° - 2 * 15° = 150°
?NOY = 180° - 2 * 50° = 80°
?MON = ?MOY - ?NOY = 150° - 80° = 70°
8. A, B and C are the three points on a circle such that the angles subtended by the chords AB and AC at the center O are 90 ° and 110 ° respectively. ?BAC is equal to

Answer - Option B
Explanation -

We know that,
?BOA + ?AOC + ?BOC = 360
90° + 110° + ?BOC = 360°
?BOC = 360° - 200° = 160°
Note: The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.
?BAC = [latex]\frac{1}{2}[/latex] ?BOC = [latex]\frac{1}{2}[/latex] * 160° = 80°
9. In the given figure ?AOC = 120° Find the ?CBE where O is the center:

Answer - Option C
Explanation -
?CBA = [latex]\frac{1}{2}[/latex] ?COA = [latex]\frac{1}{2}[/latex] * 120 = 60°
?CBE = 180° - 60° = 120°
10. ABC is a right angled at C. What is the in radius of the in circle shown in the figure?

Answer - Option B
Explanation -
In radius of the right-angled triangle ABC = [latex]\frac{AC + BC - AB}{2} = \frac{9 + 40 - 41}{2} = 4cm[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 36

1. At least two pairs of consecutive angles are congruent in a _________.

Answer - Option B
Explanation -
At least two pairs of consecutive angles are congruent in an isosceles trapezium
2. A solid has 12 vertices and 30 edges. How many faces does it have?

Answer - Option D
Explanation -
We know that F + V - E = 2
Here F = number of faces, V = number of vertices and E = number of edges
F + 12 - 30 = 2
F = 2 + 18 = 20
So required number of faces = 20
3. If the sum of the measures of all the inter for angles of polygon is 1440° , find the number of sides of the polygon.

Answer - Option C
Explanation -
Sum of interior angles of a polygon = (N - 2) × 180 °
[latex]\rightarrow[/latex] 1440 ° = (N - 2) × 180 °
[latex]\rightarrow[/latex] [latex]\frac {1440 °}{180 °} [/latex] = (N - 2)
[latex]\rightarrow[/latex] 8 + 2 = N
[latex]\rightarrow[/latex] N = 10
Number of sides of the polygon = 10
4. Dodecahedron has 12 faces and 20 vertices . How many edges does it have?

Answer - Option A
Explanation -
According to Euler's formula
V + F = E +2
where,
V= number of vertices
E= Number of edges
F= number of faces
20 + 12 = E + 2
E = 30
5. The measure of each interior angle of a regular hexagon is how much?

Answer - Option C
Explanation -
We have each interior angle of a polygon = [latex]\frac {(n - 2)180°}{n}[/latex]
[latex]\frac {(6 - 2)180°}{6}[/latex](n = 6 of hexagon)
[latex]180° * \frac {4}{6} = 120°[/latex]
6. If the interior angle of a regular polygon is double the measure of exterior angle, then the number of slides of the polygon is

Answer - Option A
Explanation -
Sum of interior +exterior angles = 180°
interior angle = 2*exterior angle
3 * exterior angle = 180
exterior angle = 60
60 = [latex]\frac {360}{n}[/latex]
n= [latex]\frac {360}{60}[/latex]
n = 6
7. The sum of interior angles of a regular polygon is 1440°. The number of sides of the polygon is how much?

Answer - Option A
Explanation -
If the number of sides regular polygon be n, then
(2n – 4) × 90 = 1440
2n - 4 = [latex]\frac {1440}{90}[/latex] = 16
n =10
8. The area of an isosceles trapezium is 176 [latex]{cm}^{2}[/latex]cm2 and the height is [latex]\frac {2}{{11}^{th}}[/latex] of the sum of its parallel sides. If the ratio of the length of the parallel sides be 4 : 7, then the length of a diagonal is

Answer - Option D
Explanation -
Let ABCD be a trapezium
Given : the ratio of the length of the parallel sides be 4 : 7
Let length of parallel sides are 4x and 7x respectively.
height is [latex]\frac {2}{{11}^{th}}[/latex] of the sum of its parallel sides

hence height = [latex]\frac {2}{11}[/latex] * (11x) = 2x
area of trapezium = [latex]\frac {1}{2}[/latex] (sum of parallel sides)×height
area = [latex]\frac {1}{2}[/latex] (11x)×2x = 11x2
given: The area of an isosceles trapezium is 176 cm2
hence, 11x2 = 176 [latex]\rightarrrow[/latex] x2 = 16 [latex]\rightarrrow[/latex] x = 4cm
hence sides of trapezium = 16, 28 cm
height = 8cm
Diagonal of trapezium ABCD is AC.
QC = AB+[latex]\frac {(CD – AB)}{2}[/latex]
QC = 16 + [latex]\frac {12}{2}[/latex]= 16+6 = 22cm
In triangle AQC
AC2 = AQ2 + QC2
AC2 = 82+222 [latex]\rightarrow[/latex] 64+ 484 = 548
AC = [latex]\sqrt{548} = 2 \sqrt{137} cm[/latex]
9. If the difference between an exterior angle of a regular polygon of n sides and an exterior angle of another regular polygon of (n + 1) sides is equal to 5 °, then the value of n is

Answer - Option D
Explanation -
Exterior angle of regular polygon = 360o/ number of sides
ATQ
[latex]\frac {360°}{n} - \frac {360°}{n + 1}[/latex] = 5
360 (n+1 – n ) = 5n(n+1)
[latex]\rightarrow[/latex] 72 = [latex]{n}^{2}[/latex] + n
[latex]\rightarrow[/latex] [latex]{n}^{2}[/latex] + n – 72 = 0
[latex]\rightarrow[/latex] (n – 8)(n + 9)= 0
[latex]\rightarrow[/latex] n = 8, -9
Hence number of sides = 8
10. The number of sides in two regular polygons are in the ratio 5 : 4 and the difference between each interior angle of the polygons is 6 °. Then the numbers of sides are

Answer - Option A
Explanation -
If the difference between Interior angles = 6o then the difference between Exterior angles will also be 6o. Since sides are in the ratio 5:4, take them 5n and 4n.
[latex]\frac {360°}{4n} - \frac {360°}{5n}[/latex] = 6
[latex]\frac {90}{n} - \frac {72}{n}[/latex] = 6
18 = 6n
n = 3
So sides are 5n, 4n = 15, 12

RRB JE 2019 Quantitative Aptitude Quiz - Set 37

1. If the sum of the interior angles of a regular polygon be 1080°, the number of sides of the polygon is:

Answer - Option B
Explanation -
Sum of the interior angles of a regular polygon of n sides
= (2n -4) * 90 °
i.e, (2n -4) * 90 ° = 1080°
(2n -4) = 1080° ÷ 90 ° = 12
i.e, 2n = 12 + 4 = 16
n = [latex] \frac{16}{2} = 8[/latex]
2. If each interior angle of a regular polygon is 150 ° the number of sides of the polygon is

Answer - Option D
Explanation -
The relationship between angle and side of a polygon can be given as:
Angle = [latex] \frac{(2 × number of sides - 4) × 90°}{number of sides} [/latex]
Hence 150 = [latex] \frac{(2n - 4) ×90}{n}[/latex]
[latex]\rightarrow[/latex] 150n = 180n – 360
[latex]\rightarrow[/latex] 30n = 360
[latex]\rightarrow[/latex] n = 12
Hence number of sides = 12
3. If the measure of each interior angle of a regular Polygon be 144° the number of sides of the Polygon is

Answer - Option A
Explanation -
Interior angle = 144
Exterior anlge = 180 - 144 = 36
36 = [latex] \frac{360}{n}[/latex]
n = [latex] \frac{360}{36}[/latex]
n = 10
4. The number of diagonals of a polygon is 54. The number of sides of the polygon is how much?

Answer - Option C
Explanation -
Number of diagonals of a polygon = [latex] \frac{n(n-3)}{2}[/latex]
54 = [latex] \frac{n(n-3)}{2}[/latex]
108 = n(n-3)
12*9 = n(n-3)
n = 12
5. In the given figure, O is the center of the circle, PQR = 100° and STR = 105°. What is the value (in degrees) of OSP?

Answer - Option D
Explanation -

Angle PQR=100 °
Angle STR=105 °
Refex angle of SOR= 2 × STR = 2 × 105 = 210 °
Angle SOR = 360 - 210 = 150
Angle PQR = 100
So angle formed by chord PR at the centre must be double of PQR
Hence angle POR = 2 × PQR = 200 °
Angle SOP+angle SOR = 200 °
Angle
So angle SOP = 180 - 150 = 50 °
In tringle SOP OS = OP = radius of angle
So angle OSP = angle OPS = x
2x + 50 = 180
X = 65 °
So angle SOP = 65°
6, The diagonal of a cube is [latex]\sqrt {192}[/latex] Its volume (in [latex]{cm}^{3}[/latex]) will be

Answer - Option C
Explanation -
Diagonal of a cube is = [latex]\sqrt {3}[/latex] x = [latex]\sqrt {192}[/latex]
x= [latex]\sqrt {\frac {192}{3}}[/latex] = 8cm
Now, volume of the cube = [latex]{a}^{3}[/latex] = ([latex]{8}^{3}[/latex]) = 512 [latex]{cm}^{3}[/latex]
7. Find the ratio of internal angle to external angle in a regular polygon which has 54 diagonals.

Answer - Option C
Explanation -
We know that,
no of diaonals = [latex]\frac {n(n - 3)}{2}[/latex]
54 = [latex]\frac {n(n - 3)}{2}[/latex]
n (n - 3) = 108
[latex]{n}^{2} - 3n - 108 = 0[/latex]
[latex]{n}^{2} - 12n + 9n - 108 = 0[/latex]
n(n - 12) + 9(n - 12) = 0
(n - 12)(n + 9) = 0
n = 12
Each interior angle
[latex]\frac {(n - 2)180}{n}[/latex]
[latex]\frac {10 * 180}{12} = 150[/latex]
Each exterior angle = 360/12 = 30°
Required
Interior angle : Exterior angle = 5 : 1
8. What is the area (in sq cm) of a rectangle if its diagonal is 26 cm and one of its sides is 10 cm?

Answer - Option B
Explanation -
Given: diagonal of a rectangle = 26 cm and length, l = 10 cm
Then, breadth, b = [latex]\sqrt {{26}^{2} - {10}^{2}}[/latex]
= [latex]\sqrt {576}[/latex]
= 24 cm
Now, area of the rectangle = l × b
= 10 × 24
240 [latex]{cm}^{2}[/latex]
9. If ABCDEF is a regular hexagon, then what is the value (in degrees) of ADB?

Answer - Option B
Explanation -
Since ABCDEF is a regular hexagon
Measure of each interior angle = [latex]\frac {(n - 2)180}{n}[/latex] = [latex]\frac {(6 - 2)180}{6}[/latex] = 120
ADB = [latex]\frac {120}{4}[/latex] = 30
10. An equilateral triangle of area 300 [latex]{cm}^{2}[/latex] is cut from its three vertices's to form a regular hexagon. Area of hexagon is what percent of the area of triangle?

Answer - Option A
Explanation -
Area of [latex]\triangle[/latex] = [latex]\frac {\sqrt {3}}{4}{a}^{2} = 300[/latex]
[latex]{a}^{2} = \frac {4}{\sqrt {3}} * 300[/latex]
a = 20[latex]{\sqrt {3}}{4}[/latex]
Side of Hexagon = [latex]\frac {20}{3}\sqrt 4{3}[/latex]
Area of Hexagon = [latex]6 * \frac {\sqrt {3}}{4} {\frac {20}{3}\sqrt 4{3}}^{2}[/latex]
= 6 * [latex]\frac {\sqrt {3}}{4}\frac {400}{9}\sqrt {3}[/latex]
= 200
Required
% = [latex]\frac {200}{300} * 100[/latex]
= 66.66%

RRB JE 2019 Quantitative Aptitude Quiz - Set 38

1. If [latex]{2}^{x} = {3}^{y} = {6}^{-z} [/latex] then [latex]\frac {1}{x} + \frac {1}{y} + \frac {1}{z}[/latex] is equal to

Answer Option A
Explanation :
[latex]{2}^{x} = {3}^{y} = {6}^{-z} [/latex] = k
[latex]2 = {k}^{\frac {1}{x}}: 3 = {k}^{\frac {1}{y}}; 6 = {k}^{- \frac {1}{z}} [/latex]
2 * 3 = 6
[latex]{k}^{\frac {1}{x}} \times {k}^{\frac {1}{y}}; = {k}^{- \frac {1}{z}} [/latex]
[latex]{k}^{\frac {1}{x} + \frac {1}{y}} = {k}^{- \frac {1}{z}} [/latex]
[latex]\frac {1}{x} + \frac {1}{y} = - \frac {1}{z}[/latex]
[latex]\frac {1}{x} + \frac {1}{y} + \frac {1}{z}[/latex] = 0
2. If the selling price is Rs 1680 after getting a discount of 16%, what was the marked price?

Answer Option A
Explanation :
We know that,
MP = SP [latex]\times \frac {100}{100 - discount%}[/latex]
= 1680 [latex]\times \frac {100}{84}[/latex]
= Rs 2000
3. A number consists of two digits and the digit in the ten’s place exceeds the digit in the units place by 5. If 5 times the sum of the digits be subtracted from the number, the digits of the number are reversed. Then the sum of digits of the number is

Answer Option C
Explanation :
Let unit’s digit be x.
Ten’s digit = x + 5
Number = 10 (x + 5) + x = 11x+ 50
Again,
= 11x + 50 - 5 (2x + 5) = 10x + x + 5
[latex]\Rightarrow [/latex] 11x + 50 - 10x - 25 = 11x + 5
[latex]\Rightarrow [/latex] 10x = 20 [latex]\Rightarrow [/latex] x = 2
i.e, Required sum
= 2x + 5 = 2 [latex]\times [/latex] 2 + 5 = 9
4. A discount of 20% on one article is the same as a discount of25% on another article. The cost price of the two articles can be:

Answer Option B
Explanation :
20% of x = 25% of y
[latex]\Rightarrow \frac {x}{y} = \frac {25}{20} = \frac {5}{4}[/latex]
So ,x and y must be in the ratio of 5 : 4.
Rs. 50 and Rs. 40
5. The bus fare between two cities is increased in the ratio 1:2. Find the increase in the fare, if the original fare is Rs. 175.

Answer Option D
Explanation :
Let the fare before and after increase be x and 2x respectively, then
Increase in fare = 2x – x = x
i.e, Increase in fare = x = Rs 175
6. If the diameter of a circle is increased by 8% then its area is increased by:

Answer Option A
Explanation :
8 + 8 + [latex] \frac {8 \times 8}{100}[/latex] = 16.64%
7. The cross section of a canal is in the shape of an isosceles trapezium which is 4 m wide at the bottom and 5 m wide at the top. If the depth of the canal is 2 m and it is 120 m long, what is the maximum capacity of this canal?

Answer Option D
Explanation :
Width of canal = 4m
Length of canal = 120m
Depth of canal = 2m
Crosse sectional area of canal = [latex] \frac {1}{2} 120 (4 + 5)[/latex] = 540sqcm
Maximum capacity of canal= cross sectional area × height = 540× 2 = 1080 cubic mts
8. A hemispherical bowl of internal radius 15 cm contains a liquid. The liquid is to be filled into cylindrical shaped bottles of diameter 5 cm and height 8 cm. The number of bottles required to empty the bowl is

Answer Option A
Explanation :
Number of bottles = [latex] \frac {Volume of hemisphere}{Volume of cylinder} = \frac { \frac {2}{3}\pi {R}^{3}}{\pi {r}^{2} h }[/latex]
[latex] \frac {\frac {2}{3} \times 15 \times 15 \times 15}{{(2.5)}^{2}\times 8}[/latex] = 45
9. Marked price of an item is Rs 200. On purchase of 1 item discount is 22%, on purchase of 4 items discount is 33%. Rabia buys 5 items, what is the effective discount?

Answer Option B
Explanation :
Marked price of 4 items = 800
Discount on 4 items= 33 %
Then dicount= 33 % 800 = 264
Discount on one item = 22 % 200 = 44
Total discount on 5 items = 264 + 44 = 308
Effective discount = 308 [latex]\times \frac {100}{1000}[/latex] = 30.8
10. The average of seven consecutive positive integers is 26. The smallest of these integers is:

Answer Option B
Explanation :
x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 26 * 7
x = 182 - 21 = 161
x = [latex] \frac {161}{7}[/latex] = 23

RRB JE 2019 Quantitative Aptitude Quiz - Set 39

1. The distance between the centres of two equal circles each of radius 3 cm is 10 cm. The length of a transverse tangent is

Answer Option C
Explanation :
Transverse tangent formula
PQ = [latex]\sqrt {{d}^{2} - {(r + r)}^{2}} = \sqrt {{10}^{2} - {(3 + 3)}^{2}}[/latex]
[latex]\sqrt {64}[/latex] = 8
2.If [latex]{x}^{2} + {y}^{2} - 4x - 4y + 8 = 0,[/latex] then the value of x - y is

Answer Option C
Explanation :
[latex]{x}^{2} + {y}^{2} - 4x - 4y + 8 = 0,[/latex]
[latex]\Rightarrow {x}^{2} -4x + 4 + {y}^{2} - 4y + 4 = 0,[/latex]
[latex]\rightarrow[/latex][latex]{(x - 2)}^{2} + {(y - 2)}^{2} = 0,[/latex]
[latex]\rightarrow[/latex](x - 2) = 0 and (y - 2) = 0
[latex]\rightarrow[/latex] = 2, y = 2
Now (x - y) = (2 - 2) = 0
3. Two trains start from station A and B and travel towards each other at speeds of 16 miles/hour and 21 miles/hour respectively. At the time of their meeting, the second train has travelled 60 miles more than the first. The distance between A and B (in miles) is:

Answer Option A
Explanation :
Here speeds of two trains = 16 miles/hour and 21 miles/hour respectively.
So I can say in one-hour second train runs 5 miles more than the first train.
Now at the time of their meeting, the second train has travelled 60 miles more than the first. So I can say trains meet after 12 hours([latex] \frac {60}{5}[/latex])
Hence distance between A & B = (16 + 21)(12) = 444 miles
Hence Option A is correct
4. By selling a fan for Rs. 600, a man loses 10%. To make a gain of 20%, the selling price of the fan should be

Answer Option D
Explanation :
CP of fan = [latex] \frac {600 \times 100}{90}[/latex]
Required SP = [latex] \frac {600 \times 100}{90} \times \frac {120}{100}[/latex] = Rs 800
Hence Option D is correct
5. Find the value of 3 + [latex]\frac {1}{\sqrt {3}} + \frac {1}{\sqrt {3} + 3} + \frac {1}{\sqrt {3} - 3}[/latex]

Answer Option A
Explanation :
3 + [latex]\frac {1}{\sqrt {3}} + \frac {1}{\sqrt {3} + 3} + \frac {1}{\sqrt {3} - 3}[/latex]
= 3 + [latex]\frac {1}{\sqrt {3}} + \frac {\sqrt {3} - 3 + \sqrt {3} + 3}{(\sqrt {3} + 3) (\sqrt {3} - 3)} + \frac {1}{\sqrt {3} - 3}[/latex]
3 + [latex]\frac {1}{\sqrt {3}} + \frac {2 \sqrt {3}}{3 - 9}[/latex]
3 + [latex]\frac {1}{\sqrt {3}} + \frac { \sqrt {3}}{3}[/latex]
3 + [latex]\frac {1}{\sqrt {3}} + \frac {1}{\sqrt {3}}[/latex] = 3
6. The area of the triangle formed by the graphs of 3x + 4y = 12. x-axis and y-axis (in sq. units)'

Answer Option C
Explanation :
Putting y=0 in 3x + 4y = 12, we get x = 4
Co-ordinates of point of intersection on x-axis = (4, 0)
Similarly for x =0 we get y =3
Co-ordinates of point of intersection on y-axis= (0, 3)
Area of Triangle formed = [latex]\frac {1}{2}[/latex] × 4× 3 = 6 sq units
Hence option C is correct
7. The greatest among the numbers [latex]\sqrt {0.09}[/latex], [latex]\sqrt {0.064}[3][/latex], 0.5 and [latex]\frac {3}{5}[/latex] is

Answer Option D
Explanation :
[latex]\sqrt {0.09}[/latex] = 0.3
[latex]\sqrt [3] {0.064}[/latex] = 0.4
0.5
[latex]\frac {3}{5}[/latex] = 0.6 Hence, it is clear that [latex]\sqrt {0.09}[/latex] < [latex]\sqrt {0.064}[3][/latex] < 0.5 < [latex]\frac {3}{5}[/latex]
8. Simplify : [latex]\sqrt {3 \frac {33}{64}} \div \sqrt {9 \frac {1}{7}} \times 2 \sqrt {3 \frac {1}{9}}[/latex]

Answer Option D
Explanation :
[latex]\sqrt {3 \frac {33}{64}} \div \sqrt {9 \frac {1}{7}} \times 2 /sqrt {3 \frac {1}{9}}[/latex]
[latex]= \sqrt {\frac {225}{64}} \div \sqrt {\frac {64}{7}} \times 2 \sqrt { \frac {28}{9}}[/latex]
[latex]= \frac {15}{8} \div \frac {8}{\sqrt {7}} \times 2 \times 2 \times \frac {/sqrt {7}}{9}[/latex]
[latex]= \frac {5 \times 7}{16} = \frac {35}{16} = 2 \frac {3}{16}[/latex]
9. If [latex]= \frac {a}{1 - a}+ \frac {b}{1 - b} + \frac {c}{1 - c}[/latex] = 1 then the value of [latex] = \frac {1}{1 - a} + \frac {1}{1 - b} + \frac {1}{1 - c}[/latex] is

Answer Option D
Explanation :
[latex]= \frac {a}{1 - a} = \frac {b}{1 - b} + \frac {c}{1 - c}[/latex] = 1
[latex]= ( \frac {a}{1 - a} + 1) = (\frac {b}{1 - b} + 1) + ( \frac {c}{1 - c} + 1)[/latex] = 3 + 1 = 4
[latex]= \frac {a + 1 - a}{1 - a} = \frac {b + 1 - b}{1 - b} + \frac {c + 1 - c}{1 - c}[/latex] = 4
[latex] = \frac {1}{1 - a} + \frac {1}{1 - b} + \frac {1}{1 - c}[/latex] = 4
10. [latex] = 3 \frac {1}{2}[/latex] - ([latex]2 \frac {1}{4}[/latex] + ([latex] 1 \frac {1}{4}[/latex] - [latex] \frac {1}{2}[/latex]([latex] = 1 \frac {1}{2} - \frac {1}{3} - \frac {1}{6}[/latex]))) is

Answer Option A
Explanation :
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5}{4}[/latex] - [latex] \frac {1}{2}[/latex]([latex] \frac {3}{2} - \frac {1}{3} - \frac {1}{6}[/latex])))
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5}{4}[/latex] - [latex] \frac {1}{2}[/latex]([latex] \frac {9 - 2 - 1}{6}[/latex])))
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5}{4} - \frac {1}{2}[/latex]))
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5 - 2}{4}[/latex]))
[latex] = \frac {7}{2}[/latex] - ([latex] \frac {9}{4} + \frac {3}{2}[/latex])
[latex] = \frac {7}{2} - 3[/latex] = [latex] \frac {7 - 6}{2} = \frac {1}{2}[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 40

1. If sin[latex]\theta[/latex] - cos [latex]\theta[/latex] = [latex]\frac {7}{13}[/latex] and 0 < [latex]\theta[/latex] < 90, then the value of sin[latex]\theta[/latex] + cos [latex]\theta[/latex] is

Answer Option A
Explanation : sin[latex]\theta[/latex] - cos [latex]\theta[/latex] = [latex] \frac {7}{13}[/latex] ....(i)
sin[latex]\theta[/latex] + cos [latex]\theta[/latex] = x ....(ii)
On squaring both equations and adding,
2( [latex]{sin}^{2} \theta + {cos}^{2}\theta[/latex]) = [latex] \frac {49}{169} + {x}^{2}[/latex]
[latex] {x}^{2} = 2 \frac{338 - 49} {169} = \frac{289} {169}[/latex]
x = [latex] \sqrt {\frac{289} {169}} = \frac{17} {13}[/latex]
2. Successive discounts of 50% and 50% is equivalent to

Answer Option B
Explanation :
Let X be the price
Price after first 50% discount = x - [latex] \frac {50 x}{100}[/latex] = - [latex] \frac {x}{2}[/latex] = [latex] \frac {x}{2}[/latex]
Price after second 50% discount
[latex] \frac {x}{2}[/latex] - [latex] \frac {50 \times\frac{x}{2}}{100}[/latex] = [latex] \frac {x}{2} - \frac {x}{4}[/latex] = [latex] \frac {x}{4}[/latex]
Hence final price = [latex] \frac {x}{4}[/latex]
Total discount = x - [latex] \frac { x}{4}[/latex] = [latex] \frac {3 x}{4}[/latex]
% total discount = ([latex] \frac {3 x}{4}[/latex]) × [latex] \frac {100}{x}[/latex] = 75%
3. The numerical value of ([latex] \frac {1}{cos \theta} + \frac {1}{cos \theta}[/latex])([latex] \frac {1}{cos \theta} - \frac {1}{cos \theta}[/latex]) is?

Answer Option C
Explanation : ([latex] \frac {1}{cos \theta} + \frac {1}{cos \theta}[/latex])([latex] \frac {1}{cos \theta} - \frac {1}{cos \theta}[/latex])
([latex] \frac {1}{cos \theta} + \frac {sin \theta}{cos \theta}[/latex])([latex] \frac {1}{cos \theta} - \frac {sin \theta}{cos \theta}[/latex])
[latex] \frac {1 + sin \theta}{cos \theta} + \frac {1 - sin \theta}{cos \theta}[/latex]
[latex] \frac {1 - {sin}^{2} \theta}{cos \theta} = \frac {{cos}^{2} \theta}{{cos}^{2} \theta}[/latex] = 1
4. The value of tan4°× tan43°× tan47° × tan86° is

Answer Option B
Explanation : = tan 4°× tan 43°× tan 47° × tan 86°
= (tan 4°× tan 86°)× (tan 47° × tan 43°)
= (tan 4°× cot 4°)× (tan 43° × cot43°)
= 1
5. There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge of this pyramid is of length [latex]\frac {5 a}{2}[/latex] cm, then the volume of this pyramid is

Answer Option C
6. [latex]\frac {sin A}{(1 + cosA)} + \frac {(1 + cosA)}{sin A}[/latex] is equal to

Answer Option B
Explanation : [latex]\frac {sin A}{(1 + cosA)} + \frac {(1 + cosA)}{sin A} [/latex]
[latex]\frac {sin A}{(1 + cosA)} \times \frac {1 - cosA}{1 - cosA} [/latex] + [latex]\frac {1 + cosA}{sinA)} \times \frac {sin A}{sin A} [/latex]
[latex]\frac {sin A - sin AcosA + sin A + sin AcosA}{{sin}^{2}A}[/latex]
= 2cosecA
7. If the ratio of volumes of two cones is 2 : 3 and the ratio of the radius of their bases is 1 : 2 then the ratio of their heights will be

Answer Option D
Explanation : Volume of cone = [latex]\frac {1}{3 \pi {r}^{2} h} [/latex]
Required ratio = [latex]\frac {1 \pi {{r}_{1}}^{2} {h}_{1}}{3} [/latex] : [latex]\frac {1 \pi {{r}_{2}}^{2} {h}_{2}}{3} [/latex] = 2 : 3
[latex]\frac {1}{4} \times \frac {{h}_{1}}{{h}_{2}} = \frac {2}{3}[/latex]
[latex]\frac {{h}_{1}}{{h}_{2}} = \frac {2}{3} \times = \frac {8}{3}[/latex]
Hence option D is correct
8. The ratio of radii of two cones is 3 : 4 and the ratio of their heights is 4 : 3. Then the ratio of their volumes will be

Answer Option A
Explanation : [latex]\frac {{V}_{1}}{{V}_{2}} [/latex] = [latex]\frac {\pi {{r}_{1}}^{2} {h}_{1}}{\pi {{r}_{2}}^{2} {h}_{2}}\frac {{V}_{1}}{{V}_{2}} [/latex] = [latex]\frac {\pi {{r}_{1}}^{2} {h}_{1}}{\pi {{r}_{2}}^{2} {h}_{2}}[/latex]
= [latex]{\frac {3}{4}}^{2} \times \frac {4}{3}[/latex]
= [latex] \frac {3}{4} \times \frac {3}{4} \times \frac {4}{3} = \frac {3}{4}[/latex]
9. Two numbers are respectively 25% and 40% less than a third number. What per cent of the first is the second?

Answer Option D
Explanation : Let third no = 100
[latex]{1}^{st}[/latex] and [latex]{2}^{nd}[/latex] numbers will be then 75 and 60
Required Percentage = 60 [latex]\times \frac {100}{75}[/latex] = 80 %
10. From each of the two given unequal numbers, half the smaller number is subtracted. Then of the resulting numbers, the larger one is five times than the smaller one. Then the ratio of the larger to smaller one is

Answer Option C
Explanation : Let the numbers be x and y where x > y
i.e, x - [latex] \frac {y}{2} = 5 (y - \frac {y}{2})[/latex]
x - [latex] \frac {y}{2} = 5 \frac {y}{2}[/latex]
x = [latex] \frac {5y}{2} + \frac {y}{2} = \frac {6y}{2}[/latex]
x = 3y
[latex]\frac {x}{y} = \frac {3}{1}[/latex]
i.e, x : y = 3 : 1

RRB JE 2019 Quantitative Aptitude Quiz - Set 41

1. Marks of two candidates P and Q are in the ratio of 2 : 5. If the marks of P are 120, marks of Q are

Answer Option C
Explanation :
Let marks of P and Q be 2x and 5x
2x = 120 x = 60
5x = 300
Marks of Q = 300
2. If (a + b) : (b + c) : (c + a) = 6 : 7 : 8 and (a + b + c) = 14, then the value of c is

Answer Option A
Explanation :
[latex]\frac {a + b}{6} = \frac {b + c}{7} = \frac {c + a}{8}[/latex] = k
[latex]\Rightarrow a + b = 6k;b + c = 7k;[/latex]
c + a = 8k
i.e, a + b + b + c + c + a
[latex]\Rightarrow 6k + 7k + 8k [/latex]
[latex]\Rightarrow 2 \times 14 = 21 k \Rightarrow k = \frac {4}{3} [/latex]
i.e, c = (a + b+ c)- (a + b)
= 14 - 6 [latex]\times \frac {4}{3} = 14 - 8 = 6[/latex]
3. If the price of rice be raised by 25%, the percent by which a house-holder must reduce his consumption of rice so as not to increase his expenditure on rice is

Answer Option D
Explanation :
Price ratio = 100 : 125
Consumption ratio = 125:100
So,reduction = [latex]25 \times \frac {100}{125}[/latex] =20 %
4. If x = a(b - c), y = b(c - a) and z = c (a - b), then [latex]{\frac {x}{a}}^{3} + {\frac {y}{b}}^{3} + {\frac {z}{c}}^{3}[/latex] =

Answer Option C
Explanation :
[latex] \frac {x}{a} = b - c; \frac {y}{b} = c - a; \frac {z}{c} = a - b[/latex]
Again, b - c + c - a + a - b = 0
Using identity If a + b +c = 0, then [latex]{a}^{3} + {b}^{3} + {c}^{3} [/latex] = 3abc
i.e, [latex]{\frac {x}{a}}^{3} + {\frac {y}{b}}^{3} + {\frac {z}{c}}^{3} [/latex] = [latex]{b - c}^{3} + {c - a}^{3} + {a - b}^{3} [/latex]
= 3 (b - c)(c - a)(a - b)
= [latex] \frac {3 xyz}{abc}[/latex]
5. A team played 40 games in a season and won in 24 of them. What percent of games played did the team win?

Answer Option C
Explanation :
Total game played= 40
Number of won games = 24
% of won games = [latex]24 \times \frac {100}{40}[/latex] = 60 %
6. If the total surface area of a cube is 96 [latex]{cm}^{2}[/latex], its volume is

Answer Option C
Explanation :
Edge of cube = x cm
i.e, [latex]6{x}^{2} \Rightarrow 96 {x}^{2} = \frac {96}{6}[/latex] = 16
[latex]\Rightarrow x = \sqrt {16} cm = 4 [/latex]
Volume of cube = [latex]{(edge)}^{2}[/latex] (edge)3 = (4)3 = 64 cu. Cm
7. The radius of cross section of a solid cylindrical rod of iron is 50 cm. The cylinder is melted down and formed into 6 solid spherical balls of the same radius as that of the cylinder. The length of the rod (in metres) is

Answer Option D
Explanation :
Length of rod = h cm
i.e, Volume of cylinder = Volume of 6 spheres
[latex]\Rightarrow \pi {r}^{2} h = 6 \times \frac {4}{3} \pi {r}^{3} [/latex]
[latex]\Rightarrow h = 6 \times \frac {4}{3} \times r = 8 \times 50[/latex]
= 400 cm = 4 metre
8. In a division sum, the divisor is 3 times the quotient and 6 times the remainder. If the remainder is 2, then the dividend is

Answer Option A
Explanation :
Divisor = 6 × 2 = 12
Again divisor = 3 × quotient
Quotient = [latex] \frac {12}{3}[/latex] = 4
Dividend = 12 × 4 + 2 = 50
Hence Option A is correct
9. If [latex] \frac {a}{b} + \frac {b}{a}[/latex] = 2, then the value of a - b is :

Answer Option C
Explanation :
[latex] \frac {a}{b} + \frac {b}{a}[/latex] = 2
[latex] \frac {{a}^{2} + {b}^{2}}{ab} = 2 [/latex]
[latex]{a}^{2} + {b}^{2} - 2ab = 0[/latex]
[latex]{(a - b)}^{2} = 0[/latex]
a - b = 0
[latex]\Rightarrow 12 x - 1[/latex]
New Number = 10(2x - 1) + x = 0
10. The length of the diagonal of a rectangle is 10 cm and that of one side is 8 cm. What is the area of this rectangle?

Answer Option B
Explanation :
In rectangle we know that ,
[latex]{D}^{2} = {L}^{2} + {B}^{2}[/latex]
100= 64 + [latex]{B}^{2}[/latex]
[latex]{B}^{2}[/latex] = 100 - 64 = 36
B = 6
Area = L × B = 6 × 8 = 48 sq cm

RRB JE 2019 Quantitative Aptitude Quiz - Set 42

1. In a two-digit number, the digit at the unit's place is 1 less than twice the digit at the ten's place. If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is

Answer Option D
Explanation :
Ten digit = x
Unit digit = 2x - 1
Original number = 10 + 2x - 1
21x - 10
New Number = 10 (3x 1) + x
Difference able new number and original number = 21x - 10 - 12x + 1
9x- 9
12x - 1 = 9x +9 = 20
3x = 12
x = 4
Hence original number = 12 * 4 - 1 = 47
2.A dishonest grocer sells rice at a profit of 10% and also uses weights which are 20% less than the marked weight. The total gain earned by him will be

Answer Option A
Explanation :
Let CP = 100
SP = 110
but he sells rice worth 80 only.
So profit = 110-80=30
Profit % = ([latex] \frac {3O}{80}\times [/latex]) 100 = 37.5%
Hence Option A is correct
3. [latex]\sqrt{ \frac {9.5 \times 0.085}{0.0017 \times 0.19}} [/latex] equals

Answer Option A
Explanation :
[latex]\sqrt{ \frac {9.5 \times 0.085}{0.0017 \times 0.19}} [/latex] = [latex]\sqrt{ \frac {95 \times 85 \times 100}{17 \times 19}} = 5 \times 10 = 50[/latex]
4. A does 30% of a work in 30 days. He then calls in B and they together finish the remaining work in 20 days. How long B alone would take to do the whole work?

Answer Option A
Explanation :
let total work be 100 units
Given: A does 30 % of a work in 30 days
Then, efficiency of A = [latex] \frac {3O}{30} [/latex]
= 1 unit/day
He then calls in B and they together finish the remaining work in 20 days.
So, remaining work = 70 units
Let the efficiency of B be x units/day, then
[latex] \frac {7O}{1 + x} = 20[/latex]
[latex]\Rightarrow[/latex] 20 + 20x = 70
[latex]\Rightarrow[/latex] 20x = 50
[latex]\Rightarrow[/latex] x = 2.5 units/day
Then, working alone B would take = ([latex] \frac {100}{2.5} [/latex])
= 40 days
5. If [latex]{a}^{2} + {b}^{2} + {c}^{2}[/latex] -ab -bc - ca then a b: c is :

Answer Option B
Explanation :
[latex]{a}^{2} + {b}^{2} + {c}^{2}[/latex] -ab -bc - ca = 0
therefore, a = b = c
thus, a : b : c = 1 : 1 : 1
6. If each of the length and breadth of a rectangle is increased by 50% by what percent does its area increase ?

Answer Option A
Explanation :
Resultant increase per cent
= (50 + 50 + [latex]\frac {50 \times 50}{100}[/latex])%
= 125 %
7. A chair listed at ? 350 is available at successive discounts of 25% and 10%. The selling price of the chair is

Answer Option A
Explanation :
Single equivalent discount = ([latex]25 + 10 - \frac {25 \times 10}{100}[/latex]) = 32.5 %
i.e, S.P of chair = [latex]350 \frac {(100 - 32.5)}{100}[/latex]
= [latex]\frac {350 \times 67.5}{100}[/latex]
= 236.25
8. Two right circular cones of equal height of radii of bases 3 cm and 4 cm are melted together and made to a solid sphere of radius 5 cm. The height of a cone is

Answer Option B
Explanation :
Volume of cones = Volume of sphere
[latex] \frac {1}{3}\pi {{r}_{1}}^{2}h + \frac {1}{3}\pi {{r}_{1}}^{2}h = \frac {4}{3}\pi {r}^{3} [/latex]
[latex]\Rightarrow({{r}_{1}}^{2} + {{r}_{2}}^{2}) h = 4 {r}^{3}[/latex]
[latex]\Rightarrow (9 + 16)h = 4 {5}^{3}[/latex]
[latex]\Rightarrow 25 h = 4 \times 125 [/latex]
[latex]\Rightarrow h = 4 \times 5 = 20 cm[/latex]
9. If a - b = -5 and [latex]{a}^{2} +{b}^{2} = 73[/latex], then find ab.

Answer Option D
Explanation :
We know that,
[latex]{(a - b)}^{2} = {a}^{2} +{b}^{2} -2ab[/latex]
[latex]\Rightarrow[/latex] 25 = 73 – 2ab
[latex]\Rightarrow[/latex] ab = [latex]\frac {48}{2}[/latex]
[latex]\Rightarrow[/latex] ab = 24
10. A man bought some eggs of which 10% are rotten. He gives 80% of the remainder to his neighbours. Now he is left out with 36 eggs. How many eggs he bought?

Answer Option D
Explanation :
Let the man bought x eggs.
Therefore,
x = [latex]\frac {90}{100} \times \frac {20}{100} = 36[/latex]
x = [latex] \frac {36 \times 100 \times 100}{90 \times 20}[/latex]
x = 200 eggs

RRB JE 2019 Quantitative Aptitude Quiz - Set 43

1. If [latex]{x}^{2} + 9{y}^{2} = 6xy[/latex] then x : y is

Answer Option C
Explanation :
[latex]{x}^{2} + 9{y}^{2} = 6xy[/latex]
[latex]{(x - 3y)}^{2} = 0[/latex]
x = 3y
[latex]\frac {x}{y} = \frac {3}{1}[/latex]
2. The diameter of a cylinder s 7 cm and its height is 16 cm. Using the value of [latex]\pi = \frac {22}{7}[/latex], the lateral surface area of the cylinder is

Answer Option A
Explanation :
Lateral surface area of the cylinder = [latex]2 \pi r h[/latex]
= 2 [latex]\times \frac {22}{7} \times \frac {7}{2} \times 16[/latex]
= 352 sq cm.
3. If sin([latex]\theta + 18[/latex]) (0 < [latex]\theta[/latex] < 90), then the value of cos 5[latex]\theta[/latex] is

Answer Option A
Explanation :
Sin ([latex]\theta[/latex] + 18) = cos 60
or, cos(90 - ([latex]\theta[/latex] + 180)) = cos 60
or, 90 - ([latex]\theta[/latex] + 18) = 60
or, 90 -[latex]\theta[/latex] - 18 =60
or, [latex]\theta[/latex] = 12
cos 5[latex]\theta[/latex]= cos60 = [latex]\frac {1}{2}[/latex]
4. An article of cost price ? 8,000 is marked at ? 11,200. After allowing a discount of x% a profit of 12% is made. The value of x is

Answer Option B
Explanation :
S.P. for a profit of 12 % = [latex]\frac {8000 \times 112}{100}[/latex] = ? 8960
i.e, Discount = 11200 – 8960 = ? 2240
If the discount per cent be x, then
[latex]\frac {11200 \times x}{100}[/latex] = 2240
x = [latex]\frac {2240 \times 100}{11200}[/latex] = 20 %
5. The average age of a husband and his wife was 23 years at the beginning of their marriage. After five years they have a one-year old child. The average age of the family of three, when the child was born, was

Answer Option C
Explanation :
At the time of their marriage age of
H + W = 23 [latex]\times[/latex] 2 = 46 years
After 5 years the age of Child was 1 year, so the child was born 4 years after marriage. Total age at that time of child's birth
H + W + C = 46 + 8 (4 years added to both H and W) = 54 years
Average age = [latex]\frac {54}{3}[/latex] = 18 years
6, The ratio of the number of boys and girls In a school is 3 : 2. If 20% of the boys and 25% of the girls are scholarship holders, then the percentage of the students, who do not get the scholarship, is:

Answer Option A
Explanation :
Boys = 30, Girls = 20 (let)
Boys getting no scholarship = 24
Girls getting no scholarship = 15
Sum of students with no scholarship = 24 + 15 = 39
i.e, Required percentage = [latex]\frac {39}{50} \times 100[/latex] = 78
7. If a + b + c = 0, then the value of [latex]\frac {1}{(a + b)(b + c)} + \frac {1}{(a + c)(b + a)} + \frac {1}{(c + a)(c + b)}[/latex] is

Answer Option B
Explanation :
[latex]\frac {1}{(a + b)(b + c)} + \frac {1}{(a + c)(b + a)} + \frac {1}{(c + a)(c + b)}[/latex]
[latex]\frac {(a + b) + (b + c) + (c + a)}{(a + b)(b + c)(c + a)} + \frac {1}{(a + c)(b + a)} + \frac {1}{(c + a)(c + b)}[/latex]
[latex]\frac {2 (a + b + c)}{(a + b)(b + c)(c + a)}[/latex] = 0
i.e, (a + b + c) = 0
8. The average age of 80 boys in a class is 15 years. The average age of a group of 15 boys in a class is 16 years and the average age of another 25 boys in the class is 14 year. What is the average age of remaining boys in the class?

Answer Option A
Explanation :
Let the required average age be x years.
i.e, 80 × 15 = 15 × 16 + 25 × 14 + 40 × x
=> 40x = 1200 - 240 - 350 = 610
i.e, x = [latex] \frac {610}{40}[/latex] = 15.25 years
9. The cost price of a book is ? 150. At what price should it be sold to gain 20% ?

Answer Option B
Explanation :
S.P. of book
= [latex] \frac {150 \times 120}{100}[/latex]
= 180
10. A shopkeeper by selling 20 Timex watches, earns a profit equal to the selling price of 4 Timex watches. His profit percentage is

Answer Option B
Explanation :
Given: A shopkeeper by selling 20 Timex watches, earns a profit equal to the selling price of 4 Timex watches, then
20 SP = 20CP + Profit
20SP = 20 CP + 4SP
[latex] \frac {SP}{CP}[/latex] = [latex] \frac {20}{16}[/latex]
Profit % = [latex] \frac {4}{16}\times 100[/latex]
= 25 %

RRB JE 2019 Quantitative Aptitude Quiz - Set 44

1. The ratio of the length of the parallel sides of a trapezium is 3:2. The shortest distance between them is 15 cm. If the area of the trapezium is 450 [latex]{cm}^{2}[/latex] the sum of the lengths of the parallel sides is

Answer Option D
Explanation :
Area of the trapezium = [latex] \frac {1}{2}[/latex] (Sum of parallel sides)×Altitude
450 = [latex] \frac {1}{2}[/latex] (3x + 2x) × 15
5x = 450 × [latex] \frac {2}{15}[/latex] = 60 cm
Hence Option D is correct
2. A right pyramid stands on a square base of 16 cm and its height is 15 cm. The area (in [latex]{cm}^{2}[/latex]) of its slant surface is

Answer Option B
Explanation :
Lateral surface area= [latex] \frac {1}{2}[/latex] × perimeter of base × slant height
Slant height = [latex]\sqrt {{8}^{2} + {15}^{2}} [/latex] = 17cm
Required area = [latex] \frac {1}{2}[/latex] × 17 ×64 = 544 sq.cm
Hence option B is correct
3. A solid spherical copper ball, whose diameter is 14 cm, is melted and converted into a wire having diameter equal to 14 cm. The length of the wire is

Answer Option D
Explanation :
Volume of spherical copper ball = Volume of wire
[latex] \frac {4}{3} \pi {7}^{3} = \pi {7}^{2} h[/latex]
h = [latex] \frac {28}{3}[/latex] cm
4. If the cost price of 15 tables be equal to the selling price of 20 tables, the loss percent is :

Answer Option C
Explanation : Let C. P. of 1 table = X Rs
Hence C. P. of 15 tables = 15X Rs
S.P. of one table = Y Rs
Hence S. P. of 20 tables = 20Y Rs
Given: 15X = 20Y
Y = [latex] \frac {3}{4}[/latex]X
Loss in one table = X -Y = X -([latex] \frac {3}{4}[/latex]) X = [latex] \frac {X}{4}[/latex]
Hence loss % = [latex] \frac {(\frac {X}{4}) \times 100}{X} = (\frac {100}{4})[/latex] = 25%
5. If a shirt costs Rs. 64, after 20% discount is allowed, what was its original price in Rs. ?

Answer Option B
Explanation :
Discount = 20 %
After discount price of shirt= 64
Hence selling price = market price – discount
Let Market price = X ?
Hence X - 20% of X = 64
64 = X – ([latex] \frac {20X}{100}[/latex]) = X – ([latex] \frac {X}{5}[/latex]) = 4X/5
X = ([latex] \frac {64 \times 5}{4}[/latex]) = 16 × 5 = 80
Hence original price of shirt = ? 80
6. The price of an article is discounted by 25%. To restore to its original value, the new price must be increased by

Answer Option D
Explanation : Given: discount = 25 %
Let initial price = Rs 100
Then, discounted price = Rs 75
Now, percentage increase to make discounted price equal to the initial price = [latex] \frac {100 - 75}{75} \times 100[/latex]
= 33.33%
7. Rs, 800 at 5% per annum compounded annually will amount to Rs. 882 in

Answer Option B
8. The sum of two numbers is 75 and their difference is 25. The product of the two numbers is

Answer Option B
Explanation :
Let first number be x. then, second number = 75 – x
Since, difference is 25.
Therefore,
5 - x - x = 25
75- 2x = 25
2x = 50
x = 25
Therefore, other number = 75 – 25 = 50
Thus, the of two numbers = 25 × 50 = 1250
9. A parallelepiped whose sides are in ratio 2 : 4 : 8 have the same volume as a cube. The ratio of their surface area is :

Answer Option D
Explanation :
Volume of cube = Volume of parallelepiped (v is basically a cuboid)
Volume of cube= 2 x 4 x 8
[latex]{(a)}^{3}[/latex] = 64
a = 4
Surface area of parallelepiped : Surface area of cube = 2 (2 * 4 + 4 * 8 + 2 * 8) : 6 * 4 * 4 = 7 : 6
10.If the number 583_437 is completely divisible by 9, then the smallest whole number in the place of the blank digit will be

Answer Option D
Explanation :
583_437 sum of all the digits of this number should be a multiple of 9 according to the divisibility rule.
Let the missing number be x
30 + x should be a multiple of 9
To check the which digit will make the number divisible by 9 we will substitute each value from the options given
For x = 4, 30 + 4 = 34; not a multiple of 9
For x = 5, 30 + 5 = 35; not a multiple of 9
For x = 3, 30 + 3 = 33; not a multiple of 9
For x = 6, 30 + 6 = 36; is a multiple of 9

RRB JE 2019 Quantitative Aptitude Quiz - Set 45

1. If x + [latex] \frac {1}{4x}[/latex], find the value of [latex]8{x}^{3} + \frac {1}{8 {x}^{3}}[/latex]

Answer Option A
Explanation :
x + [latex] \frac {1}{4x}[/latex] = [latex] \frac {3}{2}[/latex]
2x + [latex] \frac {2}{4x}[/latex] = 3
2x + [latex] \frac {1}{2x}[/latex] = 3
Now taking cube of both sides
[latex]{(2x + \frac {2}{4x})}^{3} = {3}^{3}[/latex]
[latex]2{x}^{3} + {(\frac {1}{2x})}^{3} + 3 (2x + \frac {1}{2x})[/latex] = 27
[latex]8{x}^{3} + (\frac {1}{{8x}^{3}}) + 9[/latex] = 27
[latex]8{x}^{3} + (\frac {1}{{8x}^{3}}) [/latex] = 18
2. The LCM of two numbers is 4 times their HCF. The sum of LCM and HCF is 125.If one of the numbers is 100, then the other number is?

Answer Option B
Explanation :
If LCM be L and HCF be H, then
L = 4H
i.e, H + 4H = 125
[latex]\rightarrow[/latex] 5H = 125
[latex]\rightarrow[/latex] H = 25
[latex]\rightarrow[/latex] L = 100
Product of two numbers = Product of their LCM and HCF
100 * Second Number = 25 * 100
Second Number = 25
3. The radius of the base of a right circular cone is doubled. To keep the volume fixed, the height of the cone will be

Answer Option D
Explanation :
Original volume = [latex] \frac {1}{3}\pi {r}^{2}h[/latex]
New volume = [latex] \frac {1}{3}\pi {2r}^{2}{h}^{!}[/latex]
i.e, [latex] \frac {1}{3}\pi {r}^{2}{h}^{!}[/latex] = [latex] \frac {1}{3}\pi {r}^{2}h[/latex]
4[latex] {h}^{!}[/latex] = h
[latex] {h}^{!}[/latex] = [latex]\frac {h}{4}[/latex]
4. The cost price of 400 lemons is equal to the selling price of 320 lemons. Then the profit percent is

Answer Option C
Explanation :
400 CP = 320 SP
SP/CP = [latex]\frac {400}{320}[/latex]
Profit percent
= [latex]\frac {400 - 320}{320} \times 100[/latex]
[latex]\frac {80}{320} \times 100[/latex] = 25 %
5. A cone of radius 3.5 cm and height 12 cm is completely ?lled with water. This water is emptied into an empty cylindrical vessel of radius 7 cm. What will be the height of water in this vessel?

Answer Option D
Explanation :
Volume of cone = [latex] \frac {1}{3}\pi {r}^{2}h[/latex] = [latex] \frac {1}{3}\pi {3.5}^{2}\times 12[/latex]
Volume of cylinder = p r2h = p 72 × h
[latex]\Rightarrow[/latex] [latex] \frac {1}{3}\pi {3.5}^{2} \times 12[/latex] = [latex] \pi {}^{2}h[/latex]
[latex]\Rightarrow[/latex] h = 1 cm
6. 5% discount is offered on an item. By applying a promo code the customer wins 20% cash back. What is the effective discount?

Answer Option B
Explanation :
We know that,
Effective discount % = x + y – [latex] \frac {xy}{100}[/latex]
i.e, Effective discount % = 5 + 20 – [latex] \frac {(5 × 20)}{100}[/latex]
= 24%
Cashback is given on discount price or the amount buyer pays
7. If the base of right pyramid is a triangle of sides 21 cm, 20 cm, 13 cm and its altitude is 30 cm, then its volume (in [latex] {cm}^{3}[/latex]) will be

Answer Option D
Explanation :
the base of right pyramid is a triangle of sides 21 cm, 20 cm, 13 cm
hence perimeter of base(2S) = 21 + 20 + 13 = 54 cm
hence S = 27 cm
area of base
= [latex]\sqrt {S (S - c)(S - b)(S - c)}[/latex]
= [latex]\sqrt {27 (27 - 21)(27 - 20)(27 - 13)}[/latex]
[latex]\sqrt {27 \times 6 \times 7 \times 14}[/latex]
= 126 [latex] {cm}^{2}[/latex]
Volume of right pyramid = [latex] \frac {1}{3}[/latex] (area of base × height)
= 126 × [latex] \frac {30}{3}[/latex] = [latex] \frac {3780}{3}[/latex] = 1260 [latex] {cm}^{3}[/latex]
8. The height of a solid right circular cylinder is 6 metres and three times the sum of the areas of its base is equal to two times the curved surface area. The radius of the base, in metre, is

Answer Option A
Explanation :
Given that,
3 × sum of the areas of base of the cylinders
= 2 × Area of curved surface
Let r be the radius of the cylinder
Therefore,
3 [latex]\times 2 \pi {r}^{2} = 2 \times 2 \pi r h[/latex]
3 [latex]\times 2 \pi {r}^{2} = 2 \times 2 \pi r \times 6[/latex]
[latex]r = \frac {12}{3} = 4 m[/latex]
9. If books bought at prices from Rs. 150 to Rs. 300 are sold at prices ranging from ?250 to Rs. 350, what is the greatest possible profit that might be made in selling 15 books ?

Answer Option D
Explanation :
Minimum cost price = 150 [latex]\times [/latex] 15 = Rs. 2250
Maximum selling price = 350 [latex]\times [/latex] 15 = Rs. 5250
Gain = 5250 – 2250 = Rs. 3000
10. A certain amount of money at r %, compounded annually after two and three years becomes Rs. 1440 and Rs. 1728 respectively, r is

Answer Option D
Explanation :
I the principal be Rs. P, then
A = P [latex]{(1 + \frac {r}{100})}^{T}[/latex]
[latex]\Rightarrow 1440 = P {(1 + \frac {r}{100})}^{2}[/latex]...(i)
[latex]\Rightarrow 1728 = P {(1 + \frac {r}{100})}^{3}[/latex]...(ii)
On dividing equation (ii) by (i),
[latex] \frac {1728}{1440} = 1 + \frac {r}{100}[/latex]
i.e, [latex]\frac {r}{100} = \frac {1728}{1440} - 1[/latex]
= [latex] \frac {1728 - 1440}{1440} = \frac {288}{1440}[/latex]
[latex]\Rightarrow r = \frac {288 \times 100}{1440}[/latex]
= 20 % per annum

RRB JE 2019 Quantitative Aptitude Quiz - Set 46

1. If x = a + [latex]\frac {1}{a}[/latex] and y = a - [latex]\frac {1}{a}[/latex], then the value of [latex]{x}^{4}[/latex] + [latex]{y}^{4}[/latex] - [latex]2{x}^{2}{y}^{2}[/latex] is

Answer Option C
Explanation :
x + y = a + [latex]\frac {1}{a}[/latex] - a - [latex]\frac {1}{a}[/latex] = 2a
and x - y = (a + [latex]\frac {1}{a}[/latex]) - ( a - [latex]\frac {1}{a}[/latex]) = [latex]\frac {2}{a}[/latex]
i.e, (x + y)(x - y) = 2a [latex]\times \frac {2}{a}[/latex] = 4
[latex]{({x}^{2} -{y}^{2})}^{2} = {4}^{2} = 16[/latex]
[latex]{x}^{4} + {y}^{4} - 2{x}^{2}{y}^{2}[/latex] = 16
2. A radio set, marked 750 is sold at 570. The rate of discount offered is

Answer Option C
Explanation :
Market rate of radio = ? 750
Selling price of Radio = ? 570
Hence discount = Market Price – selling price
Discount = 750 – 570 = 180
Discount % = [latex] \frac {(discount×100)}{market price}[/latex] = [latex] \frac {180 \times 100}{750}[/latex]
= [latex] \frac {360}{15}[/latex] = 24 %
3. A sells an article to B at a gain of 10 %, B sells it to C at a gain of 5 %. If C pays 462 for it, what did it cost to A?

Answer Option D
Explanation : Let price of an article = X
A sells an article to B at a gain of 10 %
Hence A’s profit = [latex] \frac {10 x}{100}[/latex] = [latex] \frac {x}{10}[/latex]
S. P. for A = x + [latex] \frac {x}{10}[/latex] = [latex] \frac {11 x}{10}[/latex]
[latex] \frac {11 x}{10}[/latex] Rs. Is C. P. for B
Now B sells it to C at a gain of 5%.
Hence B’s profit = [latex]5 \times \frac {\frac {11 x}{10}}{100} = \frac {11 x}{200}[/latex]
Hence S. P. for B =[latex]\frac {11x}{10} + \frac {11x}{200} = 220 x + \frac {11x}{200} = \frac {231x}{200}[/latex]Rs
This [latex]\frac {231X}{200}[/latex] Rs is C. P. for C
But given that C pays 462
Therefore
[latex]\Rightarrow X = 200 \times \frac {462}{231}[/latex]
= 200 [latex]\times[/latex] 2 = 400
Hence A cost Rs 400 for article.
4. A sold an article sold for Rs.8,750 with 25% profit on the price he bought. And he bought the article at 30% concession on the marked price. What was the marked price?

Answer Option D
Explanation :
Let marked Price = 100 unit
CP = 70 unit (30 % discount)
SP = 70 *125/100 = 70 * 5/4 = 350/4 = 87.5 unit
87.5 unit = 8750 Rs
100 unit = 10000 Rs
5. Is 999 [latex] \frac {998}{999}\times 999[/latex] equal to:

Answer Option A
Explanation :
999 [latex] \frac {998}{999}\times 999[/latex]
[latex](999 + \frac {998}{999})\times 999[/latex]
[latex] {999}^{2} + 998[/latex]
[latex] {(1000 - 1)}^{2} + 998[/latex]
= 1000000 - 2000 + 1 + 998
= 998999
6. The whole surface of a cube is 150 sq. cm. Then the volume of the cube is

Answer Option A
Explanation :
Whole surface of a cube is =150
or, 6[latex]{a}^{2}[/latex] = 150
a = 5
So, Volume of cube = [latex]{a}^{3}[/latex]= 125
7. Acid and water are mixed in a vessel A in the ratio of 5:2 and in the vessel B in the ratio 8:5. In what proportion should quantities be taken out from the two vessels so as to form a mixture in which the acid and water will be in the ratio of 9:4?

Answer Option A
Explanation :
Let the amount of mixture in vessels A and B are X and Y Liters respectively. [latex] \frac {5X}{7}[/latex]
Hence amount of acid in vessels A = [latex] \frac {5X}{7}[/latex]
Amount of water in vessel A = [latex] \frac {2X}{7}[/latex]
Hence amount of acid in vessels B = [latex] \frac {8Y}{13}[/latex]
Amount of water in vessel B = [latex] \frac {5Y}{13}[/latex]
Therefore amount of acid in vessel C = ([latex] \frac {5X}{7}[/latex]) + ([latex] \frac {8Y}{13}[/latex]) = [latex] \frac {(65X + 56Y)}{91}[/latex]
Therefore amount of water in vessel C = (2X/7) + (5Y/13) = (26X + 35)/91
Given: [latex] \frac {(65X + 56Y)}{91}[/latex] : [latex] \frac {(26X + 35Y)}{91}[/latex] = 9 : 4
4(65X + 56Y) = 9(26X + 35Y)
260X + 224Y = 234X + 315Y
26X = 91Y [latex]\rightarrow \frac {X}{Y}[/latex]
= [latex] \frac {91}{26}[/latex]
= [latex] \frac {7}{2}[/latex]
X : Y = 7 : 2 [latex] \frac {(26X + 35Y)}{91}[/latex]
8. The greatest value among the fractions [latex] \frac {2}{7}, \frac {1}{3}, \frac {5}{6}, \frac {3}{4} [/latex] is:

Answer Option B
Explanation :
[latex] \frac {5}{6}, > \frac {3}{4}, > \frac {1}{3}, > \frac {2}{7} [/latex]
The greatest is [latex] \frac {5}{6}[/latex]
9. A number is greater than thrice its reciprocal by [latex] \frac {26}{3} [/latex]. What is the number?

Answer Option C
Explanation :
Let the number be x
Then its reciprocal will be [latex] \frac {1}{x} [/latex]
[latex]\Rightarrow x - \frac {3}{x} = \frac {26}{3} [/latex]
[latex]3{x}^{2} - 9 - 26x = 0 [/latex]
[latex]3{x}^{2} - 26x - 9 = 0 [/latex]
[latex]3{x}^{2} - 27x + x - 9 = 0 [/latex]
3x (x - 9) + 1 (x - 9) = 0
(x - 9)(3x + 1) = 0
x = 9
10. The printed price of a book is 320. A retailer pays 244.80 for it. He gets successive discounts of 10% and an another rate. His second rate is :

Answer Option A
Explanation :
The printed price of a book = 320
A retailer pays 244.80 for it
First discount = 10%
Hence discount price =320 [latex]\times \frac {10}{100} [/latex] = 32 Rs
Hence price after first discount = 320 – 32 = ? 288
Let second discount = x%
Price after second discount = 244.80
Hence
[latex]288 \times \frac {x}{100} = 288 - 244.80 = 43.20 [/latex]
[latex]\Rightarrow = 43.20 \times \frac {100}{288} = 15 [/latex]
Hence second % discount = 15%

RRB JE 2019 Quantitative Aptitude Quiz - Set 47

1. The length of a road is one kilometer. The number of plants required for plantation at a gap of 20 meters in both sides of the road is

Answer Option A
Explanation :
No. of trees to be planted at one side of the road = [latex] \frac {1000}{20} [/latex] + 1 = 51 (There will be 1 more tree due to the inclusion of corners)
Thus number of trees required to be planted on both sides = 51 + 51 = 102
Hence Option A is correct
2. P and Q can nish a work in 24 and 12 days respectively. In how many days can they complete 50% of the work if they work together?

Answer Option C
Explanation :
It is given here that, P and Q can finish a work in 24 and 12 days respectively. Let the total work be 24 units (LCM of 24 and 12), then
Efficiency of P = [latex] \frac {24}{24} [/latex] = 1 units/day
And that of Q = [latex] \frac {24}{12} [/latex] = 2 units/days
Number of days needed to finish half of the work together = [latex] \frac {12}{1 + 2} [/latex]
= 4 days
3. If x - y = [latex]\frac {x + y}{7} = \frac {xy}{4}[/latex] the numerical value of xy is

Answer Option A
Explanation :
let all equal to k
x - y = k
x + y =7k
solving both x = 4k and y = 3k
Using in xy = 3k
12[latex]{k}^{2}[/latex] = 3k, k = [latex]\frac {1}{3}[/latex]
So, xy = [latex]\frac {4}{3}[/latex]
4. If the cost price of 25 articles is equal to selling price of 20 articles, then the gain percentage i

Answer Option C
Explanation :
Let Cost price of 1 article be Re.1.
Therefore, Cost price of 25 articles = ?25.
Selling price of 20 articles = 25
Hence S.P. of 1 article = 25/20 = 5/4
Hence gain = [latex]\frac {5}{4} - 1 = \frac {1}{4}[/latex]
% gain = [latex] \frac {\frac {1}{4} \times 100}{1}[/latex] = 25%
5. If diagonals of a rhombus are 24 cm and 32 cm, then perimeter of that rhombus is

Answer Option A
Explanation :
Each side of the rhombus can be found out by using pythagorus theorem & we know that perimeter will be 4 times of the side.
[latex]{12}^{2} + {16}^{2}[/latex] = 20
P = 4 x 20
p = 80 cm
6. The units digit of the expression [latex]{25}^{6251} + {36}^{528} + {73}^{54}[/latex]is

Answer Option D
Explanation :
Unit digit of powers of 5 is always 5
Unit digit of powers of 6 is always 6
Now unit digit of powers of 3
[latex]{3}^{1}[/latex] = 3
[latex]{3}^{2}[/latex]= 9
[latex]{3}^{3}[/latex]= 9 x 3 = 27
[latex]{3}^{4}[/latex] =7 x 3 =21
[latex]{3}^{5}[/latex] = 1 x 3 =3
Hence unit digit repeats after every 4 powers, so the cyclicity of powers of 3 is 4
So the unit digit of [latex]{73}^{54} = {3}^{54} = {3}^{2}[/latex] = 9
Required unit digit= sum of unit digits = 5 + 6 + 9 = 20 = 0
7. A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm. The height of the cone is

Answer Option B
Explanation :
Volume of ball = [latex]\frac {4}{3}\pi {r}^{3}[/latex] = [latex]\frac {4}{3} \times \frac {22}{7} {3}^{3} = \frac {792}{7}[/latex]
Volume of cone = Volume of ball
or, [latex]\frac {4}{3}\pi {R}^{3} h = \frac {792}{7}[/latex]
or, [latex]\frac {1}{3} \times \frac {22}{7} {6}^{2} \times h = \frac {792}{7}[/latex]
or, h = 3cm
8. The cost of an article worth Rs 100 is increased by 10% first and again increased by 10%. The total increase in rupees is

Answer Option B
Explanation :
Percentage effect = 10 + 10 + [latex]\frac {10 \times 10}{100}[/latex] = 21%
Therefore increase = Rs 21
Hence Option B is correct
9. If [latex](\frac {(1 - cosA)}{2})[/latex] , then the value of x is

Answer Option D
Explanation :
[latex](\frac {(1 - cosA)}{2})[/latex] = x
cosA = 1 - [latex]2{sin}^{2}(\frac {A}{2})[/latex]
i.e, = [latex]\frac {1}{2}[1 - (1 - 2{sin}^{2}(\frac {A}{2})[/latex]
x = [latex]{sin}^{2}(\frac {A}{2})[/latex]
10; Painter ‘A’ can paint a house in 50 days and ‘B’ can do it in 25 days. With help of ‘C’, they did the job in 10 days only. Then, ‘C’ alone can do the job in

Answer Option A

RRB JE 2019 Quantitative Aptitude Quiz - Set 48

1. If 3x + [latex]\frac {1}{2x}[/latex] = 5 then the value of is [latex]8 {x}^{3} + \frac {1}{27 {x}^{3}}[/latex] :

Answer Option B
Explanation :
Multiply this by [latex]\frac {2}{3} [/latex]

3x + [latex]\frac {1}{2x} = 5[/latex]
2x + [latex]\frac {1}{3x} = \frac {10}{3}[/latex]
Cubing it [latex]{(\frac {10}{3})}^{3} -3 \times \frac {2}{3}[ \frac {10}{3} ][/latex]
[latex]\frac {1000}{27} - \frac {20}{3} = \frac {820}{2} = 30\frac {10}{27}[/latex]
2. Value of ([latex]\frac {(4{a}^{2} + 12ab + 9{b}^{2})}{(2a + 3b)}[/latex]) is

Answer Option B
Explanation :
[latex]\frac {(4{a}^{2} + 12ab + 9{b}^{2})}{(2a + 3b)}[/latex]
[latex]\frac {{2a + 3b}^{2})}{(2a + 3b)}[/latex] = 2a + 3b
3. If [latex]{(a – 1)}^{2} + { (b + 2)}^{2} + {(c + 1)}^{2}[/latex] = 0, then the value of 2a - 3b + 7c is

Answer Option
Explanation :
[latex]{(a – 1)}^{2} + { (b + 2)}^{2} + {(c + 1)}^{2}[/latex] = 0
Hence [latex]{(a – 1)}^{2} = 0; { (b + 2)}^{2} = 0; {(c + 1)}^{2} = 0;[/latex]
a = 1; b= -2; c = -1
i.e, 2a - 3b + 7c = 2 - 3 (-2) -7 = 2 + 6 - 7 = 8 - 7 = 1
4. A number is greater than five times its reciprocal by [latex]\frac {19}{2}[/latex]. Find the number.

Answer Option C
5. The length of the diagonal of a square is 14 cm. What is area of this square?

Answer Option C
Explanation :
Diagonal of a square = a[latex]\sqrt {2}[/latex] = 14
a = 7[latex]\sqrt {2}[/latex] cm
Area of the square = [latex]{a}^{2}[/latex]
= [latex]{7(\sqrt {2})}^{2}[/latex]
= 98 sq. cm
6. The price of an article has been reduced by 25%. In order to restore the original price, the new price must be increased by

Answer Option C
Explanation :
100 - 25 = 75
In 75 the increase to be 25
i.e, 100, the increase to be = [latex]\frac {25}{75}\times 100 = 33 \frac {1}{3}[/latex]
7. In an examination 80% of the boys passed in English and 85% passed in Mathematics, while 75% passed in both. If 45 boys failed in both, the number of boys who sat for the examination was

Answer Option B
Explanation :
Let X be the total number of students sat for the examination
80% of the boys passed in English, 85% passed in Mathematics and 75% passed in both subjects.
Hence % of failed students in both subjects = 100 – (85+80 – 75) = 100 – 90 = 10%
{Using the formula n(A or B ) = n(A)+n(B) – n(A and B)}
If 45 boys failed in both
Therefore 10% of students = 45
[latex]\frac {10X} {100}[/latex] = 45 ? X = [latex]\frac {4500} {10}[/latex] = 450
8. A double bed is marked at Rs. 7,500. The shopkeeper allows successive discounts of 8%, 5% and 2% on it. What is the net selling price?

Answer Option C
Explanation :
SP = 7500 [latex]\times \frac {92} {100} \times \frac {95} {100}\times \frac {98} {100}[/latex]
SP = 6,423.90
9. By selling a table for Rs. 1140, a man loses 5%, in order to the gain 5%, the table must be sold for

Answer Option A
Explanation :
5% = [latex]\frac {1} {20}[/latex]
in case of loss 19 unit = 1140
1 unit = 60
in case of profit 21 unit = 1260
10. What are the roots of the quadratic equation 12[latex]{x}^ {2}[/latex] - 37x - 28 = 0 ?

Answer Option C
Explanation :
12[latex]{x}^ {2}[/latex] - 37x 28 = 0
12[latex]{x}^ {2}[/latex] - 49x + 12x - 28 = 0
7x (3x - 7) + 4 (3x - 7) = 0
(7 + 4)(3x - 7) = 0
roots of the quadratic equation = [latex]\frac {7} {3}[/latex],[latex]\frac {-4} {7}[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 49

1. Successive discounts of 20% and 10% are equivalent to a single discount of :

Answer Option A
Explanation :
We have two successive discount be x% and y%, then equivalent discount
= (X + Y - [latex]\frac {xy}{100}[/latex]) %
= (20 + 10 - [latex]\frac {20 \times 10}{100}[/latex])
= (30 - 2)%
= 28%
2. What is the distance between the points (4,7) and ( - 1, - 5)?

Answer Option B
Explanation :
Distance between two points = [latex]\sqrt {{(x" - x')}^{2} + {(y" -y')}^{2}}[/latex]
= [latex]\sqrt {{(-1 -4)}^{2} + {(-5 -7)}^{2}}[/latex]
= 13 units
3. Profit of Rs 144000 has to be divided among three partners Akram, Bipin and Chintan in the ratio 3:2:7. How much Rs. Chintan get?

Answer Option A
Explanation :
Profit = 144000
Ratio in which profit is to divided between Akram, Bipin, chintan = 3:2:7
Chintan profit = 144000 [latex]\times \frac {7}{12}[/latex] = 84000
4. Two years ago the average age of a family of 8 members was 18 years. After the addition of a baby, the average age of the family remains the same today. What is the age of the baby?

Answer Option A
Explanation :
Two year ago the average age of a family with 8 members = 18
Hence 2 years ago sum of ages of these 8 members = 18 × 8 = 144
After 2 year sum of their ages = 144 + (2 × 8) = 144 + 16 = 160
Let an age (X years)of a baby is also added with these members, now sum of ages = 160 + X
Now average = [latex] \frac {(160 + X)}{9}[/latex]
Given that this new average is same age average age of 8 family members 2 years ago
Hence [latex] \frac {(160 + X)}{9}[/latex]= 18 ? 160 + X = 162 ? X = 2 years
Therefore age of baby = 2 years
5. The reciprocal of the sum of the reciprocals of [latex] \frac {3}{8}[/latex] and [latex] \frac {8}{11}[/latex] is:

Answer Option B
Explanation :
Sum of the reciprocals of [latex] \frac {3}{8}[/latex] and [latex] \frac {8}{11}[/latex] = [latex] \frac {64 + 33}{24}[/latex]
= [latex] \frac {97}{24}[/latex]
Thus, the reciprocal of the sum of the reciprocals of[latex] \frac {3}{8}[/latex] and [latex] \frac {8}{11}[/latex] is [latex] \frac {24}{97}[/latex]
6. A fraction becomes [latex] \frac {1}{3}[/latex] when 1 subtracted from both the numerator and the denominator. The same fraction becomes [latex] \frac {1}{2}[/latex] when 1 is added to both the numerator and the denominator. The sum of numerator and denominator of the fraction is

Answer Option A
Explanation :
Let the original fraction be [latex] \frac {x}{y}[/latex]
[latex] \frac {x - 1}{y - 1}[/latex] = [latex] \frac {1}{3}[/latex]
3x - 3 = y - 1
3x - y = 2 .………… (i)
Again, [latex] \frac {x + 1}{y + 1}[/latex] = [latex] \frac {1}{2}[/latex]
2x + 2 = y + 1
2x - y = -1 .………… (ii)
From equation (i) – (ii)
3x - y - 2x + y = 2 + 1
x = 3
From equation (i)
3 [latex]\times [/latex] 3 - y = 2
y = 9 - 2 = 7
x + y = 3 + 7 = 10
7. A godown is 15 m long and 12 m broad. The sum of the areas of the floor and the celling is equal to the sum of areas of the four walls. The volume (in m3) of the godown is:

Answer Option B
Explanation :
If the height of the godown be h meter, then
2 (15 [latex]\times [/latex] 12) = 2 [latex]\times [/latex] h (15 + 12)
27h = 15 [latex]\times [/latex] 12
h = [latex]\frac {15 \times 12}{27} = \frac {20}{3}[/latex]
Volume of the godown
[latex]\frac {15 \times 12 \times 20}{3} [/latex]
= 1200 cu. m
8. The total number of prime factors in [latex]{4}^{10}\times{7}^{3}\times{16}^{2}\times 11 \times{10}^{2}[/latex] is

Answer Option C
Explanation :
[latex]{4}^{10}\times{7}^{3}\times{16}^{2}\times 11 \times{10}^{2}[/latex]
[latex]{2}^{20}\times{7}^{3}\times{2}^{8}\times 11 \times{2}^{2}\times{5}^{2}[/latex]
[latex]{2}^{30}\times{5}^{2}\times{16}^{2}\times {7}^{3}\times {11}^{1}[/latex]
Total number of prime factors = 30 + 2 + 3 + 1 = 36
9. If 18, x and 50 are in continued proportion then the value of x will be how much?

Answer Option A
Explanation :
[latex]\frac{18}{x} = \frac{x}{50}[/latex]
[latex]{x}^{2} = 18 \times 50[/latex]
= 900
x = [latex]\sqrt{900}[/latex] = 30
10. Anil started a business with an investment of Rs. 25,000. after 3 months, vishal joined his business with a capital of Rs. 30,000 At the end of the year, they have made a profit of Rs. 19,000. what will be anil’s share in the profit?

Answer Option A
Explanation :
Ratio of the equivalent capitals of Anil and Vishal for 1 month
= 25000 x 12 : 30000 x 9
= 25 x 12 : 30 x 9
= 10 : 9
sum of the terms of ratio =10 + 9 = 19
Anil’s share = [latex]\frac {10}{9}[/latex] x 19000
= Rs. 10000

RRB JE 2019 Quantitative Aptitude Quiz - Set 50

1. Parallel sides of a trapezium are 26 cm and 40 cm and the area is 792 [latex]{cm}^{2}[/latex]. What is the value of the distance (in cm) between parallel sides?

Answer Option A
Explanation :
Area of trapezium = [latex]\frac {1}{2}[/latex] × (sum of parallel sides) × h
[latex]\frac {1}{2}[/latex] × (26 + 40) × h = 792
[latex]\frac {1}{2}[/latex] × 66 × h = 792
33 h = 792
h = 24
Hence, the correct option is A
2. What is the length of the longest rod that can be placed in a room which is 3 metres long, 4 metres broad and 5 metres high?

Answer Option C
Explanation :
Length of room=3m
And breadth of room = 4m and height ='5'm
Then length of rod = [latex]\sqrt {{3}^{2}+ {4}^{2}+ {5}^{2}}[/latex] = [latex]\sqrt {9 + 16 + 25}[/latex] = [latex]\sqrt {50}[/latex] = 5 [latex]\sqrt {2}[/latex]
3. A is 20% more efficient than B. If B alone can complete a piece of work in 12 days, then A alone can complete the same work in how many days?

Answer Option A
Explanation :
Given: A is 20% more efficient than B.
Let work done by B in one day = 5 units
Total work = efficiency of B* no. of days taken by B to complete the work = 5 * 12 = 60
Since A is 20 % more efficient than B, Efficiency of A = 5* ([latex]\frac {120}{100}[/latex]) = 6 units
No. of days taken by A to complete the work = Total work/ Efficiency of A = 60/6 = 10 days
4. Which one among [latex]\sqrt [3]{6}[/latex], [latex]\sqrt [2]{5}[/latex] and [latex]\sqrt [6]{12}[/latex] is the largest?

Answer Option B
Explanation :
LCM of 3, 2 and 6 = 6
[latex]{(6)}^{\frac {1}{3}}[/latex] = [latex]{{(6)}^{2}}^{\frac {1}{6}}[/latex]
[latex]{(5)}^{\frac {1}{2}}[/latex] = [latex]{{(5)}^{3}}^{\frac {1}{6}}[/latex]
[latex]{12}^{\frac {1}{6}}[/latex]
5. The value of [latex]\sqrt {{10}^{-5}\times 0.25}[/latex] is

Answer Option B
Explanation :
Expression = [latex]\sqrt {{10}^{-5}\times 0.25}[/latex]
= [latex]\sqrt {\frac {0.25} {{10}^{5}}}= \sqrt {\frac {25} {{10}^{5} \times {10}^{2} }}[/latex]
= [latex]\sqrt {\frac {25} {{10}^{8}}}= \sqrt {\frac {5} {{10}^{4}}}[/latex] = 0.005
6. A sofa-set listed at `800 is sold to a retailer at successive discounts of 25% and 15% by the wholesaler. What is the selling price of the sofa-set?

Answer Option B
Explanation :
After 25% discount, new price = 800 - 200 = 600
After 15% more discount, price = 600 - 90 = 510
7. A sector is formed by opening out a cone of base radius 8 cm and height 6 cm. Then the radius of the sector is (in cm)

Answer Option C
Explanation :
Radius of sector = Slant height of cone
=[latex]\sqrt {{h}^{2} + {r}^{2}}[/latex]
= [latex]\sqrt {{6}^{2} + {8}^{2}}[/latex] = [latex]\sqrt {36 + 64}[/latex] = [latex]\sqrt {100}[/latex] = 10 cm
8. 3 men and 7 women can do a job in 5 days, 5 days, while 4 men and 6 women can do it in 4 days. The number of days required for a group of 10 women working together, at the same rate as before, to finish the same job is :

Answer Option D
Explanation :
3 [latex]\times[/latex] 5 men + 7 [latex]\times[/latex] 5 women
= 4 [latex]\times[/latex] 4en + 6 [latex]\times[/latex] 4 women
16 men - 15 men = 35 women - 24 women
1man = 11women
3men + women = 40 women
i.e, [latex]{M}_{1} {D}_{1}[/latex] = [latex]{M}_{2} {D}_{2}[/latex]
40 [latex]\times[/latex] 5 = 10 [latex]\times[/latex] [latex]{D}_{2}[/latex]
[latex]{D}_{2}[/latex] = 20 days
9. A boy added all natural numbers from 1 to 12, however he added one number twice due to which the sum becomes 80. What is the number which he added twice?

Answer Option B
Explanation :
Let the number added twice be x, then
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + x = 80
x = 2
So, the number added twice is 2
10, A bus starts running with some initial speed and increases every hour by 9 km.hr. If it takes 11 hours to cover a distance of 572 km, then what was the initial speed (in km/hr) of the bus?

Answer Option B
Explanation :
Let the initial speed of the car be x
Speed is increasing by 9km per hour, speed will form an AP with cd = 9 and 1st term = x
Then the final speed of car=x + (11 - 1) × 9 = x + 90
Since for each hour sped is different but constant, so distance covered every hour will also be the same AP
Distance covered =572 = Sum of AP = no of terms/2(Sum of first and last term)
Then 572 = [latex]\frac {11}{2}[/latex] (x +x +90)
11x + 495 = 572
11X = 77
X = 7
So initial speed = 7km/hr

RRB JE 2019 Quantitative Aptitude Quiz - Set 51

1. The ratio of two numbers is 4 : 5. If both numbers are increased by 4, the ratio becomes 5 : 6. What is the sum of the two numbers?

Answer Option D
Explanation :
Let the initial numbers be 4x and 5x
As per question [latex]\frac {4x + 4}{5x + 4} = \frac {5}{6}[/latex]
X = 4
So the sum of two numbers=4x + 5x = 9x = 9 × 4 = 36
2. Average of 36 results is 18. If 2 is subtracted from each result, then what will be the new average of the results?

Answer Option B
Explanation :
Average of 36 results = 18
Since ‘2’ is subtracted from each result, new average will simply be:
18 (Old average) – 2 = 16
Hence, the correct option is B
Alternate Method:
average of 36 results = 18
sum of 36 results = 36*18
since 2 is subtracted from each result, the new sum will be = 36 * 18 - 36 * 2 = 36 * (18-2) = 36 * (16)
new average = [latex]\frac {36 * (16)}{36}[/latex] = 16
3. The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 13m. The area of the field is

Answer Option A
4. What is the distance between the points (4,7) and ( - 1, - 5)?

Answer Option B
Explanation :
Distance between two points = [latex]\sqrt {{(x" - x')}^{2} + {(y" -y')}^{2}}[/latex]
= [latex]\sqrt {{(-1 -4)}^{2} + {(-5 -7)}^{2}}[/latex]
= 13 units
5. Profit of Rs 144000 has to be divided among three partners Akram, Bipin and Chintan in the ratio 3:2:7. How much Rs. Chintan get?

Answer Option A
Explanation :
Profit = 144000
Ratio in which profit is to divided between Akram, Bipin, chintan = 3:2:7
Chintan profit = 144000 [latex]\times \frac {7}{12}[/latex] = 84000
6. Two years ago the average age of a family of 8 members was 18 years. After the addition of a baby, the average age of the family remains the same today. What is the age of the baby?

Answer Option A
Explanation :
Two year ago the average age of a family with 8 members = 18
Hence 2 years ago sum of ages of these 8 members = 18 × 8 = 144
After 2 year sum of their ages = 144 + (2 × 8) = 144 + 16 = 160
Let an age (X years)of a baby is also added with these members, now sum of ages = 160 + X
Now average = [latex] \frac {(160 + X)}{9}[/latex]
Given that this new average is same age average age of 8 family members 2 years ago
Hence [latex] \frac {(160 + X)}{9}[/latex]= 18 ? 160 + X = 162 ? X = 2 years
Therefore age of baby = 2 years
7. The reciprocal of the sum of the reciprocals of [latex] \frac {3}{8}[/latex] and [latex] \frac {8}{11}[/latex] is:

Answer Option B
Explanation :
Sum of the reciprocals of [latex] \frac {3}{8}[/latex] and [latex] \frac {8}{11}[/latex] = [latex] \frac {64 + 33}{24}[/latex]
= [latex] \frac {97}{24}[/latex]
Thus, the reciprocal of the sum of the reciprocals of[latex] \frac {3}{8}[/latex] and [latex] \frac {8}{11}[/latex] is [latex] \frac {24}{97}[/latex]
8. A fraction becomes [latex] \frac {1}{3}[/latex] when 1 subtracted from both the numerator and the denominator. The same fraction becomes [latex] \frac {1}{2}[/latex] when 1 is added to both the numerator and the denominator. The sum of numerator and denominator of the fraction is

Answer Option A
Explanation :
Let the original fraction be [latex] \frac {x}{y}[/latex]
[latex] \frac {x - 1}{y - 1}[/latex] = [latex] \frac {1}{3}[/latex]
3x - 3 = y - 1
3x - y = 2 .………… (i)
Again, [latex] \frac {x + 1}{y + 1}[/latex] = [latex] \frac {1}{2}[/latex]
2x + 2 = y + 1
2x - y = -1 .………… (ii)
From equation (i) – (ii)
3x - y - 2x + y = 2 + 1
x = 3
From equation (i)
3 [latex]\times [/latex] 3 - y = 2
y = 9 - 2 = 7
x + y = 3 + 7 =10
9. A godown is 15 m long and 12 m broad. The sum of the areas of the floor and the celling is equal to the sum of areas of the four walls. The volume (in m3) of the godown is:

Answer Option B
Explanation :
If the height of the godown be h meter, then
2 (15 [latex]\times [/latex] 12) = 2 [latex]\times [/latex] h (15 + 12)
27h = 15 [latex]\times [/latex] 12
h = [latex]\frac {15 \times 12}{27} = \frac {20}{3}[/latex]
Volume of the godown
[latex]\frac {15 \times 12 \times 20}{3} [/latex]
= 1200 cu. m
10. The total number of prime factors in [latex]{4}^{10}\times{7}^{3}\times{16}^{2}\times 11 \times{10}^{2}[/latex] is

RRB JE 2019 Quantitative Aptitude Quiz - Set 52

1.Find the missing number in the following series: 4320, 720, ?, 36, 12, 6

Answer Option A
Explanation :
[latex]\frac {4320}{6}[/latex] = 720
[latex]\frac {720}{5}[/latex] = 144
[latex]\frac {144}{4}[/latex] = 36
[latex]\frac {36}{3}[/latex] = 12
[latex]\frac {12}{2}[/latex] = 6
Thus 144 is the required number.
Hence Option A is correct
2. In each of the following questions, choose the correct alternative from the given ones that will complete the series. 480, ?, 24, 8, 4

Answer Option B
Explanation :
If we look the series in reverse order then,
8 = 4 × 2;
24 = 8 × 3;
[latex]\rightarrow[/latex] Multiplier increased by ‘1’ in every next step.
Therefore ?= 24 × 4 = 96
[latex]\rightarrow[/latex] 480 = 96 × 5
Hence ‘?’ = 96
3. A train 300 metres long is running at a speed of 90 km/hr. How many seconds will it take to cross a 200 metres long train running in the same direction at a speed of 60 km/hr?

Answer Option A
4. Without any stoppage a person travels a certain distance at an average speed of 42 km/hr and with stoppages he covers the same distance at an average speed of 28 km/hr. How many minutes per hour does he stop?

Answer Option D
Explanation :
Here [latex]{S}_{1}[/latex] = 42 and [latex]{S}_{2}[/latex] = 28
i.e, Stoppage time/hr = [latex]\frac{{S}_{1} - {S}_{2}}{{S}_{1}}[/latex]
= [latex]\frac{42 - 28}{42}[/latex] = 20 min
5. In a kilometre race, A beats B by 100 meters. B beats C by 100 meters. By how much meters does A beat C in the same race?

Answer Option C
Explanation :
While A covers 1000 meters, B can cover 900 meters
[latex]\Rightarrow[/latex] While B covers 1000 meters, C can cover 900 meters
[latex]\Rightarrow[/latex] Lets assume that all three of them are running same race. So, when
B runs 900 meters, C can run 900 × 9/10 = 810
[latex]\Rightarrow[/latex] So A can beat C by 190 meters.
6. In a 200 m race A runs at a speed of 1.66 m/s. If A gives a start of 8m to B and still beats him by 24 seconds. What is the speed of B?

Answer Option B
Explanation :
Time taken by A to cover 200 meters = 120 seconds (Approx.)
[latex]\Rightarrow[/latex] time taken by B = 144 seconds
[latex]\Rightarrow[/latex] B takes 144 seconds to cover 192 meters (since A gives a start of 8m to B)
[latex]\Rightarrow[/latex] Speed of B = 192/144 = 1.33 m/s
7. Ram walks at a speed of 24 km/h. Today the day was very hot so walked at [latex]\frac {5}{6}[/latex] of his average speed. He arrived his school 20 minutes late. Find the usual time he takes to cover the distance between his school and home?

Answer Option C
Explanation :
If Ram is walking at [latex]\frac {5}{6}[/latex] of his usual speed that means he is taking 6/5 of usual time.
[latex]\Rightarrow[/latex] [latex]\frac {6}{5}[/latex] of usual time - usual time = 20 mins
[latex]\Rightarrow[/latex] [latex]\frac {1}{5}[/latex] of usual time = 20 mins
[latex]\Rightarrow[/latex] Usual time = 100 mins
8. John travelled from his town to city. John went to city by bicycle at the speed of 25 km/h and came back at the speed of 4 km/h. If John took 5 hours 48 min to complete his journey, what is the distance between town and city?

Answer Option C
Explanation :
Average speed of John = [latex]\frac {2xy}{(x + y)}[/latex] = 2 × 25 ×[latex]\frac {4}{25}[/latex] + 4 = [latex]\frac {200}{29}[/latex] km/h
[latex]\Rightarrow[/latex] Distance travelled = Speed × Time = [latex]\frac {200}{29}[/latex] × [latex]\frac {29}{5}[/latex] = 40 Km
[latex]\Rightarrow[/latex] Distance between city and town = [latex]\frac {40}{2}[/latex] = 20 km
9. Average weight of 20 girls in a class is 25 kg, when the age of a teacher is included the average weight increased by 2 then find out the weight of the teacher.

Answer Option C
Explanation :
It is given that average weight of 20 girls is 25 kg, let the teacher’s weight is X kg
25 = [latex]\frac {Sum of the weights of 20 girls}{20}[/latex]
Sum of the weights of 20 girls = 500 kg
25 + 2 = [latex]\frac {500 + x}{21}[/latex]
X = 67 kg
Hence (III) is the correct answer.
10. [latex]\frac {2sin68°}{cos22°} - \frac {2cot15°}{5tan75°} - \frac {3tan45°, tan20°, tan40°, tan50°, tan70°}{5}[/latex] is equal to

Answer Option C

RRB JE 2019 Quantitative Aptitude Quiz - Set 53

1. If [latex]x= \frac {4 \sqrt {15}}{\sqrt {5} + \sqrt {3}}[/latex], the value of [latex]\frac {x + \sqrt {20}}{x - \sqrt {20}} + \frac {x + \sqrt {12}}{x - \sqrt {12}}[/latex] is

Answer Option B
2. Each interior angle of regular polygon is 144o. The number of sides of the polygon is

Answer Option C
Explanation :
If the number of sides of the polygon be n, then
[latex]\frac {2n - 4}{n}\times 90[/latex] = 144
[latex]\frac {(2n - 4) 5}{n}[/latex]
10n - 20 = 8n
2n = 20 [latex]\Rightarrow[/latex] n = 10
3. The value of [latex]\frac{1}{\sqrt{2} + 2} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} +.... + \frac{1}{\sqrt{100} + \sqrt{99}}[/latex] is

Answer Option B
4. A fraction having denominator 30 and lying between and is :

Answer Option B
[latex]\frac {5}{8}[/latex] = 0.625; [latex]\frac {7}{11}[/latex] = 0.636
[latex]\frac {20}{30}[/latex] = 0.666; [latex]\frac {19}{30}[/latex] = 0.633
5. Kamala bought a bicycle for Rs 1,650. She had to sell it at a loss of 8%. She sold it for

Answer Option A
Explanation :
To have a loss of 8%, she should sell it at 92% of cost price ,
Then SP = [latex]\frac {1650 \times 92}{100}[/latex] = 1518 Rs.
6. A boat travels 24 km upstream in 6 h and 20 km downstream in 4 h. then, the speed of a boat in still water and the speed of current are respectively.

Answer Option B
Explanation :
distance travelled by boat in upstream = 24 km
Time taken = 6 h
Speed of the boat in upstream = [latex]\frac {24}{6}[/latex] = 4 km/h
And distance travelled
By boat in downstream = 20 km
Time taken = 4h
Speed of the boat in downstream = [latex]\frac {20}{4}[/latex] km/h = 5 km/h
Now, speed of the boat in still water = [latex]\frac {1}{2}[/latex] [ speed of the boat in still water +speed of the boat in downstream]
[latex]\frac {1}{2}[/latex] [4 + 5] = [latex]\frac {1}{2}[/latex] × 9 = 4.5 km/h
And speed of the current = [latex]\frac {1}{2}[/latex]
[speed of the boat in downstream – speed of the boat in upstream]
[latex]\frac {1}{2}[/latex] [5 – 4] = [latex]\frac {1}{2}[/latex] × 1 = 0.5 km/h
7. Pipe A can fill a tank in 10 hours, when it works along with pipe B, it can fill the tank in 6 hours. In how many hours can pipe B alone fill the same tank?

Answer Option D
Explanation :
Pipe A can fill the tank in 10 hours, so in one hour A fills = 1/10
Now A and B can fill together it in 6 hours,
Let B can fill the tank alone in X hours.
So, in one hour B fills = [latex]\frac {1}{X}[/latex]
[latex]\frac {1}{X}[/latex] + [latex]\frac {1}{10}[/latex] = [latex]\frac {1}{6}[/latex]
X = 15 hours.
8. Ratio between the monthly incomes of A and B is 9 : 8 and the ratio between their expenditures is 8 : 7. If they save ? 500 each, find A’s monthly income.

Answer Option C
Explanation :
If the ratio of the incomes of A and B be a ; b and that of their expenses be c : d and each saves Rs. x, then,
A’s income = [latex]\frac {ax (d - c)}{ad - bc}[/latex]
= [latex]\frac {9 \times 500 (7 - 8)}{9 \times 7 - 8 \times 8}[/latex]
= 9 [latex]\times[/latex] 500 = Rs. 4500
9. Price of milk has increased by 20%. To keep the expenditure unchanged, the present consumption is to be reduced by:

Answer Option D
Explanation :
Required reduction in percentage = [latex](\frac {100 \times 20}{100 + 20})[/latex]% = [latex]\frac {200}{12}[/latex] %
[latex]\frac {50}{3}[/latex] % = 16 [latex]\frac {2}{3}[/latex] %
Hence Option D is correct
10. The parallel sides of a trapezium are in a ratio 2 :3 and their shortest distance is 12 cm. If the area of the trapezium is 480 sq. cm., the longer of the parallel sides is of length :

Answer Option D
Explanation :
A = 480
[latex]\Rightarrow[/latex] [latex]\frac {1}{2}[/latex] * 12 * (3x + 2x) = 480
[latex]\Rightarrow[/latex] x = 16
The longer side = 16 X 3 = 48

RRB JE 2019 Quantitative Aptitude Quiz - Set 54

1. The average of nine consecutive numbers is n. If the next two numbers are also included the new average will

Answer Option D
Explanation : Fifth number = n
Tenth number = n + 5
Eleventh number = n + 6
New average = [latex]\frac {9n + n + 5 + n + 6}{11}[/latex]
= [latex]\frac {11n + 11}{11}[/latex] = [latex]\frac {(n + 1)\times 11}{11}[/latex] = n + 1
2. If 120 is 20% of a number, then 120% of that number will be

Answer Option B
Explanation :
Let the number be x
Therefore
x × [latex]\frac {20}{100}[/latex] = 120
=>x = 600
120% of 600 = 720
Hence Option B is correct
3. What is to be added to 15% of 160 so that the sum may be equal to 25% of 240?

Answer Option D
Explanation :
15 % of 160 = 24
Let the number be added be x
24 + x = 25 % of 240
24 + x = 60
x = 36
Hence Option D is correct
4. [latex] \frac {4 \frac {2}{7} - \frac {1}{2}}{3 \frac {1}{2} + 1 \frac {1}{7}} = \frac {1}{2 + \frac {1}{2 + \frac {1}{5 - \frac {1}{5}}}}[/latex]

Answer Option C
5. [latex] \sqrt{64}[/latex] - [latex] \sqrt{36}[/latex] is equal to

Answer Option C
Explanation : [latex] \sqrt{64}[/latex] - [latex] \sqrt{36}[/latex] = 8 - 6 = 2
6. The average salary of all staff of a school is ?10,000. The average salary of 20 teaching staff is ?12,000 and that of non-teaching staff is ?5000, the number of non-teaching staff will be

Answer Option B
Explanation : Number of non-teaching staff = x
i.e, 20 [latex] \times[/latex] 12000 + x [latex] \times[/latex] 5000
= (x + 20 ) [latex] \times[/latex] 10000
[latex] \Rightarrow[/latex] 240000 + 5000x = (x +20) [latex] \times[/latex] 10
[latex] \Rightarrow 240 + 5x = (x + 20) \times 10 [/latex]
[latex] \Rightarrow[/latex]10x = 5x = 240 - 200
[latex] \Rightarrow[/latex] 5x = 40
[latex] \Rightarrow[/latex] x = [latex] \frac {40}{5}[/latex] = 8
7. If [latex]3{a}^{2} = {b}^{2}\neq[/latex] then the value of [latex] \frac {{(a + b)}^{3} - {(a - b)}^{3}}{{(a + b)}^{2} - {(a - b)}^{2}}[/latex] is

Answer Option A
Explanation :
[latex]3{a}^{2} = {b}^{2}\neq[/latex] then the value of [latex] \frac {{(a + b)}^{3} - {(a - b)}^{3}}{{(a + b)}^{2} - {(a - b)}^{2}}[/latex] = [latex] \frac {b({b}^{2} + 3{a}^{3})}{{a}^{3} + {b}^{3}}[/latex] = [latex] \frac {3 \times 2{b}^{3}}{4{b}^{2}} = \frac {3b}{2}[/latex]
8. If 12x = [latex]{19}^{2} - {11}^{2}[/latex], what is the value of x?

Answer Option A
Explanation :
12x = [latex]{19}^{2} - {11}^{2}[/latex]
[latex]\Rightarrow 12x = 361 - 121[/latex]
x = [latex] \frac {240}{12}[/latex]
= 20
9. If [latex]\sqrt {y} = 4x, then \frac {{x}^{2}}{y} is:[/latex]

Answer Option A
Explanation :
[latex]\sqrt {y}[/latex] = 4x
y = 16 [latex]{x}^{2}[/latex]
[latex] \frac {{x}^{2}}{y} = \frac {1}{16}[/latex]
10, From a solid cylinder of height 10 cm and base radius of 6 cm, a cone of same height and same base is removed. The volume of the remaining solid is :

Answer Option A
Explanation :
Let ‘r’ be the radius of base of cylinder: r = 6 cm.
As volume of a cylinder of height ‘h’ with radius of base ‘r’= [latex]\pi {r}^{2} h[/latex]
Hence volume of the given cylinder = [latex]\pi {r}^{2} \times 10 = 10 \pi {r}^{2} [/latex]
A cone of same height and same base is removed from cylinder hence volume of that cylinder = [latex] \frac {1}{3} = 10 \pi {r}^{2} [/latex]
Therefore volume of remaining solid
= volume of cylinder - volume of cone
= [latex]\pi {r}^{2} h - \frac {10}{3} \pi {r}^{2} h[/latex]
= [latex]\frac {20}{3} \pi {6}^{2} [/latex]
[latex]h = 240 \pi cu. cm[/latex]

RRB JE 2019 Quantitative Aptitude Quiz - Set 55

1. ABCD is a cyclic trapezium whose sides AD and BC are parallel to each other. If ?ABC = 72 °, then the measure of the ?BCD is

Answer Option D
Explanation :
?ABC + ?CDA = 180 °
?CDA = 180 ° – 72 °
= 108 °
So, ?ADE = 180 ° – 108 °= 72 °
AD || BC
?BCD = ?ADE = 72 ° (corresponding angles)
Hence, option D is correct.
2. The base of a right prism is a trapezium. The lengths of the parallel sides are 8 cm and 14 cm and the distance between the parallel sides is 8 cm. If the volume of the prism is 1056 cm3, then the height of the prism is

Answer Option C
Explanation :
Area of trapezium = [latex]\frac {8(14 + 8)}{2}[/latex]
= 88 square cm
Area of prism= area of base * height
1056 = 88 * h
h= 12 cm
Hence, option C is correct.
3. Mahesh purchased a radio at 9/10 of its selling price and sold it at 8% more than its original selling price. His gain per cent is :

Answer Option
Explanation :
Let the original selling price of radio = Rs. 100
C.P. of ratio = Rs. 90
New selling price = Rs. 108
Gain per cent = 18/90 * 100 = 20%
4. The ratio of the age of a father to that of his son is 5 : 2. If the product of their ages in years is 1000, then the father’s age (in year) after 10year will be :

Answer Option B
Explanation :
Let father’s age be 5x years.
Son’s age = 2x years
5X × 2X = 1000
[latex]{x}^{2} = 100 \Rightarrow x = 10[/latex]
Father’s age after 10l years
= 5x + 10
= 5 × 10 + 10 = 60 years
5. Three solid iron cubes of edges 4 cm, 5 cm and 6 cm are melted together to make a new cube. 62 [latex]{cm}^{3}[/latex] of the melted material is lost due to improper handling. The area (in [latex]{cm}^{2}[/latex] ) of the whole surface of the newly formed cube is

Answer Option A
Explanation :
Volume of all three cubes = [latex]{4}^{3}[/latex] + [latex]{5}^{3}[/latex] + [latex]{6}^{3}[/latex] = 405 cu.cm.
Volume of new cube = 405 – 62 = 343 cu.cm
Edge of cube = [latex]\sqrt[3]{343}[/latex] = 7 cm
Surface Area = 6 × 7 × 7 = 294 sq.cm
6. The marked price is 20% higher than cost price. A discount of 20% is given on the marked price. By this type of sale, there is

Answer Option A
Explanation :
Cost price = 100
Marked price = 120
Selling price = [latex]\frac {120 \times 80}{100} [/latex]
= 96
i.e, Loss = 4 and loss per cent = 4%
7. If in ?ABC, ?ABC = 5 ?ACB and ?BAC = 3 ?ACB, then ?ABC =?

Answer Option C
8. 360 sq. cm and 250 sq.cm are the areas of two similar triangles. If the length of one of the sides of the first triangle be 8cm, then the length of the corresponding side of the second triangle is

Answer Option C
Explanation :
Let AB = 8cm
As the triangle are similar
So area of trinagle ABC/ ARea of trinagle DEF= [latex]\frac {8 \times 8}{DE \times DE} [/latex]
[latex]\frac {360}{250} [/latex] = [latex]\frac {64}{DE \times DE} [/latex]
DE =[latex]\frac {5 \times 8}{6} [/latex] = [latex]\frac {40}{6} [/latex] = 6 [latex]\frac {2}{3} [/latex] cm
9. A fan is listed at `150 and a discount of 20% is given. Then the selling price is

Answer Option C
Explanation :
MP = Rs 150
Discount = 20%
SP = 150 - ([latex]\frac {150 \times 20}{100} [/latex]) = Rs 120
Hence Option C is correct
10. The radius of two concentric circles are 17 cm and 10 cm. A straight line ABCD intersects the larger circle at the point A and D and intersect the smaller circle at the points B and C. If BC = 12 cm, then the length of AD (in cm) is:

Answer Option C

RRB JE 2019 Quantitative Aptitude Quiz - Set 56

1. A man invests a certain sum of money at 6% p.a. simple interest and another sun at 7% p.a. simple interest. His income from interest after 2 years was Rs.354. One-fourth of the first sum is equal to one-fifth of the second sum. The total sum invested was:

Answer Option b
Explanation :
Let the sums be x and y.
[latex]\frac {x \times 6 \times 2}{100} [/latex] + [latex]\frac {y \times 7 \times 2}{100} [/latex] = 354 or 6x + 7y = 17700
Also, [latex]\frac {x}{4} [/latex] = [latex]\frac {y}{5} [/latex] or 6x + 7y = 17700....{i}
Solving (i) and (ii), we get : x =1200 and y =1500
Total sum = Rs.2700
2. If the simple interest on a certain sum for 15 months at 7 [latex]\frac {1}{2} [/latex] % per annum exceeds the simple interest on the same sum for 8 months at 12 [latex]\frac {1}{2} [/latex]% per annum by Rs.32.50, then the sum (in Rs.) is :

Answer Option C
Explanation :
Let the sum be Rs. x. then, (x [latex]\times \frac {15}{2} \times \frac {5}{4} \times \frac {1}{100}[/latex]) - [latex]\frac {25}{2} \times \frac {2}{3} \times \frac {1}{100}[/latex] = 32.50
[latex]\Rightarrow \frac {75 x}{8} - \frac {25x}{3}[/latex] = 3250
x = [latex]\frac {3250 \times 24}{25}[/latex] = 3120
3. Mr Scindia after selling 5.5% stock at Rs. 92 realizes Rs 32200. Then he invested 1/3 of the amount in 4.5% stock at Rs. 92, 2/5 of the amount at Rs. 115 in 5% stock and the remaining in 6% stock at Rs. 56. The change in his income is :

Answer Option C
Explanation :
Let Mr. Scindia has x shares of 5.5%
x * 92 = 32,200
x = 350 shares
Income = 350 * 5.5 = 1925
Now, after investment his income is
([latex] \frac {1}{2} \times \frac {32200}{92} \times 4.5[/latex]) + ([latex]\times \frac {2}{5} \times \frac {32200}{92} \times 5[/latex]) + ([latex]\times \frac {4}{15} \times \frac {32200}{56} \times 6[/latex])
= 525 + 560 + 920 = 2005
Profit = 2005 – 1925 = Rs. 80
4. The length of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, the distance of the other chord from the centre is

Answer Option D
Explanation :
The length of two parallel chords of a circle are PQ = 6 cm and AB = 8 cm
CP and CA are the radius of circle
CD = 4cm, PD = DQ = [latex]\frac {6}{2}[/latex] = 3cm
AK = KB = [latex]\frac {8}{2}[/latex] = 4cm
In triangle CPD
[latex]{CP}^{2} = {CD}^{2} + {PD}^{2}[/latex]
= [latex]{4}^{2} + {2}^{2} [/latex] = 16 + 9 = 25 [latex]\rightarrow[/latex] CP = 5cm
Now in triangle CAK
[latex]{CA}^{2} = {AK}^{2} + {CK}^{2}[/latex]
[latex]{5}^{2} = {4}^{2} + {CK}^{2} \rightarrow 25 = 16 + {CK}^{2} \rightarrow {CK}^{2} = 25 – 16 = 9 \rightarrow[/latex] CK = 3cm
Hence the distance of the chord AB from the centre is 3cm
5. In triangle ABC, AD is the internal bisector of â A, meeting the side BC at D. If BD = 5 cm, BC = 7.5 cm, then AB : AC is

Answer Option A
Explanation :
In triangle ABC, AD is the internal bisector of â A
Given: BD = 5 cm, BC = 7.5 cm [latex]\rightarrow[/latex] DC = 7.5 – 5 = 2.5 cm
Now by the theorem of internal bisector of angle
[latex]\frac {BD}{DC}[/latex] = [latex]\frac {AB}{AC}[/latex]
[latex]\frac {5}{2.5}[/latex] = [latex]\frac {AB}{AC}\rightarrow[/latex][latex]\frac {AB}{AC} = \frac {2}{1}[/latex]
[latex]\rightarrow[/latex] AB : AC = 2 : 1
6. If = b +Cc – 2a, y = c +Ca – 2b, z = a + b – 2c, then the value of [latex]{x}^{2} + {y}^{2} - {z}^{2} + 2xy[/latex]is

Answer Option C
7. The average age of a class of 35 students is 15 years. If the teacher’s age is also included the average age increases by one year. Furthermore, if the average age of the teacher’s family is 40, having a wife and a son’s age is 80% less than his mother’s age, then the age of the teacher’s wife is

Answer Option A
Explanation :
The average age of a class of 35 students is 15 years.
Hence sum of ages of students = 35 × 15 = 525
If the teacher’s age is also included the average age increases by one year ? average = 16 years
Now sum of ages = 16 × 36 = 576
Hence age of teacher = 576 - 525 = 51 years
Now average age of teachers family (teacher, his wife and his son) is 40
Hence sum of ages of teacher’s family = 40 × 3 = 120
Let wife’s age = A years [latex]\frac {A}{5}[/latex]
Given son’s age is 80% less mother’s age, hence son’s age = 20% of mother’s age = [latex]\frac {20A}{100}[/latex] = [latex]\frac {A}{5}[/latex]
Now sum of their ages = 51 + A + [latex]\frac {A}{5}[/latex] = 120 [latex]\rightarrow[/latex] [latex]\frac {6A}{5}[/latex] = 120 - 51 [latex]\rightarrow[/latex] 6A = 5 × 69 [latex]\rightarrow[/latex] X = [latex]\frac {345}{6}[/latex] = 57.5 years
Hence teacher’s wife age = 57.5 years.
8. In a triangle the length of the side opposite the angle which measures 60 ° is 6[latex]\sqrt {3}[/latex] cm. What is the length of the side opposite to the angle which measures 90 °?

Answer Option C
9. If (36 - 16x) - (4x - 8) = 4, then the value of x is

Answer Option b
Explanation :
(36 - 16x) - (4x - 8) = 4
[latex]\rightarrow[/latex] 36 – 16x – 4x + 8 = 4
[latex]\rightarrow[/latex] 20x = 40
[latex]\rightarrow[/latex] x = 2
10. The area of an equilateral triangle is 9[latex]\sqrt {3}[/latex] sq cm, nd height of the triangle ?

Answer Option C
Explanation :
Area of equilateral triangle = [latex]\frac {\sqrt {3}}{4} {(side)}^{2}[/latex]
Let the side of triangle = a
Hence [latex]\frac {\sqrt {3}}{4} {(a)}^{2}[/latex] = 9 [latex]\sqrt {3}[/latex]
A = 6ccm
Height of equilateral triangle= = 3[latex]\sqrt {3}[/latex] cm

RRB JE 2019 Quantitative Aptitude Quiz - Set 57

1. What length of the side of an equilateral triangle. If its area is 64[latex]\sqrt {3}[/latex] sq cm ?

Answer Option B
Explanation :
Area of an equilateral triangle = [latex]\frac {\sqrt {3}}{4} {(side)}^{2}[/latex]
Let the side equilateral triangle= a
[latex]\frac {\sqrt {3}}{4} {(a)}^{2}[/latex] = 64[latex]\sqrt {3}[/latex]
[latex]{(side)}^{2}[/latex] = 256
a = 16 cm
2. An angle is four times its complementary angle. What is the measure of the angle?

Answer Option A
Explanation :
Let the complementary angle be x
The angle =4x
We know that sum of complementary angle =90 °
X +4x= 90 °
X = 18 °
Hence the required angle = 18 ° × 4=72 °
3. What is the equation of the line passing through the point (1,3) and making an angle of -45 ° with the positive x-axis?

Answer Option A
Explanation :
Slope of the given line= tan-45 ° = -1
Given point are (1,3)
y - y1 = m(x - x1)
y -3 = -1(x - 1)
y -3 = -x + 1
x + y = 4
hence option A is the correct answer
4. If x + y = 13 and [latex] {x}^{2} + {y}^{2} = 89, [/latex] then find xy is

Answer Option D
Explanation :
x + y= 13
[latex] {x}^{2} + {y}^{2} = 89 [/latex]
[latex] {(x + y)}^{2}[/latex] = [latex] {x}^{2} + {y}^{2}[/latex] + 2xy
[latex] {(13)}^{2}[/latex] = 89 + 2xy
2xy = 169 - 89 = 80
xy = 40
5. Reflection of the point (3,-6) in the y-axis is

Answer Option C
Explanation :
Given coordinates (3,-6) are in 4th quadrant. When it is reflected through y axis than the following coordinate will be in third quadrant.
In third quadrant x coordinate is also negative
So reflection is (-3-6)
6. In an isosceles trapezium _____.

Answer Option A
Explanation :
In an isosceles trapezium diagonals are congruent.
7. The price of an article is cut by 31%, to restore to its original value, the new price must be increased by

Answer Option B
Explanation :
Let the original price be 100
Then cut price = 100 - 31% 100 = 69
To restore to its original price = [latex]\frac {31}{69} \times 100[/latex] = 44.93 percent
8. Deepanshu lent Rs. 8400 to Jaipal for 15 years and Rs. 5100 to Kareem for 14 years on simple interest at the same rate of interest and received Rs. 17766 in all from both of them as interest. The rate of interest per annum is

Answer Option D
Explanation :
SI = [latex]\frac { P R T }{100}[/latex] Here R1 = R2 = R , P1 =
8400, P2 = 5100, T1 = 15YRS , T2 = 14 yrs
Total SI = 17766
So, [latex]\frac {8400 \times 15 \times R }{100} \frac {5100 \times 14 \times R }{100}[/latex]
17766= ( 84 × R × 15 ) + ( 51 × 14 × R)
17766 = 1260 R + 714 R
17766 = 1974 R
R= 9
9. Three consecutive natural numbers are such that the square of the greatest is greater than the product of the other two by 19. The smallest of these numbers is

Answer Option A
Explanation :
Let the numbers are x, x + 1, x + 2
[latex] {(x+2)}^{2} [/latex] = x(x + 1) + 19
[latex] {x}^{2} [/latex] + 4x + 4 = x2 + x + 19
3x = 15
X = 5
Smallest no = 5
10. If [latex]\sqrt { {(cosec)}^{2}A - 1}[/latex] = x, then the value of x is

Answer Option D
Explanation :
[latex]\sqrt { {(cosec)}^{2}A - 1}[/latex] = x
[latex]\sqrt { {(cot)}^{2}A} = x[/latex] as [latex]1 + \sqrt { {(cot)}^{2}A }[/latex] = [latex]\sqrt { {(cosec)}^{2}A}[/latex]
Cot A = x

RRB JE 2019 Quantitative Aptitude Quiz - Set 58

1. Rajat sells a machine for Rs 53 lakhs at a loss. Had he sold it for Rs 64 lakh, his gain would have been 10 times the former loss. Find the cost price of the machine.

Answer Option D
Explanation :
Let cost price = x lakhs
Selling price = 53 lakhs
Loss = x - 53
If selling price is 64 lakhs then
Gain = 64 - x = 10 (x - 53)
64 - x = 10x - 530
11x = 594
X = 54 lakhs
2. Point Q (-2,b) is the midpoint of segment EF. Co-ordinates of E are (-7,-6) and F are (a,0). What is the value of a and b?

Answer Option A
Explanation :
As Q (-2, b ) is the mid point of segment EF . So mid point coordinate will be
-2 = [latex]\frac {(-7 + a)}{2}[/latex] ; b = [latex]\frac {(-6 + 0)}{2}[/latex]
[latex]\rightarrow[/latex] -7 + a = -4 [latex]\rightarrow[/latex] b =-3
[latex]\rightarrow[/latex] a = 3
3. Slope of the side DA of the rectangle ABCD is [latex]\frac {-3}{4}[/latex]. What is the slope of the side AB?

Answer Option B
Explanation :
As AD is perpendicular to AB , slope of AB= -1 /(slope of AD)
= [latex]\frac {-1}{\frac {-3}{4}}[/latex]
= 4/3
4. If the length of the side of an equilateral triangle is 6 cm, what is its area?

Answer Option D
Explanation :
We know that,
Area of an equilateral triangle = [latex]\frac {\sqrt {3}}{4}[/latex] [latex]{(side)}^{2}[/latex]
= [latex]\frac {\sqrt {3}}{4}[/latex] × 62
= 9[latex]\sqrt {3}[/latex] [latex]{cm}^{2}[/latex]
5. The circumference of a circle is equal to the perimeter of an equilateral triangle. If the radius of the circle is 7 cm what is the length of the side of the equilateral triangle?

Answer Option B
Explanation :
Given that: perimeter of an equilateral triangle = circumference of a circle
Let a side of equilateral triangle be s, then
3s = 2 p r
s = [latex]\frac {2pr}{3}[/latex]
= ([latex]\frac {44}{3}[/latex]) cm
6. What will be the roots of the quadratic equation [latex]{x}^{2} - 25x + 156 = 0[/latex]?

Answer Option A
Explanation :
[latex]\rightarrow {x}^{2} - 25x + 156 = 0 [/latex]
[latex]\rightarrow {x}^{2}[/latex] -13x-12x + 156 = 0
[latex]\rightarrow [/latex] x(x-13)-12(x-13) = 0
[latex]\rightarrow [/latex] (x-12)(x-13) = 0
[latex]\rightarrow [/latex] x = 12, 13
7. If the amount received at the end of 2nd and 3rd year at Compound Interest on a certain Principal is Rs 32448 and Rs 33745.92 respectively, what is the rate of interest?

Answer Option D
Explanation :
We know that,
A = P [latex]{(1 + \frac {r}{100})}^{n}[/latex]
Given that, for 2 years
32448 = P [latex]{(1 + \frac {r}{100})}^{2}[/latex]------(i)
And for 3 years,
33745.92 = P [latex]{(1 + \frac {r}{100})}^{3}[/latex] -------(ii)
Dividing eq. (ii) by eq. (i), we get
= [latex](1 + \frac {r}{100}) = \frac {33745.92}{32448} \times 100[/latex]
= 4%
8. At what point does the line 3x + y = -6 intercept the x-axis?

Answer Option B
Explanation :
At x axis y coordinate will be zero
So put y = 0 in th given equation
3x + 0 = -6
x = -2
hence the required coordinate will be (-2,0)
9. If Giri's salary is [latex]\frac {9}{5}[/latex] times of Hariprasad's and Shaukat's is [latex]\frac {9}{2}[/latex] times of Hariprasad's, what is the ratio of Giri's salary to Shaukat's?

Answer Option D
Explanation :
Let Hariprasad'ssalary = x
So Giri’s salary = [latex]\frac {9}{5}[/latex] (x)
Shaukat’s salary = [latex]\frac {9}{2}[/latex](x)
= [latex]\frac {Giri salary}{Shaukat salary}[/latex] = [latex]\frac {\frac {9x}{5}}{\frac {9x}{2}}[/latex] = [latex]\frac {2}{5}[/latex]
= 2 : 5
10. sin(A+B) + sin(A-B) is equal to

Answer Option
Explanation :
sin(A + B) + sin(A - B)
= sin A cos B + cos A sin B + sin A cos B – cos A sin B
= 2sin A cos B

RRB JE 2019 Quantitative Aptitude Quiz - Set 59

1. Between 200 and 400 how many numbers are divisible by 7?

Answer Option B
Explanation :
First and last number between 200 and 400 which is divisible by 7 are 203 and 399
Let there are n numbers divisible by 7 between 200 and 400, then
Tn = a + (n – 1)d
399 = 203 + (n – 1)7
[latex]\rightarrow[/latex] n = [latex]\frac {196}{7}[/latex] + 1
= 29
2. On dividing 221[latex]{a}^{2}{b}^{2}[/latex] by 13[latex]{b}^{2}[/latex], we get _____.

Answer Option A
Explanation :
[latex]\frac{221{a}^{2}{b}^{2}}{13{b}^{2}}[/latex]
= 17[latex]{a}^{2}[/latex]
3. The point P(3,-2) divides the segment joining the points (x,0) and (0,y) in the ratio 1:3. Find x and y?

Answer Option A
Explanation :
Point P(3,-2) divides the segment in ratio of 1 : 3
We know that the coordinate of point dividing the line segment is given by x = [latex]\frac {m {x}^{2} + nx1}{m + n}[/latex] and y = [latex]\frac {m {y}^{2} + ny1}{m + n}[/latex]
Here m=1 and n=3 and x=3 and y=-2
So 3 = [latex]\frac {1 \times 0\times 3 \times x}{1 + 3}[/latex] and -2 = [latex]\frac {1 \times y\times 3 \times 0}{1 + 3}[/latex]
12 = 3x and -8 =y
X = 4 and y = -8
Hence option A is the correct answer
4. What should be the missing digit so that the number 347_547 becomes exactly divisible by 11?

Answer Option B
Explanation :
Let the missing number be x
Given number is 347 x 547
We know that a number is divisible by 11 if the difference between the sum of digit at even place and odd place is either 0 or multiple of 11
So sum of digit at odd place= 7 + 5 + 7 + 3= 22
Sum of digit at even place= 4 + x + 4 = 8 + x
22 - 8 - x = 14 -x
The number is divisible if x= 14 or x= 3,, 14-14=0 or 14 - 3 = 11 both are multiple of 11
Hence the required number is 3
5. What should be added to 8(3x - 4y) to obtain 18x - 18y?

Answer Option C
Explanation :
Let k should be added to obtain the required polynomial
[latex]\rightarrow[/latex] 8(3x - 4y) + k = 18x - 18y
[latex]\rightarrow[/latex] 24x - 32y + k = 18x - 18y
[latex]\rightarrow[/latex] k = 14y - 6x
6. tan([latex]\frac {A}{2} [/latex]) is equal to

Answer Option C
Explanation :
tan([latex]\frac {A}{2} [/latex])
= [latex]\frac {sin \frac {A}{2}}{cos \frac {A}{2}} [/latex]
use the formula of cos 2A form, to get the next line.
cos A = 2[latex]{sin}^{2}[/latex]([latex]\frac {A}{2} [/latex]) - 1
cos([latex]\frac {A}{2} [/latex]) = [latex]\sqrt{1}[/latex] + cos A
and cos A = 1 - 2[latex]{sin}^{2}[/latex]([latex]\frac {A}{2} [/latex])
sin([latex]\frac {A}{2} [/latex]) = [latex]\sqrt{1}[/latex] - cos A
[latex]\sqrt{\frac {(1 - cos A)}{(1 + cos A)}}[/latex]
= [latex]\sqrt{\frac {(1 - cos A)(1 - cos A)}{(1 + cos A)(1 - cos A)}}[/latex]
= [latex]\sqrt{\frac {{(1 - cos A)}^{2}}{(1 - {cos }^{2}A)}}[/latex]
= [latex]\frac {1 - cos A}{sin A}[/latex]
= [latex]\frac {1}{sin A}[/latex] - [latex]\frac {cos A}{sin A}[/latex]
= cosec A – cot A
7. Two cars travel from city A to city B at a speed of 24 and 32 km/hr respectively. If one car takes 2.5 hours lesser time than the other car for the journey, then the distance between City A and City B is

Answer Option C
Explanation :
Let the distance between two cities be x km, then according to the question
[latex]\frac{x}{24} [/latex] - [latex]\frac{x}{32} [/latex] - 2.5
[latex]\frac{4x - 3x}{96} [/latex] = 2.5
[latex]\rightarrow[/latex] x = 240 km
8. If and then value of x and y is

Answer Option C
Explanation :
7x + 6y = 5xy
[latex]\Rightarrow \frac{7}{y} - \frac{6}{x} = 5[/latex]....(i)
10x - 4y = 4xy
[latex]\Rightarrow \frac{5}{y} - \frac{2}{x} = 2[/latex]....(ii)
Now, let [latex]\frac{1}{x} [/latex] = u and [latex]\frac{1}{y} [/latex] = v, then
Eq. (i) and (ii) will be written as:
7v + 6u = 5
5v – 2u = 2
Solving these two eq., we get
v = [latex]\frac{1}{2} [/latex] and u = [latex]\frac{1}{4} [/latex]
Therefore, x = 4 and y = 2
9. Sister can bake 50 cakes in 25 hours, Sister and Mummy together can bake 75 cakes in 15 hours. How many cakes Mummy can bake in 15 hours?

Answer Option B
Explanation :
Given that, sister can bake 50 cakes in 25 hours, Sister and Mummy together can bake 75 cakes in 15 hours.
Efficiency of Sister = [latex]\frac{50}{25} [/latex]
= 2 cakes/hour
And efficiency of Sister and Mummy together = [latex]\frac{75}{15} [/latex]
= 5 cakes/hour
Efficiency of Mummy = 5 – 2
= 3 cakes/hour
Number of cakes Mummy can make in 15 hours = 15 × 3
= 45 cakes
10. If a - b = 11 and ab = 24,then value of [latex]{a}^{2} + {b}^{2} [/latex] is

Answer Option A
Explanation :
We know that,
[latex]{(a – b)}^{2} [/latex] = [latex]{a}^{2} [/latex] + [latex]{b}^{2} [/latex] – 2ab
[latex]\rightarrow[/latex] [latex]{a}^{2} [/latex] + [latex]{b}^{2} = {(a – b)}^{2} [/latex] + 2ab
= [latex]{(11)}^{2} [/latex] + 2 × 24
= 121 + 48
= 169

RRB JE 2019 Quantitative Aptitude Quiz - Set 60

1. Shopkeeper, sold cashew nuts at the rate Rs 1,260 a kg and bears a loss of 8%. Now if he decides to sell it at Rs 1,386 per kg, what will be the result?

Answer Option A
Explanation :
If SP = Rs 1260 per kg, loss = 8%
Then, CP = 1260 × ([latex]\frac {100}{92} [/latex])
= Rs 1370 per kg
percentage gain on selling nuts at Rs1386 per kg = [latex]\frac {1386 - 1370}{1370}\times 100 [/latex]
1.2%
2. A cone of radius 7 cm and height 12 cm is completely filled with water. This water is emptied into an empty cylindrical vessel of radius 3.5 cm. What will be the height of water in this vessel ?

Answer Option A
Explanation :
Radius of cone = 7cm
Height of cone = 12cm
Volume of cone = [latex]\frac{1}{3}\pi {r}^{2} h [/latex] = [latex]\frac{1}{3}\pi \times 49 \times 12 [/latex]
Volume of cylinder = [latex]\pi {r}^{2} h [/latex] = [latex]\frac{7}{2}\times \pi \times \frac{7}{2} \times h [/latex]
Volume of cone must be equal to volume of cylinder
Hence [latex]\frac{1}{3}\pi \times 49 \times 12 [/latex] = [latex]\pi {r}^{2} h [/latex] = [latex]\frac{7}{2}\times \pi \times \frac{7}{2} \times h [/latex]
[latex]\Rightarrow \frac{12 \times 4}{3} [/latex]
h = 16cm
3. When a discount of 20% is given on an African safari ticket, the prot is 43%. If the discount is 28%, then the prot is

Answer Option D
Explanation :
Let the MP of the ticket be Rs100, then after giving a discount of 20%, SP = Rs 80
Given that, profit after giving this discount = 43%
CP = 80 × [latex]\frac{100}{143} [/latex]
Rs56
Also, If the discount is 28%, then SP = Rs72
Hence, the required profit is [latex]\frac{72 - 56}{56} \times 100[/latex]
28.7%
4. What is the equation of the line which passes through the points (-1,2) and (4,-3)?

Answer Option B
Explanation :
Equation of a line passing through two given points is given by:
y – y’ = [latex]\frac{Y" - Y'}{X" - X'} [/latex] (x – x’)
Here, x’ = -1, x’’ = 4, y’ = 2 and y’’ = -3
Therefore, equation of the line passing through these two points is
y – 2 = [latex]\frac{- 3 - 2}{4 - (-1)} [/latex] (x – (-1))
y – 2 = -1(x + 1)
x + y = 1
5. What will be the sum of the measures all the interior angles of a polygon having 14 sides?

Answer Option B
Explanation :
Sum of interior angle = (n - 2)180 °
[latex]\Rightarrow[/latex]n = number of sides
[latex]\Rightarrow[/latex]so required sum = (14 - 2)180 ° = 2160 °
6. What is the value of tan [latex]\frac{7 \pi}{6} [/latex] ?

Answer Option A
Explanation :
[latex]\Rightarrow[/latex] tan [latex]\frac{7 \pi}{6} [/latex] = tan [latex]\pi + \frac{\pi}{6} [/latex]
[latex]\Rightarrow[/latex] tan [latex]\frac{\pi}{6} [/latex] = tan30° = [latex]\frac{1}{\sqrt {3}} [/latex]
7. Coefcient of [latex]{x}^{2} [/latex] in (x + 9)(6 - 4x)(4x - 7) is

Answer Option C
Explanation :
[latex]\Rightarrow[/latex] (x + 9)(6 - 4x)(4x – 7)
[latex]\Rightarrow[/latex] (6x - 4[latex]{x}^{2} [/latex] + 54 - 36x)(4x - 7)
[latex]\Rightarrow[/latex] ( - 4[latex]{x}^{2} [/latex] - 30x + 54)(4x - 7)
[latex]\Rightarrow[/latex] ( - 16[latex]{x}^{3} [/latex] + 28[latex]{x}^{2} [/latex] - 120[latex]{x}^{2} [/latex] + 210x + 216x - 378)
[latex]\Rightarrow[/latex] -16[latex]{x}^{3} [/latex] - 92[latex]{x}^{2} [/latex] + 426x - 378
Hence coeff of [latex]{x}^{2} [/latex] is -92
8. Curved surface area of a cylinder is 1232 sq cm. If circumference of its base is 154 cm, then what will be the height of the cylinder? (Take p = [latex]\frac{22}{7} [/latex])

Answer Option C
Explanation :
Let r be the radius and h be the height of cylinder
Circumference of base = 2pr = 154
[latex]\Rightarrow 2 \times \frac{22}{7} \times r[/latex] = 154
[latex]\Rightarrow r = \frac{154 \times 7}{44}[/latex] = 24.5 cm
Curved surface are of cylinder = 2prh = 1232
h = [latex]\frac{1232}{2pr} [/latex] = [latex]\frac{1232}{154} [/latex] = 8 cm
9. If 18, x and 50 are in continued proportion then the value of x will be how much?

Answer Option A
Explanation :
[latex]\frac{18}{x} = \frac{x}{50}[/latex]
[latex]{x}^{2} = 18 \times 50[/latex]
= 900
x = [latex]\sqrt{900}[/latex] = 30

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