# RRC Group D Mathematics – CBT

5 Steps - 3 Clicks

# RRC Group D Mathematics – CBT

### Introduction

• The RRC Group D Mathematics section in Computer Based Test (CBT) contains 25 questions with 25 Maximum Marks (Each question carry 1 Mark).

• There will be negative marking and 1/3 marks shall be deducted for each wrong answer.

### Pattern

RRC Group D Level-1 – Computer Based Test (CBT) Pattern:

• The examination duration and number of questions for CBT are indicated below:

Exam
Duration
in Minutes
No of Questions (each of 1 mark) from Total No of
Questions
General
Science
Mathematics General
Intelligence
and Reasoning
General
Awareness
and Current
Affairs
90 25 25 30 20 100

• The examination duration will be 120 Minutes for eligible PwBD candidates accompanied with Scribe.

• The section wise distribution given in the above table is only indicative and there may be some variation in the actual question paper.

### Samples

Number system

1. The weight of the LSB as a binary number is:

A. 1
B. 2
C. 3
D. 4

2. What is the difference between binary coding and binary coded decimal?

A. Binary coding is pure binary.
B. BCD is pure binary.
C. Binary coding has a decimal format.
D. BCD has no decimal format.

3. Convert the binary number 1001.0010 to decimal.

A. 125
B. 12.5
C. 90.125
D. 9.125

4. What is the decimal value of the hexadecimal number 777?

A. 191
B. 1911
C. 19
D. 19111

5. What is the resultant binary of the decimal problem 49 + 1 =?

A. 01010101
B. 00110101
C. 00110010
D. 00110001

BODMAS, Decimals

1. 8 – 4.2 ÷ 6 + 0.3 × 0.4

A. 4.32
B. 7.42
C. 5.42
D. 6.42

Explanation:
8 – 4.2 ÷ 6 + 0.3 × 0.4
= 8 – 0.7 + 0.3 × 0.4, (Simplifying ‘division’ 4.2 ÷ 6 = 0.7)
= 8 – 0.7 + 0.12, (Simplifying ‘multiplication’ 0.3 × 0.4 = 0.12)
= 7.3 + 0.12, (Simplifying ‘subtraction’ 8 – 0.7 = 7.3)
= 7.42, (Simplifying ‘addition’ 7.3 + 0.12 = 7.42)
=7.42

2. 12.8 – 0.4 of (7.2 – 3.7) + 2.4 × 3.02

A. 18.648
B. 17.648
C. 16.648
D. 12.648

Explanation:
12.8 – 0.4 of (7.2 – 3.7) + 2.4 × 3.02
= 12.8 – 0.4 of 3.5 + 2.4 × 3.02, (Simplifying 7.2 – 3.7 = 3.5)
= 12.8 – 0.4 × 3.5 + 2.4 × 3.02, (Simplifying ‘of’)
= 12.8 – 1.4 + 2.4 × 3.02, (Simplifying ‘multiplication’ 0.4 × 3.5 = 1.4)
= 12.8 – 1.4 + 7.248, (Simplifying ‘multiplication’ 2.4 × 3.02 = 7.248)
= 11.4 + 7.248, (Simplifying ‘subtraction’ 12.8 – 1.4 = 11.4)
= 18.648, (Simplifying ‘addition’ 11.4 + 7.248 = 18.648)
= 18.648.

3. 7.6 – [3 + 0.5 of (3.1 – 2.3 × 1.02)]

A. 3.223
B. 5.323
C. 4.223
D. 1.223

Explanation:
7.6 – [3 + 0.5 of (3.1 – 2.3 × 1.02)]
= 7.6 – [3 + 0.5 of (3.1 – 2.346)], (Simplifying 2.3 × 1.02 = 2.346)
= 7.6 – [3 + 0.5 of 0.754], (Simplifying ‘subtraction’ 3.1 – 2.346 = 0.754)
= 7.6 – [3 + 0.5 × 0.754], (Simplifying ‘of’)
= 7.6 – [3 + 0.377], (Simplifying ‘multiplication’ 0.5 × 0.754 = 0.377)
= 7.6 – 3.377, (Simplifying ‘addition’ 3 + 0.377 = 3.377)
= 4.223, (Simplifying ‘subtraction’ 7.6 – 3.377 = 4.223)
= 4.223.

4. If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?

A. 0.172
B. 1.72
C. 17.2
D. 172

Explanation:
$$\frac {29.94}{1.45}$$ = $$\frac {299.4}{14.5}$$

= ($$\frac {2994}{14.5}$$ x $$\frac {1}{10}$$)[ Here, Substitute 172 in the place of 2994/14.5]
= $$\frac {172}{10}$$
= 17.2

5. The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:

A. 0.02
B. 0.2
C. 0.04
D. 0.4

Explanation:
Given expression = (11.98)$$^{2}$$ + (0.02)$$^{2}$$ + 11.98 x x.
For the given expression to be a perfect square, we must have
11.98 x x = 2 x 11.98 x 0.02 or x = 0.04

Fractions

1. 138.009 + 341.981 – 146.305 = 123.6 + ?

A. 120.o85
B. 199.57
C. 295.05
D. None of these

Explanation:
Let 138.009 + 341.981 – 146.305 = 123.6 + z
Then, z = (138.09+341.981) – (146.305+123.6)= 479.99 – 269.905 = 210.085

2. When 0.46 is written in the simplest form, the sum of the numerator and the denominator is?

A. 69
B. 73
C. 96
D. 41

Explanation:
0.46 = 46/100 = 23/50.
Sum of the numerator and denominator is 23 + 50 = 73.

3. How many digits will be there to the right of the decimal point in the product of 95.75 and 0.2554?

A. 6
B. 7
C. 8
D. 9

Explanation:
Sum of the decimal places = 7.
Since the last digit to the extreme right will be zero ( 5 * 4 = 20), so there will be 6 significant digits to the right of the decimal point.

4. What is the difference between the biggest and the smallest fraction among 2/3, 3/4, 4/5 and 5/6?

A. 1/6
B. 1/12
C. 1/20
D. 1/30

Explanation:
Converting each of the given fractions into decimal form, we get
2/3 = 0.66, 3/4 = 0.75, 4/5 = 0.8, 5/6 = 0.833
Since 0.833>0.8>0.75>0.66
So, 5/6 > 4/5 > 3/4 > 2/3
Therefore, Required Difference = 5/6 – 2/3 = 1/6

5. If 1.5x = 0.04y, then the value of ( (y – x) / (y+x) ) is :

A. 73/77
B. 7.3/77
C. 730/77
D. 7300/77

Explanation:
x / y = 0.04 / 1.5 = 4 / 150 = 2 / 75.
= > ( (y – x) / (y+x) ) = (1 – (x / y)) / ( 1 + ( x / y)) = (1 – 2/75) / (1 + 2/75) = 73/77.

LCM

1. The least number which when add 3 to it, completely divided 8, 12 and 18 is?

A. 71
B. 70
C. 69
D. 68

Explanation:
LCM = 72
72 – 3 = 69

2. Four bells begin to toll together respectively at the intervals of 8, 10, 12 and 16 seconds. After how many seconds will they toll together again?

A. 246 seconds
B. 242 seconds
C. 240 seconds
D. 243 seconds

Explanation:
LCM = 240

3. The least number which when diminished by 7 is divisible by 21, 28, 36 and 45 is?

A. 1267
B. 1265
C. 1261
D. 68

Explanation:
LCM = 1260
1260 + 7 = 1267

4. The greatest number of five digits which is divisible by 32, 36, 40, 42 and 48 is:

A. 90730
B. 90725
C. 90715
D. 90720

5. LCM of 87 and 145 are:

A. 1305
B. 435
C. 875
D. 48

HCF

1. A man was employed on the promise that he will be paid the highest wages per day. The contract money to be paid was Rs. 1189. Finally, he was paid only Rs. 1073. For how many days did he actually work?

A. 39
B. 40
C. 37
D. 35

Explanation:

HCF of 1189, 1073 = 29
1073/29 = 37

2. An officer was appointed on maximum daily wages on contract money of Rs. 4956. But on being absent for some days, he was paid only Rs. 3894. For how many days was he absent?

A. 3
B. 4
C. 5
D. 6

Explanation:

HCF of 4956, 3894 = 354
(4956 – 3894)/354 = 3

3. A merchant has three different types of milk: 435 litres, 493 litres and 551 litres. Find the least number of casks of equal size required to store all the milk without mixing.

A. 51
B. 61
C. 47
D. 45

Explanation:

HCF of 435, 493, 551 = 29
(453/29) + (493/29) + (551/29) = 51

4. A wholesale tea dealer 408 kilograms, 468 kilograms and 516 kilograms of three different qualities of tea. He wants it all to be packed into boxes of equal size without mixing. Find the capacity of the largest possible box?

A. 50
B. 36
C. 24
D. 12

Explanation:

HCF of 408, 468, 516 = 12

5. Find the size of the largest square slabs which can be saved on the floor of a room 5 meters 44 cm long and 3 meters 74 cm broad?

A. 56
B. 42
C. 38
D. 34

Explanation:

HCF of 374, 544 = 34

Ratio and Proportion

1. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?

A. Rs. 500
B. Rs. 1500
C. Rs. 2000
D. None of these

Explanation:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x – 3x = 1000
x = 1000.
B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

2. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40: 57. What is Sumit’s salary?

A. Rs. 17,000
B. Rs. 20,000
C. Rs. 25,500
D. Rs. 38,000

Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,$$\frac {2x + 4000}{3x + 4000}$$ = $$\frac {40}{57}$$
57(2x + 4000) = 40(3x + 4000)
6x = 68,000
3x = 34,000
Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

3. If 0.75 : x :: 5 : 8, then x is equal to:

A. 1.12
B. 1.2
C. 1.25
D. 1.30

Explanation:
(x x 5) = (0.75 x 8) x = $$\frac {6}{5}$$= 1.20

4. The fourth proportional to 5, 8, 15 is:

A. 18
B. 24
C. 19
D. 20

Explanation:
Let the fourth proportional to 5, 8, 15 be x.
Then, 5: 8: 15: x
5x = (8 x 15)
x = $$\frac {(8 x 15)}{5}$$ = 24.

5. Two number are in the ratio 3: 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:

A. 27
B. 33
C. 49
D. 55

Explanation:
Let the numbers be 3x and 5x.
Then,$$\frac {3x – 9}{5x – 9}$$ = $$\frac {12}{23}$$
23(3x – 9) = 12(5x – 9)
9x = 99
x = 11.
The smaller number = (3 x 11) = 33.

Percentages

1. After decreasing 24% in the price of an article costs Rs.912. Find the actual cost of an article?

A. 1400
B. 1300
C. 1200
D. 1100

Explanation:
CP* (76/100) = 912
CP= 12 * 100 => CP = 1200

2. 40% of a number is more than 20% of 650 by 190. Find the number?

A. 600
B. 700
C. 800
D. 900

Explanation:
(40/100) * X – (20/100) * 650 = 190
2/5 X = 320
X = 800

3. 60% of a number is added to 120, the result is the same number. Find the number?

A. 300
B. 200
C. 400
D. 500

Explanation:
(60/100) * X + 120 = X
2X = 600
X = 300

4. 96% of the population of a village is 23040. The total population of the village is?

A. 32256
B. 24000
C. 24936
D. 25640

Explanation:
X * (96/100) = 23040
X = 240 * 100
X = 24000

5. If the price has fallen by 10% what percent of its consumption be: increased so that the expenditure may be the same as before?

A. 11%
B. 10%
C. 11 1/9 %
D. 9 1/11 %

Explanation:
100 – 10 = 90
90——10
100——? => 11 1/9%

Mensuration

1. The parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 14 cm. Find the circumference of a semicircle whose diameter is equal to the side of the square. (Round off your answer to two decimal places)

A. 77.14 cm
B. 47.14 cm
C. 84.92 cm
D. 94.94 cm
E. 23.57 cm

Explanation:
Let the side of the square be a cm.
Parameter of the rectangle = 2(16 + 14) = 60 cm Parameter of the square = 60 cm
i.e. 4a = 60
A = 15
Diameter of the semicircle = 15 cm
Circimference of the semicircle
= 1/2(∏)(15)
= 1/2(22/7)(15) = 330/14 = 23.57 cm to two decimal places.

2. A cube of side one-meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?

A. 10
B. 100
C. 1000
D. 10000
E. None of these

Explanation:
Along one edge, the number of small cubes that can be cut
= 100/10 = 10
Along each edge, 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000

3. The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square?

A. 600 cm
B. 800 cm
C. 400 cm
D. 1000 cm
E. None of these

Explanation:
Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.

4. The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square.

A. 200 sq cm
B. 72 sq cm
C. 162 sq cm
D. Cannot be determined
E. None of these

Explanation:
Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l. b. Therefore a cannot be found.
Area of the square cannot be found.

5. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?

A. Rs. 3642.40
B. Rs. 3868.80
C. Rs. 4216.20
D. Rs. 4082.40
E. None of these

Explanation:
Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 * 7) = 315 (13 – 0.04) = 4095 – 12.6 = Rs. 4082.40.

Time and Work

1. A can do a piece of work in 4 days. B can do it in 5 days. With the assistance of C they completed the work in 2 days. Find in how many days can C alone do it?

A. 10 days
B. 20 days
C. 5 days
D. 4 days

Explanation:

C = 1/2 – 1/4 – 1/5 = 1/20 => 20 days.

2. A can do a piece of work in 30 days. He works at it for 5 days and then B finishes it in 20 days. In what time can A and B together it?

A. 16 2/3 days
B. 13 1/3 days
C. 17 1/3 days
D. 16 1/2 days

Explanation:

5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 days

3. A and B can do a piece of work in 12 days and 16 days respectively. Both work for 3 days and then A goes away. Find how long will B take to complete the remaining work?

A. 15 days
B. 12 days
C. 10 days
D. 9 days

Explanation:

3/12 + (3 + x)/16 = 1
x = 9 days

4. A can do a piece of work in 40 days; B can do the same in 30 days. A started alone but left the work after 10 days, then B worked at it for 10 days. C finished the remaining work in 10 days. C alone can do the whole work in?

A. 24 days
B. 30 days
C. 44 days
D. 17 1/2 days

Explanation:

10/40 + 10/30 + 10/x = 1
x = 24 days

5. A work which could be finished in 9 days was finished 3 days earlier after 10 more men joined. The number of men employed was?

A. 18
B. 20
C. 22
D. 24

Explanation:

x ——- 9
(x + 10) —- 6
x * 9 = (x + 10)6
x = 20

Time and Distance

1. The speed of a train is 90 kmph. What is the distance covered by it in 10 minutes?

A. 15 kmph
B. 12 kmph
C. 10 kmph
D. 5 kmph

Explanation:

90 * 10/60 = 15 kmph

2. In what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way?

A. 9 sec
B. 8 sec
C. 7 sec
D. 6 sec

Explanation:

T = 60/36 * 18/5 = 6 sec

3. A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?

A. 50 kmph
B. 52 kmph
C. 54 kmph
D. 56 kmph

Explanation:

S = 240/16 * 18/5 = 54 kmph

4. If a man walks to his office at ¾ of his usual rate, he reaches office 1/3 of an hour late than usual. What is his usual time to reach office?

A. 1 hour
B. 2 hour
C. 3 hour
D. 4 hour

Explanation:

Speed Ratio = 1:3/4 = 4:3
Time Ratio = 3:4
1 ——– 1/3
3 ——— ? è 1 hour.

5. Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance travelled by them if the slower car takes 1 hour more than the faster car.

A. 906 km
B. 960 m
C. 960 km
D. 966 km

Explanation:
60(x + 1) = 64x
X = 15
60 * 16 = 960 km

Simple and Compound Interest

1. On a sum of money, simple interest for 2 years is Rs 660 and compound interest is Rs 696.30, the rate of interest being the same in both cases.

A. 8%
B. 9%
C. 10%
D. 11%

Explanation:
Difference between C.I and S.I for 2 years = 36.30
S.I. for one year = 330.
S.I. on Rs 330 for one year = 36.30
So R% = $$\frac{100 \times 36.30}{330 \times 1}$$ = 11%

2. A sum of money invested at compound interest to Rs. 800 in 3 years and to Rs 840 in 4 years. The rate on interest per annum is.

A. 4%
B. 5%
C. 6%
D. 7%

Explanation:
S.I. on Rs 800 for 1 year = 40
Rate = (100*40)/(800*1) = 5%

3. Mr. Thomas invested an amount of 13,900 divided into two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be 3508, what was the amount invested in Scheme B?

A. 6400
B. 6500
C. 7200
D. 7500

Explanation:
Let the sum invested in Scheme A be x and that in Scheme B be (13900 – x).
Then, (x x 14 x 2)/100 + ((13900 – x) x 11 x 2)/100 = 3508
28x – 22x = 350800 – (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = (13900 – 7500) = 6400.

4. A sum fetched a total simple interest of 4016.25 at the rate of 9 %.p.a. in 5 years. What is the sum?

A. 4462.50
B. 8032.50
C. 8925
D. None of these

Explanation:
Principal = (100 x 4016.25)/(9 x 5)
= 401625/45
= 8925.

5. How much time will it take for an amount of 450 to yield 81 as interest at 4.5% per annum of simple interest?

A. 3.5 years
B. 4 years
C. 4.5 years
D. 5 years

Explanation:
Time = (100 x 81)/(450 x 4.5) years = 4 years.

Profit and Loss

1. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

A. 30%
B. 70%
C. 100%
D. 250%

Explanation:
Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 – 125) = Rs. 295.
Required percentage = ($$\frac {295}{420}$$ x 100)% = $$\frac {1475}{21}$$% = 70% (approximately).

2. A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

A. 3
B. 4
C. 5
D. 6

Explanation:
C.P. of 6 toffees = Re. 1
S.P. of 6 toffees = 120% of Re. 1 = Rs. $$\frac {6}{5}$$
For Rs. $$\frac {6}{5}$$, toffees sold = 6.
For Re. 1, toffees sold = (6 x $$\frac {5}{6}$$ = 5.

3. A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?

A. Rs. 18.20
B. Rs. 70
C. Rs. 72
D. Rs. 88.25

Explanation:
C.P. = Rs. ($$\frac{100}{122.5}$$ x 392) = Rs. $$\frac {1000}{1225}$$ x 392 = Rs. 320
Profit = Rs. (392 – 320) = Rs. 72.

4. A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?

A. Rs. 1090
B. Rs. 1160
C. Rs. 1190
D. Rs. 1202

Explanation:
S.P. = 85% of Rs. 1400 = Rs. ($$\frac {85}{100}$$ x 1400) = Rs. 1190

5. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:

A. Rs. 45
B. Rs. 50
C. Rs. 55
D. Rs. 60

Explanation:
(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)
C.P. of 12 balls = S.P. of 17 balls = Rs.720.
C.P. of 1 ball = Rs. $$\frac {720}{12}$$ = Rs. 60.

Algebra

1. If a * b = 2a – 3b + ab, then 3 * 5 + 5 * 3 is equal to :

A. 22
B. 24
C. 26
D. 28

Explanation:

a * b = 2a – 3b + ab
Þ 3 * 5 = 2 × 3 – 3 × 5 + 3 × 5 = 6
5 * 3 = 2 × 5 – 3 × 3 + 3 × 5
= 10 – 9 + 15 = 16
Therefore, 3 * 5 + 5 *3
= 6 + 16 = 22

2. p and q are positive numbers satisfying 3p + 2pq = 4 and 5q + pq = 3. Find the value of p.

A. 1 or $$\frac {9}{5}$$
B. $$\frac {1}{2}$$ or $$\frac {20}{3}$$
C. 1 or –$$\frac {20}{3}$$
D. $$\frac {1}{2}$$ or –$$\frac {9}{5}$$

3. For what value of ‘a’, the polynomial 2×3+ ax2+ 11x + a + 3, is exactly divisible by (2x – 1) ?

A. 7
B. –7
C. 5
D. –5

4. If a + b + c = 15 and a2+ b2+ c2 = 83, then a3 + b3 + c3 – 3abc = ?

A. 160
B. 175
C. 180
D. 100

5. What will be the value of (x – a)3+ (x – b)3+ (x – c)3 – 3 (x –a) (x – b) (x – c) if a + b+c = 3x ?

A. 1
B. 3
C. 0
D. 5

Geometry and Trigonometry

1. ABC is a right angled triangle with a right angle at A. Points D, E are the middle points of AB and AC respectively. Which of the following relations is correct ?

A. 3 (BE2 + CD2) = 4 BC2
B. 4(BE2 + CD2) = 5 BC2
C. 4(BE2 + CD2) = 3 BC2
D. None of these

2. Two circles whose radii are 10 cm and 8 cm, intersect each other and their common chord is 12 cm long. What is the distance between their centres?

A. 11.27 cm
B. 12.29 cm
C. 12.27 cm
D. 13.29 cm

3. In D ABC, AB = 6 cms, BC = 10 cms, AC = 8cm and AD ^ BC. Find the value of the ratio of BD : DC.

A. 3: 4
B. 9: 16
C. 4: 5
D. 16: 25

4. If cosq + secq = 2, then cos5q + sec5q= ?

A. 1
B. 2
C. –1
D. –2

5. 2 (sin6q + cos6q) – 3 (sin4q + cos4q) + 1 = ?

A. 1
B. 0
C. –1
D. 2

Elementary Statistics

1. A numerical value used as a summary measure for a sample, such as sample mean, is known as a

A. population parameter
B. sample parameter
C. sample statistic
D. the population mean
E. None of the above answers is correct.

Explanation:

If it pertains to sample it is called a statistic, if it pertains to the population it is called a parameter.

2. Since the population size is always larger than the sample size, then the sample statistic

A. can never be larger than the population parameter
B. can never be equal to the population parameter
C. can never be zero
D. can never be smaller than the population parameter
E. None of the above answers is correct.

Explanation:

Sample statistic will depend upon the sample chosen. It can be less than, greater than, equal to the population parameter. It can assume the value of zero.

3. The mean of a sample is

A. always equal to the mean of the population
B. always smaller than the mean of the population
C. computed by summing the data values and dividing the sum by (n – 1)
D. computed by summing all the data values and dividing the sum by the number of items
E. None of the above answers is correct.

Explanation:

Mean= Total of sample values/ sample size

4. The sum of the percent frequencies for all classes will always equal

A. one
B. the number of classes
C. the number of items in the study
D. 100
E. None of the above answers is correct.

Explanation:

If we count the total frequency it is equal to the sample size n. n/n *100= 100

5. Since the mode is the most frequently occurring data value, it

A. can never be larger than the mean
B. is always larger than the median
C. is always larger than the mean
D. must have a value of at least two
E. None of the above answers is correct.

Explanation:

The mean, median and mode values will be distributed according to the skewness of the distribution. Accordingly, the mode can be greater than or less than mean or mode.

Square root

1. If (89)$$^{2}$$ is added to the square of a number, the answer so obtained is 16202. What is the (1/26) of that number?

A. 5.65
B. 2.7
C. 3.5
D. 6.66

Explanation:
Let the number is = x
(89)$$^{2}$$ + x$$^{2}$$ = 16202
x$$^{2}$$ = 8281
x = 91
=> (1/26) of 91 = 3.5

2. The largest perfect square between 4 and 50 is

A. 25
B. 36
C. 49
D. 45

3. 6$$^{3}$$ is equal to

A. 249
B. 241
C. 216
D. 316

4. Cube root of 729 then square it

A. 9
B. 36
C. 81
D. 144

Explanation:
729 = 9 x 9 x 9
=> Cube root of 729 = 9
Now, required square of 9 = 9 x 9 = 81.

5. $$\sqrt{0.0169 \times ?}$$ = 1.3

A. 10
B. 100
C. 1000
D. None of these

Explanation:
$$sqrt {0.0169 \times x}$$ = 1.3
Then, 0.0169x = (1.3)$$^{2}$$ = 1.69
x = $$\frac {1.69}{0.0169}$$ = 100

Age Calculations

1. Raju age after 15 years will be 5 times his age 5 years back, What is the present age of Raju

A. 15
B. 14
C. 10
D. 8

Explanation:
Clearly,
x+15 = 5(x-5)
4x = 40 => x = 10

2. Sachin is younger than Rahul by 7 years. If the ratio of their ages is 7:9, find the age of Sachin

A. 23.5
B. 24.5
C. 12.5
D. 14.5

Explanation:
If Rahul age is x, then Sachin age is x-7,
so (x-7)/x = 7/9
=> 9x-63 = 7x
=> 2x = 63
=> x = 31.5

So Sachin age is 31.5 – 7 = 24.5

3. The ratio between the present ages of P and Q is 6:7. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years

A. 7:8
B. 7:9
C. 3:8
D. 5:8

Explanation:
Let P age and Q age is 6x years and 7x years.
Then 7x – 6x = 4 x = 4
So required ratio will be (6x+4): (7x+4) => 28:32 => 7:8

4. Ages of two persons differ by 16 years. If 6 year ago, the elder one be 3 times as old the younger one, find their present age

A. 12,28
B. 14,30
C. 16,32
D. 18,34

Explanation:
Let the age of younger person is x,
Then elder person age is (x+16)
=> 3(x-6) = (x+16-6) [6 years before]
=> 3x-18 = x+10
=> x = 14.
So other person age is x + 16 = 30

5. The sum of the ages of a father and son is 45 years. Five years ago, the product of their ages was four times the father’s age at that time. The present age of father and son

A. 34,11
B. 35,10
C. 36,9
D. 40,5

Explanation:
Let sons age = x years. Then fathers age = (45 – x)years.
(x—5)(45—x—5) = 4(45- x – 5) hence (x—5) = 4 so x = 9
Their ages are 36 years and 9 years.

Calendar & Clock

1. It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

A. Sunday
B. Saturday
C. Friday
D. Wednesday

Explanation:
On 31$$^{st}$$ December 2005, it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31$$^{st}$$ December 2009, it was Thursday.
Thus, on 1$$^{st}$$ Jan 2010, it is Friday.

2. Today is Monday. After 61 days, it will be:

A. Wednesday
B. Saturday
C. Tuesday
D. Thursday

Explanation:
Each day of the week is repeated after 7 days.
So, after 63 days, it will be on Monday.
After 61 days, it will be on Saturday.

3. On 8$$^{th}$$ Feb 2005, it was Tuesday. What was the day of the week on 8$$^{th}$$ Feb 2004?

A. Tuesday
B. Monday
C. Sunday
D. Wednesday

Explanation:
The year 2004 is a leap year. It has 2 odd days.
The day on 8$$^{th}$$ Feb 2004 is 2 days before the day on 8$$^{th}$$ Feb 2005.
Hence, this day is Sunday.

4. Find the time between 2 and 3 o’clock when the hands of a clock will be together.

A. 10 9/11 minutes
B. 10 10/11 minutes
C. 8 minutes
D. 10 6/11 minutes

5. A watch which gains uniformly is 5 minutes slow at 8 o’clock in the morning on Monday and it is 5 min. 48 sec fast at 8 pm the following Monday. When was it correct?

A. 20 min past 6 on Tuesday
B. 20 min past 6 on Wednesday
C. 20 min past 7 on Wednesday
D. 20 min past 7 on Thursday

Pipes & Cistern

1. Taps X and Y can fill a tank in 30 and 40 minutes respectively.Tap Z can empty the filled tank in 60 minutes.If all the three taps are kept open for one minute each,how much time will the taps take to fill the tank?

A. 48min
B. 72min
C. 24min
D. None of these

Explanation:
Given taps, X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,
part filled by tap X in 1 minute = 1/30
part filled by tap Y in 1 minute = 1/40
Tap Z can empty the tank in 60 minutes. Therefore,
part emptied by tap Z in 1 minute = 1/60
Net part filled by Pipes X,Y,Z together in 1 minute = [1/30 +1/40 – 1/60] = 5/120 = 1/24
i.e., the tank can be filled in 24 minutes.

2. One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 min, then the slower alone will be able to fill the tank in:

A. 81 min
B. 108 min
C. 144 min
D. 192 min

Explanation:
Let the slower pipe alone fill the tank in x minutes.
Then, the faster pipe will fill it in x/3 minutes.
= > $$\frac {1}{x}$$ + $$\frac {3}{x}$$ = $$\frac {1}{36}$$
=> $$\frac {4}{x}$$ = $$\frac {1}{36}$$ => x = 144mins.

3. 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?

A. 8
B. 15
C. 16
D. 18

Explanation:
Capacity of the tank =(12 x 13.5) liters =162 liters.
The capacity of each bucket =9 litres
Number of buckets needed = 162/9 =18.

4. Two pipes A and B can separately fill a cistern in 10 and 15 minutes respectively. A person opens both the pipes together when the cistern should have been was full he finds the waste pipe open. He then closes the waste pipe and in another 3 minutes, the cistern was full. In what time can the waste pipe empty the cistern when fill?

A. 8.21 min
B. 8 min
C. 8.57 min
D. 8.49 min

Explanation:
1/10 + 1/15 = 1/6 x 3 = 1/2
1 – 1/2 = 1/2
1/10 + 1/15 – 1/x = 1/2
x = 8.57 min

5. Pipe A can fill a tank in 16 minutes and pipe B cam empty it in 24 minutes. If both the pipes are opened together after how many minutes should pipe B be closed so that the tank is filled in 30 minutes?

A. 21 min
B. 24 min
C. 20 min
D. 22 min