Q1. Deepa decided to donate 8% of her salary to an orphanage, On the day of donation she changed her mind and donated Rs. 2240 was 80% of what she had decided earlier. How much is Deepa’s salary?

A. Rs. 36000 B. Rs. 42000 C. Rs. 35000 D. Rs. 45000

Answer: C
Solution: Salary [latex]\rightarrow[/latex]100
What She decided to donated = 8
She donated = 8 [latex]\times \frac {80}{100}[/latex] = 6.4
6.4 r [latex]\rightarrow[/latex] 2240
1 r [latex]\rightarrow[/latex] 350
100 r [latex]\rightarrow[/latex] 35000 RS
Q2. When the price of the radio was reduced by 20%, its sale increased by 80%. What was the net effect on the sale?

A. 44% increase B. 44% decrease C. 66% increase D. 75% increase

Answer: A
Solution: Let Price of Radio [latex]\rightarrow[/latex] 100
Sale of Radio [latex]\rightarrow[/latex] 100
Total Sale [latex]\Rightarrow[/latex] 10000
Reduced Price = 100 [latex]\times \frac {80}{100}[/latex] = 80
Increased Sale = 100 [latex]\times \frac {180}{100}[/latex] = 180
Total Sale = 80 [latex]\times[/latex] 180 = 14400
% increase in Sale = [latex]\frac {4400}{10000} \times[/latex] 100 = 44 %
Q3. If the price of sugar is increased by 7%, then by how much percent should a housewife reduce her consumption of sugar, to have no extra expenditure?

A. 7 over 107% B. 107 over 100% C. 100 over 107% D. 7%

Answer: A
Solution: Price Ratio [latex]\Rightarrow[/latex] 100 [latex]\colon[/latex] 107
Consumption [latex](\alpha \frac {1} {Price})[/latex] Ratio
[latex]\Rightarrow[/latex] 107 [latex]\colon[/latex] 100
Reduction In Consumption = [latex]\frac {7} {107} \times[/latex] 100
[latex]\Rightarrow \frac {7} {107}[/latex] %
Q4. A sum of Rs. 4558 is divided among A, B and C such that A receives 20% more than C, and C receives 25% less than B. What is A’s share in the amount?

A. Rs. 1548 B. Rs. 1720 C. Rs. 1290 D. Rs. 1345

Answer: A
Solution: A = [latex]\frac {120} {100}[/latex] C
C = [latex]\frac {75} {100}[/latex] B
B =

A [latex]\colon[/latex] C [latex]\Rightarrow[/latex] 6 [latex]\colon[/latex] 5
C [latex]\colon[/latex] B [latex]\Rightarrow[/latex] 3 [latex]\colon[/latex] 4 ------------------------- A [latex]\colon[/latex] C [latex]\colon[/latex] B [latex]\Rightarrow[/latex] 18 : 15 : 20

A'S Amount = 4558 [latex]\times \frac {18}{53}[/latex]
= 86 [latex]\times[/latex] 18
= 1548
Q5. A spider climbed 62(1/2)% of the height of the pole in one hour and in the next hour it covered 12(1/2)% of the remaining height. If the height of the pole is 192 m, then distance climbed in second hour is:

A. 3 m B. 5 m C. 7 m D. 9 m

Answer: D
Solution: Hight Climbed is [latex]{1}^{st}[/latex] = 192 [latex]\times \frac {5}{8}[/latex] = 24 [latex]\times[/latex] 5 = 120
Remaing Height = 192 - 120 = 72
Height Climbed is Second hour = 72 [latex]\times \frac {25}{200}[/latex] 9 m
Q6. The difference between the value of a number increased by 25% and the value of the original number decreased by 30% is 22. What is the original number?

A. 70 B. 65 C. 40 D. 90

Answer: C
Solution: Let original number 100
125r - 70r [latex]\Rightarrow[/latex] 22
55r [latex]\Rightarrow[/latex] 2
1r = [latex]\frac {2}{5} \times 100[/latex] = 40
Original Number = 40
Q7. If 12% of 75% of a number is greater than 5% of a number by 75, the number is

A. 1875 B. 1890 C. 1845 D. 1860

Answer: A
Solution: [latex]\frac {12}{100} \times {75}{100} \times X - \frac{5X}{100}[/latex] = 75
[latex]X \times (\frac {9}{100} - \frac {5}{100}) [/latex] = 75
[latex]X \times \frac {4}{100} [/latex] = 75
X = 25 [latex] \times [/latex] 75 = 1875
Q8. The salary of Raju and Ram is 20% and 30% less than the salary of Saroj respectively. By what % is the salary of Raju is more than the salary of Ram?

A. 33.33% B. 50% C. 15.18% D. 14.28%

Answer: D
Solution: Let Salary of Saroj = 100
Salary of Raju = 80
Salary of Ram = 70
Required % = [latex]\frac {10}{70} \times 100[/latex]
= [latex]\frac {100}{7} [/latex] = 14.28 %
Q9. A fraction is such that if the double of the numerator and the triple of the denominator is changed by +10% and –30% respectively then we get 11% of 16/21. Find the fraction.

A. [latex]\frac {4}{25} [/latex] B. [latex]\frac {2}{25} [/latex] C. [latex]\frac {3}{25} [/latex] D. None of these

Answer: B
Solution: Let traction [latex]\rightarrow \frac {X}{Y}[/latex]
ATQ
[latex]\frac {2x \times \frac {110}{100}}{3y \times \frac {70}{100}}[/latex] = [latex]\frac {11}{100}[/latex]
[latex]\times [/latex] [latex]\frac {16}{21}[/latex]
[latex]\frac {{2X} \times {11}}{{3Y} \times {7}}[/latex] = [latex]\frac {{11} \times {16}} {{100} \times {21}}[/latex]
[latex]\frac{X}{Y}[/latex] = [latex]\frac {80}{100} [/latex] = [latex]\frac {2}{25}[/latex]
Q10. In class, 65% of the students are boys. On a particular day, 80% of girl students were present. What was the fraction of boys who were present that day if the total number of students present that day was 70%?

A. [latex]\frac {2}{3}[/latex] B. [latex]\frac {28}{65}[/latex] C. [latex]\frac {5}{6}[/latex] D. [latex]\frac {42}{65}[/latex]

Answer: D
Solution: Let total no. of students in school [latex]\rightarrow[/latex] 100
Boys = 65
Girls = 35
Girls Present = [latex]\frac {{35} \times {80}} {{100}}[/latex] = [latex]\frac {280}{10}[/latex] = 28
Total No. of Students Present = 70
No. of Boys Present = 70 - 28 = 42
Fraction of boys Present = [latex]\frac {42}{65}[/latex]

Q1. A T.V was sold at a profit of 5%. If it had been sold at a profit of 10% the profit would have been Rs. 1000 more. What is its cost price?

A. Rs. 20000 B. Rs. 5000 C. Rs. 10000 D. Rs. 15000

Answer: A
Solution: + 10 % - 5 % [latex]\rightarrow[/latex] 1000 RS.
5 % [latex]\rightarrow[/latex] 1000 RS.
1 % [latex]\rightarrow[/latex] 200 RS.
100 % [latex]\rightarrow[/latex] 20,000 RS.
Q2. A man sells two chairs at Rs. 120 each and by doing so he gains 25% on one chair and loses 25% on the other. His loss on the whole in Rs. Is

A. 20 B. 16 C. 25 D. 30

Answer: B
Solution: Loss % = [latex] (\frac {x}{10})^{2} [/latex]
= [latex]({2.5})^{2} [/latex]
= 6.25 %
C.P = 240 [latex] \times \frac {100}{93.75}[/latex] = 256
S.P = 240
Loss = RS. 16
Q3. A man sold two articles at Rs. 375 each. On one, he gains 25% and on the other, he loses 25%. The gain or loss% on the whole transaction is

A. 6% loss B. 416% Profit C. Rs. 50 profit D. 614% loss

Answer: D
Solution: Shortcut = [latex] (\frac {x}{10})^{2} [/latex] = 6.25 %
= 6 [latex]\frac {1}{4} [/latex] %
Q4. A man wanted to sell an article with 20% profit: but he acutually sold at 20% loss for Rs. 480, At what price he wanted to sell it to earn the profit?

A. Rs. 720 B. Rs. 840 C. Rs. 600 D. Rs. 750

Answer: A
Solution: C.P [latex] \times \frac {80}{100} [/latex] = 480
C.P = 600
S.P that he want to sell = 600 [latex] \times \frac {120}{100} [/latex] = 720 RS.
Q5. If an article is sold at 5% gain instead of 5% loss, the man gains Rs. 5 more. Find the cost price of that article

A. Rs. 100 B. Rs. 105 C. Rs. 50 D. Rs. 110

Answer: C
Solution: + 5 % - (- 5 %) [latex]\Rightarrow[/latex] RS. 5
10 % = RS. 5
1 % = [latex]\frac {1}{2}[/latex]
100 % = RS. 50
Q6. On selling an article for Rs. 105 a trader loses 9%. To gain 30% he should sell the article at

A. Rs. 126 B. Rs. 144 C. Rs. 150 D. Rs. 139

Answer: C
Solution: C.P [latex] \times \frac {91}{100} [/latex] = 105
C. P = [latex]\frac {10500}{91} [/latex]
S.P = [latex]\frac {10500}{91} \times \frac {130}{100} [/latex]
= 150 RS.
Q7. An article is sold at a loss of 10%. Had it been sold for Rs. 9 more there would have been a gain of 12 1/2% on it. The cost price of the article is:

A. Rs. 40 B. Rs. 45 C. Rs. 50 D. Rs. 35

Answer: A
Solution: Let C.P [latex]\rightarrow[/latex] 100
C.P S.P 100 90 100 112.5
22.5 [latex]\rightarrow[/latex] RS. 9
1 r [latex]\rightarrow \frac {9}{22.5} [/latex]
100 r [latex]\rightarrow \frac {9}{22.5} \times [/latex] 100
= [latex] \frac {9}{225} \times [/latex] 1000
= 40 RS.
Q8. By selling a table for Rs. 350 instead of Rs. 400, loss percent increases by 5%. The cost price of table is:

A. Rs. 1,050 B. Rs. 417.50 C. Rs. 435 D. Rs. 1,000

Answer: D
Solution: 5 % [latex]\rightarrow[/latex] RS. 50
1 % [latex]\Rightarrow[/latex] RS. 10
100 % [latex]\Rightarrow[/latex] RS. 1000
Q9. The marked price of an article is Rs. 500. It is sold at successive discounts of 20% and 10%. The selling price of the article (Rs.) is:

A. Rs. 350 B. Rs. 375 C. Rs. 360 D. Rs. 400

Answer: C
Solution: Single discount = - 20 - 10 + 2 = -28 %
S.P = 500 [latex]\times \frac {72}{100}[/latex]
= 360 RS.
Q10. The marked price of an article is 10% higher than the cost price. A discount of 10% is given on the marked price. In this kind of sale, the seller bears

A. B. 5% C. 1% D. 1%

Answer: D
Solution: P = 10 - 10 - 1
= -1 %
[latex]\Rightarrow[/latex] Loss [latex]\rightarrow[/latex] 1 %

Q1. There are 2 teams-A and B. If 3 people are shifted from Team A to Team B, then Team B has thrice the number of members than Team A. If 2 people are shifted from Team B to Team A, then Team B has double the number of members than Team A. How many members does Team B have originally?

A. 15 B. 18 C. 42 D. 45

Answer: C
Solution: Member in Team A [latex]\Rightarrow[/latex] x
Member in Team B [latex]\Rightarrow[/latex] y
ATQ
3(x - 3) = y + 3
y + 3 = 3x - 9
3x - y = 12 ------------------1
(y - 2) = 2 (x + 2)
y - 2 = 2x + 4
2x - y = -6 ------------------2
From 1 & 2
x = 18
y = 42
Q2. (6² + 7² + 8² + 9² + 10²)/(√(7 + 4√3) – √(4 + 2√3) ) is equal to

A. 330 B. 355 C. 305 D. 366

Answer: A
Solution: [latex]\frac {{6}^{2} + {7}^{2} + {8}^{2} + {9}^{2} + {10}^{2}} {\sqrt {7 + 4 \sqrt 3} - \sqrt {4 + 2 \sqrt 3}}[/latex]
[latex]\frac {{36} + {49} + {64} + {81} + {100}} {\sqrt {4 + 3 + 2 \times 2 \sqrt 3} - \sqrt {1 + 3 + 2 \sqrt 3}}[/latex]
[latex]\frac {200 + 130} {\sqrt {(2 + \sqrt 3})^{2} - \sqrt {(1 + \sqrt 3})^{2}}[/latex]
[latex]\frac {330}{ 2 + \sqrt {3} - 1 - \sqrt {3}}[/latex]
= 330
Q3. There are 1400 students in a school, 25% of those wear spectacles and 2/7 of those wearing spectacles are boys. How many girls in the school wear spectacles?

A. 250 B. 100 C. 200 D. 300

Answer: A
Solution: Student Wearing Spectacles = 1400 [latex]\times \frac {1}{4}[/latex] = 350
Boys Wearing Spectacles = 350 [latex]\times \frac {2}{7}[/latex] = 100
Girls Wearing Spectacles = 350 - 100 = 250
Q4. If [latex] {x}^{2} + \frac {1}{{x}{2}} = 98 (x > 0) [/latex], then the value of [latex] {x}^{3} + \frac {1}{{x}^{3}}[/latex]is

A. 970 B. 1030 C. -970 D. -1030

Answer: A
Solution: [latex]{x}^{2} + \frac {1}{{x}^{2}}[/latex] = 98
Adding 2 both sides
[latex]{x}^{2} + \frac {1}{{x}^{2}} + 2[/latex] = 98+2
[latex](x + \frac {1}{x})^{2}[/latex] = 100
[latex]x + \frac {1}{x}[/latex] = 10
Cubing both sides
[latex](x + \frac {1}{x})^{3}[/latex] = [latex]{10}^{3}[/latex]
[latex]{x}^{3} + \frac {1}{{x}^{3}} + 3 (x + \frac {1}{x})[/latex] = 1000
[latex]{x}^{3} + \frac {1}{{x}^{3}}[/latex] = 1000 - 30 = 970
Q5. The marked price of an article is Rs. 5000 but due to festive offer a certain percent of discount is declared. Mr. X availed this opportunity and bought the article at reduced price. He then sold it at Rs. 5000 and thereby made a profit of 11(1/9)%. The percentage of discount allowed was?

A. 10 B. 3(1/3 ) C. 7(1/2) D. 11(1/9)

Answer: A
Solution: S.P for x [latex]\Rightarrow[/latex] 5000
Profit % = 11 [latex]\frac {1}{9}[/latex] %
= [latex]\frac {100}{9}[/latex] %
[latex]\frac {1 \rightarrow {profit}}{9 \rightarrow {C.P}}[/latex]
S.P = 9+1 = 10
10 r [latex] \rightarrow [/latex] 5000
1 r [latex] \rightarrow [/latex] 500
9 r [latex] \rightarrow [/latex] 4500 RS.
Discount %= [latex] \frac {500}{5000} \times 100[/latex]
= 10 %
Q6. Arun buys one kilogram of apples for Rs. 120 and sells it to Swati gaining 25%. Swati sells it to Divya and Divya again sells it for Rs. 198, making a profit of 10%. What is the profit percentage made by swati?

A. 25% B. 20% C. 16.67% D. 15%

Answer: B
Solution: C.P of Arun = 120
S.P of Arun and C.P of Swati = 120 [latex]\times \frac {125}{100}[/latex] = 150 RS.
S.P of Divya = 198
C.P of Divya and S.P of Swati = 198 [latex]\times \frac {100}{110}[/latex] = 180 RS.
Proft of Swati = [latex]\frac {180 - 150}{150} \times 100[/latex]
= [latex]\frac {30}{150} \times 100[/latex] = 20 %
Q7. If 7 sin α = 24 cos α; 0 < α < π/2, then the value of 14 tan α – 75 cos α – 7 sec α is equal to

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

Answer: D
Solution: 7 Sin [latex]\alpha = 24 Cos \alpha[/latex]
Tan = [latex]\frac {24 \rightarrow {Perpendicular}}{7 \rightarrow {Base}}[/latex]
H = [latex]\sqrt {576 + 49}[/latex]
= [latex]\sqrt {625}[/latex]
= 25
Cos [latex]\alpha = \frac {B}{H} = \frac {7}{25}[/latex]
Sec [latex]\alpha = \frac {1}{Cos \alpha} = \frac {25}{7}[/latex]
14 Tan [latex]\alpha - 75 {Cos \alpha} - 7 {Sec \alpha}[/latex]
= 14 [latex]\times \frac {24}{7} - 75 \times \frac {7}{25} - 7 \times \frac {25}{7}[/latex]
= 48 - 21 - 25
= 48 - 46
= 2
Q8. A tower is observed from a point on the horizontal through the foot of the tower. The distance of this point from the foot of the tower is equal to the height of the tower. The angle of elevation of the top of the tower is

A. 60° B. 45° C. 40° D. 30°

Answer: B
Solution: Tan [latex]\theta = \frac {h}{h}[/latex]
[latex]\theta = {45}^{0}[/latex]
Q9. At present ages of a father and son are in the ratio of 7 : 3 and they will be in the ratio 2 : 1 after 10 years. What is the present age of father?

A. 70 years B. 65 years C. 60 years D. 50 years

Answer: A
Solution: (8 - 7) r = 10
1 r = 10
Present Age of Father = 7 [latex]\times 10[/latex]
= 70 years
Q10. John cycling at a constant speed of 10 km/hr., reaches his school in time. If he cycles at a constant speed of 15 km/hr., he reaches his school 12 minutes early. Number of km he has to cycle for his school is :

A. 4 B. 6 C. 9 D. 12

Answer: B
Solution: Ratio of Speed

10 : 5 2 : 3

Ratio of Time [latex]\rightarrow 3 : 2[/latex]
(3 - 2) r [latex]\Rightarrow [/latex] 12 minutes
3 r [latex]\rightarrow [/latex] 36 minutes
Distance = 10 km/hr [latex]\times [/latex] 36/ 60
= 6 km

Q1 Simplify: [latex](\frac {2}{7} + \frac {3}{5} ) \div (\frac {2}{5} + \frac {2}{7})[/latex]

A. [latex]\frac {31}{24}[/latex] B. [latex]\frac {13}{45}[/latex] C. [latex]\frac {26}{45}[/latex] D. [latex]\frac {45}{13}[/latex]

Answer: A
Solution: [latex]\frac {31}{35} ÷ \frac {24}{35}[/latex]
[latex]\frac {31}{35} \times \frac {35}{24}[/latex] = [latex]\frac {31}{24}[/latex]
Q2 Simplify: Simplify (–4.6) × (–4.6) ÷ (–4.6 + 0.6)

A. –5.29 B. –0.529 C. –4.06 D. 5.01

Answer: A
Solution: [latex]\Rightarrow 4.6 \times 4.6 \times \frac {-1}{4}[/latex]
[latex]\Rightarrow [/latex] - 5.9
Q3 Simplify: [latex]\frac {9}{13} ÷ \frac {18}{26} ÷ \frac {90}{52}[/latex]

A. [latex]\frac {45}{26}[/latex] B. [latex]\frac {13}{45}[/latex] C. [latex]\frac {26}{45}[/latex] D. [latex]\frac {45}{13}[/latex]

Answer: C
Solution:
[latex]\frac {9}{13} \times \frac {26}{8} \times \frac {52}{90}[/latex]
[latex]\Rightarrow [/latex][latex]\frac {26}{45}[/latex]
Q4 Solve: 12 – [26 – {2 + 5 × (6 – 3)}]

A. 2 B. 3 C. 7 D. 8

Answer: B
Solution: 12 - [26 - {2 + 5 [latex]\times [/latex] 3}]
= 12 - [26 - 17]
= 3
Q5 Solve: [latex]{10}^{9} \times {10}^{7} ÷ {10}^{-3} \times {10}^{-6} ÷ {10}^{-4} ÷ {10}^{2}[/latex]

A. [latex]{10}^{15}[/latex] B. [latex]{10}^{25}[/latex] C. [latex]{10}^{5}[/latex] D. [latex]{10}^{35}[/latex]

Answer: A
Solution: [latex]\Rightarrow {10}^{9 + 7 + 3 - 6 + 4 - 2}[/latex]
= [latex]{10}^{15}[/latex]
Q6. [latex]\frac {12}{13} \times \frac {285}{96} ÷ \frac {171}{169} \times \frac{485}{81} ÷ \frac {291}{162}[/latex]

A. 9[latex]\frac {1}{36}[/latex] B. 10[latex]\frac {1}{36}[/latex] C. 10[latex]\frac {1}{24}[/latex] D. 10[latex]\frac {1}{48}[/latex]

Answer: A
Solution: [latex]{12}{13} \times \frac {285}{96} \times \frac {169}{171} \times \frac {481}{81} \times \frac {162}{291} [/latex]
[latex]\Rightarrow \frac {325}{36}[/latex]
= 9[latex]\frac {1}{36}[/latex]
Q7 Solve: [latex]\frac {{(998)}^{2} - {(997)}^{2} - 45}{{(98)}^{2} - {(97)}^{2}}[/latex]

A. 10 B. 12 C. 14 D. 15

Answer: A
Solution: [latex]\Rightarrow [/latex][latex]\frac {[{(998)}^{2} - {(997)}^{2}] - 45} {({98} - {97}) ({98} + {97})} [/latex]
[latex]\Rightarrow [/latex][latex]\frac {({998} - {997}) ({998} + {997}) - 45} {{1} \times {195}}[/latex]
[latex]\Rightarrow [/latex][latex]\frac {{1995} - {45}} {195}[/latex]
[latex]\Rightarrow [/latex] [latex]\frac {1950}{195}[/latex] = 10
Q8 Solve: [latex]\frac {{0.8 \bar {3}} ÷ {7.5}}{{2.3 \bar {21}} - {0.0 \bar {98}}}[/latex]

A. 0.05 B. 0.04 C. 0.03 D. 0.02

Answer: A
Solution: [latex]\Rightarrow [/latex][latex]\frac {\frac {83 - 8}{90} \div 7.5}{\frac {321 - 3}{990} - \frac {98}{990}}[/latex]
[latex]\Rightarrow [/latex] [latex]\frac {\frac {75}{90} \times \frac {1}{7.5}} {2 \frac {318}{990} - \frac {98}{990}}[/latex]
= [latex]\frac {\frac {1}{9}} {2 \frac {220}{990}}[/latex]
= [latex]\frac {\frac {\frac {1}{9}}{2200}} {990}[/latex] = [latex]\frac {1}{220}[/latex] = [latex]\frac {1}{20}[/latex]
= 0.05
Q9 Simplify: b – [b – (a + b) – {b – (b – a + b)} + 2a]

A. a B. 2a C. 4a D. 0

Answer: D
Solution: b – [b – a – b – b + b – a + b + 2a]
= b – b = 0
Q10 Solve: [latex]{(6A)}^{4} ÷ {(36)}^{3} \times 216[/latex] = [latex]{(6)}^{(? - 5)}[/latex]

A. 6 B. 7 C. 4 D. 1

Answer: A
Solution: [latex]{6}^{1} = {6}^{? - 5}[/latex]
1 = ? - 5
? = 6

Q1. If (6x – 1) – (8x – 5) = 7, then the value of x is ______.

A. –3/2 B. 3/2 C. 11/2 D. –11/2

Answer: A
Solution: (6x - 1) - (8x - 5) = 7
6x - 1 - 8x + 5 = 7
- 2x = 3
x = [latex]\frac {- 3}{2}[/latex]
Q2. If 7/8th of 5/4th of a number is 315, then 5/9th of that number is _____.

A. 123 B. 81 C. 140 D. 160

Answer: D
Solution: [latex]\frac {7}{8} \times \frac {5}{4} \times x[/latex] = 315
x = 288
[latex]\frac {5}{9} x = 288 \times \frac {5}{9}[/latex] = 160
Q3. The average marks of 56 students is shown as 60. It includes a wrong entry of 92 marks instead of 29 marks. The correct average is _____.

A. 58.875 marks B. 61.125 marks C. 63.375 marks D. 56.625 marks

Answer: A
Solution: [latex]\frac {^{s} {56}}{56}[/latex] =60
[latex]{^{s}{56}}[/latex] = 3360
[latex]{^{s}{55}} + 92[/latex] = 3360
[latex]{^{s}{55}}[/latex] = 3360 - 92 = 3268
[latex]{^{s}{55}} + 29[/latex] = 3268 + 29
[latex]{^{s}{56}}[/latex] = 3297
Average = [latex]\frac {3297}{56}[/latex] = 58.875
Q4. If 13x² = 17² – 9², find the value of x?

A. 16 B. 12 C. 8 D. 4

Answer: D
Solution:
[latex]13 {x}^{2} = {17}^{2} - {9}^{2}[/latex]
[latex]13 {x}^{2}[/latex] = 289 - 81
[latex]13 {x}^{2}[/latex] = 208
[latex]{x}^{2}[/latex] = 16
x = 4
Q5. If 2 + 2x < 3 + 5x and 3(x – 2) < 5 – x; then x can take which of the following values?

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

Answer: A
Solution: 2 + 2x < 3 + 5x
- 1 < 3x
x > [latex]\frac {- 1}{3}[/latex] ------------------1
And,
3(3 - 2) < 5 - x
3x - 6 < 5 - x
4x < 11
x < [latex]\frac {11}{4}[/latex] ------------------2
Form 1 and 2
[latex]\frac {- 1}{3} < x < \frac {11}{4}[/latex]
- 0.33 < x <2.75
x = 1
Q6. If the radius of a circle is increased by 17% its area increases by ______.

A. 34 percent B. 36.89 percent C. 17 percent D. 18.445 percent

Answer: B
Solution: If r [latex]\uparrow 17[/latex] %
Area [latex]\uparrow = 17 + 17 + \frac {17 \times 17}{100}[/latex]
= 36.89 %
Q7. A bank offers 20% compound interest per half year. A customer deposits Rs 2800 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is ______.

A. Rs 3584 B. Rs 896 C. Rs 1792 D. Rs 448

Answer: C
Solution: R= 20 %p, Half yearly
Principal on Jan = RS. 2800
Interest on [latex]{1}^{st} July[/latex]
= 2800 [latex]\times \frac {20}{100}[/latex]
= 560 RS.
Principal on [latex]{1}^{st} July[/latex] = 2800 + 560 + 2800 = 6160
Interest at the end of the year = 6160 [latex]\times \frac {20}{100}[/latex] = 1232 RS.
Total Interest = 560 + 1232 = 1792 RS.
Q8. 4 hrs after a goods train passed a station, another train traveling at a speed of 60 km/hr following that goods train passed through that station. If after passing the station the train overtakes the goods train in 8 hours. What is the speed of the goods train?

A. 40 km/hr B. 48 km/hr C. 60 km/hr D. 32 km/hr

Answer: A
Solution: Let the speed of goods train be x
Speed of another train = 60 km/hr
Distance between teo trains = 4x
8 = [latex]\frac {4x}{60 - x}[/latex]
480 - 8x = 4x
x = 40 km/hr
Q9. Ticket for an adult is Rs 1000 and a child is Rs 500. One child goes free with two adults. If a group has 17 adults and 5 children what is the discount the group gets?

A. 14.7 percent B. 32 percent C. 12.82 percent D. 22 percent

Answer: B
Solution: 17 Adults and 5 Children
5 Children will go free 10 Adults
So, discount = 5 [latex]\times 500[/latex]
= 2500 RS.
Total cost of money without discount = 17000 + 2500
= 19500 RS.
Discount % = [latex] \frac {2500}{19500} \times 100[/latex]
= 12.82 %
Q10. Find the value of p if –3x – 11, x + p and 5x + 7 are in arithmetic progression.

A. 9 B. 2 C. - 9 D. - 2

Answer: A
Solution: If A, B, C are in arithematic Progression
Than,
2B = A + C
B = x + P
A = - 3x - 11
C = 5x + 7
2x + 2P = - 3x - 1 + 5x + 7
2x + 2P = 2x - 4
2P = -4
P = - 2