Number System
1. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

A. 9 B. 11 C. 13 D. 15

Answer: Option (D)
Explanation: Let the three integers be x, x + 2 and x + 4. Then, 3x = 2(x + 4) + 3 => x = 11. Third integer = x + 4 = 15.
2. Which one of the following is not a prime number?

A. 31 B. 61 C. 71 D. 91

Answer: Option (D)
Explanation: 91 is divisible by 7. So, it is not a prime number.
3. 5358 x 51 = ?

A. 273258 B. 273268 C. 273348 D. 273358

Answer: Option (A)
Explanation: 5358 x 51 = 5358 x (50 + 1) = 5358 x 50 + 5358 x 1 = 267900 + 5358 = 273258.
4. 72519 x 9999 = ?

A. 725117481 B. 674217481 C. 685126481 D. 696217481 E. None of these

Answer: Option (A)
Explanation: 72519 x 9999 = 72519 x (10000 - 1) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481.
5. Convert the decimal number 151.75 to binary.

A. 10000111.11 B. 11010011.01 C. 00111100.00 D. 10010111.11

Answer: Option (D)
Percentages
1. What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?

A. 1 B. 14 C. 20 D. 21

Answer: Option (C)
Explanation: Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69. Number of such numbers =14 Required percentage = [[latex]\frac{14}{70}[/latex] x 100 ]% = 20%.
2. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

A. 2700 B. 2900 C. 3000 D. 3100

Answer: Option (A)
Explanation: Number of valid votes = 80% of 7500 = 6000. Valid votes polled by another candidate = 45% of 6000 = [[latex]\frac{45}{100}[/latex]x 6000] = 2700.
3. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

A. 57% B. 60% C. 65% D. 90%

Answer: Option (A)
Explanation: Total number of votes polled = (1136 + 7636 + 11628) = 20400. Required percentage =[[latex]\frac{11628}{20400}[/latex] x 100]% = 57%.
4. The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

A. 4.37% B. 5% C. 6% D. 8.75%

Answer: Option (B)
Explanation: Increase in 10 years = (262500 - 175000) = 87500. Increase% = [[latex]\frac{87500}{175000}[/latex] x 100]% = 50%. Required average = [latex]\frac{50}{10}[/latex]% = 5%.
5. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

A. 39, 30 B. 41, 32 C. 42, 33 D. 43, 34

Answer: Option (C)
Explanation: Let their marks be (x + 9) and x. Then, x + 9 = [latex]\frac{56}{100}[/latex](x + 9 + x) 25(x + 9) = 14(2x + 9) 3x = 99 x = 33 So, their marks are 42 and 33.
Decimal & Fractions
1. 54.327 x 357.2 x 0.0057 is same as

A. 5.4327 x 3.572 x5.7 B. 5.4327 x 3.572 x0.57 C. 54327 x 3572 x 0.0000057 D. None of these

Answer: Option (A)
Explanation: Number of decimal places in the given expression = 8 Number of decimal places in (A) = 8 Number of decimal places in (B) = 9 Number of decimal places in (C)= 7 Clearly, the expresssion in (A) is the same as the given Expression.
2. (0.2 * 0.2 + 0.01) ( 0.1 * 0.1 + 0.02)^(-1) is equal to :

A. 5/3 B. 9/3 C. 11/3 D. 13/3

Answer: Option (A)
Explanation: Given = (0.2 * 0.2 + 0.01) / ( 0.1 * 0.1 + 0.02) = (0.04 + 0.01) / (0.01 + 0.02) = 0.05 / 0.03 = 5/3...
3. 138.009 + 341.981 - 146.305 = 123.6 + ?

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

Answer: Option (D)
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.
4. 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

Answer: Option (B)
Explanation: 0.46 = 46/100 = 23/50. Sum of the numerator and denominator is 23 + 50 = 73
5. 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

Answer: Option (A)
Explanation: Sum of the decimal places = 7. Since the last digit to the extreme right wil be zero ( 5 * 4 = 20), so there will be 6 significant digits to the right of the decimal point.
Average
1. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?

A. Rs. 4991 B. Rs. 5991 C. Rs. 6001 D. Rs. 6991

Answer: Option (A)
Explanation: Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009. Required sale = Rs. [ (6500 x 6) - 34009 ] = Rs. (39000 - 34009) = Rs. 4991.
2. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

A. 0 B. 1 C. 10 D. 19

Answer: Option (D)
Explanation: Average of 20 numbers = 0. Sum of 20 numbers (0 x 20) = 0. It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
3. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?

A. 76 kg B. 76.5 kg C. 85 kg D. Data inadequate E. None of these

Answer: Option (C)
Explanation: Total weight increased = (8 x 2.5) kg = 20 kg. Weight of new person = (65 + 20) kg = 85 kg.
4. The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:

A. 3500 B. 4000 C. 4050 D. 5000

Answer: Option (B)
Explanation: Let P, Q and R represent their respective monthly incomes. Then, we have: P + Q = (5050 x 2) = 10100 .... (i) Q + R = (6250 x 2) = 12500 .... (ii) P + R = (5200 x 2) = 10400 .... (iii) Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv) Subtracting (ii) from (iv), we get P = 4000. P's monthly income = Rs. 4000.
5. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

A. 17 kg B. 20 kg C. 26 kg D. 31 kg

Answer: Option (D)
Explanation: Let A, B, C represent their respective weights. Then, we have: A + B + C = (45 x 3) = 135 .... (i) A + B = (40 x 2) = 80 .... (ii) B + C = (43 x 2) = 86 ....(iii) Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv) Subtracting (i) from (iv), we get : B = 31. B's weight = 31 kg.
Simplification
1. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for?

A. 160 B. 175 C. 180 D. 195

Answer: Option (B)
Explanation: Suppose the man works overtime for x hours. Now, working hours in 4 weeks = (5 x 8 x 4) = 160. 160 x 2.40 + x x 3.20 = 432 3.20x = 432 - 384 = 48 x = 15. Hence, total hours of work = (160 + 15) = 175.
2. 60 + 5 * 12 / (180/3) = ?

A. 60 B. 120 C. 13 D. 61 E. None of these

Answer: Option (D)
Explanation: 60 + 5 * 12 / (180/3) = 60 + 5 * 12 / (60) = 60 + (5 * 12)/60 = 60 + 1 = 61.
3. 9000 + 16 2/3 % of ? = 10500

A. 1500 B. 1750 C. 9000 D. 7500 E. None of these

Answer: Option (C)
Explanation: 9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500 50/(3 * 100) of ? = 1500 => ? = 1500 * 6 ? = 9000
4. 6 2/3 * 9 3/5 * 2 1/3 + 1 1/3 = ?

A. 150 1/3 B. 150 2/3 C. 149 2/3 D. 149 1/3 E. None of these

Answer: Option (B)
Explanation: 6 2/3 * 9 3/5 * 2 1/3 + 1 1/3 = 20/3 * 48/5 * 7/3 + 4/3 = 448/3 + 4/3 = 452/3 = 150 2/3
5. 0.003 * ? * 0.0003 = 0.00000027

A. 9 B. 3 C. 0.3 D. 0.03 E. None of these

Answer: Option (C)
Explanation: 0.003 * ? * 0.0003 = 0.00000027 3/1000 * ? * 3/10000 = 3/1000 * 3/1000 * 3/100 ? = 3/10 = 0.3
Problems on Ages
1. A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was:

A. 14 years B. 19 years C. 33 years D. 38 years

Answer: Option (A)
Explanation: Let the son's present age be x years. Then, (38 - x) = x 2x = 38. x = 19. Son's age 5 years back (19 - 5) = 14 years.
2. A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?

A. 7 B. 8 C. 9 D. 10 E. 11

Answer: Option (D)
Explanation: Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years. (2x + 2) + 2x + x = 27 5x = 25 x = 5. Hence, B's age = 2x = 10 years.
3. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:

A. 14 years B. 18 years C. 20 years D. 22 years

Answer: Option (D)
Explanation: Let the son's present age be x years. Then, man's present age = (x + 24) years. (x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 x = 22.
4. Sachin is younger than Rahul by 7 years. If the ratio of their ages is 7:9, find the age of Sachin

A. 24.5 B. 25.5 C. 26.5 D. 27.5

Answer: Option (A)
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
5. I am five times as old as you were when I was as old as you are", said a man to his son. Find out their present ages, if the sum of their ages is 64 years?

A. Father = 50; Son =14 B. Father = 40; Son =24 C. Father = 60; Son =4 D. Father = 48; Son =16

Answer: Option (B)
Explanation: Let the present age of the man be 'P' and son be 'Q', Given, P + Q = 64 or Q = (64 - P) Now the man says "I am five times as old as you were, when I was as old as you are", So, P = 5[B - (P - Q)] We get 6P = 10Q, Substitute value for Q, 6P = 10(64 - P), Therefore P = 40, Q = 24.
Profit and Loss
1. same article for Rs. 1280. At what price should the article be sold to make 25% profit?

A. Rs. 2000 B. Rs. 2200 C. Rs. 2400 D. Data inadequate

Answer: Option (A)
Explanation: Let C.P. be Rs. x. Then, [latex]\frac{1920 - x}{x}[/latex] x 100 = [latex]\frac{x - 1280}{x}[/latex]x 100 1920 - x = x - 1280 2x = 3200 x = 1600 Required S.P. = 125% of Rs. 1600 = Rs. [latex]\frac{125}{100}[/latex] x 1600 = Rs 2000.
2. 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

Answer: Option (C)
Explanation: C.P. = Rs.([latex]\frac{100}{122.5}[/latex] x 392) = Rs. ([latex]\frac{1000}{1225}[/latex] x 392) = Rs. 320 Profit = Rs. (392 - 320) = Rs. 72.
3. 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

Answer: Option (C)
Explanation: S.P. = 85% of Rs. 1400 = Rs.([latex]\frac{85}{100}[/latex]x 1400) = Rs. 1190
4. 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

Answer: Option (D)
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.([latex]\frac{720}{12}[/latex]) = Rs. 60.
5. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

A. No profit, no loss B. 5% C. 8% D. 10% E. None of these

Answer: Option (B)
Explanation: C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600. S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680. Gain = ([latex]\frac{80}{1600}[/latex] x 100)% = 5%.
Time and Work
1. A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?

A. 8 hours B. 10 hours C. 12 hours D. 24 hours

Answer: Option (C)
Explanation: A's 1 hour's work =[latex]\frac{1}{4}[/latex]; (B + C)'s 1 hour's work =[latex]\frac{1}{3}[/latex]; (A + C)'s 1 hour's work =[latex]\frac{1}{2}[/latex] . (A + B + C)'s 1 hour's work = ([latex]\frac{1}{4}[/latex]+[latex]\frac{1}{3}[/latex]) =[latex]\frac{7}{12}[/latex]. B's 1 hour's work =([latex]\frac{7}{12}[/latex]-[latex]\frac{1}{2}[/latex])= [latex]\frac{1}{12}[/latex]. Therefore B alone will take 12 hours to do the work.
2. A and B complete a work in 6 days. A alone can do it in 10 days. If both together can do the work in how many days?

A. 3.75 days B. 4 days C. 5 days D. 6 days

Answer: Option(A)
Explanation: 1/6 + 1/10 = 8/30 = 4/15 15/4 = 3.75 days.
3. A and B together can do a piece of work in 8 days. If A alone can do the same work in 12 days, then B alone can do the same work in?

A. 20 days B. 16 days C. 24 days D. 28 days

Answer: Option (C)
Explanation: B = 1/8 – 1/2 = 1/24 => 24 days.
4. 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

Answer: Option (B)
Explanation: 5/30 + 20/x = 1 x = 24 1/30 + 1/24 = 3/40 40/3 = 13 1/3 days.
5. 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

Answer: Option (D)
Explanation: 3/12 + (3 + x)/16 = 1 x = 9 days.
Ratio and Proportions
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

Answer: Option (C)
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

Answer: Option (D)
Explanation: Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively. Then,[latex]\frac{2x + 4000}{3x + 4000}[/latex] = [latex]\frac{40}{57}[/latex] 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

Answer: Option (B)
Explanation: (x x 5) = (0.75 x 8) x = [latex]\frac{6}{5}[/latex] = 1.20
4. Length and width of a field are in the ratio 5 : 3. If the width of the field is 42 m then its length is:

A. 100 m B. 80 m C. 50 m D. 70 m

Answer: Option (D)
5. If 12, 21, 72, 126 are in proportion, then:

A. 12 × 21 = 72 × 126 B. 12 × 72 = 21 × 126 C. 12 × 126 = 21 × 72 D. none of these

Answer: Option (C)
Time and Distance
1. Express a speed of 36 kmph in meters per second?

A. 10 mps B. 12 mps C. 14 mps D. 17 mps

Answer: Option (A)
Explanation: 36 * 5/18 = 10 mps
2. A car covers a distance of 624 km in 6 ½ hours. Find its speed?

A. 104 kmph B. 140 kmph C. 104 mph D. 10.4 kmph

Answer: Option (A)
Explanation: 624/6 = 104 kmph
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

Answer: Option (C)
Explanation: S = 240/16 * 18/5 = 54 kmph
4. Walking 7/6 of his usual rate, a boy reaches his school 4 min early. Find his usual time to reach the school?

A. 25 min B. 26 min C. 27 min D. 28 min

Answer: Option (D)
Explanation: Speed Ratio = 1:7/6 = 6:7 Time Ratio = 7:6 1 -------- 7 4 --------- ? -> 28 m
5. Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance traveled 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

Answer: Option (C)
Explanation: 60(x + 1) = 64x X = 15 60 * 16 = 960 km.
HCF & LCM
1. LCM of 87 and 145 is:

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

Answer: Option (B)
2. HCF of 3/16, 5/12, 7/8 is:

A. 2/47 B. 3/47 C. 1/48 D. 5/48

Answer: Option (C)
Explanation: HCF of numerators = 1 LCM of denominators = 48 => 1/48
3. Find the H.C.F of 54/9, 3 9/17 and 36/51?

A. 17/6 B. 6/153 C. 6 D. 17

Answer: Option (B)
4. What will be the LCM of 8, 24, 36 and 54

A. 54 B. 108 C. 216 D. 432

Answer: Option (C)
Explanation: LCM of 8-24-36-54 will be 2*2*2*3*3*3 = 216.
5. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A. 123 B. 127 C. 235 D. 305

Answer: Option (B)
Explanation: Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127.
Mixtures & Allegations
1. For a car there are 5 tyres including one spare tyre(4+1). All tyres are equally used. If the total distance travelled by the car is 40000km then what is the average distance travelled by each tyre?

A. 10000 B. 40000 C. 32000 D. 8000

Answer: Option (C)
Explanation: total distance travelled by the car=40000km so total distance travelled by 4 wheels=4*40000=160000 as all tyres(4+1) are equally used so average distance travelled by the each tyre=160000/5=32000
2. Average marks of a,b,c, are 48. When d joins average becomes 47. E has 3 more marks than d. Average marks of b,c,d,e is 48. What are the marks of a?

A. 46 B. 43 C. 49 D. 52

Answer: Option (B)
Explanation: a+b+c=48*3=144 a+b+c+d=47*4=188 b+c+d+e=48*4=192 d=188-144=44 e=3+44=47 b+c+d=192-47=145 a=188-145=43.
3. A milkman mixed 1: 4 solution of milk and water with another 1: 2 solution of milk and water in the volume of ratio 3: 2. If the profit earned by selling the first solution was 20% and the mixture was sold at the same price, what is the profit or loss percentage? You have to assume that water comes free of cost.

A. 5.26% B. 5.25% C. 6.25% D. None of these

Answer: Option (B)
Explanation: Consider 90 litre of solution1 and 60 litre of solution2 90 litre of solution1 contains 90*1/5=18 litre of milk and (90-18)=72 litre of water 60 litre of solution2 contains 60*1/3=20 litre of milk and (60-20)=40 litre of water The mixture will have 18+20=38 litre of milk 72+40=112 litre of water i.e., Quantity of milk : Quantity of water = 38 : 112 = 19 : 56 Take 75 litres of solution1. The quantity of milk = 75*1/5=15 litre. Profit earned in selling solution is 20% assume cost price of 1 litre of milk = Rs.1 Cost price of 75 litre of solution1 = Rs.15. Selling price of 75 litre of solution1 = 15*120/100 = Rs.18 Take 75 litres of mixture. The quantity of milk = 75*19/75=19 litre. Cost price is Rs.19 Selling price will be the same and is Rs.18 so there will be a loss Loss = Rs.1 in selling 75 litres of the mixture Percentage loss = 1*100/19 = 5.25% loss
4. A solution of 25 litres contain 20% sugar, if 10 litres of the solution is evaporated, find the percentage of sugar in the final solution?

A. 27.5% B. 25% C. 30% D. 33.33%

Answer: Option (D)
Explanation: it is 33.33% given 25 litres of sol has 20% sugar. qty of sugar =(20%/100%)*25 =5 lit of sugar (assume sugar qty in a litre for easy calculation ). therefore 25 lit sol contains 20 lit of water and 5 litres of sugar. now the sol boils losing 10 litres of water. therefore the new 15-litre sol contains 10-litre water and 5-litre sugar. there new % of sugar is 5/15 * 100 =33.33%
5. A mixture of a certain quantity of milk with 16 litres of water is worth 90p per litre. if pure milk is worth Rs.1.08 per litre, how much milk is there in the mixture?

A. 75 B. 80 C. 60 D. 65

Answer: Option (B)
Explanation: let the mixture contain 'x' lit of milk and already given 16lit of water as water don't cost (x+16) lit of mixture = x lit of pure milk (x+16)90=(x*1.08*100) /*since given in paise*/ x=80lit.
Simple & Compound Interest
1. If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.

A. 1261 B. 1234 C. 1256 D. 1287

Answer: Option (A)
Explanation: Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. So principal=RS [100*1200]/3*5=RS 8000 Amount = Rs. 8000 x [1 +5/100]^3 - = Rs. 9261. C.I. = Rs. (9261 - 8000) = Rs. 1261.
2. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

A. Rs. 650 B. Rs. 690 C. Rs. 698 D. Rs. 700

Answer: Option (C)
Explanation: S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698.
3. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

A. 3% B. 4% C. 5% D. 6% E. None of these

Answer: Option (D)
Explanation: S.I. = Rs. (15500 - 12500) = Rs. 3000. Rate = ([latex]\frac{100 \times 3000}{12500 \times 4}[/latex])% = 6%
4. The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:

A. Rs. 400 B. Rs. 500 C. Rs. 600 D. Rs. 800

Answer: Option (B)
Explanation: Let the sum be Rs. P. Then, P ((1 + [latex]\frac{10}{100})^{2}[/latex] - P) = 525 P (([latex]\frac{11}{10})^{2}[/latex] - 1) = 525 P = [latex]\frac{525 \times 100}{21}[/latex] = 2500. Sum = Rs . 2500. So, S.I. = Rs. [latex]\frac{2500 \times 5 \times 4}{100}[/latex] = Rs. 500
5. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging interest of 10%, the effective rate of interest becomes:

A. 10% B. 10.25% C. 10.5% D. None of these

Answer: Option (B)
Explanation: Let the sum be Rs. 100. Then, S.I. for first 6 months = Rs.[latex]\frac{100 \times 10 \times 1}{100 \times 2}[/latex] = Rs. 5 S.I. for last 6 months = Rs. [latex]\frac{105 \times 10 \times 1}{100 \times 2}[/latex] = Rs. 5.25 So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25 Effective rate = (110.25 - 100) = 10.25%
Data Interpretation etc

1. What was the ratio of investment in 1997 over the investment in 1992?

A. 5.50 B. 5.36 C. 5.64 D. 5.75

Answer: Option (A)
Explanation: The 1997 figure of investment as a factor of 1992 investment = (31.36/5.70) = 5.50
2. What was the absolute difference in the FDI to India in between 1996 and 1997?

A. 7.29 B. 7.13 C. 8.13 D. None of these

Answer: Option (B)
Explanation: The difference in investments over 1996-1997 was 31.36 - 24.23 = € 7.13 millions.
3. If India FDI from OPEC countries was proportionately the same in 1992 and 1997 as the total FDI from all over the world and if the FDI in 1992 from the OPEC countries was Euro 2 million. What was the amount of FDI from the OPEC countries in 1997?

A. 11 B. 10.72 C. 11.28 D. 11.5

Answer: Option (A)
Explanation: Let x be the FDI in 1997. Then: (2/5.7) = (x/31.36) x = (2/5.7) x 31.36 x = 11
4. Which year exhibited the highest growth in FDI in India over the period has shown?

A. 1993 B. 1994 C. 1995 D. 1996

Answer: Option (D)
Explanation: It can be seen that the FDI in 1996 more than doubles over that of 1995. No other year is close to that rate of growth.
5. What was India's total FDI for the period shown in the figure?

A. 93.82 B. 93.22 C. 93.19 D. None of these

Answer: Option (A)
Explanation: Total FDI investment in the figure shown is = 5.7 + 10.15 + 12.16 + 10.22 + 24.23 + 31.36 = 93.82 billion.