Percentage
1. The tax on a commodity is diminished by 20% but its consumption is increased by 10%. Find the decrease percent in the revenue derived from it?

A. 20% B. 18% C. 15% D. 12%

Answer: Option (D)
Explanation: 100 * 100 = 10000 80 * 110 = 8800 10000------- 1200 100 ------- ? = 12%
2. At an examination in which full marks were 500. A got 10% less than B, B got 25% more than C and C got 20% less than D. If A got 360marks, what percentage of full marks was obtained by D?

A. 70% B. 90% C. 80% D. 75%

Answer: Option (C)
Explanation: A B C D 90 100 80 100 A D 90 ----- 100 360 ------ ? = 400 500 ------ 400 100 ------- ? => 80%
3. If the population of a certain city increases at a rate of 5%. If the population in 1981 was 138915, then the population in 1978 was?

A. 1,20,000 B. 1,10,000 C. 1,00,000 D. 90,000

Answer: Option (A)
Explanation: X * (105/100) * (105/100) * (105/100) = 138915 X = 138915/1.157625 X = 120000
4. There were two candidates in an election. Winner candidate received 62% of votes and won the election by 288 votes. Find the number of votes casted to the winning candidate?

A. 456 B. 744 C. 912 D. 1200

Answer: Option (B)
Explanation: W = 62% L = 38% 62% - 38% = 24% 24% -------- 288 62% -------- ? => 744
5. A man spends 10% of his income in house rent, 20% of the rest on his children’s education, 25% of the rest miscellaneous causes. If he now posses Rs. 1944 then his income is?

A. Rs.3600 B. Rs.4000 C. Rs.4500 D. Rs.3000

Answer: Option (A)
Explanation: X * (90/100) * (80/100) * (75/100) = 1944 X * 0.9 * 0.8 * 0.75 X = 1944/0.54 X = 3600
Number System
1. When writing numbers from 1 to 10,000, how many times is the digit 9 written?

A. 3200 B. 3600 C. 4000 D. 4200

Answer: Option(C)
Explanation: The digits 9 occurs in the thousands place in 1000 numbers. It occurs in the hundreds place in 1000 numbers and so on The digit occurs 4000 times.
2. Which digits should come in place of @ and # if the number 62684@# is divisible by both 8 and 5?

A. 4, 0 B. 0, 4 C. 4, 4 D. 1, 1

Answer: Option (A)
Explanation: Since the given number is divisible by 5, so 0 or 5 must come in place of #. But, a number ending with 5 is never divisible by 8. So, 0 will replace #. Now, the number formed by the last three digits is 4@0, which becomes divisible by 8, if @ is replaced by 4. Hence, digits in place of @ and # are 4 and 0 respectively.
3. How many keystrokes are needed to type numbers from 1 to 1000 on a standard keyboard?

A. 3001 B. 2893 C. 2704 D. 2890

Answer: Option (B)
Explanation: While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them requires one keystroke. That is 9 keystrokes. There are 90 two-digit numbers, from 10 to 99. Each of these numbers requires 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers. There are 900 three-digit numbers, from 100 to 999. Each of these numbers requires 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers. Then 1000 is a four-digit number which requires 4 keystrokes. Totally, therefore, one requires 9+180+2700+4= 2893 keystrokes.
4. How many natural numbers below 660 are divisible by 5 and 11 but not by 3?

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

Answer: Option (A)
Explanation: If the number is divisible by 5 and 11 it must be divisible by 55. The numbers are less than 660. Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part). The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3. The numbers of such, multiples are the quotient of 11 by 3. Quotient of [latex]\frac {11}{3}[/latex] = 3. Out of 11 multiples of 55, 3 are multiples of 3. Hence, numbers less than 660 and divisible by 5 and 11 but not by 3=11−3=8
5. What is the maximum value of m such that 7m divides into 14! evenly?

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

Answer: Option(B)
Explanation: The term 14! equals the product of the numbers 1,2,3,4,5,6,7,8,9,10,11,12,13 and 14. Only two of these numbers are divisible by 7. The numbers are 7 and 14. Hence, 14! can be expressed as the product of k×7×14, where k is not divisible by 7. Now, since there are two 7s in 14!, the numbers 7 and 7[latex]^{2}[/latex] divides 14! evenly. 7[latex]^{3}[/latex] and further powers of 7 leave a remainder when divided into 14!. Hence, the maximum value of m is 2.
Average
1. The captain of a cricket team of 11 members is 26 years old and the wicketkeeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

A. 23 years B. 24 years C. 25 years D. None of these

Answer: Option (A)
Explanation: Let the average age of the whole team by x years. 11x - (26 + 29) = 9(x -1) 11x - 9x = 46 2x = 46 x = 23. So, the average age of the team is 23 years.
2. 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 are 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.
3. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

A. 35 years B. 40 years C. 50 years D. None of these

Answer: Option (B)
Explanation: Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years. Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years. Husband's present age = (90 - 50) years = 40 years.
4. 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.
5. 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).
Number Series
1. 42 40 38 35 33 31 28

A. 25 22 B. 26 23 C. 26 24 D. 25 23 E. 26 22

Answer: Option (C)
Explanation: This is an alternating subtraction series in which 2 is subtracted twice, then 3 is subtracted once, then 2 is subtracted twice, and so on.
2. 6 10 14 18 22 26 30

A. 36 40 B. 33 37 C. 38 42 D. 34 36 E. 34 38

Answer: Option (E)
Explanation: This simple addition series adds 4 to each number to arrive at the next.
3. 8 12 9 13 10 14 11

A. 14 11 B. 15 12 C. 8 15 D. 15 19 E. 8 5

Answer: Option (B)
Explanation: This is an alternating addition and subtraction series, in which the addition of 4 is alternated with the subtraction of 3.
4. 36 31 29 24 22 17 15

A. 13 11 B. 10 5 C. 13 8 D. 12 7 E. 10 8

Answer: Option (E)
Explanation: This is an alternating subtraction series, which subtracts 5, then 2, then 5, and so on.
5. 3 5 35 10 12 35 17

A. 22 35 B. 35 19 C. 19 35 D. 19 24 E. 22 24

Answer: Option (C)
Explanation: This is an alternating addition series, with a random number, 35, interpolated as every third number. The pattern of addition is to add 2, add 5, add 2, and so on. The number 35 comes after each "add 2" step.
Ratio and Percentage
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 40% of a number is equal to two-thirds of another number, what is the ratio of the first number to the second number?

A. 2: 5 B. 3: 7 C. 5: 3 D. 7: 3

Answer: Option (C)
Explanation: Let 40% of A = [latex]\frac {2}{3}[/latex] B Then,[latex]\frac {40A}{100}[/latex] = [latex]\frac {2B}{3}[/latex] [latex] \frac{2A}{5}[/latex] = [latex]\frac {2B}{3}[/latex] [latex]\frac {A}{B}[/latex] = [latex]\frac{2}{3}[/latex] x [latex]\frac {5}{2}[/latex] = [latex]\frac {5}{3}[/latex] A : B = 5 : 3.
4. 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

Answer: Option (B)
Explanation: Let the numbers be 3x and 5x. Then, [latex]\frac {3x - 9}{5x - 9}[/latex] = [latex]\frac {12}{23}[/latex] 23(3x - 9) = 12(5x - 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.
5. The greatest ratio out of 2:3, 5:4, 3:2 and 4:5 is?

A. 4:5 B. 3:2 C. 5:4 D. 2:3

Answer: Option (B)

Mensuration and Geometry
1. The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?

A. 1: 96 B. 1: 48 C. 1: 84 D. 1: 68 E. None of these

Answer: Option (A)
Explanation: Let the length and the breadth of the rectangle be 4x cm and 3x respectively. (4x)(3x) = 6912 12x[latex]^{2}[/latex] = 6912 x[latex]^{2}[/latex] = 576 = 4 * 144 = 2[latex]^{2}[/latex] * 12[latex]^{2}[/latex] (x > 0) => x = 2 * 12 = 24 Ratio of the breadth and the areas = 3x : 12x[latex]^{2}[/latex] = 1 : 4x = 1: 96.
2. The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?

A. 8.5 m B. 17 m C. 34 m D. 51 m E. None of these

Answer: Option (B)
Explanation: Let the breadth of the plot be b m. Length of the plot = 3 b m (3b)(b) = 867 3b[latex]^{2}[/latex] = 867 b[latex]^{2}[/latex] = 289 = 17[latex]^{2}[/latex] (b > 0) b = 17 m.
3. The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?

A. 27 m B. 24 m C. 18 m D. 21 m E. None of these

Answer: Option (C)
Explanation: Let the length and the breadth of the floor be l m and b m respectively. l = b + 200% of b = l + 2b = 3b Area of the floor = 324/3 = 108 sq m l b = 108 i.e., l * l/3 = 108 l[latex]^{2}[/latex] = 324 => l = 18.
4. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What is 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

Answer: Option (D)
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.
5. What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58?

A. Rs. 3944 B. Rs. 3828 C. Rs. 4176 D. Cannot be determined E. None of these

Answer: Option (A)
Explanation: Let the side of the square plot be an ft. a[latex]^{2}[/latex] = 289 => a = 17 Length of the fence = Perimeter of the plot = 4a = 68 ft. Cost of building the fence = 68 * 58 = Rs. 3944.
Quadratic Equation
1. The sum and the product of the roots of the quadratic equation x[latex]^{2}[/latex] + 20x + 3 = 0 are?

A. 10, 3 B. -10, 3 C. 20, -3 D. -10, -3 E. None of these

Answer: Option (E)
Explanation: Some of the roots and the product of the roots are -20 and 3 respectively.
2. If the roots of the equation 2x[latex]^{2}[/latex] - 5x + b = 0 are in the ratio of 2:3, then find the value of b?

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

Answer: Option (A)
Explanation: Let the roots of the equation 2a and 3a respectively. 2a + 3a = 5a = -(- 5/2) = 5/2 => a = 1/2 Product of the roots: 6a[latex]^{2}[/latex] = b/2 => b = 12a[latex]^{2}[/latex] a = 1/2, b = 3.
3. The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?

A. 9, 10 B. 10, 11 C. 11, 12 D. 12, 13 E. None of these

Answer: Option (A)
Explanation: Let the two consecutive positive integers be x and x + 1 x[latex]^{2}[/latex] + (x + 1)[latex]^{2}[/latex] - x(x + 1) = 91 x[latex]^{2}[/latex] + x - 90 = 0 (x + 10)(x - 9) = 0 => x = -10 or 9. As x is positive x = 9 Hence the two consecutive positive integers are 9 and 10.
4. One root of the quadratic equation x[latex]^{2}[/latex] - 12x + a = 0, is thrice the other. Find the value of a?

A. 29 B. -27 C. 28 D. 7 E. None of these

Answer: Option (E)
Explanation: Let the roots of the quadratic equation be x and 3x. Sum of roots = -(-12) = 12 a + 3a = 4a = 12 => a = 3 Product of the roots = 3a[latex]^{2}[/latex] = 3(3)[latex]^{2}[/latex] = 27.
5. A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?

A. 10 B. 8 C. 15 D. 7.50 E. None of these

Answer: Option (A)
Explanation: Let the price of each notebook be Rs.x. Let the number of notebooks which can be brought for Rs.300 each at a price of Rs.x be y. Hence xy = 300 => y = 300/x (x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy =>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0 multiplying both sides by -1/10x => x[latex]^{2}[/latex] + 15x - 10x - 150 = 0 => x(x + 15) - 10(x + 15) = 0 => x = 10 or -15 As x>0, x = 10.
Simple Interest
1. Find the simple interest on Rs.500 for 9 months at 6 paisa per month?

A. Rs.345 B. Rs.270 C. Rs.275 D. Rs.324

Answer: Option (B)
Explanation: I = (500*9*6)/100 = 270.
2. A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum?

A. Rs.500 B. Rs.515 C. Rs.525 D. None

Answer: Option (D)
Explanation: 840 = P [1 + (10*3)/100] P = 646.
3. Rs.2500 is divided into two parts such that if one part be put out at 5% simple interest and the other at 6%, the yearly annual income may be Rs.140. How much was lent at 5%?

A. Rs.1500 B. Rs.1300 C. Rs.1200 D. Rs.1000

Answer: Option (D)
Explanation: (x*5*1)/100 + [(2500 - x)*6*1]/100 = 140 X = 1000
4. 4000 was divided into two parts such a way that when the first part was invested at 3% and the second at 5%, the whole annual interest from both the investments is Rs.144, how much was put at 3%?

A. Rs.2500 B. Rs.2700 C. Rs.2800 D. Rs.5000

Answer: Option (C)
Explanation: (x*3*1)/100 + [(4000 - x)*5*1]/100 = 144 3x/100 + 200 – 5x/100 = 144 2x/100 = 56 è x = 2800.
5. If rupee one produces rupees nine over a period of 40 years, find the rate of simple interest?

A. 20 % B. 10 % C. 15 % D. 22 1/2%

Answer: Option (D)
Explanation: 9 = (1*40*R)/100 R = 22 1/2 %.
Compound Interest
1. Find the C.I. on a sum of Rs.1600 for 9 months at 20% per annum, interest being compounded quarterly?

A. Rs.17684 B. Rs.1684 C. Rs.2522 D. Rs.3408

Answer: Option (C)
2. Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?

A. Rs.81.60 B. Rs.160 C. Rs.1081.60 D. Rs.99

Answer: Option (A)
Explanation: SI = 40 + 40 CI = 40 + 40 + 1.6 = 81.6
3. The C.I. on a certain sum for 2 years Rs.41 and the simple interest is Rs.40. What is the rate percent?

A. 4% B. 5% C. 6% D. 8%

Answer: Option (B)
Explanation: SI = 20 + 20 CI = 20 + 21 20 ---- 1 100 ---- ? => 5%
4. Find the least number of complete years in which a sum of money put out at 25% compound interest will be more than double of itself?

A. 6 years B. 1 year C. 2 years D. 4 years

Answer: Option (D)
5. The difference between compound and simple interest on a certain sum of money for 3 years at 6 2/3% p.a is Rs.184. Find the sum?

A. Rs.12000 B. Rs.14200 C. Rs.17520 D. Rs.13500

Answer: Option (D)
Explanation: P = (184*106) / [6 2/3 * 6 2/3 *(300*6 2/3)] P = 13500.
Profit and Loss
1. 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.
2. When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?

A. Rs. 21,000 B. Rs. 22,500 C. Rs. 25,300 D. Rs. 25,800

Answer: Option (C)
Explanation: 85 : 18700 = 115 : x => x =[latex]\frac {18700 \times 115}{85}[/latex] = 25300. Hence, S.P. = Rs. 25,300.
3. 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%.
4. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:

A. 15 B. 16 C. 18 D. 25

Answer: Option (B)
Explanation: Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x. S.P. of x articles = Rs. 20. Profit = Rs. (20 - x). ([latex]\frac {20 - x}{x}[/latex] x 100 = 25) 2000 - 100x = 25x 125x = 2000 x = 16.
5. 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%

Answer: Option (B)
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 =[latex]\frac{295}{420}[/latex]x 100% = [latex]\frac {1475}{21}[/latex]% = 70% (approximately).

Problems on Ages
1. The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. what is the age of the youngest child ?

A. 4 B. 8 C. 10 D. None

Answer: Option (A)
Explanation: Let x = the youngest child. Each of the other four children will then be x+3, x+6, x+9, x+12. We know that the sum of their ages is 50. so, x+(x+3)+(x+6)+(x+9)+(x+12) = 50 => x= 4 The youngest child is 4 years old.
2. If Raj was one-third as old as Rahim 5 years back and Raj is 17 years old now, How old is Rahim now?

A. 40 B. 41 C. 36 D. 48

Answer: Option (B)
Explanation: Raj’s age today = 17 decades, Hence, 5 decades back, he must be 12 years old. Rahim must be 36 years old, Because (3×12). 5 years back Rahim must be 41 years old today. Because (36+5).
3. The ages of Krish and Vaibhav are in the proportion of 3: 5. After 9 years, the proportion of their ages will be 3: 4. Then the current age of Vaibhav is:

A. 10 B. 13 C. 15 D. 18

Answer: Option (C)
Explanation: Krish’s age = 3A and Vaibhav’s age = 5A (3A+9)/(5A+9) = 3/4 => 4 (3A + 9) = 3 (5A + 9) => A = 3 Therefore, Vaibhav’s age = 15 years.
4. If two times of the daughter’s age in years is included to the mother’s age, the total is 70 and if two times of the mother’s age is included to the daughter’s age, the total is 95. So the Mother’s age is,

A. 30 B. 38 C. 40 D. 41

Answer: Option (C)
Explanation: Let daughter’s age = A and mother’s age = B Given: 2A+B = 70 and A+2B = 95 Solving B, we will get B = 40.
5. The ratio of the ages of Mahesh and Nilesh is 5: x. Mahesh is 18 years younger to Ramesh. After nine years Ramesh will be 47 years old. If the difference between the ages of Mahesh and Nilesh is the same as the age of Ramesh, what is the value of x?

A. 11.8 B. 12.9 C. 13.7 D. 14.5

Answer: Option (D)
Explanation: Let the present ages of Mahesh, Nilesh and Ramesh be k, l and m respectively. k/l = 5/x ------ (1) k = m - 18 ------ (2) m + 9 = 47 ------ (3) k - l = m ----- (4) (3) => m = 47 - 9 = 38 years (2) => k = 38 -18 = 20 years (1) => 20/l = 5/x => l = 4x (4) => 4x - 20 = 38 => 4x = 58 => x = 14.5.
Speed, Distance, and Time
1. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

A. 9 B. 10 C. 12 D. 20

Answer: Option (B)
Explanation: Due to stoppages, it covers 9 km less. Time taken to cover 9 km =[latex]\frac {9}{54}[/latex] x 60 min = 10 min.
2. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:

A. 5 kmph B. 6 kmph C. 6.25 kmph D. 7.5 kmph

Answer: Option (A)
Explanation: Let Abhay's speed be x km/hr. Then,[latex]\frac {30}{x}[/latex] - [latex]\frac {30}{2x}[/latex] = 3 6x = 30 x = 5 km/hr.
3. A farmer traveled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance traveled on foot is:

A. 14 km B. 15 km C. 16 km D. 17 km

Answer: Option (C)
Explanation: Let the distance traveled on foot be x km. Then, distance travelled on bicycle = (61 -x) km. So, [latex]\frac{x}{4}[/latex] + [latex]\frac {(61 -x)}{9}[/latex] = 9 9x + 4(61 -x) = 9 x 36 5x = 80 x = 16 km.
4. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

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

Answer: Option (A)
Explanation: Let the duration of the flight be x hours. Then, [latex]\frac{600}{x}[/latex] - [latex]\frac{600}{x + (1/2}[/latex] = 200 [latex]\frac{600}{x}[/latex] - [latex]\frac{1200}{2x + 1}[/latex] = 200 x(2x + 1) = 3 2x[latex]^{2}[/latex] + x - 3 = 0 (2x + 3)(x - 1) = 0 x = 1 hr. [neglecting the -ve value of x]
5. A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:

A. 35.55 km/hr B. 36 km/hr C. 71.11 km/hr D. 71 km/hr

Answer: Option (C)
Explanation: Total time taken = [latex]\frac{160}{64}[/latex] + [latex]\frac{160}{80}[/latex]hrs = [latex]\frac{9}{2}[/latex]hrs. Average speed = 320 x [latex]\frac{2}{9}[/latex] km/hr = 71.11 km/hr.
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]1) = [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. 4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

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

Answer: Option (B)
Explanation: Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y. Then, 4x + 6y = [latex]\frac{1}{8}[/latex] and 3x + 7y = [latex]\frac{1}{10}[/latex]. Solving the two equations, we get: x =[latex]\frac{11}{400}[/latex], y =[latex]\frac{1}{400}[/latex] 1 woman's 1 day's work =[latex]\frac{1}{400}[/latex]. 10 women's 1 day's work =[latex]\frac{1}{400}[/latex] x 10 =[latex]\frac{1}{40}[/latex]. Hence, 10 women will complete the work in 40 days.
3. A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the work?

A. 40 B. 50 C. 54 D. 60

Answer: Option (D)
Explanation: (A + B)'s 20 day's work = ([latex]\frac{1}{30}[/latex] x 20)= [latex]\frac {2}{3}[/latex]. Remaining work = (1 - [latex]\frac{2}{3}[/latex] ) = [latex]\frac {1}{3}[/latex]. Now,[latex]\frac {1}{3}[/latex] work is done by A in 20 days. Therefore, the whole work will be done by A in (20 x 3) = 60 days.
4. Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:

A. 15 B. 16 C. 18 D. 25

Answer: Option (B)
Explanation: The ratio of times taken by Sakshi and Tanya = 125: 100 = 5: 4. Suppose Tanya takes x days to do the work. 5 : 4 :: 20 : x x = [latex]\frac{4 \times 20}{5}[/latex] x = 16 days. Hence, Tanya takes 16 days to complete the work.
5. Twenty women can do work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?

A. 3: 4 B. 4: 3 C. 5: 3 D. Data inadequate

Answer: Option (B)
Mixture and Allegations
1. Two varieties of wheat - A and B costing Rs. 9 per kg and Rs. 15 per kg were mixed in the ratio 3 : 7. If 5 kg of the mixture is sold at 25% profit, find the profit made?

A. Rs. 13.50 B. Rs. 14.50 C. Rs. 15.50 D. Rs. 16.50 E. None of these

Answer: Option (D)
Explanation: Let the quantities of A and B mixed be 3x kg and 7x kg. Cost of 3x kg of A = 9(3x) = Rs. 27x Cost of 7x kg of B = 15(7x) = Rs. 105x Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66 Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50
2. In a mixture of milk and water, the proportion of milk by weight was 80%. If, in a 180 gm mixture, 36 gms of pure milk is added, what would be the percentage of milk in the mixture formed?

A. 80% B. 100% C. 84% D. 87.5% E. None of these

Answer: Option (E)
Explanation: Percentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.
3. In a can, there is a mixture of milk and water in the ratio 4: 5. If it is filled with an additional 8 liters of milk the can would be full and the ratio of milk and water would become 6: 5. Find the capacity of the can?

A. 40 B. 44 C. 48 D. 52 E. None of these

Answer: Option (B)
Explanation: Let the capacity of the can be T liters. Quantity of milk in the mixture before adding milk = 4/9 (T - 8) After adding milk, quantity of milk in the mixture = 6/11 T. 6T/11 - 8 = 4/9(T - 8) 10T = 792 - 352 => T = 44.
4. In what ratio should a variety of rice costing Rs. 6 per kg be mixed with another variety of rice costing Rs. 8.75 per kg to obtain a mixture of costing Rs. 7.50 per kg?

A. 5: 6 B. 3: 4 C. 7: 8 D. 8: 9 E. None of these

Answer: Option (A)
Explanation: Let us say the ratio of the quantities of cheaper and dearer varieties = x: y By the rule of allegation, x/y = (87.5 - 7.50) / (7.50 - 6) = 5/6.
5. A mixture of 70 liters of milk and water contains 10% water. How many liters of water should be added to the mixture so that the mixture contains 12 1/2% water?

A. 2 B. 8 C. 4 D. 5 E. None of these

Answer: Option (A)
Explanation: Quantity of milk in the mixture = 90/100 (70) = 63 litres. After adding water, milk would form 87 1/2% of the mixture. Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72 Hence 72 - 70 = 2 litres of water must be added.
Probability
1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

A. 1/2 B. 2/5 C. 8/15 D. 9/20

Answer: Option (D)
Explanation: Here, S = {1, 2, 3, 4, ...., 19, 20}. Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. P(E) = n(E)/n(S) = 9/20.
2. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A. 1/10 B. 2/5 C. 2/7 D. 5/7

Answer: Option (C)
Explanation: P (getting a prize) = 10/(10+25) = 10/35 = 2/7.
3. A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of heart is:

A. 1/13 B. 2/13 C. 1/26 D. 1/52

Answer: Option (C)
Explanation: Here, n(S) = 52. Let E = event of getting a queen of the club or a king of heart. Then, n(E) = 2. P(E) = n(E)/n(S) = 2/52 = 1/26.
4. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen, and King only)?

A. 1/13 B. 3/13 C. 1/4 D. 9/52

Answer: Option (B)
Explanation: Clearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card) = 12/52 = 3/13.
5. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A. 3/4 B. 4/7 C. 1/8 D. 3/7

Answer: Option (B)
Explanation: Let number of balls = (6 + 8) = 14. The number of white balls = 8. P (drawing a white ball) = 8/14 = 4/7.

Linear Equation
1. Value of ‘x’ in 6x - 4 = 3x + 8 should be

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

Answer: Option (D)
2. The sum of two digits and the number formed by interchanging its digit is 110. If ten is subtracted from the first number, the new number is 4 more than 5 times of the sum of the digits in the first number. Find the first number.

A. 46 B. 48 C. 64 D. 84

Answer: Option (C)
3. A fraction becomes. when subtracted from the numerator and it becomes. when 8 is added to its

denominator. Find the fraction. A. 4/12 B. 3/13 C. 5/12 D. 11/7

Answer: Option (C)
4. Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What is the present age of A.

A. 20 B. 50 C. 60 D. 40

Answer: Option (B)
5. What will be the solution of these equations ax+by=a-b, bx-ay=a+b

A. x=1, y=2 B. x=2,y=-1 C. x=-2, y=-2 D. x=1, y=-1

Answer: Option (D)
Data Sufficiency
Direction (1-5): In each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and
Give answer
(A) If the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question (B) If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question (C) If the data either in statement I alone or in statement II alone are sufficient to answer the question (D) If the data given in both statements I and II together are not sufficient to answer the question and (E) If the data in both statements I and II together are necessary to answer the question.
1. Question: How is 'No' coded in the code language? Statements: I. 'Ne Pa Sic Lo' means 'But No None And' and 'Pa Lo Le Ne' means 'If None And But'. II. 'Le Se Ne Sic' means 'If No None Will' and 'Le Pi Se Be' means 'Not None If All'.

A. I alone is sufficient while II alone is not sufficient B. II alone is sufficient while I alone is not sufficient C. Either I or II is sufficient D. Neither I nor II is sufficient E. Both I and II are sufficient

Answer: Option (A)
Explanation: In the two statements given in I, the common words are 'But', 'None', 'And' and the common codewords are 'Ne', 'Pa', 'Lo'. So, 'Ne', 'Pa' and 'Lo' are codes for 'But', 'None' and 'And'. Thus, in the first statement, 'Sic' is the code for 'No'.
2. Question: Who among P, Q, T, V, and M is exactly in the middle when they are arranged in ascending order of their heights? Statements: I. V is taller than Q but shorter than M. II. T is taller than Q and M but shorter than P.

A. I alone is sufficient while II alone is not sufficient B. II alone is sufficient while I alone is not sufficient C. Either I or II is sufficient D. Neither I nor II is sufficient E. Both I and II are sufficient

Answer: Option (E)
Explanation: From I, we have: M > V > Q. From II, we have: T > Q, T > M, P > T. Combining the above two, we have: P>T>M>V>Q i.e. Q Clearly, M is in the middle.
3. Question: Which code word stands for 'good' in the coded sentence 'sin co bye' which means 'He is good'? Statements: I. In the same code language, 'co mot det' means 'They are good'. II. In the same code language, 'sin mic bye' means 'He is honest'.

A. I alone is sufficient while II alone is not sufficient B. II alone is sufficient while I alone is not sufficient C. Either I or II is sufficient D. Neither I nor II is sufficient E. Both I and II are sufficient

Answer: Option (C)
Explanation: In the given statement and I, the common word is 'good' and the common code word is 'co'. So, 'co' is the code for 'good'. In the given statement and II, the common words are 'He' and 'is' and the common codewords are 'sin' and 'bye'. So 'sin' and 'bye' are the codes for 'He' and 'is'. Thus, in the given statement, 'co' is the code for 'good'.
4. Question: What is the numerical code for 'water' in a certain code? Statements: I. The code for 'give me water' is '719'. II. The code for 'you can bring water for me' is written as '574186'.

A. I alone is sufficient while II alone is not sufficient B. II alone is sufficient while I alone is not sufficient C. Either I or II is sufficient D. Neither I nor II is sufficient E. Both I and II are sufficient

Answer: Option (D)
Explanation: In I and II, the common words are 'me' and 'water' and the common code numbers are '7' and '1'. So, the code for 'water' is either '7' or '1'.
5. Question: How many visitors saw the exhibition yesterday? Statements: I. Each entry pass holder can take up to three persons with him/her. II. In all, 243 passes were sold yesterday.

A. I alone is sufficient while II alone is not sufficient B. II alone is sufficient while I alone is not sufficient C. Either I or II is sufficient D. Neither I nor II is sufficient E. Both I and II are sufficient

Answer: Option (D)
Explanation: From I and II, we find that maximum (243 x 3) i.e. 729 visitors saw the exhibition. But the exact number cannot be determined.
Simplification
1. A man has Rs.480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?

A. 45 B. 60 C. 75 D. 90

Answer: Option (D)
Explanation: Let the number of notes of each denomination be x. Then x + 5x + 10x = 480 16x = 480 x = 30. Hence, total number of notes = 3x = 90.
2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:

A. 20 B. 80 C. 100 D. 200

Answer: Option (C)
Explanation: Let the number of students in rooms A and B be x and y respectively. Then, x - 10 = y + 10 x - y = 20 .... (i) and x + 20 = 2(y - 20) x - 2y = -60 .... (ii) Solving (i) and (ii) we get: x = 100 , y = 80. The required answer A = 100.
3. 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.
4. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:

A. 22 B. 23 C. 24 D. 26

Answer: Option (D)
Explanation: Let the number of hens be x and the number of cows be y. Then, x + y = 48 .... (i) and 2x + 4y = 140 x + 2y = 70 .... (ii) Solving (i) and (ii) we get: x = 26, y = 22. The required answer = 26.
5. Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:

A. [latex]\frac{1}{7}[/latex] B. [latex]\frac{1}{8}[/latex] C. [latex]\frac{1}{9}[/latex] D. [latex]\frac{7}{8}[/latex]

Answer: Option (A)
Explanation: Original share of 1 person =[latex]\frac{1}{8}[/latex] New share of 1 person = [latex]\frac{1}{7}[/latex] Increase = [latex]\frac{1}{7}[/latex] - [latex]\frac{1}{8}[/latex] = [latex]\frac{1}{56}[/latex] Required fraction = [latex]\frac{(1/56)}{(1/8)}[/latex] = [latex]\frac{1}{56}[/latex] x [latex]\frac{8}{1}[/latex] = [latex]\frac{1}{7}[/latex]
Permutation and Combination
1. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A. 360 B. 480 C. 720 D. 5040 E. None of these

Answer: Option (C)
Explanation: The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.
2. If a five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition, the total number of ways this can be done is

A. 216 B. 240 C. 600 D. 3125

Answer: Option (A)
3. In the above problem, if the first digit cannot be 0, then

A. 42100 B. 421200 C. 42400 D. 421600

Answer: Option (B)
Explanation: n= 26x 25x 9x 9x 8=421200.
4. In how many ways can 10 people be seated in a row so that a given pair is not next to each other?

A. 10 ! - 9 ! B. 10! - 2(9!) C. 9! D. none of these

Answer: Option (B)
5. Suppose a license plate contains two letters followed by three digits with the first digit not zero. How many different license plates can be printed?

A. 608000 B. 608200 C. 608400 D. 608600

Answer: Option (C)
Explanation: Each letter can be printed in twenty-six different ways, the first digit in nine ways and each of the other two digits in ten ways. Hence 26 x 26 x 9 x 10 x 10 = 608400 different plates can be printed.
Data Interpretation (Pie Chart)
Dirction (1-5) : Go through the following pie chart and answer the questions given below that.

1. Approximately how many degrees should be there in the central angle of the sector for agriculture expenditure?

A. 220 B. 213 C. 210 D. 208

Answer: Option (B)
Explanation: In the pie chart, 100% is spread over 360°. Therefore 1 % = 3.6°. Agriculture expenditure = 59 %. Therefore 3.6 59 = 212.4°.
2. Approximately what is the ratio of expenditure on agriculture to that on dairy?

A. 8:3 B. 9:2 C. 70:1 D. 10:1

Answer: Option (D)
Explanation: Over here, one common mistake is that students calculate the actual values of agriculture and dairy. Since budget expenditure is proportional to % of the area covered, the ratio of agriculture to dairy expenditure would be the ratio of corresponding % allocations. Therefore, Agriculture/Dairy = 59/6 = 10/1.
3. In Haryana, in 2000, a total expenditure of Rs. 120mn was incurred. Approximately how many millions did the Haryana government spend on roads?

A. 15.4 B. 20 C. 10.8 D. 12

Answer: Option (C)
Explanation: Total expenditure = 100% = Rs. 120 mn. Expenditure on roads = 9% = 9/100 × 120 = Rs. 10.8mn.
4. If Rs. 9mn was spent in 2000 on Dairy, what would have been the total expenses in that year in million?

A. 150 B. 140 C. 160 D. 130

Answer: Option (A)
Explanation: 9mn were spent on dairy. This amount represents 6% of total expenditure in the year 2000. 6 = (Dairy expenditure/Total expenditure ) × 100 6 = (9/Total expenditure ) × 100 Total Exp = 100 × (9/6) = Rs. 150mn.
5. In 1999, Haryana spent 11% of all its expenses on Roads. “Did Haryana spend more on Roads expenditure in 1999 than in 2000?” To answer the question, we

A. Do not require any additional data. B. Require to know the total amount spent in each year. C. Require the exact %age break-up of the various items of expenses in 1999. D. Require the expenditure on dairy in both the years.

Answer: Option (B)
Explanation: % gives us just a proportional indication of the total quantity. Unless total expenditure is known, % of that is meaningless. Hence, to compare expenditure on roads in 1999 and 2000, the total amount spent must be known.