The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: A.10 B.14 C.23 D.30 E.None of these

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: A.10 B.14 C.23 D.30 E.None of these
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1 Answer

Answer :

Answer – C (23) Explanation – L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 – 37) = 23
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Related questions

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: A.30 B.22 C.40 D.60 E.None of these

Description : The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: A.30 B.22 C.40 D.60 E.None of these

Last Answer : Answer – C (40) Explanation – Let the numbers be 2x and 3x. Then, their L.C.M. = 6x. So, 6x = 48 or x = 8. The numbers are 16 and 24. Hence, required sum = (16 + 24) = 40.

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of theseSix bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of these

Description : Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of these

Last Answer : Answer – D (16) Explanation – L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds, i.e, 2 minutes.In 30 minutes, they will toll together 30/2 + 1 = 16

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: A.68 B.98 C.180 D.364 E.None of these

Description : The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: A.68 B.98 C.180 D.364 E.None of these

Last Answer : Answer – D (364) Explanation – L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. Required number = (90 x 4) + 4 = 364.

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: A.534 B.486 C.544 D.548 E.None of these

Description : The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: A.534 B.486 C.544 D.548 E.None of these

Last Answer : Answer – D (548) Explanation – Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548.

The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is: A.11115 B.15110 C.15130 D.15310 E.None of these

Description : The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is: A.11115 B.15110 C.15130 D.15310 E.None of these

Last Answer : Answer – B (15110) Explanation – Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10. Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10 = 15120 – 10 = 15110

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point? A.15 minutes 15 seconds B.42 minutes 30 seconds C.42 minutes D.46 minutes 12 seconds E.None of these

Description : A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point ... A.15 minutes 15 seconds B.42 minutes 30 seconds C.42 minutes D.46 minutes 12 seconds E.None of these

Last Answer : Answer – D (46 minutes 12 seconds) Explanation – L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 see i.e., 46 min. 12 sec

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: A.1 B.2 C.3 D.5 E.None of these

Description : The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: A.1 B.2 C.3 D.5 E.None of these

Last Answer : Answer – B (2) Explanation – Let the numbers 13a and 13b. Then, 13a x 13b = 2028 ab = 12. Now, the co-primes with product 12 are (1, 12) and (3, 4).

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A.124 B.100 C.111 D.175 E.None of these

Description : The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A.124 B.100 C.111 D.175 E.None of these

Last Answer : Answer – C (111) Explanation – Let the numbers be 37a and 37b. Then, 37a x 37b = 4107 ab = 3. Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: A.269 B.275 C.308 D.310 E.None of these

Description : The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: A.269 B.275 C.308 D.310 E.None of these

Last Answer : Answer – C (308) Explanation – Other number =[11 x 7700]/275 = 308

The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 years hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Description : The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 ... hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Last Answer : 5 : 30 : 31

What least number would be subtracted from 427398 so that the remaining number is divisible by 15? 1. 13 2. 3 3. 16 4. 11 5. 14

Description : What least number would be subtracted from 427398 so that the remaining number is divisible by 15? 1. 13 2. 3 3. 16 4. 11 5. 14

Last Answer : Answer- 2 ( 3) Explanation:- On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted 

A two digit number is such that the sum of the digits is 11.When the number with the same digits is reversed is subtracted from this number, the difference is 9.What is the number? 1. 23 2. 24 3. 65 4. 14 5. 32

Description : A two digit number is such that the sum of the digits is 11.When the number with the same digits is reversed is  subtracted from this number, the difference is 9.What is the number? 1. 23 2. 24 3. 65 4. 14 5. 32

Last Answer : Answer- 3(65) Explanation:- let the number be written as xy x-ten’s place y units place x+y=11…………………1 (10x+y)-(10y+x)=9 10x+y-10y-x=9 9x-9y=9 x-y=1………………….2 Add (1) & (2) 2x=12 x=6 y=5 the number is 65

A two digit number exceeds forty percent of another two digit number by 16. If the sum of these two digit numbers is 72, which of the following is the difference between these two digit numbers ? a) 10 b) 6 c) 8 d) 14 e) None of these

Description : A two digit number exceeds forty percent of another two digit number by 16. If the sum of these two digit numbers is 72, which of the following is the difference between these two digit numbers ? a) 10 b) 6 c) 8 d) 14 e) None of these

Last Answer : c) 8 Let first two digit no. and second two digit no. be x and y respectively. x = y × 40/100 +16 x - 2y/5 = 16....(i) x + y = 72...(ii) From equation (i) and (ii) x = 32, y = 40 Difference = 8 Answer: c)

8 1/4 ×5 2/3 – 4 1/6 × 4 1/5 = ? a) 31 b) 18 c) 14 d) 27 e) 23

Description : 8 1/4 ×5 2/3 – 4 1/6 × 4 1/5 = ? a) 31 b) 18 c) 14 d) 27 e) 23

Last Answer : ? = (33/4) × (27/5) – (25/6) × (21/5) =891/20 -35/2 ≈ 44.5 – 17.5 =27 Answer: d)

Ashwin borrowed a sum of Rs.6, 300 from Mishra at the rate of 14% for 3 years. He then added some more money to the borrowed sum and lent it to Ramesh at the rate of 16% of simple interest for the same time. If Ashwin gained Rs.618/- in the whole transaction, then what sum did he lend to Ramesh? a) Rs.6600 b) Rs.6800 c) Rs.4800 d) Rs.9800 e) Rs.8500

Description : Ashwin borrowed a sum of Rs.6, 300 from Mishra at the rate of 14% for 3 years. He then added some more money to the borrowed sum and lent it to Ramesh at the rate of 16% of simple interest for the same time. If ... sum did he lend to Ramesh? a) Rs.6600 b) Rs.6800 c) Rs.4800 d) Rs.9800 e) Rs.8500

Last Answer : Suppose Ashwin lend Rs. x to Ramesh then, Xx16x3/100 - (X – 6300) × 14 × 3/100 = 618......(1) From equation I , x = Rs. 6800 Answer: b)

The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number a) 135 b) 105 c) 155 d) 165 e) None of these

Description : The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number a) 135 b) 105 c) 155 d) 165 e) None of these

Last Answer : Let the two numbers be 3x and 4x. When they are increased by 9 they become 3x + 9 and 4x + 9. It is given that the ratio is 18:23 Thus, 3x + 9/4x + 9 = 18:23 23(3x + 9) = 18(4x + 9) 69x + 207 = 72x ... = -45 X = 15 Thus two numbers are 3x15 = 45 and 4 x 15 = 60 And the sum is 60+45 = 105 Answer: b)

Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021? a) 17 b) 19 c) 21 d) 23 e) 25

Description : Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021? a) 17 b) 19 c) 21 d) 23 e) 25

Last Answer : Let the correct ages of Ramesh and Suresh be X in 2015 (since they are twins, they will have the same age) Given: 2x + 9 = 43 Or x = 17 years in 2015. Age in 2021 = 17 + 6 = 23 Note: year of birth = 2015 – 17 = 1998. Reversing, Suresh = 1989 or 9 year older. Answer: d)

Sum of present age of A, B and C is 72 years. If 4 years ago, their ages were in the ratio 1 : 2 : 3, find A’s present age. a) 7 years b) 10 years c) 12 years d) 14 years e) None of these

Description : Sum of present age of A, B and C is 72 years. If 4 years ago, their ages were in the ratio 1 : 2 : 3, find A’s present age. a) 7 years b) 10 years c) 12 years d) 14 years e) None of these

Last Answer : Sum of present ages of A, B and C is 72 years. sum of their ages (4 years ago) = 72 – (3 × 4) = 60 years 4 years ago, ratio of ages of A, B and C was 1 : 2 :3. A’s age, 4 years ago = (1/6) ×60 = 10 years A’s present age = 10 + 4 = 14 years Answer: d)

2 3 5 8 14 23 41 69 a) 5 b) 8 c) 14 d) 41 e) 69

Description : 2 3 5 8 14 23 41 69 a) 5 b) 8 c) 14 d) 41 e) 69

Last Answer : The series is an alternate series, having S 1 = 2 5 14 41; ×3 – 1 in each term S 2 = 3 8 23 69; ×3 – 1 in each term Answer: e)

Is X divisible by 12? a. X leaves a remainder 2 when divided by 8 b. X is divisible by 3. c. X is divisible by 6. a) If the data in statement a is sufficient to answer the question, while the data in statement b and c are not required to answer the question. b) If the data in statement b is sufficient to answer the question, while the data in statement a and c are not required to answer the question c) If the data in statement a and b are sufficient to answer the question, while the data in statement c is not required to answer the question. d) If the data in statement a, b and c together are necessary to answer the question. e) If the data in statement a, b and c together are not sufficient to answer the question.

Description : Is X divisible by 12? a. X leaves a remainder 2 when divided by 8 b. X is divisible by 3. c. X is divisible by 6. a) If the data in statement a is sufficient to answer the question, while ... the question. e) If the data in statement a, b and c together are not sufficient to answer the question.

Last Answer : From (a), we notice that the number is of the form 8n+2, which means it also leaves a remainder 2 on being divided by 4. So, its not divisible by 12. From (b) and (c) together, the number can be any multiple of 6, from which we cant conclusively say if number is divisible by 12. Answer: a)

The sum of 5 natural numbers A, B, C, D, E is 126. It is known that A:B = 7:8, B:C = 2:3, C:D = 4:5, D:E = 5:7.What is the difference between A and E? a) 14 b) 22 c) 24 d) 28 e) 32

Description : The sum of 5 natural numbers A, B, C, D, E is 126. It is known that A:B = 7:8, B:C = 2:3, C:D = 4:5, D:E = 5:7.What is the difference between A and E? a) 14 b) 22 c) 24 d) 28 e) 32

Last Answer : A:B = 7:8 B:C = 2:3 = 8:12 SO, A:B:C = 7:8:12 C:D = 4:5 = 12:15 A:B:C:D = 7:8:12:15 D:E = 5:7 = 15:21 SO, A:B:C:D:E = 7:8:12:15:21 given x(7+8+12+15+21) = 126 or 63x = 126 or x =2 E – A = 21X – 7X = 14X = 28 Answer: d)

One year ago the ratio between Sam and Ash’s age was 4:3. One year hence the ratio of their ages will be 5:4. What is the sum of their present ages in years? A.14 B.16 C.18 D.12 E.None of these

Description : One year ago the ratio between Sam and Ash’s age was 4:3. One year hence the ratio of their ages will be 5:4. What is the sum of their present ages in years? A.14 B.16 C.18 D.12 E.None of these

Last Answer : Answer - B (16) Explanation - Let one year ago Sam's age be 4X years And, Ash's age be 3X years Present age of Sam = (4X + 1) years Present age of Ash= (3X + 1) years One year hence Sam's age = (4X + 2) ... = 2 Sum of their present ages = 4X + 1 + 3X + 1 = 7X + 2 = 7 x 2 + 2 = 16 years.

A is 2 years older than B who is twice as old as C. If sum of ages of A, B and C is 37 years, find the age of A. a) 7 years b) 9 years c) 14 years d) 16 years e) None of these

Description : A is 2 years older than B who is twice as old as C. If sum of ages of A, B and C is 37 years, find the age of A. a) 7 years b) 9 years c) 14 years d) 16 years e) None of these

Last Answer : Let the age of C be x years. Then age of B = 2x and age of A = 2x + 2 sum of ages of A, B and C = (2x + 2) + (2x) + x = 37 (5x + 2) = 37  x = 7 Age of A = 2x + 2 = 16 years Answer: d)

There are two sets of five numbers each – set P and set Q. P is a set of consecutive even numbers and Q is a set of consecutive numbers. The sum of the numbers of P is 230. The second least number in Q is 33 less than twice the least number in P. Find the sum of the numbers in Q. 1 : 260 2 : 250 3 : 240 4 : 270 5 : None of these

Description : There are two sets of five numbers each - set P and set Q. P is a set of consecutive even numbers and Q is a set of consecutive numbers. The sum of the numbers of P is 230. The second least number in Q is 33 less ... the sum of the numbers in Q. 1 : 260 2 : 250 3 : 240 4 : 270 5 : None of these

Last Answer : 1 : 260

Kathir and Anand can do a work in 20 days and 30 days respectively. Kathir started the work and left after 4 days. Anand took over and completed the work. In how many days was the total work completed? a) 28 days b) 20 days c) 23 days d) 25 days e) 26 days

Description : Kathir and Anand can do a work in 20 days and 30 days respectively. Kathir started the work and left after 4 days. Anand took over and completed the work. In how many days was the total work completed? a) 28 days b) 20 days c) 23 days d) 25 days e) 26 days 

Last Answer : a Kathir’s one day’s work= 1/20 Kathir’s 4 day’s work =4* (1/20) = 1/5 Work left= 1-1/5 = 4 /5 Anand’s one day’s work= 1/30 Anand can do work in = (4/5) *(30/ 1) = 24 days So total days = 24+4 = 28 days

Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Description : Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Last Answer : Five years ago, x – 5/y – 5 = 4/1 x – 4y = – 15 —-(i) After 10 years, x + 10/y + 10 = 2/1 x – 2y = 10 —–(ii) From eqn (i) and (ii) => x = 35, y = 12.5 Answer: d)

The average age of 30 students in a class is 12 years. The average age of a group of 5 of the students is 10 years and that of another group of 5 of them is 14 years. The average of the remaining students is : a) 8 years b) 10 years c) 12 years d) 14 years e) 15 years

Description : The average age of 30 students in a class is 12 years. The average age of a group of 5 of the students is 10 years and that of another group of 5 of them is 14 years. The average of the remaining students is : a) 8 years b) 10 years c) 12 years d) 14 years e) 15 years

Last Answer : Let it be x . Then : 5×10+5×14+20×x=30×12 20x=360-120 or 20x=240 or x = 12 Answer: c)

Sum of present ages of A and B is 60 years. 5 years hence, their ages will be in the ratio 3:4. Find A’s present age. a) 25 years b) 30 years c) 35 years d) 40 years e) None of these

Description : Sum of present ages of A and B is 60 years. 5 years hence, their ages will be in the ratio 3:4. Find A’s present age. a) 25 years b) 30 years c) 35 years d) 40 years e) None of these

Last Answer : Sum of their present ages = 60 years sum of their ages, 5 years hence = 60 + (5 × 2) = 70 years. after 5 years, ratio of ages of A and B will be 3 : 4. A’s age after 5 years = (3/7) × 70 = 30 years A’s present age = 30 – 5 = 25 years Answer: a)

The sum of ages of Mohan and his father is 35 years. When Mohan’s age will be equal to present age of his father, then sum of their ages will be 85 years. Find present age of the father. a) 20 years b) 25 years c) 30 years d) 32 years e) None of these

Description : The sum of ages of Mohan and his father is 35 years. When Mohan’s age will be equal to present age of his father, then sum of their ages will be 85 years. Find present age of the father. a) 20 years b) 25 years c) 30 years d) 32 years e) None of these

Last Answer : Sum of present ages = 35 years sum of their ages will be 85 years after (85-35) / 2 = 25 years from now. the father is 25 years older than his son. Present age of father = (35 + 25)/2 = 30 years. Answer: c)

Sum of Rs.118 was shared among 50 boys and girls,each girl receive Rs.2.60 and boy receive Rs.1.80.Find the number of girls. A) 15 B)25 C)30 D)35

Description : Sum of Rs.118 was shared among 50 boys and girls,each girl receive Rs.2.60 and boy receive Rs.1.80.Find the number of girls. A) 15 B)25 C)30 D)35

Last Answer : D)35

A mixture contains milk and water in the ratio of 3:2 litter of water is added to the mixture, milk, milk and water in the mixture become equal find the quantities of milk and water in the mixture . a) 12, 8 litter b) 4,3 litter c) 12, 6 litres d) 10,8 litres e) None of these

Description : A mixture contains milk and water in the ratio of 3:2 litter of water is added to the mixture, milk, milk and water in the mixture become equal find the quantities of milk and water in the mixture . a) 12, 8 litter b) 4,3 litter c) 12, 6 litres d) 10,8 litres e) None of these

Last Answer : Let quantities of milk and water in the mixture be 3x and 2x. Then if 4 litres of water is added to the mixture the ratio of milk and water become 1:1. It can be written as (3x): (2x + 4) = 1 ... Therefore, the milk in the mixture is 4x3 = 12 litres and quantity of water = 4x2 = 8 liters Answer: a)

Reshma purchased 120 chairs at price of Rs. 110 per chair. He sold 30 chairsat a profit of Rs. 12 per chair and 75 chairs at profit of Rs. 14 per chairs. The remaining chairs were sold at a loss of Rs. 7 per chairs. What is the average profit per table? A. Rs 10.56 B. Rs 10.87 C. Rs 12.123 D. Rs 12.67

Description : Reshma purchased 120 chairs at price of Rs. 110 per chair. He sold 30 chairsat a profit of Rs. 12 per chair and 75 chairs at profit of Rs. 14 per chairs. The remaining chairs were sold at a loss of Rs. 7 per chairs ... the average profit per table? A. Rs 10.56 B. Rs 10.87 C. Rs 12.123 D. Rs 12.67 

Last Answer : Answer – B (Rs 10.875) Explanation – Total C.P = Rs. (120 x 110) =Rs. 13200. Total S.P = Rs.[(30 x 110 + 30 x 12) + (75 x 110 + 75 x 14) + (15 x 110 – 15 x 7) =Rs..14505 Average profit = Rs. (14505 – 13200) /120 = Rs. 1305/120= 10.875

A sum of money is divided among Suresh, Ganesh, Vignesh and Mahesh in the ratio of 3 : 4 : 9 : 10 respectively. If the share of Vignesh is Rs.2,580/-more than the share of Ganesh, then what is the total amount of money of Suresh and Mahesh together ? a) Rs. 6,985/- b) Rs. 6,487/- c) Rs. 6,708/- d) Rs. 7,156/- e) Rs. 8,457/-

Description : A sum of money is divided among Suresh, Ganesh, Vignesh and Mahesh in the ratio of 3 : 4 : 9 : 10 respectively. If the share of Vignesh is Rs.2,580/-more than the share of Ganesh, then what is the total amount of money of Suresh ... ,985/- b) Rs. 6,487/- c) Rs. 6,708/- d) Rs. 7,156/- e) Rs. 8,457/-

Last Answer : According to the question, difference of ratio of Ganesh and Vignesh is 5 so, 1 ratio = 2580/5 = 516 Now, total ratio of Suresh + Mahesh = 13 So, total money = 13 × 516 = Rs. 6708 Answer: c)

400 gm spirit solution has 30 % spirit in it , what is the ratio of spirit should be added to make it 80 % in the solution ? A)2:5 B)4:3 C)5:2 D)2:7

Description : 400 gm spirit solution has 30 % spirit in it , what is the ratio of spirit should be added to make it 80 % in the solution ? A)2:5 B)4:3 C)5:2 D)2:7

Last Answer : A)2:5

In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Description : In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage ... mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Last Answer : 2) 28. 57%

A shopkeeper declares that he sells rice at the cost price. However he uses a weight of 425 grams. Instead of 500 grams, what is his percentage of profit? a) a) 16 5/6 b) 15 c) c) 20 7/23 d) 25 e) None of these

Description : A shopkeeper declares that he sells rice at the cost price. However he uses a weight of 425 grams. Instead of 500 grams, what is his percentage of profit? a) a) 16 5/6 b) 15 c) c) 20 7/23 d) 25 e) None of these

Last Answer : % profit = Loss/Actual Weight x 100 = 500 – 425/500 × 100 = 75/500 x 100 = 15% Answer: b)

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: a) 10 b) 14 c) 12 d) 13 e) 15

Description : Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: a) 10 b) 14 c) 12 d) 13 e) 15

Last Answer : (A+B+C) - (A+B) (A+B+C) =1/6 and (A+B+C) in 2hrs=2/6 and remaining part 1-2/6=2/3. So (A+B) in 7 hrs is 2/ (3*7) =2/21. 1/6-2/21=1/14 so answer is 14. Answer: b)

In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Description : In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Last Answer : A)11:77:7

13.5 × 14 + 22.5 × 18 – 4.7 × 30 = ? a) 405 b) 423 c) 435 d) 453 e) 481

Description : 13.5 × 14 + 22.5 × 18 – 4.7 × 30 = ? a) 405 b) 423 c) 435 d) 453 e) 481

Last Answer : ?=189+405-141 = 453 Answer: d)

What is the difference between money spent by both companies on tax ( in lakh Rs.)? a) 26.8 b) 24.4 c) 24.6 d) 23.8 e) None of these

Description : What is the difference between money spent by both companies on tax ( in lakh Rs.)? a) 26.8 b) 24.4 c) 24.6 d) 23.8 e) None of these

Last Answer : e) None of these

6 pens and 14 exercise books cost Rs.162. 5pens and 8 exercise books cost Rs.102. The ratio of cost of 1 pen to that of 1 exercise book is : a) 10 : 3 b) 2 : 3 c) 3 : 2 d) 5 : 6 e) None of these

Description : 6 pens and 14 exercise books cost Rs.162. 5pens and 8 exercise books cost Rs.102. The ratio of cost of 1 pen to that of 1 exercise book is : a) 10 : 3 b) 2 : 3 c) 3 : 2 d) 5 : 6 e) None of these

Last Answer : Let cost of 1 pen be Rs.x and that of 1 exercise book be Rs.y. Then 6x +14 y = 162 =>3x + 7y =81……..(1) 5x + 8y =102…………………………………………..(2) (1)× 5 – (2) × 3 =>11y=99=>y=9.Putting y=9 in equation (1),we get : x = 6 x : y =6 : 9 =2 : 3 Answer: b)

24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Description : 24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Last Answer : The series consists of two series 1 and 2: Series 1 : 24 26 28 30 +2 +2 +2 Series 2 : 14 15 16 +1 +1 Answer: c)

The average age of 30 boys of a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher? A.50 B.44 C.45 D.42 E.None of these

Description : The average age of 30 boys of a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher? A.50 B.44 C.45 D.42 E.None of these

Last Answer : Answer – C (45) Explanation – Total ages of 30 boys = 14 x 30 = 420 years Total age when class teacher is included = 15 x 31 = 465 years Age of class teacher = 465 – 420 = 45 years

In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Last Answer : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

Which of the following represents the largest 4-digit number which can be added to 7249 in order to make the derived number divisible by each of 12,14,21,33 and 54? a) 9123 b) 9383 c) 8727 d) 8673

Description : Which of the following represents the largest 4-digit number which can be added to 7249 in order to make the derived number divisible by each of 12,14,21,33 and 54? a) 9123 b) 9383 c) 8727 d) 8673

Last Answer : b) 9383

Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Description : Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Last Answer : Here, p(x) = x4 + 2x3 - 4x2 + 6x - 3, g(x) = x2 - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).

Find the least number of five digits which is exactly divisible by 12, 15 and 18. (a) 1080 (b) 10080 (c) 10025 (d) 11080

Description : Find the least number of five digits which is exactly divisible by 12, 15 and 18. (a) 1080 (b) 10080 (c) 10025 (d) 11080

Last Answer : 10080 Hint : The least number of 5 digits is 10000. L.C.M. of 12, 15 and 18 is 180 . On dividing 10000 is 100. => 10000 + 180- 100 = 10080 is divisible by 180.

In 10 years time, Ram will be 2 times older than his son. 5 years ago he was 9 times as old as his son. What will be the sum of the ages of Ram and his son 3 years from now? a) 46 b) 66 c) 86 d) 76 e) None of these

Description : In 10 years time, Ram will be 2 times older than his son. 5 years ago he was 9 times as old as his son. What will be the sum of the ages of Ram and his son 3 years from now? a) 46 b) 66 c) 86 d) 76 e) None of these

Last Answer : Let current age of Ram and his son be R,S (R + 10) = 3(S + 10) [2 times older = 3 times the age] R = 3S + 20 (R - 5) = 9(S - 5) 3S + 20 – 5= 9S – 45 3S + 15 = 9S – 45 or S = 10., R = 50 sum of ages 3 years from now = 10 + 3 + 50 + 3 = 66. Answer: b)

The ages of A and B are in the ratio 8 : 5. If the sum of their ages is 39 years, what will be the ratio of their ages after 9 years? a) 3 : 2 b) 8 : 7 c) 10: 7 d) 11 : 8 e) None of these

Description : The ages of A and B are in the ratio 8 : 5. If the sum of their ages is 39 years, what will be the ratio of their ages after 9 years? a) 3 : 2 b) 8 : 7 c) 10: 7 d) 11 : 8 e) None of these

Last Answer : Sum of ratios = 8 +5 = 13. Sum of ages = 39 years 13 ratio = 39 years  1 ratio = 3 years 9 years = 3 ratio new ratio = (8+3) : (5+3) = 11 : 8 Answer: d)