InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is: |
| Answer» N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4 | |
| 2. |
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: |
| Answer» 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). [Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ] So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4). Clearly, there are 2 such pairs. | |
| 3. |
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? |
| Answer» L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30 ---------------------------- = 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15 ---------------------------- Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5 = 630. | |
| 4. |
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is: |
| Answer» L.C.M. of 5, 6, 7, 8 = 840. Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 x 2 + 3) = 1683. | |
| 5. |
If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to: |
| Answer» Let the numbers be a and b. Then, a + b = 55 and ab = 5 x 120 = 600. The required sum = 1 + 1 = a + b = 55 = 11 a b ab 600 120 | |
| 6. |
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is: |
| Answer» Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Required number (9999 - 399) = 9600. | |
| 7. |
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: |
| Answer» 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. | |
| 8. |
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is: |
| Answer» Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm. | |
| 9. |
Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is: |
| Answer» Since the numbers are co-prime, they contain only 1 as the common factor. Also, the given two products have the middle number in common. So, middle number = H.C.F. of 551 and 1073 = 29; First number = 551 = 19; Third number = 1073 = 37. 29 29 Required sum = (19 + 29 + 37) = 85. | |
| 10. |
Find the highest common factor of 36 and 84. |
| Answer» 36 = 22 x 32 84 = 22 x 3 x 7 H.C.F. = 22 x 3 = 12. | |
| 11. |
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: |
| Answer» Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548. | |
| 12. |
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: |
| Answer» Other number = 11 x 7700 = 308. 275 | |
| 13. |
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is: |
| Answer» Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4. So, the numbers 12 and 16. L.C.M. of 12 and 16 = 48. | |
| 14. |
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: |
| Answer» Required number = (L.C.M. of 12,16, 18, 21, 28) + 7 = 1008 + 7 = 1015 | |
| 15. |
252 can be expressed as a product of primes as: |
| Answer» Clearly, 252 = 2 x 2 x 3 x 3 x 7. | |
| 16. |
Find the lowest common multiple of 24, 36 and 40. |
| Answer» 2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360. | |
| 17. |
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: |
| Answer» 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. | |
| 18. |
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. |
| Answer» Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4. | |
| 19. |
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 again at the starting point ? |
| Answer» L.C.M. of 252, 308 and 198 = 2772. So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec. | |
| 20. |
The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is: |
| Answer» Clearly, the numbers are (23 x 13) and (23 x 14). Larger number = (23 x 14) = 322. | |
| 21. |
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 ? |
| Answer» L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together 30 + 1 = 16 times. 2 | |
| 22. |
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: |
| Answer» 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. | |
| 23. |
Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: |
| Answer» Let the numbers be 3x, 4x and 5x. Then, their L.C.M. = 60x. So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40). Hence, required H.C.F. = 40. | |
| 24. |
The G.C.D. of 1.08, 0.36 and 0.9 is: |
| Answer» Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18, H.C.F. of given numbers = 0.18. | |
| 25. |
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is: |
| Answer» Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127. | |
| 26. |
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: |
| Answer» 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. | |