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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
Options- A. 504
- B. 536
- C. 544
- D. 548
- Correct Answer
- 548
Explanation
Required number = (L.C.M. of 12, 15, 20, 54) + 8= 540 + 8
= 548.
- 1. Which of the following fraction is the largest?
Options- A. 7⁄8
- B. 13⁄16
- C. 31⁄40
- D. 63⁄80 Discuss
- 2. 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:
Options- A. 75
- B. 81
- C. 85
- D. 89 Discuss
- 3. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
Options- A. 101
- B. 107
- C. 111
- D. 185 Discuss
- 4. 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:
Options- A. 55⁄601
- B. 601⁄55
- C. 11⁄120
- D. 120⁄11 Discuss
- 5. 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:
Options- A. 4
- B. 5
- C. 6
- D. 8 Discuss
- 6. Which of the following has the most number of divisors?
Options- A. 99
- B. 101
- C. 176
- D. 182 Discuss
- 7. 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:
Options- A. 276
- B. 299
- C. 322
- D. 345 Discuss
- 8. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
Options- A. 123
- B. 127
- C. 235
- D. 305 Discuss
- 9. 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:
Options- A. 279
- B. 283
- C. 308
- D. 318 Discuss
- 10.
(469 + 174)2 - (469 - 174)2 =? (469 x 174)
Options- A. 2
- B. 4
- C. 295
- D. 643 Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: 7⁄8
Explanation:
L.C.M. of 8, 16, 40 and 80 = 80.
| 7 | = | 70 | ; | 13 | = | 65 | ; | 31 | = | 62 |
| 8 | 80 | 16 | 80 | 40 | 80 |
| Since, | 70 | > | 65 | > | 63 | > | 62 | , so | 7 | > | 13 | > | 63 | > | 31 |
| 80 | 80 | 80 | 80 | 8 | 16 | 80 | 40 |
| So, | 7 | is the largest. |
| 8 |
Correct Answer: 85
Explanation:
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.
Correct Answer: 111
Explanation:
Let the numbers be 37 a and 37 b
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.
Correct Answer: 11⁄120
Explanation:
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 |
Correct Answer: 4
Explanation:
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
Correct Answer: 176
Explanation:
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
Correct Answer: 322
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
∴ Larger number = (23 x 14) = 322.
Correct Answer: 127
Explanation:
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
Correct Answer: 308
Explanation:
| Other number = | ❨ | 11 x 7700 | ❩ | = 308. |
| 275 |
Correct Answer: 4
Explanation:
| Given exp. = | (a + b)2 - (a - b)2 |
| ab |
| = | 4ab |
| ab |
= 4 (where a = 469, b = 174.)
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