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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
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
- Correct Answer
- 111
Explanation
Let the numbers be 37 a and 37 bThen, 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.
- 1. 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
- 2. 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
- 3. 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?
Options- A. 4
- B. 10
- C. 15
- D. 16 Discuss
- 4. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 5. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Options- A. 74
- B. 94
- C. 184
- D. 364 Discuss
- 6. 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
- 7. Which of the following fraction is the largest?
Options- A. 7⁄8
- B. 13⁄16
- C. 31⁄40
- D. 63⁄80 Discuss
- 8. 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 Discuss
- 9. Which of the following has the most number of divisors?
Options- A. 99
- B. 101
- C. 176
- D. 182 Discuss
- 10. 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
Problems on H.C.F and L.C.M
Search Results
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: 16
Explanation:
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 |
Correct Answer: 2
Explanation:
Let the numbers 13 a and 13 b
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.
Correct Answer: 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.
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: 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: 548
Explanation:
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
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.
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