Discussion
Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
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
- 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.
- 1. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?
Options- A. 196
- B. 630
- C. 1260
- D. 2520 Discuss
- 2. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
Options- A. 40
- B. 80
- C. 120
- D. 200 Discuss
- 3. 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:
Options- A. 1677
- B. 1683
- C. 2523
- D. 3363 Discuss
- 4. Reduce 128352/238368 to its lowest terms
Options- A. 3⁄4
- B. 5⁄13
- C. 7⁄13
- D. 9⁄13 Discuss
- 5. 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?
Options- A. 26 minutes and 18 seconds
- B. 42 minutes and 36 seconds
- C. 45 minutes
- D. 46 minutes and 12 seconds Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
Problems on H.C.F and L.C.M
Search Results
Correct Answer: 630
Explanation:
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.
Correct Answer: 40
Explanation:
Let the numbers be 3 x , 4 x and 5 x
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.
Correct Answer: 1683
Explanation:
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.
Correct Answer: 7⁄13
Explanation:
128352) 238368 ( 1
128352
---------------
110016 ) 128352 ( 1
110016
------------------
18336 ) 110016 ( 6
110016
-------
x
-------
So, H.C.F. of 128352 and 238368 = 18336.
128352 128352 ? 18336 7
Therefore, ------ = -------------- = --
238368 238368 ? 18336 13
Correct Answer: 46 minutes and 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 sec. i.e., 46 min. 12 sec.
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: 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: 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: 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: 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.
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.