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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
Reduce 128352/238368 to its lowest terms
Options- A. 3⁄4
- B. 5⁄13
- C. 7⁄13
- D. 9⁄13
- Correct Answer
- 7⁄13
Explanation128352) 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 - 1. 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
- 2. The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
Options- A. 1008
- B. 1015
- C. 1022
- D. 1032 Discuss
- 3. Find the lowest common multiple of 24, 36 and 40.
Options- A. 120
- B. 240
- C. 360
- D. 480 Discuss
- 4. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
Options- A. 12
- B. 16
- C. 24
- D. 48 Discuss
- 5. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
Options- A. 3
- B. 13
- C. 23
- D. 33 Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
Problems on H.C.F and L.C.M
Search Results
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: 1015
Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
Correct Answer: 360
Explanation:
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.
Correct Answer: 48
Explanation:
Let the numbers be 3 x and 4 x . Then, their H.C.F. = x . So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
Correct Answer: 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.
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: 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: 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: 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: 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.
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