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- Problems on H.C.F and L.C.M
1.
The L.C.M. of two numbers is 45 times their H.C.F. If one of the numbers is 125 and the sum of H.C.F. and L.C.M. is 1150, the other number is:
- A.215
- C.225
- B.220
- D.235
- Answer & Explanation
- Report
Answer : [C]
Explanation :
Explanation :
| Let H.C.F. be h and L.C.M. be I | ||||||||||
| Then, I =45h and I + h = 1150. | ||||||||||
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| So, I = (1150 - 25) = 1125. | ||||||||||
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2.
H.C.F. of 3240, 3600 and a third number is 36 and their L.C.M. is 24 x 35 x 52 x 72
The third number is:
The third number is:
- A.22 x 35 x 72
- C.25 x 52 x 72
- B.22 x 53 x 72
- D.23 x 35 x 72
- Answer & Explanation
- Report
Answer & Explanation : [A]
Explanation :
Explanation :
| 3240 = 23 x 34 x 5; 3600 = 24 x 32 x 52; H.C.F. = 36 = 22 x 32 | |||||
| Since H.C.F. is the product of lowest powers of common factors, | |||||
| so the third number must have (22 x 32) as its factors | |||||
| Since L.C.M. is the product of highest powers of common prime factors, | |||||
| so the third number must have 35 and 72 as its factors. | |||||
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3.
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| and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They | ||||||
| will again beep together at: |
- A.12 midnight
- C.6 a.m.
- B.3 a.m.
- D.9 a.m.
- Answer & Explanation
- Report
Answer : [D]
Explanation :
Explanation :
| Interval after which the devices will beep together = (L.C.M. of 30, 60, 90, 105) min. | |||
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| So, the devices will again beep together 21 hrs. after 12 noon i.e., at 9 a.m. | |||
4.
The H.C.F of 204, 1190 and 1445 is:
- A.17
- C.19
- B.18
- D.21
- Answer & Explanation
- Report
Answer : [A]
Explanation :
Explanation :
| 204 = 22 x 3 x 17; 1190 = 2 x 5 x 7 x 17; 1445 = 5 x 172 | |||
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5.
| The H.C.F. of | 2 | , | 8 | , | 64 | and | 10 | is: | ||
| 3 | 9 | 81 | 27 |
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A.
2 3 -
C.
160 3
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B.
2 81 -
D.
160 81
- Answer & Explanation
- Report
Answer : [B]
Explanation :
Explanation :
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