Exercise :: Problems on H.C.F and L.C.M - General Questions
- Problems on H.C.F and L.C.M - Important Formulas
- Problems on H.C.F and L.C.M - General Questions
| 1. | Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. |
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Answer: Option A Explanation: Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4. |
| 2. | 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: |
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Answer: Option C Explanation: Clearly, the numbers are (23 x 13) and (23 x 14).
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Larger number = (23 x 14) = 322.| 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 ? |
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Answer: Option D 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).
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| 4. | 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: |
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Answer: Option A 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 |
| 5. | The greatest number of four digits which is divisible by 15, 25, 40 and 75 is: |
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Answer: Option C Explanation: Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399.
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