Problems on H.C.F and L.C.M - General Questions (4)
| 25. | The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: |
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Answer: Option C Explanation: Let the numbers be 37a and 37b. Then, 37a x 37b = 4107
Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
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ab = 3.
Greater number = 111.| 26. | 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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| 27. | 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 |
| 28. | 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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| 29. | 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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| 30. | 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. |