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
6. | 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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7. | Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: |
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Answer: Option A Explanation: Let the numbers be 3x, 4x and 5x. Then, their L.C.M. = 60x. So, 60x = 2400 or x = 40.
Hence, required H.C.F. = 40. |
8. | The G.C.D. of 1.08, 0.36 and 0.9 is: |
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Answer: Option C Explanation: Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
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9. | The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: |
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Answer: Option B Explanation: Let the numbers 13a and 13b. Then, 13a x 13b = 2028
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. |
10. | The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: |
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Answer: Option D 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.
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