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1. The product of two numbers is 4107. If the H.C.F of this numbers is 37, then the greater number is:

  • A. 101
  • B. 101
  • C. 101
  • D. 101

Answer: Option C

Explanation:

 Let the numbers be 37a and 37b

Then, 37a × 37b = 4107

ab = 3

Now, Co primes with product 3 are (1, 3)

So, the required numbers are (37 × 1, 37 × 3) = ( 37, 111)

Areater number = 111


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