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If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

  • A.\(\dfrac{55}{601}\)
  • B.\(\dfrac{601}{55}\)
  • C.\(\dfrac{11}{120}\)
  • D.\(\dfrac{120}{11}\)

Answer: C

Let the numbers be \(a\) and \(b\).
Then,
\(a + b = 55\) and
\(ab = 5 \times 120 = 600\)
The required sum
\(=\dfrac{1}{a} + \dfrac{1}{b}\\~\\= \dfrac{a+b}{ab}\\~\\=\dfrac{55}{600}\\~\\= \dfrac{11}{120}\)

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