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If the sum of two numbers is 66 and the H.C.F and L.C.M of the numbers are 6 and 120 respectively, then the sum of the reciprocals of numbers is equal to∶
- A11/120
- B21/120
- C15/130
- D55/120
Answer not set
Discussion 1
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CCommunity
Let the numbers are x and y
Then, x + y = 66
We know that,
LCM × HCF = Product of two numbers
⇒ xy = 6 × 120 = 720
∴ required sum $(= \frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} = \frac{{\left( {{\rm{x\;}} + {\rm{\;y}}} \right)}}{{{\rm{xy}}}} = \frac{{66}}{{720}} = \frac{{11}}{{120}})$
Then, x + y = 66
We know that,
LCM × HCF = Product of two numbers
⇒ xy = 6 × 120 = 720
∴ required sum $(= \frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} = \frac{{\left( {{\rm{x\;}} + {\rm{\;y}}} \right)}}{{{\rm{xy}}}} = \frac{{66}}{{720}} = \frac{{11}}{{120}})$