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The HCF of two numbers is 16 and their product is 1024. What is sum of their reciprocals?
1. \(\frac{7}{{1024}}\)
2. \(\frac{5}{{1024}}\)
3. \(\frac{7}{{64}}\)
4. \(\frac{5}{{64}}\)

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Correct Answer - Option 4 : \(\frac{5}{{64}}\)

Given:

HCF = 16

Product of numbers = 1024

Concept used:

The reciprocal of the number is 1 divided by the number

Ex:- The reciprocal of n = 1/n

Calculation:

Let the number be 16a and 16b

16a × 16b = 1024

ab = 4

Co-prime factors of 4

(1, 4)

The number will be 16 and 64

Sum of the number = 16 + 64 = 80

Sum of their reciprocal

⇒ (1/16) + (1/64)

⇒ (80/1024) = 5/64

∴ The sum of the reciprocals of numbers is 5/64.

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