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Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 710.

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Question

Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is `7/10`.

Sum
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Solution

Let the numbers be x and x + 3.

From the given information,

`1/x + 1/(x + 3) = 7/10`

`(x + 3  + x)/(x(x + 3)) = 7/10`

`(2x + 3)/(x^2 + 3x) = 7/10`

20x + 30 = 7x2 + 21x

7x2 + x – 30 = 0

7x2 – 14x + 15x – 30 = 0

7x(x – 2) + 15(x – 2) = 0

(x – 2)(7x + 15) = 0

`x = 2, (-15)/7`

Since, x is a natural number, so x = 2.

Thus, the numbers are 2 and 5.

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Chapter 6: Solving (simple) Problems (Based on Quadratic Equations) - Exercise 6 (A) [Page 70]

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Selina Concise Mathematics [English] Class 10 ICSE
Chapter 6 Solving (simple) Problems (Based on Quadratic Equations)
Exercise 6 (A) | Q 5. | Page 70
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