Question

Divide 15 into two parts such that the sum of their reciprocals is `3/10`.

Answer 1

Let the two parts be x and 15 – x.

`1/x + 1/(15 - x) = 3/10`

`(15 - x + x)/(x(15 - x)) = 3/10`

`15/(15x - x^2) = 3/10`

`cancel(15)^5/(15x - x^2) xx 10/cancel(3)_1 = 0`

`5/(15x - x^2) xx 10 = 0`

`50 = 15x - x^2`

`x^2 - 15x + 50 = 0`

`x^2 - 10x - 5x + 50 = 0`

(x – 10)(x – 5) = 0

∴ x – 10 = 0

∴ x = 10

∴ x – 5 = 0

∴ x = 5

Thus the required two parts are 5 and 10.