#### Question

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

#### Solution

Let the two parts be x and x - 15.

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

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

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

`150 = 45x - 3x^2`

`3x^2 - 45x - 3x^2`

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

(x - 5)(x - 10) = 0

x = 5, 10

x = 5 => One part = 5 and other parts = 10

x = 10 => One parts = 10 and other part = 5

Thus the required two parts are 5 and 10.

Is there an error in this question or solution?

Solution Divide 15 into Two Parts Such that the Sum of Their Reciprocals is 3/10 Concept: Quadratic Equations.