Question

The sum of two natural numbers is 15 and the sum of their reciprocals is `3/10`. Find the numbers.

Answer 1

Let the two numbers be x and y.

According to the question,

x + y = 15

`\implies` y = 15 – x   ...(i)

And `1/x + 1/y = 3/10`

`\implies 1/x + 1/(15 - x) = 3/10`   ...(From (i))

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

`\implies` 15 × 10 = 3x(15 – x)

`\implies` 150 = 45x – 3x²

`\implies` 3x² – 45x + 150 = 0

`\implies` x² – 15x + 50 = 0

`\implies` x² – 10x – 5x + 50 = 0

`\implies` x(x – 10) – 5(x – 10) = 0

`\implies` x – 10 = 0 or x – 5 = 0

`\implies` x = 10 or x = 5

Hence, the numbers are 10, 5.

Answer 2

Let the required natural numbers be x and (15 – x) 

According to the given condition,  

`1/x + 1/(15 - x) = 3/10` 

⇒ `(15 - x + x)/(x(15 - x)) = 3/10` 

⇒ `15/(15x - x^2) = 3/10` 

⇒ 15x – x² + 50 = 0 

⇒ x² – 15x + 50 = 0  

⇒ x² – 10x – 5x + 50 = 0 

⇒ x(x – 10) – 5(x – 10) = 0 

⇒ (x – 5) (x – 10) = 0 

⇒ x – 5 = 0 or x – 10 = 0 

⇒ x = 5 or x = 10 

When x = 5,  

15 – x = 15 – 5

= 10

When x = 10,  

15 – x = 15 – 10

= 5  

Hence, the required natural numbers are 5 and 10.