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-x^2+50=0` 

⇒`x^2-15x+50=0`  

⇒`x^2-10x-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.