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A Father is Three Times as Old as His Son. After Twelve Years, His Age Will Be Twice as that of His Son Then. Find the Their Present Ages. - Mathematics

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Question

A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find the their present ages.

Derivation

Solution

Let the present age of father be x years and the present age of son be years.

Father is three times as old as his son. Thus, we have

`x=3y`

`⇒ x-3y=0`

After 12 years, father’s age will be `(x+12)` years and son’s age will be `(y +12)` years. Thus using the given information, we have

`x+12=2(y+12)`

`⇒ x+12=2y+24`

`⇒ x-2y -12=0`

So, we have two equations

`x-3y=0`

`x-2y-12=0`

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

`x/((-3)xx(-12)-(-2)xx0)=(-y)/(1xx(-12)-1xx0)=(1)/(1xx(-2)-1xx(-3))`

`⇒ x/(36-0)=(-y)/(-12-0)=1/1`

`⇒ x/36=(-y)/(-12)=1/1`

`⇒ x/36=y/12=1`

`⇒ x= 36 , y = 12`

Hence, the present age of father is 36 years and the present age of son is 12 years.

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Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.9 [Page 92]

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RD Sharma Mathematics [English] Class 10
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.9 | Q 1 | Page 92
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