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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

Derivation

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.

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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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APPEARS IN

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