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Two Years Ago, a Father Was Five Times as Old as His Son. Two Year Later, His Age Will Be 8 More than Three Times the Age of the Son. Find the Present Ages of Father and Son. - Mathematics

Definition

Two years ago, a father was five times as old as his son. Two year later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

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Solution

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

After 2 years, father’s age will be (x+2) years and the age of son will be (y+2) years. Thus using the given information, we have

`x+2 =3(y+2)+8`

`⇒ x+2 =3y +6+8`

`⇒ x-3y-12=0`

Before 2 years, the age of father was (x-2) years and the age of son was (y-2) years. Thus using the given information, we have

`x-2=5(y-2)`

`⇒ x-2 =5y-10`

`⇒ x-5y+8=0`

So, we have two equations

`x-3y-12=0`

`x-5y+8=0`

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

By using cross-multiplication, we have

`x/((-3)xx8-(-5)xx-12)=(-y)/(1xx8-1xx(-12))=1/(1xx(-5)-1xx(-3))`

`⇒ x/(-24-60)=(-y)/(8+12)=1/(-5+3)`

`⇒ x/(-84)=(-y)/20=1/(-2)`

`⇒ x/84=y/20=1/2`

`⇒ x= 84/2,y=20/2`

`⇒ x=42,y=10`

Hence, the present age of father is 42 years and the present age of son is= 10 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 9 | Page 92
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