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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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