# The Present Age of a Father is Three Years More than Three Times the Age of the Son. Three Years Hence Father'S Age Will Be 10 Years More than Twice the Age of the Son. Determine Their Present Ages. - Mathematics

Definition

The present age of a father is three years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.

#### Solution

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

The present age of father is three years more than three times the age of the son. Thus, we have

x = 3y + 3

 ⇒ x- 3y -3 =0

After 3 years, father’s age will be (x+3) years and son’s age will be (y + 3) years.

Thus using the given information, we have

 x + 3 = 2(y +3)+10

 ⇒ x +3 =2y +6+10

⇒ x - 2y -13 =0

So, we have two equations

x - 3y -3 =0

x- 2y -13 =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(-13)-(-2)xx(-3))=(-y)/(1xx(-13)-1xx(-3))=1/(1xx(-2)-1xx(-3))

⇒ x/(39-6)=(-y)/(-13+3)=1/(-2+3)

⇒x/33=(-y)/(-10)=1/1

⇒ x/33=y/10 =1

⇒ x= 33, y = 10`

Hence, the present age of father is 33 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 6 | Page 92