English

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

Advertisements
Advertisements

Question

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.

Definition

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.

shaalaa.com
  Is there an error in this question or solution?
Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.9 [Page 92]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.9 | Q 6 | Page 92
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×