Question

(9) A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his
son. What are their present ages?​

Answer 1

Answer:

Let the present age of the son be x and that to of his father which is 7 times =7x

Then, 2 years ago their ages :

(x-2) of son

(7x-2) of father

Given, 13(x−2)=7x−2

Then,

13x−26=7x−2

=13x−7x=26−2

6x=24

or x=4 and that of his father = 28 years

ageofson=x=4years

ageoffather=7x=28years

_hope helps you.._

Answer 2

GiveN:-

A father is 7 times as old as his son, two years ago the Father was 13 times as old as his son.

To FinD:-

The present age of son.

SolutioN:-

Let the present age of the son be x years.

Let the present age of the father be 7x years.

2 years ago their ages were:-

Son = (x - 2) years.

Father = (7x - 2) years.

It is said that two years ago the Father was 13 times as old as his son.

According to the question,

$\large\implies{\sf{7x-2=13(x-2)}}$

$\large\implies{\sf{7x-2=13x-26}}$

$\large\implies{\sf{-2+26=13x-7x}}$

$\large\implies{\sf{24=6x}}$

$\large\implies{\sf{\dfrac{24}{6}=x}}$

$\large\implies{\sf{\dfrac{\cancel{24}}{\cancel{6}}=x}}$

$\large\implies{\sf{4=x}}$

$\large\therefore\boxed{\bf{x=4.}}$

Their present ages:-

Son's age = x = 4 years.

Father's age = 7x = 7 × 4 = 28 years.

The present age of the son is 4 years.