Question

A father is 7 times as old as his son , two years ago the Father was 13 times as old as his son .then the present age of son in ............ years​

Answer 1

Solution:-

Let the Present age of his son be x years.

therefore , the present age of father is 7x years.

According to the Question

Father is 7 times as old as his son .

→ x = 7x

Now, again

Age After 2 years , Father was 13 times as old as his son .

→ 13(x-2) = 7x-2

→ 13x - 26 = 7x - 2

→ 13x -7x = -2+26

→ 6x = 24

→ x = 24/6

→ x = 4

Therefore,the present age of his son is 4 years.

Hence, the required answer is 4 years.

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:-

1. Son's age = x = 4 years.
2. Father's age = 7x = 7 × 4 = 28 years.

The present age of the son is 4 years.