Question

A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his
son. How old are new now? ​

Answer 1

${\underline{\underline{\huge{\mathfrak{\green{Answer:-}}}}}}$

Hey Mate!!

Let the age of his son be x years.

⇒ Age of the Father = 7x years.

Ages 2 years ago:

Son = x - 2

Father = 7x -2

According to the question,

7x - 2 = 13(x - 2)

⇒ 7x - 2 = 13x - 26

⇒ 13x - 7x = 26 - 2

⇒ 6x = 24

⇒ x = 24 ÷ 6

→ x = 4 years

${\boxed{\boxed{\huge{\sf{\red{Age\: of\: Son=x=4\: years ,\\\\Age\: of\: Father=7x=28\: years}}}}}}$

Answer 2

Let the age of son = x

and age of father = y

so, according to first condition,

y = 7x

and according to 2nd condition,

y - 2 = 13(x-2)

so,

7x - y = 0 ----(eq 1)

13x - y = 24 ----(eq 2)

equating equation 1 and 2

we get,

x = 4

and y = 7x

so, y = 28

therefore, present age of son is 4 years old and present age of father is 28 years old.

Hope it helps you!!!