Five years ago, A's age was four times the age of B. Five years hence, A's age will be twice the age of B. Find their present ages.

Let present age of A be x years and B be y years.

Five years's ago their age will be :

A = (x - 5) years

B = (y - 5) years

Given,

Five years ago, A's age was four times the age of B.

⇒ (x - 5) = 4(y - 5)

⇒ x - 5 = 4y - 20

⇒ x = 4y - 20 + 5

⇒ x = 4y - 15 ………(1)

Five years's later their age will be :

A = (x + 5) years

B = (y + 5) years

Given,

Five years hence, A's age will be twice the age of B.

⇒ (x + 5) = 2(y + 5)

⇒ x + 5 = 2y + 10

⇒ x = 2y + 10 - 5

⇒ x = 2y + 5 ………(2)

From (1) and (2), we get :

⇒ 4y - 15 = 2y + 5

⇒ 4y - 2y = 5 + 15

⇒ 2y = 20

⇒ y = $\dfrac{20}{2}$ = 10 years.

Substituting value of y in equation (2), we get :

⇒ x = 2(10) + 5 = 20 + 5 = 25 years.

Hence, present age of A = 25 years and B = 10 years.