Question

The age of a man is 4 times the sum of the ages of his two sons. Ten years hence, his age will
be double of the sum of the ages of his sons. The father's present age is
(A) 50 years
(B) 55 years
(C) 60 years
(D) 65 years​

Answer 1

Given :

• The age of a man is 4 times the sum of the ages of his two sons.
• Ten years hence, his age will
• be double of the sum of the ages of his sons.

To Find :

• The present age of Father.

Solution :

Let the present age of Father be x.

Let the sum of ages of two sons be y.

Case 1 :

Age of man x is 4 into sum of ages of the two sons, y.

Equation :

$\sf{x=4y\:\:\:(1)}$

Case 2 :

Age of Father ten years hence will be (x+10) years.

And the sum of age of two sons ten years hence will be (y+20) years.

Equation :

$\longrightarrow$ $\sf{x+10=2(y+20)}$

$\longrightarrow$ $\sf{x+10=2y+40}$

$\longrightarrow$ $\sf{x-2y=40-10}$

$\longrightarrow$ $\sf{x-2y=30}$

$\longrightarrow$ $\sf{4y-2y=30}$

$\longrightarrow$ $\sf{2y=30}$

$\longrightarrow$ $\sf{y=\dfrac{30}{2}}$

$\longrightarrow$ $\sf{y=15}$

Substitute, y = 15 in equation (1),

$\longrightarrow$ $\sf{x=4y}$

$\longrightarrow$ $\sf{x=4\:\times\:15}$

$\longrightarrow$ $\sf{x=60}$

$\large{\boxed{\sf{\purple{Present\:age\:of\:father\:=\:60\:years}}}}$