Question

A man says to his son ,"seven years ago I was seven times as old as you were , and three years hence I shall be three times as old as you." Find their ages.​

Answer 1

ANSWER:

Current ages

Father's age= 42 years

Son's age=12 years.

Step by Step Explanation:

Let the son's present age be x and father's present age be y.

Seven years ago

Son's age=x-7.

Father's age=y-7

From the question it can be concluded that

7(x-7)=y-7

7x-49=y-7

7x-y=49-7

7x-y=42. ---------(i)

Three years ahead

Son's age=x+3.

Father's age=y+3

From the question it can be concluded as,

3(x+3)=y+3

3x+9=y+3.

y-3x=9-3

y-3x=6. ----------(ii)

Solution of equations,

From equation (ii)

y-3x=6

y=6+3x

Substituting the value of x in equation (i)

7x-y=42

7x-(6+3x)=42

7x-6-3x=42

4x=42+6

4x=48

x=12

Substituting the value of x in equation (ii)

y-3x=6

y-3×12=6

y-36=6

y=6+36

y=42.

Answer 2

Answer:

let the age of son five years ago be x. then age of father was 7x.

A/Q = 7x + 10 = 3(x+10)

$7x + 10 = 3(x + 10) \\ 7x-3x = 30 - 10 \\ 4x = 20 \\ x = \frac{20}{4} \\ x = 5$

= age of father seven years ago= 7x

=(7×5) years

=35 years

= age of son seven years ago = x

= 5 years

= present age of father = ( 35+7) years

= 42 years

= present age of son = ( 5+7) years

= 12 years