Question

a man said to his son 6 years ago I was 6 times older than you and after 3 years from now my age will be only 3 times of yours find out the age of his son

Answer 1

Answer:

Step-by-step explanation:

Let the present age of son be x

6 years before age of son = x-6

6 years before age of father = 6(x-6)

3 years later age of father = 6(x-6) +3= 6x-36+3= 6x-33

3 years later age of son = 3(x+3)

A/Q

3(x+3) =6x-33

3x + 9 = 6x-33

9+33 = 6x-3x

42= 3x

3x = 42

x = 42/3

x= 14

Hence, age of son = 14

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Answer 2

Given :

• 6 years ago a man was 6 times older than his son.
• After 3 years the age of man will be 3 times the age of his son.

To Find :

• The present age of son

Solution :

Let the present age of man be x years.

Let the present age of son be y years.

Case 1 :

6 years ago, man's age was 6 times older than his son.

Age of man 6 years ago, (x-6) years.

Age of son 6 years ago, (y-6) years.

Equation :

$\longrightarrow$ $\sf{x-6=6(y-6) }$

$\longrightarrow$ $\sf{x-6=6y-36}$

$\longrightarrow$ $\sf{x-6y=-36+6}$

$\longrightarrow$ $\sf{x-6y=-30}$

$\sf{x\:=\:-30+6y\:\:\:(1)}$

Case 2 :

After 3 years, age of man will be 3 times of son's age.

Age of man after 3 years, (x+3) years.

Age of son after 3 years, (y+3) years.

Equation :

$\longrightarrow$ $\sf{x+3=3(y+3)}$

$\longrightarrow$ $\sf{x+3=3y+9}$

$\longrightarrow$ $\sf{x-3y=9-3}$

$\longrightarrow$ $\sf{x-3y=6}$

From equation (1), substitute value of x,

$\longrightarrow$ $\sf{-30+6y-3y=6}$

$\longrightarrow$ $\sf{6y-3y=6+36}$

$\longrightarrow$ $\sf{3y=42}$

$\longrightarrow$ $\sf{y=\cancel\dfrac{42}{3}}$

$\longrightarrow$ $\sf{y=14}$

$\large{\boxed{\sf{\purple{Present\:age\:of\:son\:=\:14\:years}}}}$