Question

13. A mother is 20 years older than her son. Ten years ago she was 3 times as old as her son
then. Find their present ages.
​

Answer 1

AnswEr :

Let the present age of the son be x

According to the Question,

Mother's age would be x + 20

Ten years ago,her age is 3 times her son's age

• Son's age before 10 years : x - 10

• Mother's age before 10 years : (x + 20) - 10

According to the Condition,

3(x - 10) = (x + 20) - 10

➠ 3x - 30 = x + 10

➠ 3x - x = 30 + 10

➠ 2x = 40

➠ x = 20

The present age of the son is 20 years

Since,

Mother's present age : x + 20 = 20 + 20 = 40 years

His mother's present age is 40 years

Answer 2

Answer :-

$\rule{205}{2}$

↣  Age of mother = 40 years

↣  Age of son = 20 years

$\rule{205}2$

Step - by - step explanation :-

$\rule{205}2$

Let,

$\rightarrowtail\sf{present\;age\;of\;mother=x\;\;years}$

$\rightarrowtail\sf{present\;age\;of\;son=y\;\;years}$

According to the first information given

● Mother is 20 years older than her son

$\mapsto \sf{x=y+20\:\:.....equation(1)}$

According to the second information given

● Ten years ago age of mother was 3 times as old as her son then

→ Ten years ago age of mother = x - 10

→ Ten years ago age of son = y - 10

Then,

$\mapsto\sf{x-10=3(y-10)}$

$\mapsto\sf{x-10=3y-30}$

using equation (1)

$\mapsto\sf{(y+20)-10=3y-30}$

$\mapsto\sf{y+20-10-3y+30=0}$

$\mapsto\sf{-2y+40=0}$

$\mapsto\sf{-2y=-40}$

$\boxed{\longmapsto\large{\red{\sf{y=20}}}}$

putting y = 20 in equation (1)

$\mapsto\sf{x=(20)+20=40}$

$\boxed{\longmapsto\large{\sf{\red{x=40}}}}$

Hence ,

Age of mother is 40 years and

age of son is 20 years .

$\rule{205}2$