Question

In 10yrs time a father will be twice as old as his son .10yrs àgo he was six times as old find their present age​

Answer 1

ANSWER:-

Let the present age of father and his daughter are x and y respectively.

Given that ten years ago a father was six times as old as his daughter.

(x−10)=6(y−10)

⇒x−10=6y−60

⇒x−6y+50=0⟶(1)

Also given that after 10 years, he will be twice as old as his daughter

(x+10)=2(y+10)

⇒x+10=2y+20

⇒x−2y−10=0⟶(2)

Subtracting equation (1) from (2), we have

x−2y−10−(x−6y+50)=0

⇒x−2y−10−x+6y−50=0

⇒4y−60=0

⇒y=15

Substituing the value of y in equation (2), we have

x−2×15−10=0

⇒x−30−10=0

⇒x=40

Hence, the present ages of father and his daughter is 40 and 15 respectively.

∴ sum of their present ages =x+y=40+15=55

Hence, the correct answer is 55.