Question

15. A father is 10 times as old as his son. After 6 years the father will
be 4 times the age of his son. Find the present ages of father and son.​

Answer 1

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 eq
n
(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 eq
n
(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.

Answer 2

Answer:

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