Question

ten years ago, a father's age was 7 times the age of his son and two years later, twice the father age will be equal to 5 times the age of his son. find the present age

Answer 1

Hii!!

Here is ur answer :--

Let the present age of father be x years and son age be y years

ATQ,

Ten years ago, father's age was 7times the age of son.

(x-10) = 7 (y-10)

=> x - 10 = 7y - 70

=> x = 7y - 60 .................. (1)

Two years later, twice the father age will be equal to 5 times the age of his son.

2 (x+2) = 5 (y+2)

=> 2x + 4 = 5y +10

=> 2x = 5y +6 .................. (2)

Putting the value of x in (2) we get ,

=> 2 * (7y - 60) = 5y + 6

=> 14y -120 = 5y +6

=> 14y -5y = 6 + 120

=> 9y = 126

=> y = 126/9

=> y = 14

Putting the value of y in (1) we get,

x = 7 * 14 - 60

=> x = 98 - 60

=> x = 38

Therefore,

Present age of father is 38 years

Present age of son is 14 years

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Hope it helps!!