Question

A father is 35years more than his Son's age. After 5 years the father's age will be twice of his sons age. Find the present age of both.​

Answer 1

Answer:

74

Step-by-step explanation:

Let the present age of father be f and present age of son be s

Given,

f+5=3(s+5)

=>f+5=3s+15

=>f−3s=10 --- (1)

Also

f−2=4(s−2)

=>f−2=4s−8

=>f−4s=−6 --- (2)

Subtracting eqn (2) from (1), we get

s=16

From eqn (1),f−3(16)=10=>f=58

And their sum of present ages is 58+16=74 years

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Answer 2

Answer:

Let son's age be x

Father,s age = x + 35

Given, 2(x + 5) = (x + 35) + 5

2x + 10 = x + 40

2x - x = 40 - 10

x = 30

Father age=x + 35 = 65yrs.

Son age=x+5=35years.