Question

A man is 20 years older to his son, after 10 years his age becomes twice the age of his son. Find their present age.

Answer 1

Answer: Son = 10 years, Man(father)=30 years

Step-by-step explanation:

Let the age of son = S

Age of man = S + 20

After 10 years:

Age of son = S + 10

Age of man = S + 20 + 10 = S+ 30

Given:

S+ 30 = 2 *[S+ 10]

S + 30 = 2S + 20

S = 10

Age of man = S + 20 = 30

Present age of :

Son = 10 years && Man = 30 years

Answer 2

HEYA!

let age of the son be= y

Let age of the man be=x

10 years later,

Man's age= x+10

sons age= y+10

x = 20+y

x - y = 20---------1

x + 10 = (y+10) 2

x + 10 = 2 y + 20

x - 2y = 10---------2

Subtracting equation 2 from 1, we get

x-2y=10

x-y =20

SO we get...

SON'S AGE Y=10

AND

MAN'S age X=30

REFER TO THE ATTACHMENT FOR WORKED OUT ANSWER.....

HOPE IT HELPS YA!

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