Question

two years ago A man was five times as old as his son 2 years later his age will be 8 more than three times the age of his son find their present ages​

Answer 1

Let the age of the man be x yrs.

and the age of his son be y yrs.

Before 2 years

x-2 = 5(y-2)

=>x-5y = -8..... (i)

After 2 years

x+2 = 8+3(y+2)

=>x-3y = 12.....(ii)

Subtracting i and ii, we get

(x-5y) - (x-3y) = -8-12

=> y = 10

Putting the value of y in eq. i , we get

x = 42

So the age of man = 42yrs

and the age of son = 10yrs.

Answer 2

Let the age of father is x years and that of son is y years.

Then by the given question,

x-2=5(y-2)

or, x-5y=-10+2

or, x-5y=-8

or, x=5y-8

x+2=3(y+2)+8

or, x-3y=6+8-2

or, 5y-8-3y=12

or, 2y=12+8

or, y=20/2

or, y=10

then x=5y-8=50-8=42

Then, the age of father is 42 yrs. and the age of son is 10 yrs.

Alternately,

Let t f = father's present age

let s = son's age

:

Write an equation for each statement, simplify as much as possible.

:

" two years ago , a father was five times old as his son ."

f - 2 = 5(s-2)

f - 2 = 5s - 10

f = 5s - 10 + 2

f = 5s - 8

"two years later from today his age will be 8 years more than three times the age of his son "

f + 2 = 3(s+2) + 8

f + 2 = 3s + 6 + 8

f = 3s + 14 - 2

f = 3s + 12

replace f with (5s-8), from the 1st statement

5s - 8 = 3s + 12

5s - 3s = 12 + 8

2s = 20

s = 10 yrs is the son's age

then

f = 5(10) - 8

f = 42 yrs is father's age

:

So,finally the present ages of son and father are 10 and 42 respectively....