Question

Three years hence a father will be four times as old as his son will be. Before two years he was seven times as old as his son was. Find their present ages.

Answer 1

Hi there !!

___________________________

In this problem, we can assign X to the son’s age and Y to the father’s.

Therefore, we can create the equation 3X=Y.



Now we have our two equations:

Y=3X

Y=2X+12



3X = 2X+12

Now we subtract 2X from each side to isolate the variable:

X=12

The son’s age is therefore 12 years old. Should we plug that age back into our first equation, we get Y=3(12) or Y=36.

The father’s present age is 36, while the son’s present age is 12.

Thanks.

Answer 2

let the present age of son be x .

Three years hence,

Son's age = x+3

Father's age = 4(x+3)

Before 7 years, son's age = x-7

Father's age = 4(x+3) -10 { -10 to bring him 3 years ago ie to present and -7 to bring him 7 years back}

A/Q

4(x+3)-10= 7(x-7)

4x+12-10=7x-49

4x+2=7x-49

2+49=7x-4x

51 =3x

51/3=x

17 =x

So,

Son's age =17 years

Father's age = 4(x+3) -3 {-3 to bring him to present}

=4(17+3)-3

=4×20-3

=80-3

= 77 years.

$\bf{Hope\:it\:helps}$