Question

A father is 7 times as old as his son two year ago the father was 13 times as old as his they how.

Answer 1

SOLUTION:-

Given:

A father is 7 times as old as his son two years ago the father was 13 times as old as his son.

To find:

What are their present ages.

Solution:

Let Present ages of father = x

Let present ages of son = y

=) x= 7y........(1)

Therefore,

Two years ago,

⚫Father age = x-2

⚫Son age = y -2

According to the question:

=) (x-2) = 13(y-2)

=) x-2 = 13y -26

=) 7y -2 = 13y -26 [from(1)]

=) 7y -13y= -26 +2

=) -6y = -24

=) y = -24/-6

=) y = 4 years

Now,

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

=) x = 7y

=) x= 7×4

=) x= 28 years

Thus,

The father age is, x= 28 years

Father's son age is, y= 4 years

Hope it helps ☺️