Question

A father's present age is seven years less than
30 times of what his son's age was 20 years ago.
Also, the father's present age is 31 years more
than his son's present age. Find the sum of their
present ages, in year
(a) 72 (b) 73,
(c) 74. (d) 75​

Answer 1

Let the son's present age be x.

Let the father's present age be y.

We know that father's present age is seven years less than  30 times of what his son's age was 20 years ago.

So y=30(x-20)-7

We also know that father's present age is 31 years more  than his son's present age.

So y=x+31

Combining both equations:

30(x-20)-7 = x+31

30x-600-7=x+31

30x+(-600-7) = x+31

30x-607 = x+31

30x-x=31+607

29x=638

x=638/29=22

So the son's age is 22.

So the father's age is x+31=22+31=53

Sum=53+22=75 years

So the sum of their present ages is (d)75