Question

Ten years ago, the sum of the ages of two sons was one third of their father's age.One son is two years older than the other and sum of their present ages is 14 years less than the father's present age. Find the present ages of all.

Answer 1

Let the sons be A and B and father be C. Let their current ages by a, b and c respectively.

Ten years ago, the sum of their ages was 1/3 their father's age. This will give

(a-10) + (b-10) = 1/3 (c-10)

3a - 30 + 3b - 30 = c - 10

3a + 3b - c = 50 --- (1)

Let A be the elder son. Then, a = b+2

a-b = 2 --- (2)

Since the sum of their current ages is 14 less than the current age of their father, this gives

a + b = c - 14

-a - b + c = 14 --- (3)

Let us write the equations again here.

3a + 3b - c = 50 --- (1)
1a - 1b = 02 --- (2)
- a - b + c = 14 --- (3)

Adding (1) and (3) and multiplying (2) by 2, we get

2a + 2b = 64 --- (4)
2a - 2b = 04 --- (5)

Adding (4) and (5), we get

4a = 68, or a = 17.

Since a = b+2, we get b = a - 2 = 17 - 2 = 15.

and c = 17 + 15 + 14 = 46.

Thus, the current ages of sons and father are 17, 15 and 46 respectively.