Question

Solve question no.52

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.

I hope this helps u