Question

ten years ago , the sum of the ages of two sons was one third of their fathers age. one son is two years older than the other and sum of their present ages is 14 years less than the fathers present age. find the present ages if 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.

Answer 2

Let the present age of

sons= x and (x+2)

father = y

10 years age,

age of father = y-10

son= (x-10) and (x-8)

ATQ,

(x-10)+(x-8) = 1/3(y-10)

→x-10+x-8 = 1/3y-1/3×10

→2x-18-1/3y+19/3 = 0

→2x-y/3 = -10/3+18 = (-10+54)/3

→(6x-y)/3 = 44/3

→6x-y = 44...........(A)

ATQ,

x+x+2 = y-14

→2x-y = -16.........(B)

Adding eqn. (A) and (B)

6x-y = 44

2x-y = 16

___________

4x = 60

→ x = 15 and x+2 = 17

(age of two sons)

Substituting the value of x in eqn. (A)

6x-y = 44

6×15-y = 44

y = 46 (age of father)

Hope this helps