Question

Luv is having 2 friends. Three years ago, the sum of the ages of two friends was half of Luv’s age. One of his friend is 2 years older than the other and sum of their present ages is 3 years less than the Luv’s present age. Find the present ages of all.
1. 14, 6, 4
2. 13, 9, 7
3. 15, 8, 6
4. 15, 7, 5

Answer 1

Correct Answer - Option 4 : 15, 7, 5

Given∶

Sum of the ages of friends = 3 years less than the Luv’s age

Three years ago, the sum of the ages of two friends = half of Luv’s age.

Formula Used∶

Concept of Linear Equations, Elimination Method

Calculation∶

Let Luv’s present age = x years

Sum of the ages of his two friends = y years

∴ y = x – 3

⇒ x – y = 3      ----(1)

3 years ago, Luv’s age = (x – 3) years

Sum of the ages of two friends = y - 3 - 3 = (y – 6) years

∴ (y – 6) = 1/2(x – 3)

⇒ 2y – 12 = x – 3

⇒ x – 2y = -9      ----(2)

Subtracting eq.(2) from (1), we get

y = 12

Putting y = 12 in eq. (1)

x – y = 3

⇒ x – 12 = 3

⇒ x = 3 + 12 = 15

Sum of friends age = 12 years

But they differ by 2 years, therefore,

Elder friend’s age = 7 years and Younger friend’s age = 5 years

∴ Luv’s present age is 15 years while age of his two friends is 7 years and 5 years.