Problems on Ages Quiz Set 014

Let the present ages of the three friends be a - d, a and a + d. As of today a = 37. Ten years later their ages would be a + 10 - d, a + 10, a + 10 + d. Adding these we get 3a + 30 which equals 3 × 37 + 30 = 141 years.

The ages of the calves are in an AP with d = 2, n = 4, and sum S = 52. We have to find the first term a. We know S = ${n/2} × (2a + (n-1)d)$ Putting the values 52 = ${4/2} × (2a + (4-1)×2)$. Solving, get a = 10 years.

Clearly, the age of the mother is twice her daughter's present age. So the daughter's age today is 58/2 = 29 years. And, 16 years back the age of the daughter was 29 - 16 = 13 years.

Let the ages of three friends be 3r, 19r and 7r. The youngest of these is 3r. We have been given their sum. So (3 + 19 + 7)r = 174. Solving, we get r = 6. The youngest is 3 × 6 = 18 years.

Let the present age be x. Then $√(x - 8) + √(x + 8)$ = $√{2x}$. Squaring both sides, $(√(x - 8))^2 + (√(x + 8))^2 + 2√{x^2 - 8^2}$ = $2x$ which gives $x - 8 + x + 8 + 2√{x^2 - 8^2}$ = $2x$. Cancelling, and simplifying, $2√(x^2 - 64) = 0$, which leads to x = 8 years.