Problems on Ages

Let ages be K and S. 3 years ago: (K-3) + (S-3) = 36 => K+S = 42. After 4 years, each person is 4 years older, so the sum increases by 8. 42 + 8 = 50.

A = B + 2. D = 2A = 2(B+2) = 2B + 4. B = 2C => C = B/2. D - C = 40 => (2B + 4) - B/2 = 40 => 1.5B = 36 => B = 24. A = 24 + 2 = 26.

Let M and S be current ages. (M+6) = 3(S+6) => M - 3S = 12. (M-3) = 9(S-3) => M - 9S = -24. Subtracting equations: 6S = 36 => S = 6. M = 3(6) + 12 = 30.

Let Ram's present age be R and son's be S. 20 years ago, R-20 = 5(S-20). The son will be 41 in 16 years, so S+16 = 41, meaning S = 25. Substituting S=25 into the first equation: R-20 = 5(25-20) = 25, so R = 45.

F=2E. E-Y=15 => Y=E-15. (F+10)=3(Y+10). 2E+10 = 3(E-15+10) = 3(E-5) = 3E-15. E=25. F=2E=50.

Let M and S be current ages. (M-6) = 3(S-6) => M-3S = -12. (M+6) = 2(S+6) => M-2S = 6. Subtracting the first from the second: S = 18. Then M = 42.

If the son's age is 21, the father's age is 2 * 21 + 10 = 42 + 10 = 52. The difference in their ages is 52 - 21 = 31.

Let son's age be x. Father's age = 2x + 10. Given 2x + 10 = 40, then 2x = 30, so x = 15.

Let ages be x and x+16. Six years ago, (x+16-6) = 3(x-6), which simplifies to x+10 = 3x-18. Solving gives 2x = 28, so x = 14. The ages are 14 and 30.

Let ages be 9x and 8x. (9x+5)/(8x+5) = 10/9. Cross-multiplying: 81x + 45 = 80x + 50, so x = 5. The difference is 9x - 8x = x = 5 years.

Father's age (F) is 10 more than twice the son's age (S). This translates to F = 2S + 10.

Let ages be 4x and 5x. (4x+10)/(5x+10) = 6/7. 28x + 70 = 30x + 60, so 2x = 10, x = 5. Elder brother present age = 25. After 15 years, age = 25 + 15 = 40.

Let n be the current age. 10 years ago, his age was n - 10. The problem states n - 10 = n/2. Multiplying by 2 gives 2n - 20 = n, so n = 20.

Let son's age be x, mother's be 3x. After 15 years: 3x + 15 = 2(x + 15). 3x + 15 = 2x + 30. x = 15. Mother's age = 3 * 15 = 45.