Problems on Ages

One fact drives every age sum: in n years, every person gains exactly n years. The difference between two ages never changes; only the ratio does.

The four patterns:

1. "A is k times B, and their total is T." Let B = x, A = kx, so x(k + 1) = T.
2. Ages in a ratio, with the difference given. The difference of the ratio terms equals the given difference in years. If ages are 4 : 7 and B is 12 years older, then 7 − 4 = 3 parts = 12 years, so one part = 4.
3. Ratio now, another ratio after n years. Take the ages as ak and bk, add n to both, set the new ratio equal, solve for k.
4. "k times now, m times after n years." Let the son be x: kx + n = m(x + n).

The single most common mistake is in pattern 2 — dividing the age difference by the sum of the ratio terms instead of their difference. A difference in years can only be matched against a difference in parts. If the question gives a total age, divide by the sum; if it gives a difference, divide by the difference.

Always check backwards: put your answer back and confirm the future ratio really comes out as stated.

Age difference is constant · ratio a:b with difference d → one part = d/(b−a) · ratio with total T → one part = T/(a+b) · after n years add n to both

The ages of A and B are in the ratio 4 : 7. B is 12 years older than A. Find A's age. → The difference of the ratio terms is 7 − 4 = 3 parts, and that equals 12 years, so one part = 4 years. A = 4 × 4 = 16 years, B = 7 × 4 = 28. Sanity check: 28 − 16 = 12 as given, and 16 : 28 does reduce to 4 : 7.