One year ago, the age of B was twice that of A. After 9 years, if the ratio of the ages of B and A would be 4 ∶ 3, then the sum of their present ages is _____ years.

Correct — A.

Set the unknown one year ago: A = x, B = 2x.
Present: A = x + 1, B = 2x + 1.
Nine years from now: A = x + 10, B = 2x + 10.

Ratio condition: (2x + 10) ⁄ (x + 10) = 4 ⁄ 3
3(2x + 10) = 4(x + 10)
6x + 30 = 4x + 40 → 2x = 10 → x = 5

Present ages: A = 6, B = 11.
Sum = 17 → option (a).

Check: nine years on, 15 and 20, which is 4 : 3.

Why the others are wrong

• (b)
  15 — 15 is the sum one year ago (5 + 10) — the figures the first sentence hands you. The question asks for present ages, 6 and 11.
• (c)
  21 — A sum of two ages moves by 2 a year, so 21 is their combined age two years from now. Today it stands at 17.
• (d)
  30 — The sum stays odd for ever. It starts at 17 and changes by 2 each year, so an even total like 30 cannot be these two ages at any date.

An age problem is one linear equation in costume. Fix the single unknown at the earliest date mentioned — here one year ago — and reach every other date by adding the same number of years to both people.

One year ago A = x and B = 2x. Nine years from now is ten years after that date, giving x + 10 and 2x + 10.

The ratio 4 : 3 becomes 3(2x + 10) = 4(x + 10) by cross-multiplication, one equation with one unknown. Solve it, then step back to the date the question actually asks about.

Key facts

• The gap between two people's ages never changes, and here it is fixed at 5 years.
• A ratio condition becomes an equation by cross-multiplication: B ⁄ A = 4 ⁄ 3 means 3B = 4A.
• The sum of two ages rises by 2 every year, which is a quick sanity check on any option.

Common traps

• Anchoring x at the present when the doubling condition is stated for one year ago.
• Reporting the ages one year ago or nine years hence instead of today.
• Adding 9 to the one-year-ago ages when the correct step from that date is 10.

No directly related past PYQ was found.