Reversing the digits of A's age we get B's age. One year ago A was thrice the…

Concept: In a “digit-reversal” age problem, write a two-digit age using its tens digit x and units digit y, so the age is 10x + y. Reversing the digits gives the age 10y + x. When a condition relates the two ages after shifting both by the same number of years (for example, “one person was k times the other, m years ago”), that condition becomes one linear equation in the two unknown digits x and y — and since x and y can only be whole numbers from 0 to 9, there are just a few digit pairs to check.

Application:

1. Let A's present age be 10x + y and B's present age be 10y + x, since reversing the digits of A's age gives B's age.

2. One year ago, A's age was thrice B's age one year ago: (10x + y) − 1 = 3[(10y + x) − 1].

3. Expanding: 10x + y − 1 = 30y + 3x − 3, which rearranges to 7x − 29y + 2 = 0, i.e. 7x = 29y − 2.

4. Since x and y must each be a digit from 0 to 9, check each value of y in turn: only y = 2 makes 29y − 2 (= 56) divisible by 7 with x also a valid digit (x = 8).

5. So A = 10(8) + 2 = 82 and B = 10(2) + 8 = 28.

Cross-check: one year ago A was 81 and B was 27, and 81 = 3 × 27, confirming the ratio condition; also reversing the digits of 82 gives 28, confirming the digit-reversal relationship.

So B's present age is 28 years.