Two years ago, the age of A was three times the age of B. If B is currently 9 years old, then after how many years, the age of A will be double of the age of B?
- (a)2 years
- (b)3 years
- (c)4 years
- (d)5 years
Correct — D, 5 years. Work back to the moment the first condition describes. B is 9 now, so two years ago B was 7. A was then three times that, which is 21, so A is 21 + 2 = 23 now. Let t be the number of years after which A is twice B. Then 23 + t = 2(9 + t), which expands to 23 + t = 18 + 2t, giving t = 5. Check it: in five years A will be 28 and B will be 14, and 28 is exactly twice 14. The step that decides the item is going back two years for both people before setting up the equation — the ratio in the stem belongs to the past, not to the present.
- (a)2 years — In two years A would be 25 and B 11, and 25 is not twice 11. Two is the number of years mentioned in the stem, offered as a decoy.
- (b)3 years — In three years A would be 26 and B 12, and twice 12 is 24, not 26. The gap is still closing at that point.
- (c)4 years — In four years A would be 27 and B 13, and twice 13 is 26. One year short — this is what a candidate gets by forgetting to add the two years back to A's age and starting from 22.
Age problems are linear equations with one extra discipline: every age moves by the same amount, so the difference between two ages never changes. Here A is always 14 years older than B, and A is twice B exactly when B equals that constant difference — that is, when B is 14, which happens in five years. Using the fixed difference is often quicker than setting up and solving the equation.
Two time references appear in the stem — two years ago and t years hence — and mixing them is the standard failure. Convert everything to present ages first, then move forward once. The constant-difference shortcut is worth internalising for the whole family of these questions: if A is to be k times B, then B at that moment equals the age difference divided by k − 1. With a difference of 14 and k = 2, B must be 14, and since B is 9 now that is five years away.
- Two years ago B was 7 and A was 21, so A is 23 now and B is 9.
- The age difference is 14 years and never changes.
- Setting 23 + t = 2(9 + t) gives t = 5.
- In five years A is 28 and B is 14, and 28 is twice 14.
- For A to be k times B, B's age at that moment must equal the difference divided by k − 1.
The three-to-one ratio describes the past. Moving both ages to the present before equating is the whole discipline.
- Applying the three-to-one ratio to present ages instead of to the ages two years ago.
- Forgetting to bring A's age forward by two years after computing it.
- Answering 2 because that number appears in the stem.
Asked as a two-timeframe age problem, the standard CAPF shape, where the given ratio sits in the past and the asked condition sits in the future.
The age of a man is three times the sum of the ages of his two sons. Five years hence, his age will be double the sum of the ages of his sons. The father’s present age is
- (a) 40 years
- (b) 45 years
- (c) 50 years
- (d) 55 years
Answer(b) 45 years
The same two-timeframe structure, with a three-to-one ratio now and a two-to-one ratio later. The catch there is that two sons age five years each, so the sum of their ages rises by ten and not by five.
- practice — not a real PYQ
The present ages of a father and son differ by 30 years. After how many years will the father be twice as old as the son, if the son is now 20?
- (a)5 years
- (b)10 years
- (c)15 years
- (d)20 years
Answer(b) 10 years — the father is twice the son when the son's age equals the difference of 30, which is ten years from now.
- practice — not a real PYQ
Three years ago, A's age was four times B's age. If B is now 8, what is A's present age?
- (a)20 years
- (b)23 years
- (c)26 years
- (d)32 years
Answer(b) 23 years — three years ago B was 5 and A was 20, so A is 23 now.