Problems on ages, with answers

The questions, with answers

1. 1.What is the one method that solves most problems on ages?

   In short: Choose one unknown — usually the younger person's age today — write every other age at every time in terms of it, and turn the question's one fact into an equation.

   Let the unknown be a present age, never a past or future one, because the question usually mixes times. A father is 30 years older than his son; five years ago he was four times as old as the son. Let the son be x today, so the father is x + 30. Five years ago they were x − 5 and x + 25, and the fact is about that moment: x + 25 = 4(x − 5). So x + 25 = 4x − 20, 3x = 45 and x = 15. The son is 15 and the father 45 — and five years ago they were 10 and 40, which is four times. Write the table of ages at each time before writing the equation; the table is where translation errors get caught.

2. 2.Why does the difference between two people's ages never change?

   In short: Both people age one year every year, so the gap between them is fixed forever — a constant you can use while the ratio keeps moving.

   A ratio of ages changes with time; a difference does not, which makes the difference the most useful number in the question. Two cousins' ages are in the ratio 2 : 3 today, and one is 6 years older than the other. The ratio's parts differ by 1, and that one part is the 6-year gap, so the ages are 12 and 18. In 6 years they will be 18 and 24, a ratio of 3 : 4 — the ratio has moved, the gap is still 6. The same fact removes a common wrong option: if A is 8 years older than B today, no amount of time makes A 10 years older. It also means the older age is always the younger one plus the gap, so two people never need two unknowns.

3. 3.How do you solve an ages problem that gives one ratio now and another ratio later?

   In short: Write today's ages as ax and bx, add the same number of years to both, and set the new fraction equal to the new ratio.

   The ages of A and B are in the ratio 3 : 7 today; in 6 years they will be in the ratio 5 : 9. Today they are 3x and 7x, so (3x + 6) / (7x + 6) = 5 / 9. Cross-multiplying gives 27x + 54 = 35x + 30, so 8x = 24 and x = 3: the ages are 9 and 21, and in 6 years 15 and 27, which is 5 : 9. There is a check that often replaces the algebra. When the two ratios differ by the same number of parts (7 − 3 = 4 and 9 − 5 = 4), the parts are the same size in both, because the gap is constant. Here A's share grew from 3 parts to 5, two parts in 6 years, so a part is 3 years — the same answer in one line. For 'n years ago', subtract instead of adding.

4. 4.How do you handle the total of several people's ages at different times?

   In short: A total of ages changes by the number of people times the years passed — but only for people who were alive for all of those years.

   Three people's ages total 90 today, so five years ago they totalled 90 − 15 = 75, and in five years they will total 105. Combine that with a multiple: a father and his two children have ages totalling 72, and eight years ago the father was twice as old as the two children together. Let the children total C, so the father is 72 − C. Eight years ago the father was 64 − C and the children 16 years younger in total, C − 16. So 64 − C = 2(C − 16), which gives 3C = 96 and C = 32: the father is 40. Check: eight years ago he was 32 and the children together 16. Subtracting 16 assumes both children were born eight years ago; for a younger child, subtract only the years that child has lived.

5. 5.How do you solve 'I was as old as you are now' questions?

   In short: 'When I was your age' was exactly (my age − your age) years ago, because the gap is constant — so subtract that gap from both people's ages.

   A says to B, 'I am twice as old as you were when I was as old as you are now.' Their ages total 63. Let A be a and B be b. A was B's present age a − b years ago, and B was then b − (a − b) = 2b − a. The sentence says a = 2(2b − a), so 3a = 4b and the ages are in the ratio 4 : 3. With a total of 63, A is 36 and B is 27. Check it in words: 9 years ago A was 27, which is B's age now, and B was 18; A is now 36, twice 18. The question looks circular, but it is only the constant gap used twice: once to find when, once to find B's age at that time.

6. 6.How do you solve problems on ages that involve an average age?

   In short: Turn every average into a total — n people averaging m years total n × m — and remember a fixed group's total rises by n each year.

   The average age of a family of five is 28 years, and the youngest child is 8. What was the family's average age just before the youngest was born? The total today is 5 × 28 = 140. Without the youngest, the other four total 140 − 8 = 132. Eight years ago each of those four was 8 years younger, so they totalled 132 − 32 = 100, and the family then was those four people: an average of 100 / 4 = 25. The common wrong answer is 28 − 8 = 20, which takes eight years off everyone, including a child who did not exist yet. A fixed group's average rises by exactly one a year; a group that gains or loses a member needs its totals written out.

7. 7.What are the common translation mistakes in ages problems?

   In short: 'Hence' is the future and 'ago' the past; 'older by' is a difference and 'times as old' is a multiple — applied at the time the sentence names.

   Most wrong answers are a correct equation of the wrong sentence. 'Six years hence, Ravi will be three times as old as his daughter was four years ago' puts two different times into one fact. If Ravi is 26 years older than his daughter, let her be d and him d + 26. Six years hence he is d + 32; four years ago she was d − 4. So d + 32 = 3(d − 4), 2d = 44 and d = 22: she is 22 and Ravi 48. Check: in six years he will be 54, and four years ago she was 18 — three times. Two more traps: 'A is older than B by 5 years' is a − b = 5, not a = 5b; and 'the ratio of their ages will be' refers to both ages at that future time, not one future and one present.

8. 8.How do you check an ages answer quickly, or use the options instead?

   In short: Put the answer back into every sentence of the question — no negative ages, no parent younger than a child — and, with options, test them one by one.

   A is four times as old as B; in 20 years A will be twice as old as B. With options 5, 10, 15 and 20 for B's age, test one: B = 10 makes A 40, and in 20 years they are 30 and 60 — twice. The algebra, 4b + 20 = 2(b + 20), gives 2b = 20 and the same b = 10 in about the same time. Testing options pays when the equation would need fractions or a quadratic; algebra pays when the options are close together. Either way, substitute the final ages into every clause, not just the last one used: an answer that fits the ratio but makes someone younger than zero five years ago means a sentence was read in the wrong direction.

How the diagnostic asks it

One question from the Aptitude bank, exactly as a sitting would show it. The bank has 8 on ages and 84 across Aptitude.