Skip to content
Free shipping above ₹499
Oakspine Press

Problems on ages for AFCAT: equations, the ratio method and checking with options

Age problems become simple equations once you fix one unknown. Setting up "years ago" and "years later", the constant age difference, the ratio method, ages at birth and product questions, with worked examples and a practice set.

5 Oct 2026 8 min read

In this guide
  1. The set-up routine
  2. The one fact: the age difference never changes
  3. The ratio method
  4. Worked examples
  5. Checking with options
  6. Common mistakes
  7. Practice set
  8. What to do next

Problems on ages are among the most reliable marks in AFCAT Numerical Ability. The story changes (a parent and a child, two siblings, three friends) but the maths is always a small linear equation, and the options can usually be checked in seconds. Where candidates lose marks is in the set-up: adding years to one person and forgetting the other, or reading "five years ago" as "five years hence".

This guide gives you a set-up routine that avoids those slips, the one fact that makes many questions trivial, and the ratio method that handles most AFCAT versions.

The set-up routine

  1. Choose one unknown, usually the present age of the younger person, and write everyone's present age in terms of it.
  2. Move in time for everyone. "n years ago" subtracts n from every person's age; "n years hence" adds n to every person's age.
  3. Translate the condition ("twice as old", "ratio 5 : 6", "sum is 50") into one equation.
  4. Solve and check the answer against every condition in the question.

A small table helps when there are two time points:

Personn years agoNowm years hence
Ax − nxx + m
By − nyy + m

The one fact: the age difference never changes

Two people age at the same rate, so the difference between their ages is the same at every point in time. A parent who is 30 years older than a child was 30 years older at the child's birth and will still be 30 years older in 20 years.

This is why ratio questions work. If present ages are in the ratio 5 : 3, the difference is 2 parts, and those 2 parts are a fixed number of years forever. It is also why a parent's "age at the child's birth" equals the present difference in their ages.

What does change is the ratio. As both get older, the ratio of their ages moves towards 1. A ratio of 3 : 1 today might be 2 : 1 in a few years and 3 : 2 later.

The ratio method

When ages are given as a ratio, write them as 4x and 5x, move both in time, and set up the new ratio as a fraction.

(4x + 6)/(5x + 6) = 5/6 → 24x + 36 = 25x + 30 → x = 6

So the present ages are 24 and 30. Cross-multiply, collect the x terms, and you are done. Most AFCAT ratio questions solve in three lines this way.

Worked examples

Example 1 (twice and three times). A is twice as old as B. Ten years ago, A was three times as old as B. Find their present ages.

  • Let B = x, so A = 2x.
  • Ten years ago: 2x − 10 = 3(x − 10) = 3x − 30, so x = 20.
  • B is 20 and A is 40. Check: ten years ago they were 30 and 10, and 30 is three times 10.

Example 2 (sum and a past multiple). The ages of a parent and a child add up to 50. Five years ago, the parent was 7 times as old as the child. Find their present ages.

  • Let the child be C, so the parent is 50 − C.
  • Five years ago: 50 − C − 5 = 7(C − 5), so 45 − C = 7C − 35 and 8C = 80.
  • The child is 10 and the parent is 40. Check: five years ago, 35 = 7 × 5.

Example 3 (ratio with a difference). The ages of A and B are in the ratio 5 : 3, and A is 8 years older. After how many years will the ratio be 3 : 2?

  • The difference is 2 parts = 8 years, so one part = 4. A = 20, B = 12.
  • (20 + n)/(12 + n) = 3/2, so 40 + 2n = 36 + 3n and n = 4 years.
  • Check: in 4 years they will be 24 and 16, which is 3 : 2.

Example 4 (ratio now and in the past). The present ages of A and B are in the ratio 7 : 5. Eight years ago the ratio was 3 : 2. Find the present ages.

  • (7x − 8)/(5x − 8) = 3/2, so 14x − 16 = 15x − 24 and x = 8.
  • Present ages 56 and 40. Check: eight years ago, 48 and 32, which is 3 : 2.

Example 5 (age at birth). A parent is now three times as old as their child. In 6 years, the parent will be 2.5 times as old as the child. How old was the parent when the child was born?

  • Let the child be C, the parent 3C. In 6 years: 3C + 6 = 2.5(C + 6) = 2.5C + 15.
  • 0.5C = 9, so C = 18 and the parent is 54.
  • Age at the child's birth = the difference = 54 − 18 = 36 years.

Example 6 (product of ages). A is 4 years older than B, and the product of their ages is 192. Find their ages.

  • x(x + 4) = 192. Instead of solving the quadratic, look for two numbers 4 apart whose product is 192: 12 × 16 = 192.
  • B is 12 and A is 16. (The negative root, −16, is not an age.)

Checking with options

With +3 for a right answer and −1 for a wrong one, a quick check is always worth it. Substituting options is often faster than solving:

  • Try each option in the most restrictive condition first (usually the ratio or multiple).
  • An option must satisfy every condition, not just one. Examiners often include an option that fits the first condition but not the second.
  • Rule out options quickly: ages cannot be negative, and a parent must be older than the child by a realistic gap.

In Example 4, the options might be (56, 40), (49, 35), (42, 30) and (35, 25). All four are in the ratio 7 : 5, so the ratio alone decides nothing. Only 56 and 40 give 3 : 2 eight years earlier.

Common mistakes

  • Moving only one person in time. Every person gets n years older or younger.
  • Answering the wrong question. The question may ask for the sum, the ratio or someone's age at a different time, not the present ages.
  • Assuming the ratio stays the same. The difference stays the same; the ratio does not.
  • Mixing past and future. "Five years ago" and "five years hence" are ten years apart.

Practice set

  1. Two ages are in the ratio 3 : 4. In 5 years, the ratio will be 4 : 5. Find the present ages.
  2. A is 5 years older than B, and their ages add up to 35. Find their ages.
  3. A person's age is three times a child's. In 12 years, it will be twice the child's. Find their present ages.
  4. The ages of A and B add up to 60. Six years ago, A was twice as old as B. Find their ages.
  5. The average age of A, B and C is 25, and their ages are in the ratio 3 : 5 : 7. Find the ages.
  6. Ten years ago, A was half as old as B. Their present ages are in the ratio 3 : 4. Find the sum of their present ages.
  7. Four years ago, the ages of A and B were in the ratio 3 : 4. Four years hence, the ratio will be 5 : 6. Find the present ages.
  8. A sibling is 6 years older than an aspirant. In 3 years, the sibling will be 1.5 times as old as the aspirant. Find their present ages.

Answers:

  1. 15 and 20. (3x + 5)/(4x + 5) = 4/5, so 15x + 25 = 16x + 20 and x = 5.
  2. 20 and 15. 2B + 5 = 35, so B = 15.
  3. 36 and 12. 3C + 12 = 2(C + 12), so C = 12.
  4. 38 and 22. A − 6 = 2(B − 6) gives A = 2B − 6. Then 3B − 6 = 60, so B = 22. Check: six years ago, 32 = 2 × 16.
  5. 15, 25 and 35. Total 75 over 15 parts, so one part = 5.
  6. 35. 3x − 10 = (4x − 10)/2, so 6x − 20 = 4x − 10 and x = 5. Ages 15 and 20.
  7. 16 and 20. (3x + 8)/(4x + 8) = 5/6, so 18x + 48 = 20x + 40 and x = 4. Four years ago they were 12 and 16.
  8. Aspirant 9, sibling 15. A + 6 + 3 = 1.5(A + 3), so A + 9 = 1.5A + 4.5 and A = 9. Check: in 3 years, 18 = 1.5 × 12.

What to do next

  • Use the four-step set-up routine on your next 20 age questions, writing the time table for each.
  • Practise solving five questions by substitution alone, to build the option-checking habit.
  • Revise ratio and proportion, since most age questions are ratio questions, then move on to HCF and LCM.

A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Indian Air Force website .

Get the next AFCAT guide by email

New guides every week. No spam, unsubscribe any time.

Keep reading

AFCATNumerical AbilityMixtures and alligation for AFCAT: the cross rule, milk–water ratios and replacementAlligation finds the ratio in which two things must be mixed to get a given average. The cross rule and why it works, milk and water ratios, adding to change a ratio, repeated replacement and mixing two mixtures, with worked examples and a practice set. 7 min read·4 Oct 2026AFCATNumerical AbilityTrains, boats and streams for AFCAT: lengths, relative speed and upstream–downstreamTrains add their own length to the distance; boats add or subtract the stream. Poles, platforms, people and other trains, upstream and downstream speeds, round trips and the meeting-time rule, with worked examples and a practice set. 6 min read·3 Oct 2026AFCATNumerical AbilitySpeed, time and distance for AFCAT: average speed, relative speed and late–early questionsDistance = speed × time, plus four ideas that solve most AFCAT questions — unit conversion, proportionality, average speed and relative speed. Late and early, stoppages, meeting and overtaking, with worked examples and a practice set. 6 min read·1 Oct 2026AFCATNumerical AbilityTime and work for AFCAT: the LCM method, efficiency and men–days questionsWork questions turn into simple fractions or, faster, into total units of work. The fraction and LCM methods, efficiency, three workers, someone leaving, alternate days, men–days–hours and wages, with worked examples and a practice set. 7 min read·30 Sept 2026