Age problems are fundamentally simultaneous-equation problems dressed up in story language. The "story" tells you relationships between ages at different points in time — now, some years ago, some years from now. Your job is to strip the language down to algebra and solve.
Here is the core idea: a person's age changes by the same number of years in every scenario. If today is the reference point and you move back 7 years, every person in the problem gets 7 years subtracted. If you move forward 4 years, every person gets 4 added. This single observation kills most errors.
Think of it like a timeline. Everyone is standing on the same timeline. When the problem says "seven years ago", everyone slides left by 7. When it says "four years from now", everyone slides right by 4. The difference between two people's ages never changes — it is always constant. So if a father is 22 years older than his son today, he was 22 years older 10 years ago and will be 22 years older 10 years from now.
A useful analogy: imagine two cars travelling on parallel roads at exactly the same speed. The gap between them never changes, no matter how far they travel. Ages work the same way — the absolute gap between two people's ages is fixed forever.
Age problems in SSC CGL almost always reduce to one of three structures:
The language varies — "thrice as old", "twice as old", "five years elder than", "half as old" — but the algebra is always the same: write present age = variable, apply time shift, set up equation.
One firm rule: always define variables in terms of present age. Do not define x as the age 7 years ago and then try to add and subtract. That path leads to sign errors. Define present, shift to past/future inside the equation.
Let's formalise the process before diving into cases.
Step 1 — Assign present-age variables. Use the fewest variables possible. If the problem gives you a ratio (A : B = 3 : 4), immediately substitute A = 3x, B = 4x. You've just collapsed two unknowns into one.
Step 2 — Identify the time-shift. "n years ago" → subtract n from every person's age. "n years from now" → add n to every person's age.
Step 3 — Translate each sentence into one equation. One condition = one equation. Two conditions = two equations (usually enough to solve).
Step 4 — Solve and verify. Always substitute back. A 2-minute verify saves a wrong attempt.
This is the most common type in SSC CGL.
Pattern: Present ratio = k₁ : k₂ gives you A = k₁x, B = k₂x. A second condition (past or future) gives a second equation. Solve for x.
Example structure: "Present ratio of A to B is 5:7. 18 years ago, ratio was 8:13. Find A's present age."
Set A = 5x, B = 7x. Eighteen years ago: (5x − 18)/(7x − 18) = 8/13. Cross-multiply: 13(5x − 18) = 8(7x − 18) → 65x − 234 = 56x − 144 → 9x = 90 → x = 10. Present age of A = 50.
Pattern: "x years ago ratio was p:q. After y years ratio will be r:s."
Here, use the ratio from the present as your variable if a present ratio is given. If no present ratio is given, let present ages be A and B directly, and set up two equations.
Watch out: When cross-multiplying ratios in the past/future, be careful with which term is numerator and which is denominator. Write the ratio as a fraction immediately and cross-multiply — never "guess" which way to cross-multiply.
Usually structured as: ratio of three ages three years ago (or now) + sum of current ages.
Pattern: Ages three years ago are 4k, 5k, 6k → present ages are 4k+3, 5k+3, 6k+3 → sum = given value → solve for k → answer.
The ratio-in-the-past pattern is very SSC-friendly. Convert past ages to present ages by adding the time-shift, then sum them.
This is a very frequent sub-type. Let the younger person's present age = x.
Older = 3x (from "thrice as old now"). N years ago: 3x − N = 4(x − N). Solve: 3x − N = 4x − 4N → x = 3N.
So the younger person's age is always 3 times the number of years ago mentioned. Once you see this structure, you can jump directly to x = 3N without writing out the equation every time — but only when the multiplier goes from k to (k+1) as you move back in time.
Standard setup: F = aS now, and F + t = b(S + t) in the future (or F − t = c(S − t) in the past). Substitute F = aS into the second equation, solve for S, then get F.
The key discipline: don't add a third condition without checking whether it is consistent with the first two. Some SSC CGL problems include a third condition that is either redundant or used to verify. Solve from any two consistent conditions.
Since the age difference is constant:
(Age of A) − (Age of B) = constant at all times
If you know their ratio now is p:q and you also know one of their ages, you can find both immediately. If you know their ratio now and a past ratio, you can find the actual ages without any extra info.
A/B = p/q and (A − B) = D → A = pD/(p−q), B = qD/(p−q).
This is powerful when the problem gives you the age difference directly.
When the problem says "n years ago" or "n years from now", every person's age shifts by exactly n — no exceptions. Write the shift as a signed number: past = −n, future = +n. Never mix signs within one equation.
Micro-example: "Seven years ago, P was four times D" → (P − 7) = 4(D − 7). Both P and D lose 7. If you write P − 7 = 4D (forgetting to shift D), you get P = 4D + 21, which is wrong.
Speed gain: Enforcing this as a rule eliminates the single most common error, which typically costs 1–2 minutes of re-working. Correct setup: 30 seconds.
The moment you see a present-age ratio, substitute A = kx, B = lx and work in one unknown. Don't keep two variables when the ratio is given.
Micro-example: "A : B = 5 : 7 now. 18 years ago, ratio was 8 : 13." Standard method (two variables F and S, two equations, elimination): ~90 seconds. Shortcut (substitute 5x and 7x immediately, one equation): ~35 seconds.
The equation becomes (5x − 18)/(7x − 18) = 8/13 — one unknown, one step to cross-multiply, done.
When present multiplier is k and past multiplier is (k+1), and the past is N years ago:
Younger person's present age = kN.
Specifically for k = 3: "A is thrice as old as B now; N years ago, A was four times as old as B" → B's current age = 3N, A's current age = 9N. Difference = 6N.
Micro-example: "10 years ago, Ghanshyam was four times as old as Akash. Now Ghanshyam is thrice as old." N = 10, so Akash = 30, Ghanshyam = 90. Difference = 60. Standard algebraic setup: 3 steps, ~50 seconds. Direct pattern: 1 step, ~10 seconds.
If you have both the ratio of present ages AND their sum, you don't need any time-shift at all.
Formula: If A : B = p : q and A + B = S, then A = pS/(p+q), B = qS/(p+q).
Micro-example: "A + B + C = 75. A = 3B. C = B + 5." Total sum condition + ratio + difference. Substitute B = x: A = 3x, C = x + 5. Sum: 5x + 5 = 75 → x = 14. All three ages in one step.
This avoids two-equation elimination. Standard method: ~60 seconds. Direct substitution: ~20 seconds.
Age difference never changes. If you know the difference D and the ratio now (p : q), present ages are pD/(p−q) and qD/(p−q).
Micro-example: If father is always 22 years older than son, and current ratio is 3:1, then difference = (3−1) parts = 22 → 1 part = 11 → son = 11, father = 33.
Standard method (two equations): ~60 seconds. Difference shortcut: ~15 seconds. Best used when the problem directly states an age difference.
When you see an age problem in the exam hall, run through this decision tree:
Step 1 — Is a present ratio given? Yes → immediately substitute A = px, B = qx. You now have one unknown.
Step 2 — Count the time conditions. One condition → one equation in one unknown → solve directly. Two conditions → two equations (if no ratio given upfront, you have two unknowns).
Step 3 — Identify time direction. "N years ago" → subtract N from all ages in that equation. "N years from now" → add N to all ages in that equation.
Step 4 — Is it a three-person sum problem? Assign ratio-based variables to past ages, add the time-shift to convert to present, sum = given value, solve for the ratio constant.
Step 5 — Solve and verify. Substitute your answers back into the original conditions. If a condition fails, check sign errors in the time-shift first — that is where 80% of errors live.
Target time: 90 seconds per age problem. If you are at 2+ minutes, you likely kept two variables when one would do.
Why this question: The classic two-condition, two-time-shift structure — most common age problem format in SSC CGL.
Solving path: Define present ages P (Prachi) and D (daughter). Condition 1 (seven years ago): P − 7 = 4(D − 7) → P = 4D − 21. Condition 2 (four years from now): P + 4 = 2.5(D + 4) → P = 2.5D + 6. Equate: 4D − 21 = 2.5D + 6 → 1.5D = 27 → D = 18. Then P = 51. Sum = 69.
Why this question: Three-person problem with ratio substitution — tests whether you can handle A = 3B and C = B + 5 simultaneously.
Solving path: Let B = x, so A = 3x, C = x + 5. Sum: 3x + x + x + 5 = 75 → 5x = 70 → x = 14. B five years ago = 9. A three years from now = 42 + 3 = 45. Ratio = 9 : 45 = 1 : 5.
Why this question: Simultaneous equations presented as given — tests pure equation-solving speed. No time-shift needed.
Solving path: Two equations: x − 4y = 7 and 3x − y = 65. From first: x = 4y + 7. Substitute into second: 3(4y + 7) − y = 65 → 12y + 21 − y = 65 → 11y = 44 → y = 4. Then x = 23. Mayadevi's age = 3x = 69.
Why this question: The "thrice now, four times N years ago" pattern — use the direct pattern shortcut for maximum speed.
Solving path: Pattern recognition: present multiplier = 3, past multiplier = 4, N = 10 years ago. Akash's present age = 3 × 10 = 30. Ghanshyam = 90. Difference = 60. Algebraic verification: 3x − 10 = 4(x − 10) → 3x − 10 = 4x − 40 → x = 30. Confirmed.
Why this question: Present ratio given directly — a textbook Ratio-to-Variable Collapse opportunity. This type recurs across multiple SSC CGL years.
Solving path: Let present ages be 5x and 7x. Eighteen years ago: (5x − 18)/(7x − 18) = 8/13. Cross-multiply: 65x − 234 = 56x − 144 → 9x = 90 → x = 10. A's present age = 50.
Why this question: Multi-condition father-son problem — tests discipline in using only the two consistent conditions and ignoring the third (which is extraneous in this problem).
Solving path: Let father = F, son = S. F = 3S (condition 1). In 16 years: F + 16 = 2(S + 16) → F = 2S + 16 (condition 2). Equate: 3S = 2S + 16 → S = 16, F = 48. Answer: 48 years.
Why this question: Three-time-point ratio problem — present, past, and future ratios all in one question. Tests systematic equation setup.
Solving path: Let Ravi = 3x, Karan = 4x. Five years ago: (3x − 5)/(4x − 5) = 2/3 → 9x − 15 = 8x − 10 → x = 5. Present: Ravi = 15, Karan = 20. After y years: (15 + y)/(20 + y) = 5/6 → 90 + 6y = 100 + 5y → y = 10.
Why this question: Three-person, ratio-in-the-past problem. Tests the "convert past ages to present by adding time-shift, then sum" technique.
Solving path: Three years ago: Anil = 4k, Bhuvan = 5k, Chitra = 6k. Present ages: 4k+3, 5k+3, 6k+3. Sum: 15k + 9 = 66 → 15k = 57 → k = 3.8. Bhuvan's present age = 5(3.8) + 3 = 19 + 3 = 22.
Shifting only one person's age in a past/future condition. If the problem says "seven years ago, P was four times D", you must write P − 7 = 4(D − 7) — not P − 7 = 4D. Both P and D were younger by 7. This is the single most frequent error in age problems.
Defining variables as past ages instead of present ages. If you let x = D's age seven years ago, you then have to track separate expressions for P's past age and both present ages. The algebra becomes messy and error-prone. Always define present ages as your variables.
Getting ratio direction wrong. "A is thrice as old as B" means A = 3B, not B = 3A. In the exam hall, under time pressure, this flip happens more than you'd think. Read: "A is [multiplier] times as old as B" → A is the bigger one, A = multiplier × B.
Cross-multiplying ratio equations incorrectly. When you have (5x − 18)/(7x − 18) = 8/13, cross-multiply as 13(5x − 18) = 8(7x − 18). A common error is writing 13(7x − 18) = 8(5x − 18) — i.e., flipping which denominator goes with which numerator. Always match numerator-to-denominator: left-numerator × right-denominator = right-numerator × left-denominator.
Ignoring non-integer answers as wrong. Some age problems (like the three-person ratio type) yield non-integer values for the ratio constant k, but the actual age requested is still a clean integer. Don't abandon the calculation just because k is a decimal. Finish computing the final answer.
Using all three conditions in an over-determined system without checking consistency. When a problem gives three conditions for two unknowns, any two consistent conditions are enough. Solve from the first two, verify against the third. If the third fails, the problem has an inconsistency — pick the two conditions that directly give you the asked quantity.