Age problems are word problems where you work backwards or forwards in time to find unknown ages. They are one of the most predictable question types in SSC MTS — the setters reuse the same five or six templates, year after year.
Here is the core idea: every age problem is really just a linear equation (or two) dressed up in a story. The moment you strip away the story and write the algebra, you are done.
Think of it like a timeline. Right now is "the present." The problem might take you 5 years back, or 10 years forward, or both. Every person's age shifts by the same number of years when you move along that timeline. That is the universal rule that makes these problems solvable.
A useful analogy: Imagine you and a friend are standing on a train platform. The train (time) moves at the same speed for both of you. If the train moves 10 stations forward, both of you move 10 stations. If it moves 5 stations back, both of you move 5 stations back. Your age difference never changes, no matter how far the train travels.
This means: the difference between two people's ages is always constant. Father is 25 years older than his son today, he will still be 25 years older 20 years from now. This single fact eliminates half the arithmetic in many problems.
The five problem types you will see in SSC MTS:
Each has a specific attack pattern. Learn the pattern, not the problem.
The standard approach that works for every problem:
x if the question asks for 3x.When ages are in ratio a : b, let the actual ages be ax and bx. The variable x is the common multiplier. Your job is to find x.
Example framework: Ratio is 5 : 3, sum is 96.
5x and 3x5x + 3x = 96 → 8x = 96 → x = 12When you have a product instead of a sum: Ratio 3 : 1, product is 432.
3x and x3x · x = 432 → 3x² = 432 → x² = 144 → x = 12When a problem says "after N years" or "N years ago," every person's age shifts by exactly N. This is mechanical — no thinking required, just add or subtract N for each person.
"After 20 years, Manoj's age = sum of children's ages"
If Manoj is x now and children's sum is S now:
x + 20, children's sum is S + 40 (two children, so +20 each)x + 20 = S + 40Here is the mistake most people make: they forget to add N for each person separately. Two children, 20 years each — that is +40 to their sum, not +20.
This is a beautiful template. If children are born at intervals of k years:
yy + ky + 2k3y + 3k = 3(y + k)So if sum = 30, interval = 5: 3y + 15 = 30 → y = 5. Youngest is always (sum/n) − k where n is number of children.
These are the simplest. "Retired at 60, served for 3/5 of retirement age."
(3/5) × 60 = 36 years60 − 36 = 24 yearsNo equation needed. Just fraction × retirement age = service period, then subtract.
Ratio 5 : 8 : 9, sum of first and third = 56.
5x, 8x, 9x5x + 9x = 56 → 14x = 56 → x = 48 × 4 = 32The key insight: the problem tells you which pair sums to a known value. Use only those two terms to find x, then plug in for the one asked.
These look scarier than they are. The trick is to pick the right variable.
"Father is twice elder son's age. After 10 years, father is three times younger son's age. Sons differ by 15 years."
x, so father = 2xx − 152x + 10 = 3(x − 15 + 10) → 2x + 10 = 3x − 15 → x = 2550Always pick the person with the most constraints as your base variable. Father has two constraints — but elder son connects directly to father, so use elder son as x.
When ages are in ratio a : b and their sum is S, the individual ages are S × a/(a+b) and S × b/(a+b). No need to write ax + bx = S and solve for x separately. Apply the ratio directly to the sum.
Example: Ratio 5:3, sum 96. Son's age = 96 × 3/8 = 36. Done in one step instead of three.
Standard method: 3 steps, ~30 seconds. This shortcut: 1 step, ~10 seconds.
For n children born at intervals of k years, youngest age = (sum ÷ n) − k × (n−1)/2.
For 3 children, interval 5, sum 30: youngest = (30 ÷ 3) − 5 = 10 − 5 = 5. Direct answer.
Standard method (write three terms, form equation, solve): ~45 seconds. This: ~12 seconds.
When ratio is a : b and product is P, immediately write ab·x² = P, so x² = P/(ab).
Ratio 3:1, product 432: x² = 432/3 = 144, x = 12. You skip the intermediate step of writing 3x · x = 432 and directly solve x² = 144.
Standard method: 4 steps. This: 2 steps. Saves roughly 20 seconds under exam pressure.
The difference between two people's ages is fixed forever. Use this to eliminate answer options.
If father is 25 years older than son and son is now 13, father is 38. No equation needed. In ratio problems, check: does the age difference match (a−b)x? If an option gives a son's age that makes father's age unreasonably young (negative or under 18 for an adult), eliminate immediately.
This eliminates wrong options in under 5 seconds — before you even write an equation.
When a problem says "after N years, Person A's age = f(Person B's age)," write the equation by adding N to every single age on both sides before applying the function f. Don't simplify first, then add.
Mechanical rule: write today's expressions for each person, then replace each with (today's expression + N). This prevents the common error of forgetting to shift one person.
Saves 1-2 re-do cycles (~60–90 seconds) by getting the equation right the first time.
Read the problem and classify it in 5 seconds using this decision tree:
Is there a ratio?
ab·x² = P, solve for xax, bx, form two equationsNo ratio — just ages?
xFinal check before writing your answer: Does the answer make logical sense? Is the son younger than the father? Is nobody negative or under 5 years old for a "son"? If not, recheck your equation — you probably forgot to shift one person in a past/future condition.
Target time per age problem in SSC MTS: under 60 seconds.
Why this question: This is the cleanest ratio + average combination — tests whether you can convert average to sum and then apply the ratio directly.
Solving path: Average of two people = 48, so sum = 2 × 48 = 96. Ratio is 5:3, so son's age = 96 × 3/(5+3) = 96 × 3/8 = 36. One formula application, no equation needed.
Why this question: Tests whether you correctly handle "sum of two children's ages" in a future condition — the trap is adding 20 instead of 40 to the children's sum.
Solving path: Let Manoj = x, children's sum = x/2. After 20 years: Manoj = x+20, children's sum = x/2 + 40 (two children, 20 years each). Condition: x+20 = x/2+40 → x/2 = 20 → x = 40.
Why this question: Product condition with ratio — tests whether you go through 3x·x = 432 cleanly without confusing yourself.
Solving path: Father = 3x, son = x. Product: 3x² = 432 → x² = 144 → x = 12. Sum = 3(12) + 12 = 48. Note: the question asks for sum, not individual ages — don't stop at x = 12.
Why this question: Three-variable problem — tests systematic variable assignment. The key is picking elder son as base variable, not father.
Solving path: Elder son = x, father = 2x, younger son = x−15. After 10 years: 2x+10 = 3(x−15+10) = 3x−15. Solving: x = 25, father = 50.
Why this question: Ratio problem with three people — tests whether you use only the relevant pair to find x.
Solving path: Ages = 5x, 8x, 9x. Sum of A and C: 5x + 9x = 56 → 14x = 56 → x = 4. Age of B = 8×4 = 32. Don't add all three — only use the pair given.
Why this question: Simplest service/retirement type — no equation at all, just two arithmetic steps. Fast points.
Solving path: Service = 3/5 × 60 = 36 years. Joining age = 60 − 36 = 24. If you spent more than 20 seconds on this, you over-thought it.
Why this question: Children born at intervals — direct application of the interval pattern.
Solving path: Let youngest = x. Then x + (x+5) + (x+10) = 30 → 3x + 15 = 30 → x = 5. Or use the shortcut: 30/3 − 5 = 10 − 5 = 5. Same answer, half the steps.
Why this question: Multi-step past/future — requires you to work backwards from future age to find present, then work with past condition. Tests systematic approach.
Solving path: Son's present age = 20 − 7 = 13. Father's age 3 years ago = 5 × (son's age 6 years ago) = 5 × (13−6) = 5 × 7 = 35. Father's present age = 35 + 3 = 38. Ratio = 38:13.
Forgetting to shift every person in future/past conditions. If "after 10 years" and there are two children, their combined age increases by 20, not 10. Count the number of individuals on each side carefully.
Stopping at x instead of answering the actual question. You find x = 12 but the question asks for the sum of ages, which is 4x = 48. Read the last line of the question again before writing your answer.
Treating ratio as actual ages. If the ratio is 5:3 and you write father = 5, son = 3, every calculation that follows is wrong. Ratios need a multiplier x.
Mixing up "N years ago" sign. "3 years ago" means subtract 3 from present age. "After 3 years" means add 3. Many students negate the wrong direction under time pressure.
Using the wrong base variable in three-person problems. If father, elder son, and younger son are all involved, don't let father be your base — he has the most expressions. Use the person with the most direct connection to the others (usually the elder son).
Ignoring the constraint that ages must be positive. If your answer gives a negative age for any person, your equation is wrong — go back and check the sign on your past/future shift.