A complete, exam-ready guide to age wale question patterns — covering every formula, shortcut, 40+ worked examples, 50 practice MCQs, and 20 FAQs for SSC, Railway, Bank and State-level exams. Visit Cat Mock
40+ Solved Examples, 50 Practice MCQs, 4 Reference Tables20 FAQs
01 – What are Problems on Ages?
Problems on Ages are word problems that describe relationships between the present, past, or future ages of two or more people, and ask you to find one or more unknown ages using algebra. They are also commonly searched as “age wale question,” “age word problems,” or “age calculation questions.”
Every age problem, no matter how it is phrased, reduces to this: assign a variable to an unknown age, translate each sentence into an equation, and solve. The apparent difficulty comes from the language of the problem, not the mathematics — the actual algebra rarely goes beyond linear or simple quadratic equations. Cat Mock Series
A problem on ages is an arithmetic or algebraic word problem in which the ages of two or more people are related through ratios, sums, differences, or time intervals (past/present/future), and the goal is to determine one or more unknown ages.
02 – Importance in SSC & Competitive Exams
Age-based questions consistently appear in:
- SSC CGL Tier 1 and Tier 2 (Quantitative Aptitude)
- SSC CHSL, SSC MTS, SSC GD Constable, SSC CPO
- Railway (RRB NTPC, Group D)
- Bank PO/Clerk exams (IBPS, SBI)
- State PSC and Police Recruitment exams
- Reasoning sections, disguised as puzzle-based age problems
On average, 1–2 questions on Problems on Ages appear in almost every major government exam, and because the formulas are fixed and predictable, this is one of the highest “marks per minute of preparation” topics available. Students preparing seriously for SSC exams often lean on structured mock tests and a dedicated CATMock aptitude practice track to build speed on exactly this kind of question.
03 – Types of Age Problems
Understanding the type of a question is half of solving it. Here are the recurring patterns:
- Ratio-based present age problems — ages given as a ratio (e.g., 3:4)
- Sum/difference-based problems — total or difference of ages given
- Past age problems — “X years ago, A’s age was…”
- Future age problems — “X years hence, A’s age will be…”
- Combined past-future problems — two conditions at two different time points
- Product-based age problems — product of two ages is given
- Three-or-more-person problems — parent, child, grandchild chains
- Digit-reversal age problems — reversing the digits of an age gives another age
- Average age problems — based on the average age of a group
- Multi-condition SSC-style problems — combine 2–3 of the above in one question
04 – Core Formulae (Cheat Sheet)
Tip: You do not need separate formulas for every scenario. Master these five relationships and you can derive the rest.
| Situation | Formula |
|---|---|
| Age n years ago | Present age − n |
| Age n years hence (later) | Present age + n |
| Ages in ratio a : b | Ages = a·x and b·x (x = common multiplier) |
| Sum of two ages | A + B = Total |
| Difference of two ages (constant forever) | A − B = Constant, same at any point in time |
| Average age of a group of n people | (Sum of all ages) ÷ n |
| New average after removing one person | (Old total − removed person’s age) ÷ (n − 1) |
Golden rule: The difference between any two people’s ages never changes, no matter how many years pass. Only the ratio changes over time.
05 – Age Calculation Rules
- If the present age of a person is x, then their age n years ago was (x − n) and n years hence will be (x + n).
- If two ages are in the ratio a : b, always represent them as ax and bx — never as a and b directly.
- When a question gives two conditions at two different times, you will always get two linear equations in two variables — solve them simultaneously.
- The difference in ages of two people is constant at every point in time; only their ratio changes as they grow older.
- For three or more people, assign the ratio parts as ax, bx, cx and use the sum condition to find x.
06 – Shortcut Tricks for Fast Solving
| Trick | When to Use | How It Saves Time |
|---|---|---|
| Ratio-multiplier trick: express ages as ax, bx directly | Any ratio-based question | Avoids writing two separate variables |
| Constant-difference trick: subtract the two ratio equations to isolate the age gap | Two ratios at two time points | Skips one full elimination step |
| Options back-solving: substitute MCQ options into given conditions | MCQs with 4 options | Fastest for MCQs, avoids algebra entirely |
| Sum-and-ratio shortcut: age = S × [a/(a+b)] | Sum + ratio combo questions | Solves in one line |
| Digit-reversal trick: age difference is always a multiple of 9 | Digit reversal problems | Immediately narrows down possible digits |
Sum-and-ratio shortcut example: If two ages are in ratio 3:5 and their sum is 40, the smaller age = 40 × 3/8 = 15. No equation-writing required.
07 – Common Mistakes Students Make
| Common Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Treating ratio numbers (3:4) as actual ages | The ratio only shows proportion, not real values | Always write as 3x : 4x and solve for x |
| Adding “n years” to only one person’s age | Time passes equally for everyone | Shift every person’s age in that condition |
| Confusing “ago” with “hence” | Sign error (+ vs −) breaks the equation | “Ago” = subtract, “hence” = add |
| Assuming the ratio stays the same over time | Only the difference is constant, not the ratio | Recompute the ratio at each time point |
| Skipping verification of the final answer | Small arithmetic errors go unnoticed | Substitute the answer back into every condition |
Watch out: The single most common error in SSC-style wording is applying a time shift to only one person instead of both. If A and B’s ages are compared “5 years ago,” both must be reduced by 5 — not just one. Cat Mock Series
08 – Step-by-Step Method to Solve Any Age Problem
- Read the question twice and underline every time reference (“years ago,” “years hence,” “at present”).
- Assign variables to the unknown present ages (use ratio multipliers like x, if given).
- Translate each sentence into an algebraic equation, one at a time.
- Count your equations — you need as many independent equations as unknowns.
- Solve the system using substitution or elimination.
- Verify by plugging the solved ages back into the original statements.
- Answer the exact question asked — many mistakes happen from solving for the wrong variable.
09 – Visual Examples (Age Timeline)
A simple age timeline makes past/future problems much easier to visualise:
n years ago PRESENT n years hence A: (x − n) —-> x —-> (x + n) B: (y − n) —-> y —-> (y + n)
Whenever a question gives you two points on this timeline, you are reading two snapshots of the same two moving numbers — and both snapshots must produce equations that hold true simultaneously.
10 – 40+ Solved Examples (Easy to Hard)
Each example follows the same format: question, formula used, step-by-step solution, final answer, difficulty.
Example 1 Easy
The ratio of the present ages of A and B is 3:4. After 5 years, the ratio will become 4:5. Find their present ages.
- Let A = 3x, B = 4x
- (3x + 5)/(4x + 5) = 4/5 → 5(3x+5) = 4(4x+5)
- 15x + 25 = 16x + 20 → x = 5
Answer: A = 15 years, B = 20 years
Example 2 Easy
The sum of ages of A and B is 40 years, in ratio 3:2. Find their ages.
- Let A = 3x, B = 2x → 3x + 2x = 40
- 5x = 40 → x = 8
Answer: A = 24 years, B = 16 years
Example 3 Medium
Five years ago, Sita’s age was thrice Gita’s age. Ten years hence, Sita’s age will be twice Gita’s age. Find their present ages.
- S − 5 = 3(G − 5) → S = 3G − 10
- S + 10 = 2(G + 10) → S = 2G + 10
- 3G − 10 = 2G + 10 → G = 20 → S = 50
Answer: Sita = 50 years, Gita = 20 years
Example 4 Medium
The product of the ages of A and B is 96. A is 4 years older than B. Find their ages.
- Let B = x, A = x + 4 → x(x+4) = 96
- x² + 4x − 96 = 0 → (x−8)(x+12) = 0
- x = 8
Answer: B = 8 years, A = 12 years
Example 5 Medium
The sum of the present ages of a mother and daughter is 50 years. Five years ago, the mother’s age was 7 times the daughter’s age. Find their present ages.
- M + D = 50
- M − 5 = 7(D − 5) → M = 7D − 30
- 7D − 30 + D = 50 → D = 10
Answer: Daughter = 10 years, Mother = 40 years
Example 6 Hard
Three years hence, a father’s age will be three times his son’s age. Three years ago, the father was five times as old as the son. Find their present ages.
- F + 3 = 3(S + 3) → F = 3S + 6
- F − 3 = 5(S − 3) → F = 5S − 12
- 3S + 6 = 5S − 12 → S = 9
Answer: Son = 9 years, Father = 33 years
Example 7 Medium
A is now twice as old as B was 10 years ago. The difference between their present ages is 12 years. Find their present ages.
- A = 2(B − 10)
- A − B = 12 → 2(B−10) − B = 12 → B = 32
Answer: B = 32 years, A = 44 years
Example 8 Medium
The sum of the present ages of a father and son is 60 years. Six years ago, the father’s age was five times the son’s age. Find their present ages.
- F + S = 60
- F − 6 = 5(S − 6) → F = 5S − 24
- 5S − 24 + S = 60 → S = 14
Answer: Son = 14 years, Father = 46 years
Example 9 Hard
The ratio of the ages of P and Q, four years ago, was 3:4. Eight years from now, the ratio will be 4:5. Find their present ages.
- P − 4 = 3x, Q − 4 = 4x
- (3x+12)/(4x+12) = 4/5 → 5(3x+12) = 4(4x+12)
- x = 12
Answer: P = 40 years, Q = 52 years
Example 10 — digit reversal Hard
A man’s present age, written as a two-digit number, becomes his son’s age when the digits are reversed. The difference between their ages is 36 years, and the sum of the digits of the man’s age is 8. Find both present ages.
- Let man’s age = 10a + b, son’s age = 10b + a
- (10a+b) − (10b+a) = 36 → 9(a−b) = 36 → a−b = 4
- a + b = 8 → a = 6, b = 2
Answer: Man = 62 years, Son = 26 years
Example 11 Hard
The sum of the present ages of A, B, and C is 90 years, in the ratio 2:3:4. Find each of their ages after 5 years.
- 2x + 3x + 4x = 90 → x = 10
- A = 20, B = 30, C = 40
Answer: After 5 years — A = 25, B = 35, C = 45
Example 12 Hard
The ratio of a father’s age to his son’s age is 4:1. The product of their ages is 196. Find the ratio of their ages after 5 years.
- F = 4S, F × S = 196 → 4S² = 196 → S = 7, F = 28
- After 5 years: F = 33, S = 12
Answer: Ratio = 33:12 = 11:4
Example 13 Medium
A is as much younger than B as he is older than C. The sum of B’s and C’s ages is 40 years. Find A’s age.
- B − A = A − C → B + C = 2A
- 2A = 40 → A = 20
Answer: A = 20 years
Example 14 Easy
Five children are born at intervals of 3 years each. The sum of their ages is 50 years. Find the age of the youngest child.
- Ages = x, x+3, x+6, x+9, x+12
- 5x + 30 = 50 → x = 4
Answer: Youngest child = 4 years
Example 15 Medium
A father tells his son, “I was as old as you are now when you were born.” If the father’s present age is 38, find the son’s present age.
- Father’s age at son’s birth = 38 − S
- 38 − S = S → S = 19
Answer: Son = 19 years
Example 16 — product & ratioEasy
The ratio of the ages of two sisters is 2:3, and the product of their ages is 96. Find their ages.
- 2x × 3x = 96 → 6x² = 96 → x = 4
Answer: 8 years and 12 years
Example 17 Medium
A father is now three times as old as his daughter. Twelve years ago, he was nine times as old as her. Find their present ages.
- F = 3D
- F − 12 = 9(D − 12) → 3D − 12 = 9D − 108 → D = 16
Answer: Daughter = 16 years, Father = 48 years
Example 18 Easy
The total of the ages of a couple is 92 years. The husband is 4 years older than the wife. Find their ages.
- H = W + 4 → W + 4 + W = 92 → W = 44
Answer: Wife = 44, Husband = 48
Example 19 — three peopleMedium
A is twice as old as B, and B is twice as old as C. The sum of their ages is 70. Find each age.
- A = 4C, B = 2C → 4C + 2C + C = 70 → C = 10
Answer: C = 10, B = 20, A = 40
Example 20 Easy
Ten years ago, X’s age was half of Y’s age at that time. Y’s present age is 40. Find X’s present age.
- Y ten years ago = 30 → X ten years ago = 15
Answer: X’s present age = 25
Example 21 Medium
A woman’s age is 5 years more than twice her son’s age. The sum of their ages is 50. Find their ages.
- W = 2S + 5 → 2S + 5 + S = 50 → S = 15
Answer: Son = 15, Woman = 35
Example 22 — classic trick question Hard
If 6 years are subtracted from Rahul’s present age and the result is divided by 18, his grandson’s present age is obtained. The grandson is 2 years younger than Rahul’s son, who is 30 years old. Find Rahul’s present age.
- Grandson’s age = 30 − 2 = 28
- (Rahul − 6)/18 = 28 → Rahul − 6 = 504 → Rahul = 510
Answer: 510 years (a deliberately exaggerated trick question testing careful reading, not realism)
11 – SSC & Government Exam-Wise Age Questions
Different exams favour slightly different question styles. Here are worked examples matched to each exam’s typical pattern.SSC CGL-style
Example 23 Medium
The ratio of the present ages of X and Y is 5:6. Four years ago, the ratio was 3:4. Find their present ages.
- X = 5x, Y = 6x → (5x−4)/(6x−4) = 3/4 → x = 2
Answer: X = 10 years, Y = 12 years
Example 24 Medium
The sum of the present ages of A and B is 50. Five years ago, their ages were in the ratio 2:3. Find their present ages.
- A+B=50; 3A−2B=5 → A = 21
Answer: A = 21 years, B = 29 years
Example 25 Hard
A father’s present age is 6 years more than thrice his son’s age. Five years ago, the father’s age was 4 times the son’s age. Find their present ages.
- F=3S+6; F−5=4(S−5) → S = 21
Answer: Son = 21 years, Father = 69 yearsSSC CHSL-style
Example 26 Easy
A is 6 years older than B. After 4 years, the sum of their ages will be 52. Find their present ages.
- A=B+6; A+B=44 → B=19
Answer: B = 19 years, A = 25 years
Example 27 Medium
The ratio of the ages of two sisters is 7:5. After 6 years, the ratio will be 4:3. Find their present ages.
- (7x+6)/(5x+6)=4/3 → x=6
Answer: 42 years and 30 yearsSSC MTS-style
Example 28 Easy
The sum of the ages of a man and his wife is 108. The man’s age is 7/5 times his wife’s age. Find their ages.
- 7x + 5x = 108 → x = 9
Answer: Man = 63 years, Wife = 45 years
Example 29 Easy
A boy’s age after 6 years will be three times his age 6 years ago. Find his present age.
- x+6=3(x−6) → x = 12
Answer: 12 yearsSSC GD Constable-style
Example 30 Medium
The ratio of the present ages of a father and son is 7:2. After 10 years, the ratio will be 9:4. Find their present ages.
- (7x+10)/(2x+10)=9/4 → x=5
Answer: Father = 35 years, Son = 10 years
Example 31 Easy
A mother’s present age is 4 times her daughter’s. After 16 years, the mother’s age will be twice the daughter’s. Find their present ages.
- 4D+16=2(D+16) → D=8
Answer: Daughter = 8 years, Mother = 32 yearsSSC CPO-style
Example 32 Medium
The sum of the ages of two brothers is 42, and one-third of one brother’s age equals one-fourth of the other’s age. Find their ages.
- A/3=B/4 → B=4A/3; A+4A/3=42 → A=18
Answer: 18 years and 24 yearsRailway (RRB)-style
Example 33 Easy
A man is 24 years older than his son. In 2 years, his age will be twice the age of his son. Find their present ages.
- M=S+24; M+2=2(S+2) → S=22
Answer: Son = 22 years, Father = 46 years
Example 39 Hard
The ratio of a father and son’s ages 10 years ago was 3:1. The ratio of their ages 10 years hence will be 2:1. Find their present ages.
- 3x+10, x+10 present; (3x+20)/(x+20)=2 → x=20
Answer: Father = 70 years, Son = 30 yearsBank Exam-style
Example 34 Medium
The ratio of the ages of A and B is 5:7. Six years hence, the ratio will be 3:4. Find their present ages.
- (5x+6)/(7x+6)=3/4 → x=6
Answer: A = 30 years, B = 42 years
Example 38 Hard
The sum of the present ages of a man and his wife is 84. Six years ago, the man was twice as old as his wife. Find their present ages.
- M+W=84; M−6=2(W−6) → W=30
Answer: Wife = 30 years, Man = 54 yearsState PSC-style
Example 35 Hard
The ages of two colleagues are in the ratio 8:5. Twelve years hence, the ratio will be 3:2. Find their present ages.
- (8x+12)/(5x+12)=3/2 → x=12
Answer: 96 years and 60 years
Example 40 Medium
A woman’s age, three years ago, was five times her daughter’s age at that time. The woman’s present age is 48. Find her daughter’s present age.
- W−3=45=5(D−3) → D=12
Answer: Daughter = 12 yearsPolice Recruitment-style
Example 36 Medium
A policeman is twice as old as his colleague. Fifteen years ago, he was three times as old. Find their present ages.
- P=2C; P−15=3(C−15) → C=30
Answer: Colleague = 30 years, Policeman = 60 years
Example 37 Easy
The sum of the ages of two police officers is 56. The elder is 8 years older than the younger. Find their ages.
- E=Y+8; Y+8+Y=56 → Y=24
Answer: 24 years and 32 years
12 – Exam-Oriented Tips
- In MCQ-based exams, back-solving from the options is often faster than forming equations from scratch.
- Always identify how many unknowns the question has before you start.
- For ratio questions, never skip writing the multiplier “x” — it is the single most common source of silly mistakes.
- Practise mental cross-multiplication, since almost every ratio-based age question ends in that step.
- Build speed with timed, exam-pattern practice sets, such as those on the CATMock aptitude platform.
Exam hack: If a question gives a ratio AND a future/past condition, convert the ratio into “ax, bx” first — this turns a two-unknown problem into a single-variable equation. Cat PYQ
13Practice Section
Before moving to the MCQs, try solving these unaided, then check your working against the solved examples above for method:
- The ratio of ages of two friends is 4:5. After 8 years, the ratio will be 5:6. Find their present ages.
- A mother is 26 years older than her son. In 6 years, her age will be 3 times the son’s age. Find their present ages.
- The sum of ages of three cousins in ratio 2:4:5 is 66. Find each age.
- Five years ago, a man’s age was 4 times his daughter’s age. Ten years hence, his age will be twice hers. Find their present ages.
- Two friends’ ages differ by 8 years. Eight years ago, the elder was twice as old as the younger. Find their present ages.
For a larger, continuously updated practice bank with instant scoring, the CATMock platform is a useful next step.
14 – 50 Age Aptitude MCQs with Answers
Tap any question to reveal the answer and explanation.Easy · Q1–Q20
Q1. A’s present age is 20 years. B is 5 years younger. Find B’s age.
A) 12 B) 15 C) 18 D) 20
Answer: B) 15 — 20 − 5 = 15.
Q2. A person’s present age is 25 years. What will be his age after 8 years?
A) 30 B) 31 C) 33 D) 35
Answer: C) 33 — 25 + 8 = 33.
Q3. Ravi’s age 3 years ago was 27. Find his present age.
A) 24 B) 28 C) 30 D) 33
Answer: C) 30 — 27 + 3 = 30.
Q4. The sum of the ages of two friends is 40. One of them is 22. Find the other’s age.
A) 16 B) 18 C) 20 D) 22
Answer: B) 18 — 40 − 22 = 18.
Q5. A father is 30 years older than his son, who is 10. Find the father’s age.
A) 35 B) 38 C) 40 D) 42
Answer: C) 40 — 10 + 30 = 40.
Q6. Two ages are in the ratio 2:3. If the smaller age is 16, find the larger.
A) 20 B) 22 C) 24 D) 26
Answer: C) 24 — x = 8, larger = 3×8 = 24.
Q7. A boy’s age 5 years ago was 12. Find his present age.
A) 15 B) 17 C) 18 D) 20
Answer: B) 17 — 12 + 5 = 17.
Q8. X is twice as old as Y. If Y is 9, find X’s age.
A) 16 B) 18 C) 20 D) 22
Answer: B) 18 — 2 × 9 = 18.
Q9. A person’s present age is 45. Find his age 15 years ago.
A) 25 B) 28 C) 30 D) 32
Answer: C) 30 — 45 − 15 = 30.
Q10. Three friends are aged 20, 25, and 30. Find their average age.
A) 22 B) 24 C) 25 D) 27
Answer: C) 25 — (20+25+30)/3 = 25.
Q11. A person’s present age is 33. Find his age after 7 years.
A) 38 B) 39 C) 40 D) 41
Answer: C) 40 — 33 + 7 = 40.
Q12. Twins’ combined age is 30. Find each twin’s age.
A) 12 B) 14 C) 15 D) 16
Answer: C) 15 — 30 ÷ 2 = 15.
Q13. The ratio of a father’s age to his son’s age is 5:1. If the son is 8, find the father’s age.
A) 35 B) 38 C) 40 D) 45
Answer: C) 40 — 5 × 8 = 40.
Q14. Two siblings’ ages differ by 6 years. The younger is 14. Find the elder’s age.
A) 18 B) 20 C) 22 D) 24
Answer: B) 20 — 14 + 6 = 20.
Q15. A is 4 years older than B. If B is 26, find A’s age.
A) 28 B) 29 C) 30 D) 32
Answer: C) 30 — 26 + 4 = 30.
Q16. Anita was born in 2000. Find her age in 2026.
A) 24 B) 25 C) 26 D) 27
Answer: C) 26 — 2026 − 2000 = 26.
Q17. The sum of the ages of a husband and wife is 70. If the husband is 38, find the wife’s age.
A) 30 B) 32 C) 34 D) 36
Answer: B) 32 — 70 − 38 = 32.
Q18. A grandmother is 3 times as old as her grandson, who is 12. Find the grandmother’s age.
A) 30 B) 33 C) 36 D) 40
Answer: C) 36 — 3 × 12 = 36.
Q19. A tree gains one ring per year. A tree with 45 rings is how old?
A) 40 B) 42 C) 44 D) 45
Answer: D) 45
Q20. Four cousins average 18 years each. Find the total of their ages.
A) 68 B) 70 C) 72 D) 74
Answer: C) 72 — 18 × 4 = 72.
Medium · Q21–Q35
Q21. The ratio of ages of P and Q is 4:5, and their sum is 54. Find Q’s age.
A) 24 B) 27 C) 30 D) 32
Answer: C) 30 — x = 6, Q = 5×6 = 30.
Q22. A’s age 5 years ago was 20. Find A’s age 5 years hence.
A) 28 B) 29 C) 30 D) 32
Answer: C) 30 — present = 25, hence = 30.
Q23. B’s present age is 15. A is now twice as old as B was 5 years ago. Find A’s age.
A) 18 B) 20 C) 22 D) 24
Answer: B) 20 — B 5 yrs ago = 10, A = 2×10.
Q24. Two ages are in ratio 3:4, and differ by 8. Find both ages.
A) 20, 28 B) 21, 28 C) 24, 32 D) 18, 24
Answer: C) 24, 32 — x = 8.
Q25. The sum of ages of a father and son is 50. The father’s age is 4 times the son’s. Find the son’s age.
A) 8 B) 9 C) 10 D) 12
Answer: C) 10 — 5x = 50 → x = 10.
Q26. After 6 years, the sum of two brothers’ ages will be 40. Find the current sum.
A) 24 B) 26 C) 28 D) 30
Answer: C) 28 — 40 − 12 = 28.
Q27. A person’s age 10 years hence will be twice his age 10 years ago. Find his present age.
A) 25 B) 28 C) 30 D) 32
Answer: C) 30 — x+10 = 2(x−10) → x = 30.
Q28. The ratio of a mother’s age to her daughter’s is 9:2. The mother is 45. Find the daughter’s age.
A) 8 B) 9 C) 10 D) 12
Answer: C) 10 — x = 5, daughter = 2×5.
Q29. Three sisters’ ages are in ratio 2:3:4, and sum to 45. Find the youngest’s age.
A) 8 B) 9 C) 10 D) 12
Answer: C) 10 — x = 5, youngest = 2×5.
Q30. A man’s age is 4 times his son’s. After 20 years, it will be twice the son’s age. Find the son’s present age.
A) 8 B) 10 C) 12 D) 14
Answer: B) 10 — 4x+20 = 2(x+20) → x = 10.
Q31. The ratio of present ages of A and B is 7:3. Six years ago, the ratio was 3:1. Find their present ages.
A) 36, 15 B) 40, 18 C) 42, 18 D) 45, 20
Answer: C) 42, 18 — x = 6.
Q32. Two students’ ages differ by 3, and sum to 35. Find their ages.
A) 15, 18 B) 16, 19 C) 17, 20 D) 14, 17
Answer: B) 16, 19
Q33. A father’s age is the square of his son’s age, who is 6. Find the father’s age.
A) 30 B) 32 C) 36 D) 42
Answer: C) 36 — 6² = 36.
Q34. The ratio of ages of an uncle and nephew is 5:2. The nephew is 14. Find the uncle’s age.
A) 30 B) 32 C) 35 D) 40
Answer: C) 35 — x = 7, uncle = 5×7.
Q35. X’s present age is thrice Y’s. Their ages sum to 48. Find Y’s age.
A) 10 B) 12 C) 14 D) 16
Answer: B) 12 — 4y = 48 → y = 12.Hard ·
Q36–Q50
Q36. The ratio of ages of A and B is 6:5. Four years hence, the ratio will be 7:6. Find their present ages.
A) 20, 16 B) 22, 18 C) 24, 20 D) 26, 22
Answer: C) 24, 20 — x = 4.
Q37. The sum of a man and his wife’s present ages is 84. Six years ago, the man was twice as old as his wife. Find their present ages.
A) 50, 34 B) 52, 32 C) 54, 30 D) 56, 28
Answer: C) 54, 30
Q38. A’s age 3 years hence will be 5 times B’s age 3 years ago. B’s present age is 8. Find A’s present age.
A) 18 B) 20 C) 22 D) 25
Answer: C) 22 — B 3 yrs ago = 5, A+3 = 25.
Q39. Two present ages are in ratio 5:3. Nine years ago, the ratio was 2:1. Find their present ages.
A) 40, 24 B) 42, 25 C) 45, 27 D) 48, 30
Answer: C) 45, 27 — x = 9.
Q40. The sum of the present ages of a father, mother, and son is 90, in ratio 3:2:1. Find the son’s age.
A) 12 B) 14 C) 15 D) 18
Answer: C) 15 — 6x = 90 → x = 15.
Q41. A’s present age is 5 years more than twice B’s age. Their combined age 10 years hence is 61. Find B’s present age.
A) 10 B) 11 C) 12 D) 13
Answer: C) 12 — A+B = 41, 3B+5 = 41.
Q42. Three years ago, the ratio of two ages was 4:5. Seven years hence, the ratio will be 5:6. Find their present ages.
A) 40, 50 B) 41, 51 C) 43, 53 D) 45, 55
Answer: C) 43, 53 — x = 10.
Q43. Two brothers’ ages differ by 2 years, and the sum of the squares of their ages is 100. Find their ages.
A) 5, 7 B) 6, 8 C) 7, 9 D) 8, 10
Answer: B) 6, 8 — x²+(x+2)² = 100 → x = 6.
Q44. Ten years ago, the ratio of a father’s and son’s ages was 3:1. Ten years hence, the ratio will be 2:1. Find their present ages.
A) 65, 25 B) 68, 28 C) 70, 30 D) 72, 32
Answer: C) 70, 30
Q45. A is 5 years older than B. A’s father F is twice as old as A. B is twice as old as his sister S, who is 10. Find F’s age.
A) 45 B) 48 C) 50 D) 52
Answer: C) 50 — B=20, A=25, F=50.
Q46. The average age of a family of 5 is 25. Removing the youngest raises the average to 27. Find the youngest’s age.
A) 15 B) 16 C) 17 D) 18
Answer: C) 17 — total 125, remaining 4 = 108.
Q47. A mother was 24 when her twin daughters were born. The sum of all three present ages is 60. Find each daughter’s age.
A) 10 B) 11 C) 12 D) 14
Answer: C) 12 — 3D+24 = 60 → D = 12.
Q48. Four years hence, the ratio of A and B’s ages will be 5:6. Four years ago, the ratio was 1:2. Find their present ages.
A) 4, 6 B) 5, 7 C) 6, 8 D) 8, 10
Answer: C) 6, 8
Q49. A is as old as B and C together. C is younger than B by 12 years. A’s age is 30. Find B’s age.
A) 18 B) 20 C) 21 D) 24
Answer: C) 21 — B+C=30, C=B−12.
Q50. Two present ages are in ratio 4:5. Fifteen years hence, the ratio will be 9:10. Find their present ages.
A) 10, 13 B) 12, 15 C) 14, 17 D) 16, 20
Answer: B) 12, 15
For extended, exam-simulated practice with detailed analytics, the CATMock question bank is a good next stop.
15 Frequently Asked Questions
What is the basic formula for solving problems on ages?
If a person’s present age is x, their age n years ago is (x − n) and n years hence is (x + n). Most problems combine this with ratio relationships (a:b becomes ax and bx) and are solved by forming and solving linear equations from the sentences given.
How do I solve age problems quickly for SSC exams?
Convert every ratio into “x” form immediately, write one equation per time-based condition, and solve simultaneously. For MCQs, back-solving from the given options is often faster than forming equations from scratch.
Why does the difference between two people’s ages never change?
Because both people age at exactly the same rate — one year passes for both simultaneously. While the ratio of their ages shrinks over time, the numerical difference stays fixed forever.What types of age questions are asked in SSC CGL?
SSC CGL typically asks ratio-based present age questions, combined past-future condition questions, and sum-and-ratio combination questions. Tier 2 papers sometimes include three-person or digit-reversal problems for higher difficulty.
How many age questions come in SSC exams?
Most SSC exams (CGL, CHSL, MTS, GD, CPO) include 1 to 2 questions on problems on ages in the Quantitative Aptitude section, making it a small but consistently scoring topic.
What is the shortcut for sum-and-ratio age problems?
If sum S and ratio a:b are given, one age equals S × [a/(a+b)] and the other equals S × [b/(a+b)]. This avoids writing a full equation and can be done in one line.
Can age problems have more than two unknowns?
Yes. Three-person age problems (grandparent, parent, child) are common, and are typically solved by expressing all three ages using a single ratio multiplier and one sum condition.
What is a digit-reversal age problem?
It’s a problem where reversing the digits of one person’s two-digit age gives another person’s age. Because reversing digits always changes the value by a multiple of 9, the age difference in such problems is always divisible by 9.
How is the “average age” type of question solved?
Average age = total sum of ages ÷ number of people. When a member is added or removed, first find the total age using the old average, adjust the total, and divide by the new count of people.
What is the most common mistake in age word problems?
The most frequent error is applying a time shift (“5 years ago” or “6 years hence”) to only one person’s age instead of both people’s ages in that condition. Time always passes equally for everyone mentioned.
Are problems on ages asked in bank exams too?
Yes. IBPS, SBI PO, and Clerk exams regularly include an age-based question, usually in ratio or sum-difference form, within the Quantitative Aptitude or Numerical Ability section.How do I identify how many equations I need?What’s the difference between “ago” and “hence” in age problems?Is algebra necessary, or can I do these mentally?How do I solve age problems where a product of ages is given?What is the best way to practice problems on ages before an exam?Do reasoning sections also test age-based logic?How can I verify my answer to an age problem?What age-related terms should I memorise for exams?Where can I find more solved age questions and mock tests?
16 – Final Revision Notes
Quick Summary
Problems on ages test your ability to translate time-based relationships into algebraic equations. Every question fits one of a small number of patterns — ratio, sum/difference, past-future, product, or multi-person — and each pattern has a predictable solving method.
Key Takeaways
- Present age ± n years gives past/future age.
- Ratios must always be written with a multiplier (ax, bx), never as raw numbers.
- The difference between two ages is constant; the ratio is not.
- Two unknowns require two independent equations.
- Verification by substitution catches almost all careless errors.
Exam Checklist
- Have I identified all unknowns?
- Have I applied every time shift (“ago”/”hence”) to all relevant people?
- Have I double-checked “ago” vs “hence” signs?
- Have I verified the final answer against every condition in the question?
- Am I answering the exact quantity the question asked for?
Common Confusions, Clarified
- Ratio vs. actual value: a ratio of 3:4 does not mean the ages are literally 3 and 4 — always attach the multiplier x.
- “As old as” vs. “older/younger than”: “as old as” means equal; “older/younger than” implies a numerical difference.
- Sum-of-ages vs. average age: average age = sum ÷ number of people, not the sum itself.
| Do | Don’t |
|---|---|
| Write ratios as ax, bx | Treat ratio numbers as literal ages |
| Apply time shifts to every person in that condition | Shift only one person’s age |
| Verify your final answer | Submit without checking |
| Read “ago” and “hence” carefully | Mix up past and future signs |
17 – Conclusion
Problems on Ages might look intimidating in paragraph form, but as this guide has shown, every variation — ratio-based, sum-based, past-future combinations, product-based, or multi-person — reduces to the same core skill: translating a sentence into an equation and solving it carefully. With the formulas, shortcuts, 40+ solved examples, and 50 MCQs covered here, you now have a complete, exam-ready foundation for age wale question patterns across SSC CGL, SSC CHSL, SSC MTS, SSC GD, SSC CPO, Railway, Bank, State PSC, and Police recruitment exams.
The next step is repetition under timed conditions. Revisit the formula cheat sheet before every mock test, work through fresh question sets regularly, and track which question type trips you up most often. For continued, exam-simulated practice, structured question banks such as CATMock can help keep building speed and accuracy. CAT PYQ
