Problems on Ages for SSC, RRB and Banking 2026: formulas, unit method trick, 14 question models, 56 solved questions, study plan and FAQs.
Problems on Ages for SSC, RRB & Banking 2026
Formulas, Unit Method Trick, 14 Question Models, 56 Solved Questions & FAQs
Problems on Ages is one of those chapters that looks like a word puzzle but is really just ratio wearing a different shirt. In SSC CGL, CHSL, MTS, RRB NTPC, RRB Group D and bank exams you usually get one direct age question, and it is one of the few quant questions you can finish in under a minute if you know the pattern. Many students from Telugu medium tell me the maths is fine but the English sentence (“five years hence”, “thrice as old as”) slows them down. So this post first fixes the language and the basic rules, then gives you the one trick that solves most questions (the unit method), and then 56 solved questions arranged in 14 models, each with four options and clear steps. If you have not done ratios yet, finish our Ratio and Proportion post first; ages will feel much easier after that.
✔ Read fully if: you are new to arithmetic or you lose time turning age sentences into equations.
✔ Jump to Models 7–14 if: you are comfortable with “times” and ratio questions and want averages, products, digits and family puzzles.
✔ Why it matters: it is a near-sure mark in most papers, and the same unit method also helps in ratio, partnership and mixture questions.
Why Problems on Ages Matter in Govt Exams 2026
The quant section size below is from the standard exam patterns. The share of age questions is our own rough estimate from practice papers, so read it as approx. only.
| Exam | Maths / Quant questions | Problems on Ages (approx.) |
|---|---|---|
| SSC CGL / CHSL Tier-I | 25 | Approx. 0–1 direct, sometimes mixed with ratio or average |
| SSC MTS / GD | 20 | Approx. 1 (simple “times” or ratio type) |
| RRB NTPC CBT-1 | 30 | Approx. 1–2 |
| RRB Group D CBT | 25 | Approx. 1–2 |
| IBPS / SBI PO & Clerk Prelims | 35 | Approx. 0–1 in arithmetic, sometimes inside a DI or caselet |
| TGPSC / APPSC Groups, police exams | Varies by post | Arithmetic sections regularly include age questions |
Problems on Ages Concepts and Basic Formulas
Turning English into maths
| Words in the question | What you write (present age = x) |
|---|---|
| n years ago / n years back / n years before | x − n |
| n years hence / after n years / n years from now | x + n |
| A is k times as old as B | A = k × B |
| A is twice / thrice as old as B | A = 2B / A = 3B |
| Ages are in the ratio a : b | ax and bx |
| A is n years older than B | A = B + n (difference = n) |
| Average age of m people is p | Total age = m × p |
| A was n years old when B was born | A − B = n (always) |
Three rules that never change
- The age difference between two people stays the same forever. If a father is 28 years older than his son today, he was 28 years older 10 years ago and will be 28 years older 20 years later.
- The ratio of ages changes with time (it moves closer to 1 as both grow older). So never use today’s ratio for a past or future time.
- When n years pass, the total age of m people goes up by m × n. Four people, 5 years later → total grows by 20, not 5.
The unit method (fastest trick for ratio-based age questions)
Because the difference never changes, make the difference of ratio parts equal in both ratios. Then compare how many units changed with how many years passed.
- Write the two ratios for the two times.
- Multiply so that the difference of the two parts is the same in both.
- Change in one part (in units) = gap in years → value of 1 unit.
- Multiply back to get the ages.
4 : 1 (difference 3) → 8 : 2 | 3 : 1 (difference 2) → 9 : 3. Both differences are now 6.
8 → 9 is 1 unit and it took 6 years, so 1 unit = 6 → A = 48, B = 12. Check: after 6 years 54 = 3 × 18 ✔
“k times as old” vs “k times older than”
“5 times as old as B” means A = 5B. Strictly speaking, “5 times older than B” means 5B more than B, which is 6B. Exam setters sometimes use the second wording loosely, so read the options: if 5B fits and 6B does not, the setter meant “5 times as old”.
Problems on Ages Questions: 14 Models with Solved Examples
Each question has four options and a full solution. Try it yourself first with a timer (aim for 40–60 seconds), then read the steps. The correct option is marked ✔.
Model 1: Basic Present, Past and Future Age
Take the present age as x. Then the age n years ago is x − n and the age n years from now is x + n. Write one simple equation from the sentence and solve it.
Q1. Sunil is 14 years old and his uncle is 3 times as old as Sunil. After how many years will the uncle be twice as old as Sunil?
(a) 10 (b) 12 (c) 14 ✔ (d) 16
Solution:
- Uncle now = 3 × 14 = 42.
- After n years: 42 + n = 2(14 + n) = 28 + 2n.
- n = 42 − 28 = 14 years.
Answer: (c) 14
Q2. The sum of the present ages of Arjun and Bharath is 52 years. Arjun is 8 years older than Bharath. What was Bharath’s age 6 years ago?
(a) 14 (b) 16 ✔ (c) 18 (d) 22
Solution:
- Bharath = (52 − 8) ÷ 2 = 22.
- 6 years ago = 22 − 6 = 16.
Answer: (b) 16
Q3. Five years ago Meena was 19 years old. Her brother is 6 years younger than her. What will be her brother’s age 8 years from now?
(a) 24 (b) 26 ✔ (c) 28 (d) 30
Solution:
- Meena now = 19 + 5 = 24.
- Brother now = 24 − 6 = 18.
- After 8 years = 18 + 8 = 26.
Answer: (b) 26
Q4. Teja’s age 12 years from now will be 3 times his age 4 years ago. What is Teja’s present age?
(a) 10 (b) 12 ✔ (c) 14 (d) 16
Solution:
- x + 12 = 3(x − 4) = 3x − 12.
- 2x = 24 → x = 12.
- Check: 24 = 3 × 8 ✔
Answer: (b) 12
Q5. Twenty years from now, Lakshmi’s age will be 3 times what it was 4 years ago. Find her present age.
(a) 14 (b) 16 ✔ (c) 18 (d) 20
Solution:
- x + 20 = 3(x − 4) → x + 20 = 3x − 12.
- 2x = 32 → x = 16.
- Check: 36 = 3 × 12 ✔
Answer: (b) 16
Model 2: “Times” Relation: Now and Later / Earlier
Sentences like “A is 5 times as old as B” give A = 5B. When a second time is mentioned, add or subtract the same years for both people and form a second equation. You can also use the unit method because the age difference never changes.
Q6. A father is 5 times as old as his son. After 8 years he will be 3 times as old as his son. What is the father’s present age?
(a) 36 (b) 40 ✔ (c) 44 (d) 48
Solution:
- Let son = S, father = 5S.
- After 8 years: 5S + 8 = 3(S + 8) → 2S = 16 → S = 8.
- Father = 5 × 8 = 40. Check: 48 = 3 × 16 ✔
Answer: (b) 40
Q7. Rahul is 6 times as old as Anil. Four years from now, Rahul will be 4 times as old as Anil. What is the difference between their present ages?
(a) 24 (b) 30 ✔ (c) 32 (d) 36
Solution:
- 6A + 4 = 4(A + 4) → 2A = 12 → A = 6.
- Rahul = 36, difference = 36 − 6 = 30.
Answer: (b) 30
Q8. A mother told her son, “Five years ago I was 7 times as old as you were. Five years from now I will be 3 times as old as you will be.” Find their present ages (mother, son).
(a) 40, 10 ✔ (b) 42, 12 (c) 45, 15 (d) 36, 8
Solution:
- M − 5 = 7(S − 5) → M = 7S − 30.
- M + 5 = 3(S + 5) → M = 3S + 10.
- 7S − 30 = 3S + 10 → S = 10, M = 40. Check: 35 = 7 × 5, 45 = 3 × 15 ✔
Answer: (a) 40, 10
Q9. Six years ago, Kiran was 4 times as old as Divya. Twelve years from now, Kiran will be twice as old as Divya. Find the sum of their present ages.
(a) 51 (b) 54 (c) 57 ✔ (d) 60
Solution:
- K = 4D − 18 and K = 2D + 12.
- 4D − 18 = 2D + 12 → D = 15, K = 42.
- Sum = 57. Check: 36 = 4 × 9, 54 = 2 × 27 ✔
Answer: (c) 57
Q10. Four years ago, Ravi was 3 times as old as his sister. Eight years from now, he will be twice as old as her. Find their present ages (Ravi, sister).
(a) 40, 16 ✔ (b) 36, 12 (c) 44, 18 (d) 32, 14
Solution:
- 4 years ago 3 : 1 (difference 2 units). 8 years later 2 : 1 (difference 1 unit) → 4 : 2.
- From 3 → 4 is 1 unit and the time gap is 4 + 8 = 12 years, so 1 unit = 12.
- 4 years ago: 36 and 12 → now 40 and 16.
Answer: (a) 40, 16
Model 3: Ratio of Ages Changes with Time (Unit Method)
Two ratios are given for two different times. The age difference between two people is fixed, so if both ratios have the same difference of parts, the change in each part equals the years passed.
Q11. The present ages of P and Q are in the ratio 4 : 5. After 6 years the ratio will be 5 : 6. R is 5 years older than P. What is R’s present age?
(a) 27 (b) 29 ✔ (c) 31 (d) 33
Solution:
- 4 : 5 → 5 : 6, difference is 1 unit in both.
- Each part grew by 1 unit in 6 years → 1 unit = 6.
- P = 4 × 6 = 24, R = 24 + 5 = 29.
Answer: (b) 29
Q12. Three years ago, the ratio of ages of A and B was 5 : 3. Seven years from now, it will be 7 : 5. Find the sum of their present ages.
(a) 42 (b) 44 (c) 46 ✔ (d) 48
Solution:
- 5 : 3 → 7 : 5, difference 2 units in both.
- Change = 2 units in 3 + 7 = 10 years → 1 unit = 5.
- 3 years ago: 25 and 15 → now 28 and 18 → sum 46.
Answer: (c) 46
Q13. Eight years ago, the ratio of ages of Sai and Vamsi was 5 : 4. The ratio of their present ages is 7 : 6. What will be the sum of their ages 4 years from now?
(a) 56 (b) 58 (c) 60 ✔ (d) 64
Solution:
- 5 : 4 → 7 : 6, difference 1 unit in both.
- Change = 2 units in 8 years → 1 unit = 4.
- Now 28 and 24 → after 4 years 32 + 28 = 60.
Answer: (c) 60
Q14. Twelve years ago, the ratio of ages of X and Y was 2 : 5. Four years ago, it was 1 : 2. What will be the ratio of their ages 8 years from now?
(a) 3 : 5 ✔ (b) 4 : 7 (c) 2 : 3 (d) 5 : 8
Solution:
- 2 : 5 has difference 3; 1 : 2 has difference 1, so write it as 3 : 6.
- 2 → 3 is 1 unit in 8 years → 1 unit = 8.
- 12 years ago 16 and 40 → now 28 and 52 → after 8 years 36 : 60 = 3 : 5.
Answer: (a) 3 : 5
Q15. The present ages of a husband and wife are in the ratio 7 : 6. After 9 years the ratio will be 8 : 7. At the time of their marriage the ratio was 6 : 5. How many years ago did they get married?
(a) 6 (b) 8 (c) 9 ✔ (d) 12
Solution:
- 7 : 6 → 8 : 7 (difference 1 unit) → 1 unit = 9 → now 63 and 54, difference 9.
- At marriage 6 : 5, difference 1 unit = 9 → 54 and 45.
- 63 − 54 = 9 years ago.
Answer: (c) 9
Q16. Five years ago, the ratio of ages of A and B was 3 : 4. Ten years from now, it will be 6 : 7. C is 4 years younger than A. Find C’s present age.
(a) 14 (b) 16 ✔ (c) 18 (d) 20
Solution:
- 3 : 4 → 6 : 7, difference 1 unit in both.
- Change = 3 units in 15 years → 1 unit = 5.
- 5 years ago A = 15 → now A = 20 → C = 16.
Answer: (b) 16
Model 4: Find the Ratio of Present or Future Ages
Here the question gives some clue (times relation, average or difference) and asks for a ratio instead of an age. Find the actual ages first, then simplify the ratio.
Q17. Ten years ago, a father was 3 times as old as his son. Ten years from now, the father will be twice as old as his son. What is the ratio of their present ages?
(a) 7 : 3 ✔ (b) 5 : 2 (c) 2 : 1 (d) 9 : 4
Solution:
- 10 years ago 3 : 1 (difference 2) and 10 years later 2 : 1 → 4 : 2 (difference 2).
- 3 → 4 is 1 unit in 20 years → 1 unit = 20.
- 10 years ago 60 and 20 → now 70 and 30 → 7 : 3.
Answer: (a) 7 : 3
Q18. The present ages of a man and his son are in the ratio 8 : 3, and the average of their present ages is 33 years. What will be the ratio of their ages after 6 years?
(a) 9 : 4 ✔ (b) 8 : 3 (c) 7 : 3 (d) 2 : 1
Solution:
- Sum = 2 × 33 = 66 → 11 units = 66 → 1 unit = 6.
- Man 48, son 18 → after 6 years 54 : 24 = 9 : 4.
Answer: (a) 9 : 4
Q19. The ratio of present ages of X and Y is 4 : 7, and X is 15 years younger than Y. What will be the ratio of their ages after 10 years?
(a) 2 : 3 ✔ (b) 3 : 4 (c) 4 : 7 (d) 5 : 8
Solution:
- Difference = 3 units = 15 → 1 unit = 5.
- X = 20, Y = 35 → after 10 years 30 : 45 = 2 : 3.
Answer: (a) 2 : 3
Q20. The present ages of A and B are in the ratio 5 : 7. Four years ago, the sum of their ages was 40. What will be the ratio of their ages 6 years from now?
(a) 13 : 17 ✔ (b) 5 : 7 (c) 3 : 4 (d) 11 : 15
Solution:
- Present sum = 40 + 4 + 4 = 48 → 12 units = 48 → 1 unit = 4.
- A = 20, B = 28 → after 6 years 26 : 34 = 13 : 17.
Answer: (a) 13 : 17
Model 5: Sum of Ages + a “Times” Condition
The sum gives one equation (F + S = total) and the times condition gives another. Put F = total − S into the second one and solve.
Q21. The sum of the present ages of a father and his son is 56 years. After 4 years, the father will be 3 times as old as the son. Find the son’s present age.
(a) 10 (b) 12 ✔ (c) 14 (d) 16
Solution:
- (56 − S) + 4 = 3(S + 4) → 60 − S = 3S + 12.
- 4S = 48 → S = 12. Father = 44. Check: 48 = 3 × 16 ✔
Answer: (b) 12
Q22. The sum of the present ages of a mother and daughter is 50 years. Five years ago, the mother was 4 times as old as the daughter. What will be the daughter’s age 7 years from now?
(a) 18 (b) 20 ✔ (c) 22 (d) 24
Solution:
- (50 − D) − 5 = 4(D − 5) → 45 − D = 4D − 20.
- 5D = 65 → D = 13. After 7 years = 20.
Answer: (b) 20
Q23. The sum of the present ages of two brothers is 38 years. Three years ago, the elder brother was 3 times as old as the younger one. What will be the ratio of their ages after 5 years?
(a) 2 : 1 ✔ (b) 9 : 4 (c) 5 : 2 (d) 3 : 1
Solution:
- E − 3 = 3(Y − 3) → E = 3Y − 6.
- 3Y − 6 + Y = 38 → Y = 11, E = 27.
- After 5 years: 32 : 16 = 2 : 1.
Answer: (a) 2 : 1
Q24. The sum of the present ages of a mother and daughter is 52 years. Four years from now, the daughter’s age will be one-third of the mother’s age. What is the mother’s present age?
(a) 38 (b) 40 (c) 41 ✔ (d) 44
Solution:
- 3(D + 4) = M + 4 and M = 52 − D.
- 3D + 12 = 56 − D → D = 11, M = 41.
- Check: after 4 years 15 and 45 ✔
Answer: (c) 41
Model 6: Age Difference Is Given
When the difference is given directly, it is the strongest clue, because it stays the same at every point of time. Match it with the difference of ratio parts.
Q25. The difference between the ages of a father and son is 28 years. Five years ago, the father was 5 times as old as the son. Find their present ages (father, son).
(a) 40, 12 ✔ (b) 42, 14 (c) 38, 10 (d) 44, 16
Solution:
- Five years ago 5 : 1 → difference 4 units = 28 → 1 unit = 7.
- Five years ago 35 and 7 → now 40 and 12.
Answer: (a) 40, 12
Q26. Anand is 12 years older than Bhanu. After 6 years, Anand will be twice as old as Bhanu. What is Bhanu’s present age?
(a) 6 ✔ (b) 8 (c) 10 (d) 12
Solution:
- After 6 years, 2 : 1 → difference 1 unit = 12.
- Bhanu after 6 years = 12 → now 6.
Answer: (a) 6
Q27. The present ages of two sisters are in the ratio 7 : 4 and the difference of their ages is 18 years. What will be the sum of their ages after 5 years?
(a) 66 (b) 72 (c) 76 ✔ (d) 80
Solution:
- 3 units = 18 → 1 unit = 6 → ages 42 and 24.
- After 5 years: 47 + 29 = 76.
Answer: (c) 76
Q28. The difference between the ages of two cousins is one-fourth of the sum of their ages. If the elder is 25 years old, how old is the younger one?
(a) 12 (b) 15 ✔ (c) 18 (d) 20
Solution:
- 25 − Y = (25 + Y)/4 → 100 − 4Y = 25 + Y.
- 5Y = 75 → Y = 15.
Answer: (b) 15
Model 7: Average-Based Age Questions
Average × number of people = total age. When n years pass, the total of m people goes up by m × n. A new member joining or leaving changes the total by that person’s age.
Q29. Four years ago, the average age of A, B and C was 26 years. D joins them now and the average age of all four becomes 31 years. What is D’s present age?
(a) 30 (b) 32 (c) 34 ✔ (d) 36
Solution:
- Total 4 years ago = 3 × 26 = 78 → now 78 + 3 × 4 = 90.
- Total of four now = 4 × 31 = 124.
- D = 124 − 90 = 34.
Answer: (c) 34
Q30. The average age of a family of 5 members is 24 years. The youngest member is 4 years old. What was the average age of the family just before the youngest member was born?
(a) 20 (b) 22 (c) 25 ✔ (d) 28
Solution:
- Present total = 5 × 24 = 120.
- 4 years ago the other 4 members together were 120 − 4 − 4 × 4 = 100.
- Average = 100 ÷ 4 = 25.
Answer: (c) 25
Q31. The average age of 30 students in a class is 14 years. When the teacher’s age is included, the average becomes 15 years. What is the teacher’s age?
(a) 40 (b) 44 (c) 45 ✔ (d) 48
Solution:
- Students total = 30 × 14 = 420.
- With teacher = 31 × 15 = 465.
- Teacher = 465 − 420 = 45.
Answer: (c) 45
Q32. Seven years ago, at the time of their marriage, the average age of a husband and wife was 25 years. Today the average age of the husband, wife and their child is 22 years. How old is the child?
(a) 1 (b) 2 ✔ (c) 3 (d) 4
Solution:
- At marriage total = 50 → now 50 + 2 × 7 = 64.
- Family total now = 3 × 22 = 66.
- Child = 66 − 64 = 2 years.
Answer: (b) 2
Q33. The average age of A, B and C is 30 years. If A is left out, the average age of B and C is 27 years. What is A’s age?
(a) 32 (b) 34 (c) 36 ✔ (d) 38
Solution:
- Total of three = 90, total of B and C = 54.
- A = 90 − 54 = 36.
Answer: (c) 36
Model 8: After How Many Years / How Many Years Ago
Here the ages are known (or easy to find) and the question asks for the number of years n. Put x + n or x − n for both people and solve for n.
Q34. A father is 38 years old and his son is 10. After how many years will the father be twice as old as his son?
(a) 14 (b) 16 (c) 18 ✔ (d) 20
Solution:
- 38 + n = 2(10 + n) → n = 38 − 20 = 18.
- Check: 56 = 2 × 28 ✔
Answer: (c) 18
Q35. A man is 42 and his son is 14. How many years ago was the man 5 times as old as his son?
(a) 5 (b) 6 (c) 7 ✔ (d) 8
Solution:
- 42 − n = 5(14 − n) → 4n = 28 → n = 7.
- Check: 35 = 5 × 7 ✔
Answer: (c) 7
Q36. Ravi’s father is 4 times as old as Ravi. After 6 years he will be 3 times as old as Ravi. After how many years from now will the father be twice as old as Ravi?
(a) 18 (b) 20 (c) 24 ✔ (d) 30
Solution:
- 4R + 6 = 3(R + 6) → R = 12, father = 48.
- 48 + n = 2(12 + n) → n = 24.
Answer: (c) 24
Q37. A father is 4 times as old as his son. After 4 years, he will be 3 times as old as his son. Eight years from now, how many times the son’s age will the father’s age be?
(a) 2 times (b) 2 1/4 times (c) 2 1/2 times ✔ (d) 3 times
Solution:
- 4S + 4 = 3(S + 4) → S = 8, father = 32.
- After 8 years: 40 and 16 → 40 ÷ 16 = 2 1/2 times.
Answer: (c) 2 1/2 times
Model 9: Fraction, Percentage and Marriage-Based Questions
Words like “1⅓ times”, “25% more” or “half of” turn into simple fractions. Write the age at marriage (or the base age) as x and convert the sentence into one equation.
Q38. Lata got married 8 years ago. Her present age is 1⅓ times her age at the time of marriage. Her daughter’s age is one-eighth of Lata’s present age. How old is the daughter?
(a) 3 (b) 4 ✔ (c) 5 (d) 6
Solution:
- x + 8 = 4x/3 → 3x + 24 = 4x → x = 24.
- Present age = 32 → daughter = 32 ÷ 8 = 4.
Answer: (b) 4
Q39. Raghu’s age is 25% more than his wife’s age. The sum of their ages is 81 years. What is the wife’s age?
(a) 32 (b) 34 (c) 36 ✔ (d) 40
Solution:
- Raghu = 5W/4 → W + 5W/4 = 9W/4 = 81.
- W = 81 × 4 ÷ 9 = 36 (Raghu 45).
Answer: (c) 36
Q40. Half of A’s age is equal to two-thirds of B’s age, and A is 8 years older than B. What is B’s age?
(a) 18 (b) 24 ✔ (c) 30 (d) 32
Solution:
- A/2 = 2B/3 → 3A = 4B, so A : B = 4 : 3.
- Difference 1 unit = 8 → B = 3 × 8 = 24.
Answer: (b) 24
Q41. Ten years from now, Sita’s age will be one and a half times her age ten years ago. What is Sita’s present age?
(a) 40 (b) 45 (c) 50 ✔ (d) 55
Solution:
- x + 10 = 3(x − 10)/2 → 2x + 20 = 3x − 30.
- x = 50. Check: 60 = 1.5 × 40 ✔
Answer: (c) 50
Model 10: Product of Ages
When the product is given, think of factor pairs instead of solving a quadratic. Ages are whole numbers in most exam questions, so a quick factor check finds the pair.
Q42. The present ages of two brothers differ by 4 years and the product of their ages is 252. What is the younger brother’s age?
(a) 12 (b) 14 ✔ (c) 16 (d) 18
Solution:
- Look for factors of 252 that differ by 4: 14 × 18 = 252.
- Younger = 14.
Answer: (b) 14
Q43. The ages of two friends are in the ratio 4 : 5 and the product of their ages is 980. What is the sum of their ages?
(a) 54 (b) 60 (c) 63 ✔ (d) 72
Solution:
- 4k × 5k = 20k² = 980 → k² = 49 → k = 7.
- Ages 28 and 35 → sum 63.
Answer: (c) 63
Q44. A father is 4 times as old as his son. Four years ago, the product of their ages was 216. What is the father’s present age?
(a) 36 (b) 40 ✔ (c) 44 (d) 48
Solution:
- (4S − 4)(S − 4) = 216 → (S − 1)(S − 4) = 54.
- 9 × 6 = 54 → S = 10, father = 40.
- Check: 36 × 6 = 216 ✔
Answer: (b) 40
Model 11: Age at Birth of a Child (Birth-Based)
The age of a parent when a child was born is simply the age difference between them, and it never changes. Use that fixed gap to move between people.
Q45. Ramu was 26 years old when his son was born. The son’s mother is 4 years younger than Ramu. When the son is 15 years old, what will be the mother’s age?
(a) 35 (b) 37 ✔ (c) 39 (d) 41
Solution:
- Mother was 26 − 4 = 22 when the son was born.
- When the son is 15, mother = 22 + 15 = 37.
Answer: (b) 37
Q46. A father was 30 years old when his elder son was born and 34 when his younger son was born. The mother is 3 years younger than the father. When the younger son is 10, what is the total age of the family of four?
(a) 105 (b) 107 (c) 109 ✔ (d) 111
Solution:
- Younger son 10 → father 44, mother 41.
- Elder son is 4 years older than the younger → 14.
- Total = 44 + 41 + 14 + 10 = 109.
Answer: (c) 109
Q47. Priya’s brother is 4 years older than her, and her sister is 3 years younger than her. Their father was 30 when the sister was born, and he is 52 now. How old is the brother?
(a) 25 (b) 27 (c) 29 ✔ (d) 31
Solution:
- Sister = 52 − 30 = 22.
- Priya = 22 + 3 = 25, brother = 25 + 4 = 29.
Answer: (c) 29
Model 12: Family and Multi-Person Puzzles
Three or four people are linked by “older than”, “times as old” and “sum” clues. Pick the person who appears in the most clues, call that age x, and write everyone else in terms of x.
Q48. A is 3 years older than B, and B is 3 times as old as C. The total of the ages of A, B and C is 45 years. How old is B?
(a) 15 (b) 18 ✔ (c) 21 (d) 24
Solution:
- C = x, B = 3x, A = 3x + 3.
- 7x + 3 = 45 → x = 6.
- B = 18 (A = 21, C = 6).
Answer: (b) 18
Q49. Vijay is 6 years older than his friend Arun. Arun’s mother is 3 times as old as Arun, and Vijay is 3 times as old as his sister Priya. The ages of Arun’s mother and Priya differ by 62 years. What is the age of Arun’s mother?
(a) 66 (b) 69 (c) 72 ✔ (d) 75
Solution:
- Arun = A, Vijay = A + 6, mother = 3A, Priya = (A + 6)/3.
- 3A − (A + 6)/3 = 62 → 9A − A − 6 = 186 → A = 24.
- Mother = 72 (Priya = 10, 72 − 10 = 62 ✔).
Answer: (c) 72
Q50. The combined age of a husband, wife and son is 90 years. The husband is 4 years older than the wife. Eight years from now, the wife’s age will be twice the son’s age. What is the wife’s present age?
(a) 32 (b) 34 (c) 36 ✔ (d) 38
Solution:
- H = W + 4 → 2W + S = 86. W + 8 = 2(S + 8) → W = 2S + 8.
- 2(2S + 8) + S = 86 → S = 14.
- W = 36, H = 40.
Answer: (c) 36
Q51. At present A is 3 times as old as B. C is 8 years older than B, and the sum of the ages of B and C is 44 years. What will be the ratio of A’s age to C’s age 6 years from now?
(a) 15 : 8 ✔ (b) 2 : 1 (c) 27 : 13 (d) 9 : 5
Solution:
- B + (B + 8) = 44 → B = 18, C = 26, A = 54.
- After 6 years: 60 : 32 = 15 : 8.
Answer: (a) 15 : 8
Q52. A grandfather is 9 times as old as his grandson and is 56 years older than him. What is the grandfather’s age?
(a) 56 (b) 60 (c) 63 ✔ (d) 70
Solution:
- Difference = 9g − g = 8g = 56 → g = 7.
- Grandfather = 63.
Answer: (c) 63
Model 13: Digits Reversed Ages
If one age is a two-digit number 10a + b and the other has the digits reversed (10b + a), the sum is 11(a + b) and the difference is 9(a − b). These two facts solve most such questions in seconds.
Q53. A father’s age has the same digits as his son’s age, but reversed. Two years from now, the father will be twice as old as the son. What is the father’s present age?
(a) 42 (b) 52 ✔ (c) 62 (d) 72
Solution:
- Father = 10a + b, son = 10b + a. Option check is fastest.
- 52 and 25 → after 2 years 54 and 27 → 54 = 2 × 27 ✔
- Other options fail (e.g. 62, 26 → 64 and 28).
Answer: (b) 52
Q54. The ages of two persons have the same two digits in reverse order. The sum of their ages is 99 and the difference is 27. What is the elder person’s age?
(a) 54 (b) 63 ✔ (c) 72 (d) 81
Solution:
- 11(a + b) = 99 → a + b = 9.
- 9(a − b) = 27 → a − b = 3.
- a = 6, b = 3 → ages 63 and 36.
Answer: (b) 63
Model 14: Algebra-Form and Option-Check Questions
Some questions use letters (p years, x years) instead of numbers, and some are easiest by testing the options. Treat letters like numbers and keep the same steps.
Q55. A father is 3 times as old as his son. p years ago, the father was 5 times as old as the son. What is the son’s present age?
(a) p (b) 2p ✔ (c) 3p (d) 4p
Solution:
- 3S − p = 5(S − p) → 3S − p = 5S − 5p.
- 2S = 4p → S = 2p.
- Check with p = 4: son 8, father 24; 4 years ago 20 = 5 × 4 ✔
Answer: (b) 2p
Q56. The present ages of two colleagues are in the ratio 3 : 4. Six years ago the ratio was 5 : 7. Which of these are their present ages?
(a) 30, 40 (b) 36, 48 ✔ (c) 42, 56 (d) 24, 32
Solution:
- Test options: 36 and 48 → 6 years ago 30 and 42 → 30 : 42 = 5 : 7 ✔
- Unit method: 3 : 4 → 6 : 8 (difference 2), 5 : 7 (difference 2) → 1 unit = 6 → 36 and 48.
Answer: (b) 36, 48
How Many Days to Complete Problems on Ages?
Short answer: 4 to 6 days for a beginner studying about 2 hours a day, and 2 days if you have done it before and only need practice. Ages is a small chapter, but only after ratio is clear. If ratio is still shaky, add 2–3 days for that first. The plan below covers all 14 models and leaves the last day for a timed test.
| Day | Models | What to cover | Practice target |
|---|---|---|---|
| Day 1 | Concepts + Models 1–2 | English-to-maths table, three rules, basic and “times” questions | 25 Qs |
| Day 2 | Models 3–4 | Unit method, ratio changes with time, ratio of present/future ages | 30 Qs |
| Day 3 | Models 5–8 | Sum + times, difference given, average-based, “after how many years” | 30 Qs |
| Day 4 | Models 9–12 | Fractions and marriage, product of ages, birth-based, family puzzles | 25 Qs |
| Day 5 | Models 13–14 + error notebook | Digits reversed, letter answers, option check; redo all wrong answers | 20 Qs |
| Day 6 | Timed test + revision | Previous-year pattern questions from official papers or mocks, 15 Qs in 12 minutes, then the formula box | 15–20 Qs + review |
| Your level | Time per day | Days needed |
|---|---|---|
| Complete beginner (ratio not done) | 2 hours | 7–8 days (3 days ratio + 4–5 days ages) |
| Knows ratio, new to ages | 2 hours | 4–6 days |
| Revising (done it before) | 1–1.5 hours | 2 days (Days 3–6 rows quickly) |
| Working professional | 1 hour | 8–10 days (one row every 1.5 days) |
Do not skip: After finishing, add 3–4 age questions to every weekly mixed test for a month. It is an easy mark, so do not let it go rusty.
Problems on Ages Formulas: Quick Revision Box
| Concept | Formula / Rule |
|---|---|
| Age n years ago / after n years | x − n / x + n |
| Ages in ratio a : b | ax and bx |
| Age difference | Same at every point of time |
| Unit method | Make difference of parts equal; 1 unit = years passed ÷ change in units |
| Total age of m people after n years | Old total + m × n |
| Average to total | Total = average × number of people |
| After n years A = k × B | n = (A − kB) ÷ (k − 1) |
| n years ago A = k × B | n = (kB − A) ÷ (k − 1) |
| A now = k × B, after n years A = m × B | B = n(m − 1) ÷ (k − m) |
| A now = k × B, n years ago A = m × B | B = n(m − 1) ÷ (m − k) |
| Ratio a : b and product P | k = √(P ÷ ab) |
| Reversed digits 10a + b and 10b + a | Sum = 11(a + b), difference = 9(a − b) |
| Parent’s age at child’s birth | = parent − child (never changes) |
| “x% more than B” | B × (100 + x) ÷ 100 |
Tips for Telugu-Medium Students
- Learn the words once: Age = เฐตเฐฏเฐธ్เฐธు (vayassu), years ago = เฐธంเฐตเฐค్เฐธเฐฐాเฐฒ เฐ్เฐฐిเฐคం, after / hence = เฐคเฐฐ్เฐตాเฐค, times = เฐฐెเฐ్เฐฒు (retlu), ratio = เฐจిเฐท్เฐชเฐค్เฐคి. “Thrice” = 3 เฐฐెเฐ్เฐฒు, “twice” = 2 เฐฐెเฐ్เฐฒు.
- Make a mini time table: write three columns “Past | Now | Future” and fill each person’s age. Most mistakes vanish once the times are in separate columns.
- Tick the time of every ratio. Before solving, mark whether a ratio belongs to the past, present or future. Using the wrong time is the number one error.
- Use options without fear. Age answers are whole numbers. Putting an option back into both conditions takes 15 seconds and is fully valid in the exam.
- Practise in English even if you think in Telugu. RRB and some state exams offer Telugu, but SSC CGL Tier-I is in English and Hindi only. Bilingual practice keeps both doors open.
Problems on Ages – Frequently Asked Questions
1. What are problems on ages in maths?
They are arithmetic questions where you find a person’s present, past or future age from clues like ratios, “times as old”, sums, differences or averages. They are really simple linear equations written as a story.
2. What is the basic formula for age problems?
If the present age is x, the age n years ago is x − n and the age after n years is x + n. If ages are in the ratio a : b, take them as ax and bx.
3. What is the unit method for ages?
Write the two ratios, make the difference of their parts equal, and compare the change in units with the years passed. That gives the value of 1 unit, and then the ages.
4. Why does the age difference never change?
Both people get older by the same number of years at the same time, so the gap between them stays fixed. That fixed gap is the base of the unit method.
5. Does the ratio of ages stay the same over time?
No. The ratio changes as years pass and moves closer to 1. Only the difference of ages stays the same.
6. How do you solve age problems quickly?
Use the unit method for two-ratio questions and the option check for the rest: put each option into the conditions and see which one fits. Most questions take 30–60 seconds this way.
7. How do you find how many years later someone will be twice as old?
Use n = (A − 2B) ÷ (2 − 1) = A − 2B. For a father of 38 and son of 10, n = 38 − 20 = 18 years.
8. How do you solve age problems with averages?
Change the average into a total (average × number of people). When years pass, add the years once for every person in the group.
9. What is the difference between “5 times as old” and “5 times older”?
“5 times as old” means 5x. Strictly, “5 times older” means 6x, but exams sometimes use it loosely. Check which reading matches the options.
10. How do you solve age questions when the product of ages is given?
Look for factor pairs of the product that match the other clue (difference or ratio). For ratio a : b, use k = √(P ÷ ab).
11. What are digit-reversal age problems?
One age is 10a + b and the other is 10b + a, like 52 and 25. Their sum is 11(a + b) and their difference is 9(a − b), which solves the question fast.
12. How many age questions come in SSC CGL?
Usually about 0–1 direct question in Tier-I quant, sometimes mixed with ratio or average. This is an approximate estimate, not an official figure.
13. Are age questions asked in RRB NTPC and Group D?
Yes. They appear fairly regularly in the arithmetic part, mostly as “times” or ratio-change questions that can be solved in under a minute.
14. Which chapter should I study before problems on ages?
Ratio and proportion. The unit method is just ratio logic, so finishing ratio first makes ages a two-day job.
15. How many days are needed to finish problems on ages?
About 4–6 days at 2 hours a day for a beginner who already knows ratio, and 2 days for revision. The day-wise table above shows the split.
16. Can I solve age questions using options?
Yes. Age answers are whole numbers, so substituting options is quick and safe, especially for “find their present ages” questions.
17. How much time should I spend on one age question in the exam?
Aim for 40–60 seconds. If it is a long family puzzle, skip it in the first round and come back after the easy questions.
18. Is problems on ages easy for beginners?
Yes. It needs only basic ratio and simple equations. With 40–50 practice questions across the models above, most students can solve exam-level questions confidently.
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Problems on Ages Ages Questions for SSC CGL Age Problems Tricks Problems on Ages Formulas Ages Questions RRB NTPC Unit Method Ages