Problems on Ages Questions for Placement (with Solutions)

Last Updated: March 2026

Problems on Ages is a favorite topic in quantitative aptitude sections of campus placements. This guide covers 30 practice questions with step-by-step solutions, shortcuts, and tricks to help you solve these problems efficiently.

Key Concepts and Formulas

Basic Relationships

If present age is x, then:
- Age n years ago = x - n
- Age n years hence = x + n

Important Ratios

If A is twice as old as B: A = 2B
If A is n times as old as B: A = n × B

Age Difference

Age difference between two people remains constant over time
If A - B = d today, then A - B = d after n years

Common Problem Types

Present age based on past/future conditions
Ratio problems with time shifts
Problems involving multiple people
Average age problems

30 Practice Questions with Solutions

Level 1: Basic (Questions 1-10)

Q1. The sum of ages of 5 children born at intervals of 3 years each is 50 years. What is the age of the youngest child?

Solution: Let youngest child's age = x Ages: x, x+3, x+6, x+9, x+12

Sum: x + (x+3) + (x+6) + (x+9) + (x+12) = 50 5x + 30 = 50 5x = 20 x = 4

Q2. A father said to his son, "I was as old as you are at the present at the time of your birth." If the father's age is 38 years now, what was the son's age five years back?

Solution: Let son's present age = x Father's age at son's birth = 38 - x

Given: 38 - x = x (father was son's current age at son's birth) 38 = 2x x = 19

Son's age 5 years back = 19 - 5 = 14

Q3. The present ages of three persons are in proportions 4:7:9. Eight years ago, the sum of their ages was 56. Find their present ages.

Solution: Let present ages be 4x, 7x, and 9x

Eight years ago: (4x-8) + (7x-8) + (9x-8) = 56 20x - 24 = 56 20x = 80 x = 4

Present ages: 16, 28, and 36 years

Q4. The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. What is the ratio of their present ages?

Solution: Let son's age 10 years ago = x Father's age 10 years ago = 3x

Present ages: Son = x+10, Father = 3x+10

Ten years hence: Son = x+20, Father = 3x+20

Given: 3x + 20 = 2(x + 20) 3x + 20 = 2x + 40 x = 20

Present ages: Son = 30, Father = 70 Ratio = 70:30 = 7:3

Q5. Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age?

Solution: Given: R - Q = Q - T (Q is equally distant from R and T) This means: 2Q = R + T

Given: R + T = 50 Therefore: 2Q = 50, so Q = 25

R - Q = Q - T = R - 25 = 25 - T

We know R + T = 50, but cannot determine individual values. However, R - Q = Q - T, and since 2Q = R + T: R - Q = Q - T = (R+T)/2 - T = (R-T)/2

But we cannot find exact difference without more info. Actually, R - Q = Q - T, and R + T = 50 Adding: (R-Q) + (R+T) = (Q-T) + 50? No.

Since R + T = 50 and 2Q = R + T: R + T = 2Q = 50, so Q = 25

R - Q = R - 25, and we don't know R individually.

Wait - let me reconsider. We CAN determine R - Q: Since R - Q = Q - T, let's call this difference d Then R = Q + d and T = Q - d R + T = (Q+d) + (Q-d) = 2Q = 50

But d can be any value! Data inadequate.

Actually, we need to find if there's a unique answer. Since R + T = 50 and the differences are equal, Q is the middle value = 25. But R - Q can vary: If R = 30, Q = 25, T = 20, difference = 5 If R = 40, Q = 25, T = 10, difference = 15

Q6. The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A?

Solution: Given: A + B = B + C + 12 A + B - B = C + 12 A = C + 12 A - C = 12

Q7. A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?

Solution: Let mother's age = x Person's age = 2x/5

After 8 years: Person's age = 2x/5 + 8 Mother's age = x + 8

Given: 2x/5 + 8 = (x + 8)/2 Multiply by 10: 4x + 80 = 5x + 40 x = 40

Q8. The ratio of ages of two friends is 3:5. Six years hence, the ratio will become 5:7. Find the sum of their present ages.

Shortcut Solution: Let present ages be 3x and 5x

Six years hence: (3x+6)/(5x+6) = 5/7 Cross multiply: 7(3x+6) = 5(5x+6) 21x + 42 = 25x + 30 4x = 12 x = 3

Ages: 9 and 15 Sum = 24

Q9. The ratio of present ages of P and Q is 6:7. If Q is 4 years older than P, what will be the ratio of their ages after 4 years?

Solution: Let P's age = 6x, Q's age = 7x

Given: Q - P = 4 7x - 6x = 4 x = 4

Present ages: P = 24, Q = 28

After 4 years: P = 28, Q = 32 Ratio = 28:32 = 7:8

Q10. The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be:

Solution: Let son's present age = x Father's present age = 60 - x

Six years ago: Father: 60 - x - 6 = 54 - x Son: x - 6

Given: 54 - x = 5(x - 6) 54 - x = 5x - 30 84 = 6x x = 14

Son's age after 6 years = 14 + 6 = 20 years

Level 2: Moderate (Questions 11-20)

Q11. The average age of a class of 22 students is 21 years. The average increased by 1 when the teacher's age is also included. What is the age of the teacher?

Shortcut: Teacher's age = New average + (Old average × Number of students) = 22 + (21 × 22) - This is wrong.

Correct method: Total age of students = 22 × 21 = 462 Total age with teacher = 23 × 22 = 506 Teacher's age = 506 - 462 = 44

Shortcut formula: Teacher's age = New_avg + (Number_of_students × Difference_in_avg) = 22 + (22 × 1) = 44

Q12. A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son?

Solution: Let son's age = x Man's age = x + 24

In two years: Man: x + 26 Son: x + 2

Given: x + 26 = 2(x + 2) x + 26 = 2x + 4 x = 22

Q13. The ages of two persons differ by 16 years. If 6 years ago, the elder one was 3 times as old as the younger one, find their present ages.

Solution: Let younger's age = x Elder's age = x + 16

Six years ago: Younger: x - 6 Elder: x + 10

Given: x + 10 = 3(x - 6) x + 10 = 3x - 18 28 = 2x x = 14

Ages: 14 and 30

Q14. The ratio between the present ages of A and B is 5:3. The ratio between A's age 4 years ago and B's age 4 years hence is 1:1. What is the ratio between A's age 4 years hence and B's age 4 years ago?

Solution: Let A's age = 5x, B's age = 3x

Given: (5x - 4)/(3x + 4) = 1/1 5x - 4 = 3x + 4 2x = 8 x = 4

A's age = 20, B's age = 12

A's age 4 years hence = 24 B's age 4 years ago = 8

Ratio = 24:8 = 3:1

Q15. One year ago, the ratio of Sooraj's and Vimal's age was 6:7. Four years hence, this ratio would become 7:8. How old is Vimal?

Solution: One year ago: Sooraj = 6x, Vimal = 7x

Present: Sooraj = 6x+1, Vimal = 7x+1

Four years hence: Sooraj = 6x+5, Vimal = 7x+5

Given: (6x+5)/(7x+5) = 7/8 8(6x+5) = 7(7x+5) 48x + 40 = 49x + 35 x = 5

Vimal's present age = 7(5) + 1 = 36

Q16. The sum of ages of 5 children born at intervals of 2 years each is 60 years. What is the age of the youngest child?

Solution: Let youngest = x Ages: x, x+2, x+4, x+6, x+8

Sum: 5x + 20 = 60 5x = 40 x = 8

Q17. A father is 4 times as old as his son. In 20 years, the father will be twice as old as his son. Find their present ages.

Solution: Let son's age = x Father's age = 4x

In 20 years: Son: x + 20 Father: 4x + 20

Given: 4x + 20 = 2(x + 20) 4x + 20 = 2x + 40 2x = 20 x = 10

Ages: Son = 10, Father = 40

Q18. The present ages of three persons are in the ratio 4:7:9. Eight years ago, the sum of their ages was 56. Find their present ages.

Solution: Let ages be 4x, 7x, and 9x

Eight years ago: (4x-8) + (7x-8) + (9x-8) = 56 20x - 24 = 56 20x = 80 x = 4

Present ages: 16, 28, and 36 years

Q19. The ratio of present ages of two brothers is 1:2 and 5 years back, the ratio was 1:3. What will be the ratio of their ages after 5 years?

Solution: Let present ages be x and 2x

Five years back: (x-5)/(2x-5) = 1/3 3(x-5) = 2x-5 3x - 15 = 2x - 5 x = 10

Present ages: 10 and 20

After 5 years: 15 and 25 Ratio = 15:25 = 3:5

Q20. A couple has a son and a daughter. The age of the father is four times that of the son and the age of the daughter is one-third of her mother. The wife is 6 years younger than her husband and the sister is 3 years older than her brother. Find the mother's age.

Solution: Let son's age = x Father's age = 4x

Daughter is 3 years older than son = x + 3

Daughter's age = (1/3) × mother's age So mother's age = 3(x + 3) = 3x + 9

Wife is 6 years younger than husband: Mother's age = Father's age - 6 3x + 9 = 4x - 6 15 = x

Mother's age = 3(15) + 9 = 54

Level 3: Advanced (Questions 21-30)

Q21. Ten years ago, the age of a mother was three times the age of her son. After ten years, the mother's age will be twice that of her son. Find the ratio of their present ages.

Solution: Let son's age 10 years ago = x Mother's age 10 years ago = 3x

Present: Son = x+10, Mother = 3x+10

After 10 years: Son = x+20, Mother = 3x+20

Given: 3x + 20 = 2(x + 20) 3x + 20 = 2x + 40 x = 20

Present ages: Son = 30, Mother = 70 Ratio = 30:70 = 3:7

Q22. The average age of a group of 10 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students?

Shortcut Solution: Total age of original 10 = 10 × 20 = 200 Total age of 12 students = 12 × 22 = 264 Total age of 2 new students = 64 Average of new students = 32

Shortcut formula: Average of new members = New_avg + (Original_count × Increase)/New_members = 22 + (10 × 2)/2 = 22 + 10 = 32

Q23. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.

Solution: Let son's age = x Father's age = 3x + 3

Three years hence: Son: x + 3 Father: 3x + 6

Given: 3x + 6 = 2(x + 3) + 10 3x + 6 = 2x + 6 + 10 3x + 6 = 2x + 16 x = 10

Father's age = 3(10) + 3 = 33

Q24. The ratio of the ages of a man and his wife is 4:3. After 4 years, this ratio will be 9:7. If at the time of marriage, the ratio was 5:3, then how many years ago were they married?

Solution: Let present ages be 4x and 3x

After 4 years: (4x+4)/(3x+4) = 9/7 7(4x+4) = 9(3x+4) 28x + 28 = 27x + 36 x = 8

Present ages: Man = 32, Wife = 24

At marriage, ratio was 5:3 Let they married y years ago (32-y)/(24-y) = 5/3 3(32-y) = 5(24-y) 96 - 3y = 120 - 5y 2y = 24 y = 12

Q25. The ratio between the school ages of Neelam and Shaan is 5:6. If the ratio between one-third of Neelam's age and half of Shaan's age is 5:9, then what is the school age of Shaan?

Solution: Let Neelam's age = 5x, Shaan's age = 6x

Given: (1/3 × 5x) / (1/2 × 6x) = 5/9 (5x/3) / (3x) = 5/9 5x/9x = 5/9 5/9 = 5/9 ✓

This equation is always true! We need more information.

Q26. A father said to his son, "I was as old as you are at the present at the time of your birth." If the father's age is 38 years now, the son's age five years back was:

Solution: Let son's present age = x Father's age at son's birth = 38 - x

Given: 38 - x = x (father was son's current age) 38 = 2x x = 19

Five years back = 19 - 5 = 14

Q27. The sum of the ages of a father and his son is 4 times the age of the son. If the average age of the father and son is 28 years, what is the son's age?

Solution: Average = 28, so total = 56 years

Let son's age = x Father's age = 56 - x

Given: (56 - x) + x = 4x 56 = 4x x = 14

Q28. A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. What is the present age of the mother?

Solution: Let mother's age = x Person's age = 2x/5

After 8 years: Person = 2x/5 + 8 Mother = x + 8

Given: 2x/5 + 8 = (x + 8)/2 Multiply by 10: 4x + 80 = 5x + 40 x = 40

Q29. The ages of two persons differ by 20 years. If 5 years ago, the elder one was 5 times as old as the younger one, find their present ages.

Solution: Let younger's age = x Elder's age = x + 20

Five years ago: Younger: x - 5 Elder: x + 15

Given: x + 15 = 5(x - 5) x + 15 = 5x - 25 40 = 4x x = 10

Ages: 10 and 30

Q30. The ratio of the ages of two friends is 3:5. Six years hence, this ratio will become 5:7. Find the difference of their present ages.

Solution: Let present ages be 3x and 5x

Six years hence: (3x+6)/(5x+6) = 5/7 7(3x+6) = 5(5x+6) 21x + 42 = 25x + 30 4x = 12 x = 3

Ages: 9 and 15 Difference = 6

Shortcuts and Tricks

Trick 1: Cross-Multiplication for Ratio Problems

When given ratio changes over time:

Set up equation with ratio = fraction
Cross multiply to solve

Trick 2: Constant Difference Property

Age difference never changes: If A - B = 10 today, then A - B = 10 after any years

Trick 3: Shortcut for Average Age Problems

When a new person joins and average changes:

New person's age = New_avg + (Old_count × Change_in_avg)

Trick 4: Proportion Method

If ages in ratio a:b and condition given:

Let ages be ax and bx
Set up equation from condition
Solve for x

Trick 5: Quick Check for Data Adequacy

Count unknowns and equations
If unknowns > equations → Data inadequate
If equations ≥ unknowns → Can be solved

Companies Testing This Topic

Company	Frequency	Difficulty
TCS	Frequently asked	Easy-Medium
Infosys	Very common	Easy
Wipro	Common	Easy
Cognizant	Frequently asked	Easy
Accenture	Common	Easy-Medium
Capgemini	Common	Easy
HCL	Sometimes	Medium
Tech Mahindra	Common	Easy

Frequently Asked Questions (FAQs)

Q1: How many age problems typically appear in aptitude tests?

A: Usually 1-3 questions in campus placement papers. They're considered scoring questions because they follow predictable patterns.

Q2: What is the most common mistake in age problems?

A: Confusing "years ago" and "years hence." Always verify whether you're adding or subtracting from the present age. Also, mixing up whose age was being compared at what time.

Q3: Should I use algebra or shortcuts for age problems?

A: Learn the algebraic method first to build understanding. Once comfortable, shortcuts will naturally emerge. In exams, use whichever is faster for that specific problem.

Q4: Are there any online resources to practice more age problems?

A: Yes, IndiaBIX, GeeksforGeeks, and CrackU have extensive collections. Previous year papers from TCS, Infosys, and Wipro are excellent practice sources.

Q5: Can age problems be solved without using variables?

A: Simple ones can be done mentally. For example, if the sum of ages is 50 and difference is 10: older = (50+10)/2 = 30, younger = (50-10)/2 = 20. But using variables is safer for complex problems.

Practice regularly to master Problems on Ages!

Ages Problems Questions Placement