Problem on Ages – Quant Concepts, Tricks & Solved Examples

“Problem on Ages” is a very common and high-yield topic in aptitude exams like Banking, SSC CGL, CAT, Railways, and Placement tests. Problems on Ages test your ability to form linear equations based on relationships between ages, present, past, or future. The key is to translate English statements into algebraic equations.

Basic Definitions

Term	Meaning
Present age	Age right now
Past age	Age some years ago
Future age	Age after some years

If present age is x:

• Age after n years = x + n
• Age n years ago = x – n

Common Relationships

Expression	Equation
A is n years older than B	A = B + n
A is n years younger than B	A = B – n
Age of A is twice that of B	A = 2B
Sum of their ages	A + B
Average of their ages	(A + B)/2

Typical Question Types

Type 1: Present age based on age ratio

The ratio of the ages of A and B is 3:4. After 5 years, the ratio becomes 4:5. Find their present ages.

🧩 Approach:
Let A = 3x, B = 4x
After 5 years → (3x + 5)/(4x + 5) = 4/5
→ Cross-multiply and solve for x.

Type 2: Age difference type

The difference between the ages of a father and son is 25 years. 5 years ago, the father’s age was 6 times the son’s age. Find their present ages.

🧩 Approach:
Let son’s present age = x
Father’s = x + 25
5 years ago:
x + 25 – 5 = 6(x – 5)
→ Solve for x.

Type 3: Future/Past ratio problems

Ratio of A’s age to B’s age 5 years ago was 4:5. Ratio after 5 years will be 6:7. Find present ages.

🧩 Set up two equations and solve simultaneously.

Type 4: Average age / Combined age problems

Average age of 3 persons A, B, C is 25 years. If D joins, the average becomes 30. Find D’s age.

🧩 Total age of A, B, C = 3×25 = 75
Total with D = 4×30 = 120
→ D = 120 – 75 = 45 years

Type 5: Birth or age relation with family

A father is twice as old as his son. 20 years ago, the father was 12 times the son’s age. Find their ages.

🧩 Translate and solve equations.

Step-by-Step Approach

1. Assign variables for present ages.
2. Translate statements into equations.
3. Relate past or future ages with +/– years.
4. Use ratios or differences as given.
5. Solve equations (usually 1 or 2 variables).
6. Verify your answer logically.

Key Shortcuts and Tips

• Difference in ages remains constant over time.
• Ratios change, but differences don’t.
• Use LCM method when working with ratios to quickly get present ages.
• Check units (years before, after, etc.) carefully — small mistakes cost marks.

Example Questions for Practice

1. The ratio of father’s and son’s ages is 5:2. After 10 years, the ratio becomes 3:2. Find their present ages.
2. The sum of ages of A and B is 60 years. 4 years ago, the ratio of their ages was 7:5. Find their ages.
3. Ten years hence, a man’s age will be twice the age of his son. Ten years ago, he was four times as old. Find their present ages.

Common Mistakes done by Students

❌ Confusing “years ago” with “after years”
❌ Forgetting that age difference stays constant
❌ Mixing up ratios with actual ages
❌ Ignoring logical checks (e.g., negative or unrealistic ages)

Quick Practice Formulae Recap

Concept	Formula
Future age	x + n
Past age	x – n
Average age	(Sum of ages) / Number of persons
Age ratio change	(x + added years) / (y + added years)

Practice Tip

👉 In most aptitude exams, these are 1-step or 2-step linear equation questions.
👉 Aim for 30–45 seconds per question with practice.
👉 Create your own short forms (e.g., “n yrs ago → -n”).

Refer Aptitude Concepts

Practice Problem on Ages – Aptitude Questions