Study Notes: Problems on Ages

Overview

Age problems are a staple of SSC CHSL arithmetic, appearing in 2–3 questions each year. These questions test your ability to translate family relationships and time-based conditions into algebraic equations. The core skill is converting sentences like "Five years ago, A was twice as old as B" into mathematical statements.

This topic sits at the intersection of algebra and logical reasoning. You'll deal with present ages, past ages (subtract years), and future ages (add years). Most problems involve two or three people and require you to set up simultaneous equations. The beauty of age problems is that they follow predictable patterns—master the equation-formation step, and you'll solve them in under a minute. Expect straightforward two-person problems and occasional three-person or ratio-based variants in the exam.

Success here means recognizing keywords ("ago" = subtract, "hence" = add, "times as old" = multiplication), setting variables correctly, and solving linear equations quickly. Practice converts these word puzzles into routine calculations.

Key Concepts

• **Present, past, future ages**: If someone's present age is x, their age n years ago was (x − n), and n years hence will be (x + n). The value n remains the same for all people in the problem—time moves equally for everyone.

• **Age difference is constant**: If A is 5 years older than B today, A was 5 years older than B ten years ago and will be 5 years older ten years from now. The gap between two people's ages never changes.

• **"Times as old" relations**: "A is twice as old as B" means A = 2B. "A will be three times as old as B" means A + n = 3(B + n) for some future time n. Always match the time frames on both sides.

• **Ratio format**: When given age ratios (e.g., 3:2), assign variables as 3x and 2x, not just 3 and 2. The ratio tells you the proportion, not the actual ages.

• **Sum and difference method**: If you know the sum of two ages and their ratio, find individual ages by: sum × (each part of ratio) / (total parts). For difference-based problems, use the constant-gap principle.

• **Average ages**: If the average age of a group increases or decreases when one person leaves and another joins, the difference in their ages equals (change in average) × (number of people).

Formulas / Key Facts

1. **Past age**: Age n years ago = Present age − n 2. **Future age**: Age n years hence = Present age + n 3. **Age difference**: (Age of A) − (Age of B) = constant at all times 4. **Ratio to actual ages**: If ratio is a:b and sum is S, then ages are (a/(a+b))×S and (b/(a+b))×S 5. **Twice/thrice relations**: "A is k times as old as B" → A = k×B 6. **Average age formula**: Average = Sum of all ages / Number of people 7. **Replacement in group**: If one person replaces another, Difference in ages = Change in average × Number of people

Worked Examples

**Example 1**: Ten years ago, a father was three times as old as his son. Ten years hence, the father will be twice as old as his son. Find their present ages.

*Solution*: Let present age of son = x, present age of father = y.

Ten years ago: y − 10 = 3(x − 10) → y − 10 = 3x − 30 → y = 3x − 20 … (i)

Ten years hence: y + 10 = 2(x + 10) → y + 10 = 2x + 20 → y = 2x + 10 … (ii)

Equating (i) and (ii): 3x − 20 = 2x + 10 x = 30

Substitute in (ii): y = 2(30) + 10 = 70

**Answer**: Son's present age = 30 years, Father's present age = 70 years.

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**Example 2**: The ratio of ages of A and B is 5:3. After 6 years, the ratio will be 7:5. Find their present ages.

*Solution*: Let present ages be 5x and 3x.

After 6 years: (5x + 6)/(3x + 6) = 7/5

Cross-multiply: 5(5x + 6) = 7(3x + 6) 25x + 30 = 21x + 42 4x = 12 x = 3

**Answer**: A's present age = 5×3 = 15 years, B's present age = 3×3 = 9 years.

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**Example 3**: The average age of 5 students is 20 years. If one student aged 18 leaves and a new student joins, the average becomes 21 years. Find the age of the new student.

*Solution*: Total age of 5 students = 20 × 5 = 100 years

After one leaves: Total age = 100 − 18 = 82 years (4 students remaining)

New total with 5 students = 21 × 5 = 105 years

Age of new student = 105 − 82 = 23 years

**Answer**: New student's age = 23 years.

Common Mistakes

**Mistake 1**: Confusing "ago" and "hence" directions. *Wrong thinking*: "5 years ago" means add 5. *Correct fix*: "Ago" means past, so subtract. "Hence" or "from now" means future, so add.

**Mistake 2**: Applying time shift to only one person. *Wrong thinking*: Setting up "10 years ago A was twice B's age" as A − 10 = 2B. *Correct fix*: Both ages shift: (A − 10) = 2(B − 10). Time affects everyone equally.

**Mistake 3**: Using ratio numbers as actual ages. *Wrong thinking*: If ratio is 4:3, writing ages as 4 and 3. *Correct fix*: Use 4x and 3x, then solve for x.

**Mistake 4**: Forgetting that age difference stays constant. *Wrong thinking*: Treating the gap between two ages as variable over time. *Correct fix*: If A is 7 years older than B now, A was 7 years older 20 years ago and will be 7 years older in the future.

**Mistake 5**: Misreading "sum" vs. "difference" in ratio problems. *Wrong thinking*: Using sum formula when the problem gives a difference. *Correct fix*: Carefully identify whether you're given sum, difference, or a direct ratio statement, and choose the right equation setup.

Quick Reference

• **Past**: Present age − n; **Future**: Present age + n
• **Age gap never changes** over time—use this to cross-check answers
• **Ratio a:b** → actual ages are ax and bx; solve for x
• **"k times as old"** → set up equation with multiplication: A = k×B
• **Average change** → New person's age = Old person's age + (Change in average × Number of people)
• **Check reasonableness**: Ages must be positive, and parent typically older than child