Ages

Problems on Ages involve determining the present, past, or future ages of individuals based on given relationships or algebraic conditions. They test your ability to translate word problems into equations and solve them logically.

Typical question types:

• Age relationships (Father is twice the son's age, etc.)
• Age differences remain constant
• Age ratios at different times
• Conditions involving multiple people (e.g., sums or averages)

Think of age problems as time-traveling algebra — you're tracking someone's age through time using logic.

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2. Key Formulas & Shortcuts

A. Basic Formula

If a person is $x$ years old today:

• Age $y$ years ago: $x - y$
• Age $y$ years later: $x + y$

B. Constant Age Difference

If A is $d$ years older than B, then:

$$\text{(A's age at any point)} = \text{(B's age at that point)} + d$$

This difference stays the same over time.

C. Ratios & Ages

If ratio of ages is $a:b$, and their sum is $S$:

$$\text{A's age} = \frac{a}{a + b} \times S,\quad \text{B's age} = \frac{b}{a + b} \times S$$

D. Average of Ages

$$\text{Average Age} = \frac{\text{Sum of Ages}}{\text{Number of People}}$$

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3. Conceptual Tips & Common Mistakes

• Age difference is always constant, but ratios change with time.

• Clearly define variables for each person — confusion here leads to wrong equations.

• Avoid mixing up “$x$ years ago” and “in $x$ years”.

• Don't forget units — time is always in years.

• Read carefully: if a person is currently 5 times older than someone, does that mean:

  • "He is 5 times older" → 5x
  • "He is 5 years older" → x + 5

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4. Visual Explanation

Let’s say:

• Father is 30, Son is 10
  → Age Ratio = 3:1
  → Age difference = 20

10 years later:

• Father = 40, Son = 20
  → Ratio = 2:1
  → Difference = still 20

So:
Difference stays constant, but ratios vary over time.

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5. Solved Examples

Example 1: Basic Linear Equation

Q: A father is 3 times the age of his son. After 10 years, the father will be twice as old as his son. Find their current ages.

Let: Son = $x$, Father = $3x$

After 10 years:

$$3x + 10 = 2(x + 10) \Rightarrow 3x + 10 = 2x + 20 \Rightarrow x = 10$$

Ans: Son = 10 years, Father = 30 years

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Example 2: Using Age Difference

Q: The difference in ages of A and B is 6 years. After 4 years, A will be 1.5 times as old as B. Find their present ages.

Let B = $x$, then A = $x + 6$

After 4 years:

$$x + 6 + 4 = 1.5(x + 4) \Rightarrow x + 10 = 1.5x + 6 \Rightarrow 0.5x = 4 \Rightarrow x = 8$$

Ans: B = 8, A = 14

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Example 3: Ratio to Equation

Q: The present age ratio of A to B is 4:5. After 5 years, the ratio becomes 5:6. Find their current ages.

Let A = 4x, B = 5x

After 5 years:

$$\frac{4x + 5}{5x + 5} = \frac{5}{6} \Rightarrow 6(4x + 5) = 5(5x + 5) \Rightarrow 24x + 30 = 25x + 25 \Rightarrow x = 5$$

Ans: A = 20, B = 25

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Example 4: Group Average

Q: Average age of 4 friends is 25. A new friend joins and the new average becomes 24. Find the age of the new friend.

$$\text{Total of 4} = 4 \times 25 = 100 \\ \text{Total of 5} = 5 \times 24 = 120 \\ \text{New friend’s age} = 120 - 100 = 20$$

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Example 5: Age in Future

Q: Sum of ages of A and B is 50. After 5 years, the sum becomes 60. Find their current ages.

Trick: In 5 years, total age increases by $2 \times 5 = 10$

So current sum = $60 - 10 = 50$ → Already given.

This validates the condition. Now use further info if given.

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Example 6: Three-person scenario

Q: A is twice as old as B, and B is 4 years older than C. If A is 28, what are B and C’s ages?

$$A = 2B \Rightarrow B = 14 \\ B = C + 4 \Rightarrow C = 10$$

Ans: B = 14, C = 10