4.7 Problems on Ages

The Single Principle That Matters

All age problems rest on one fact:

$$\textbf{The difference between two ages never changes; the ratio does.}$$

If A is 10 years older than B today, A was 10 years older ten years ago and will be 10 years older twenty years hence. But if their ages today are in the ratio 3 : 2, that ratio was different in the past and will be different in future, always moving closer to 1 : 1 as time passes.

Exam items are built almost entirely on candidates forgetting the second half of that sentence.

Translating Sentences into Equations

The distinction between times as old as (multiplication) and years older than (addition) is the single most common translation error.

The ratio device

When a ratio is given, introduce a single multiplier $k$ rather than two unknowns. If present ages are in the ratio 5 : 3, write them as $5k$ and $3k$. This halves the algebra and is almost always the fastest route.

Worked Example 1: Present Ages from a Ratio and a Future Condition

The present ages of two engineers are in the ratio 5 : 3. Four years hence, the ratio becomes 3 : 2. Find their present ages.

Let the ages be $5k$ and $3k$. Four years hence:

$$\frac{5k + 4}{3k + 4} = \frac{3}{2}$$

Cross-multiplying: $2(5k + 4) = 3(3k + 4)$, so $10k + 8 = 9k + 12$, giving $k = 4$.

Present ages are $5(4) = \textbf{20}$ and $3(4) = \textbf{12}$ years.

Check. Four years hence they are 24 and 16, and $24 : 16 = 3 : 2$. Correct. Note that the difference, 8 years, is the same before and after — a fast sanity check.

Worked Example 2: A Past Reference

A father is three times as old as his son. Five years ago he was four times as old. Find their present ages.

Let the son be $x$ and the father $3x$. Five years ago:

$$3x - 5 = 4(x - 5)$$

So $3x - 5 = 4x - 20$, giving $x = 15$. The son is 15 and the father is 45.

Check. Five years ago they were 10 and 40, and $40 = 4 \times 10$. Correct.

Worked Example 3: Sum and Difference Together

The sum of the ages of a mother and daughter is 60 years. Six years ago the mother's age was five times the daughter's. Find the daughter's present age.

Let the daughter be $x$, so the mother is $60 - x$. Six years ago:

$$(60 - x) - 6 = 5(x - 6)$$ $$54 - x = 5x - 30 \;\Rightarrow\; 84 = 6x \;\Rightarrow\; x = 14$$

The daughter is 14 and the mother is 46. Six years ago they were 8 and 40, and $40 = 5 \times 8$. Correct.

Worked Example 4: Two Time References

Ten years ago, A was half as old as B. If the ratio of their present ages is 3 : 4, what is the sum of their present ages?

Let present ages be $3k$ and $4k$. Ten years ago:

$$3k - 10 = \tfrac{1}{2}(4k - 10)$$

Multiply through by 2: $6k - 20 = 4k - 10$, so $2k = 10$ and $k = 5$.

Present ages are 15 and 20, and their sum is 35.

Worked Example 5: Average Age Shift

Average-age questions combine this topic with averages and appear regularly.

The average age of a team of 5 shift engineers is 32 years. A new engineer joins and the average becomes 31 years. Find the age of the new engineer.

Total before: $5 \times 32 = 160$. Total after: $6 \times 31 = 186$. The new engineer's age is $186 - 160 = \textbf{26}$ years.

The general result is worth stating: when a person joins a group of $n$ people and changes the average by $d$,

$$\text{New person's age} = \text{New average} + n \times d$$

Here $31 + 5(-1) = 26$, confirming the arithmetic.

The average age of a group of 10 falls by 3 years when a 60-year-old is replaced by a new member. Find the new member's age.

The total falls by $10 \times 3 = 30$, so the new member is $60 - 30 = \textbf{30}$ years old. For a replacement, note that the group size does not change and the shortcut is total change equals $n$ times the change in average.

Common Traps

That last trap is worth emphasising. Having solved for $k$, re-read the question before selecting an option: many items solve for the multiplier and then ask for a sum or a future age.

Method Summary

1. Assign a variable to a present age, or use $k$ with a given ratio.
2. Express every other age in terms of that variable, shifting by the stated number of years.
3. Form one equation per given relationship.
4. Solve, then back-substitute into the original English sentence rather than into your own equation, which catches translation errors as well as arithmetic ones.
5. Confirm which quantity the question actually asks for.

Since the CIL CBT has no negative marking, a partially formed equation still supports elimination — checking each option against the stated conditions is often faster than solving, especially when the options are small whole numbers.