4.7 Problems on Ages
Key Takeaways
- Age is a named bullet of the CIL Paper-I Quantitative Aptitude syllabus and is solved by translating each sentence into one linear equation.
- The difference between two people's ages is constant over time, whereas the ratio of their ages changes, and most traps exploit this asymmetry.
- Always define the present age as the variable and add or subtract years for past and future references rather than introducing separate unknowns.
- Back-substituting the answer into the original word statement is faster than re-solving and catches most sign errors.
The Single Principle That Matters
All age problems rest on one fact:
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
| English | Algebra |
|---|---|
| Present age of A | $x$ |
| A's age 5 years ago | $x - 5$ |
| A's age 8 years hence | $x + 8$ |
| A is 4 times as old as B | $A = 4B$ |
| A is 4 years older than B | $A = B + 4$ |
| The ratio of their ages is 5 : 3 | $A = 5k$, $B = 3k$ |
| Sum of ages is 60 | $A + B = 60$ |
| A is twice as old as B was 6 years ago | $A = 2(B - 6)$ |
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:
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:
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:
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:
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 $d$,
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
| Trap | Guard |
|---|---|
| Applying a present ratio to a past or future year | Ratios change with time; only differences are constant |
| Confusing "three times as old" with "three years older" | Multiplication versus addition |
| Using two unknowns when a ratio is given | Use a single multiplier $k$ |
| Adding years to only one person | Every age in the statement shifts by the same amount |
| Answering the wrong quantity | The question may ask for the sum, the difference, or one specific person |
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
- Assign a variable to a present age, or use $k$ with a given ratio.
- Express every other age in terms of that variable, shifting by the stated number of years.
- Form one equation per given relationship.
- Solve, then back-substitute into the original English sentence rather than into your own equation, which catches translation errors as well as arithmetic ones.
- 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.
The present ages of two engineers are in the ratio 5 : 3. Four years hence the ratio becomes 3 : 2. The present age of the elder is:
A father is three times as old as his son. Five years ago he was four times as old. The son's present age is:
Which quantity between two people remains constant as the years pass?
The average age of 5 engineers is 32 years. When a sixth engineer joins, the average falls to 31. The age of the new engineer is: