How to Solve Problems on Ages
Solve ages aptitude problems using ratio multipliers and the constant age-difference rule, with a worked example and practice questions with answers.
Expected Interview Answer
Problems on ages are solved by assigning variables to present ages, translating "years ago" and "years hence" into simple linear expressions, and forming equations from the given ratios or conditions.
Let the present ages be variables (often a common multiplier x for a given ratio, like 3x and 5x). "n years ago" means subtracting n from every age in the statement, and "n years hence" means adding n — critically, the age difference between two people never changes over time, only the ratio between their ages does. Translate each sentence into one linear equation, then solve the resulting system. Sanity-check the answer: ages must be positive and consistent with the original ratio.
- The constant age-difference fact simplifies many problems
- Common-multiplier ratios turn word problems into single equations
- A systematic translate-then-solve method avoids setup errors
AI Mentor Explanation
Two teammates’ age gap stays fixed no matter how many seasons pass — if one is 5 years older now, they’ll be 5 years older a decade from now too, even though the ratio of their ages shifts. Age problems lean on this: the difference is invariant, only ratios change with time. Setting present ages as variables and translating "5 years ago" as subtracting 5 from each turns the wording into solvable linear equations.
Worked example
Present
- 4x and 3x
Equation
- 9(4x+6) = 11(3x+6)
- 36x+54 = 33x+66 → x=4
Ages
- 16 and 12
Step-by-Step Explanation
Step 1
Assign variables
Use a common multiplier for ratios: present ages 4x and 3x.
Step 2
Translate time shifts
"n years ago" subtracts n; "n years hence" adds n, to every age.
Step 3
Form the equation
Use the new ratio or difference stated in the problem.
Step 4
Solve and verify
Solve for x, substitute back, and check ages are positive and consistent.
What Interviewer Expects
- Correct variable setup using ratio multipliers
- Accurate translation of "years ago/hence"
- Recognizing the age difference is invariant over time
- Verification that the final ages satisfy all given conditions
Common Mistakes
- Applying the time shift to only one person’s age
- Forgetting the age difference never changes
- Cross-multiplying the ratio equation incorrectly
- Not checking that resulting ages are positive and sensible
Best Answer (HR Friendly)
“Set the present ages as variables — often using a common multiplier for a given ratio — then translate "years ago" as subtracting and "years hence" as adding the same number to every age. The key insight is that the difference between two people’s ages never changes, only the ratio does, so build your equation from whatever ratio or condition the problem gives, then solve.”
Follow-up Questions
- Why does the difference between two ages stay constant over time?
- How do you set up a three-person ages problem?
- How does the ratio of ages change as both people get older?
- How would you solve for a birth year given today’s age and current year?
MCQ Practice
1. A father is 3 times as old as his son. In 10 years, he will be twice as old. The son’s present age is?
Let son = x, father = 3x. 3x+10 = 2(x+10) → 3x+10 = 2x+20 → x = 10.
2. The present ages of A and B are in the ratio 5:3. Six years ago, A was twice as old as B. B’s present age is?
Let ages be 5x, 3x. 5x−6 = 2(3x−6) → 5x−6 = 6x−12 → x = 6. B = 3×6 = 18.
3. Two friends have ages that differ by 8 years. Ten years from now, this difference will be?
The difference between two people’s ages is always constant, regardless of elapsed time.
Flash Cards
Key invariant in ages problems? — The difference between two people’s ages never changes over time.
How to handle "n years ago"? — Subtract n from every person’s present age in that clause.
How to set up a ratio of ages? — Use a common multiplier: ages = ax and bx for ratio a:b.
What changes with time? — The ratio of ages changes; the absolute difference does not.