Problems on Ages — The Time-Travel Rules

Problems on Ages · 20 min

Aptitude · Problems on Ages · Foundations

The gap is frozen. The ratio never stops moving.

Every question in this chapter is built on two facts that pull in opposite directions: the difference between two people's ages is the same number forever, and their ratio changes every single year. This lesson makes both facts concrete, and fixes the one reading error that costs more marks here than any algebra.

Watch one gap hold while the ratio slides
A ratio given five years ago describes the ages five years ago. It says nothing directly about today — but the difference it hides is the same today as it was then.

01 The idea

One frozen number and one moving number

Vijay is 10 and Radha is 15. Radha is five years older, and she will be five years older on every birthday either of them ever has. Wind forward eighty-five years and they are 95 and 100 — still five apart. Time runs at one speed for everybody, so a difference in ages is a permanent fact about a pair of people.

Their ratio behaves nothing like that. Five years ago they were 5 and 10, a ratio of 1 : 2. Today they are 10 and 15, a ratio of 2 : 3. In five years they will be 15 and 20, which is 3 : 4. Same two people, same five-year gap, three different ratios — and it keeps going: 4 : 5, then 7 : 8, then 19 : 20.

There is a reason the ratio always moves in that direction. The gap of five years is a third of Vijay's age when he is 15, but only a nineteenth of it when he is 95. A fixed difference is a shrinking share of a growing age, so the ratio is always crawling towards 1 : 1 and can never turn back. Once you see that, a question saying "the ratio was 3 : 1 and later it is 2 : 1" stops looking like a contradiction and starts looking like the only thing that could have happened.

That leaves one reading error to kill. "Five years ago their ages were in the ratio 1 : 2" is a statement about 5 and 10, not about 10 and 15. Students routinely write the ratio against the present ages and then add five to the answer, which is a different problem with a different answer. Pin every fact to its moment on the timeline before you write an equation.

The difference between two ages is fixed forever. The ratio of two ages drifts towards 1 : 1 with every year that passes. Read every fact as belonging to a specific moment, and this chapter has no surprises left in it.
Constant gapThe difference between two people's ages. Adding the same number of years to both leaves it untouched, so (A + t) − (B + t) = A − B for any t, past or future. Vijay and Radha are 5 apart in 2015, 2025 and 2115.
The bracket ruleWhen you move an age in time, wrap it before you multiply or divide it. Five years after 3x the age is (3x + 5), never 8x. A variable and a plain number sit side by side and never merge.
Ratio driftThe way a ratio of two ages moves as both grow. Because the gap stays put while the ages rise, the ratio always shrinks towards 1 : 1 going forward and stretches away from it going back. Vijay and Radha run 1 : 2, 2 : 3, 3 : 4, 4 : 5, 7 : 8.

02 Worked example

Two ratios, two moments, one pair of ages

The present ages of Vijay and Radha are in the ratio 2 : 3. After 5 years the ratio of their ages will be 3 : 4. Find their present ages. Two ratios of the same two people at two different moments is the standard shape of an ages question, and this is the shortest honest route through it.

1
Turn the present ratio into agesA ratio is not a pair of ages, it is a pair of ages with the size scaled out. Put the size back in with one unknown multiplier.Vijay = 2x    Radha = 3x
2
Move both of them five years forward, in bracketsThe same five years is added to each of them. The brackets stop you from folding the 5 into the 2x.Vijay = (2x + 5)    Radha = (3x + 5)
3
Write the future ratio against the future agesThe 3 : 4 belongs to the moment five years from now, so it goes against the shifted ages and not against 2x and 3x.(2x + 5) / (3x + 5) = 3 / 4
4
Cross-multiply and solveOne unknown, one equation. The x terms nearly cancel, which is why exam setters like ratios one step apart.4(2x + 5) = 3(3x + 5) → 8x + 20 = 9x + 15 → x = 5
5
Read off the ages and check both factsSubstitute back into both the present and the future statement. Checking costs ten seconds and catches a sign slip every time.Vijay = 2(5) = 10, Radha = 3(5) = 15    10 : 15 = 2 : 3 ✓    15 : 20 = 3 : 4 ✓

Look at what the answer does to the gap. Today 15 − 10 = 5; in five years 20 − 15 = 5. The gap never moved, and yet the ratio went from 2 : 3 to 3 : 4. That is the signature of a correct answer in this chapter — if your two moments give two different gaps, you have made an arithmetic error, not discovered a strange family.

03 The method

The bracket method, and the shortcut that skips it

The bracket method above always works and you should be able to run it cold. But when the two ratios have the same difference in units, there is a two-line route that gets the same answer without an equation.

Present ages A and B. A fact at time t reads (A + t) : (B + t), with t negative for the past. Cross-multiplying two such facts gives two linear equations, and A − B is the same number in both.
The unit-shift shortcut: if both ratios have the same difference in units, one unit of the ratio equals the elapsed time divided by the shift in units. Our example is 2 : 3 today and 3 : 4 in five years — both have a 1-unit difference, and Vijay's share moved from 2 units to 3 units, a shift of 1 unit, over 5 years. So 1 unit = 5 years, and Vijay is 2 × 5 = 10. If the two differences are not equal, scale one ratio up until they are.
MomentVijayRadhaDifferenceRatio
5 years ago51051 : 2
Today101552 : 3
In 5 years152053 : 4
In 10 years202554 : 5
In 25 years354057 : 8
In 85 years95100519 : 20

05 Cheat sheet

The time-travel rules on one page

Six rows and three habits. If you can apply the first four rows without thinking, the six models that follow are all variations on arithmetic you already have.

The wordsWhat you writeOn Vijay 10 and Radha 15
t years agox − t10 − 5 = 5
t years hence / later / afterx + t10 + 5 = 15
Ratio a : bax and bx2x and 3x, x = 5
Ratio a : b, t years ago(A − t) : (B − t) = a : b5 : 10 = 1 : 2
Difference of their agesA − B, at any moment5 years, always
Ratio of their ageschanges every year2 : 3 today, 3 : 4 in 5 years
"3x + 5" collapsed to "8x"never legal(2x + 5) stays as it is
Pin the ratio to its momentA ratio given for five years ago describes 5 and 10, not 10 and 15. Writing it against the present ages solves a different problem, and the wrong answer is usually one of the options.
Brackets before multipliersTwice an age five years from now is 2(x + 5), which is 2x + 10 — not 2x + 5. Wrap first, expand second, and the sign errors disappear.
Sanity-check with the gapCompute the difference at both moments in the question. If the two differences disagree, the arithmetic is wrong; no pair of people can change how far apart they are.

06 Where & why

Where these two rules earn their marks

Ages is a small chapter with a reliable one or two questions in almost every Indian placement and government paper, and it is scored on speed rather than depth.

TCS NQT · TCS Digital
Two ratios at two moments

The house style: a present ratio and a ratio some years later, four options, ninety seconds. The unit-shift shortcut answers most of them without writing an equation.

Bank PO · SSC · RRB
"Five years ago the ratio was..."

Almost always anchored in the past, because writing the past ratio against present ages is the error the setter is fishing for. Reading discipline is the whole question.

CAT · XAT data sufficiency
Is one ratio enough?

One ratio at one moment is one equation in two unknowns, so it is never sufficient alone. Two ratios at two different moments almost always are — unless they are the same ratio, in which case they still are not.

Averages, mixtures, partnerships
The same constant-versus-drifting split

A total that grows by a fixed amount while an average moves is the identical idea in a different chapter. Getting it here saves you learning it three more times.

Two habits carry the whole chapter: label every fact with its moment, and check the gap at the end. Everything in Models 1 to 6 is a variation on the arithmetic you just did.

07 Interview questions

What an interviewer asks about ages

Ten in escalating order — the two rules, then the reading traps, then the sufficiency question that separates people who understand the drift from people who have memorised a method.

What is the one fact that never changes in an ages problem?
The difference between two people's ages. Both people gain the same number of years over the same period, so (A + t) − (B + t) is always A − B. If Radha is 5 years older than Vijay today, she was 5 years older at his birth and she will be 5 years older at 100.
Then why does the ratio of their ages keep changing?
Because the gap is a fixed number but the ages are not. Five years is a third of Vijay's age when he is 15 and a nineteenth of it when he is 95, so the same difference is a shrinking share of a growing age. The ratio therefore drifts towards 1 : 1 as time runs forward and away from it going back.
A question says the ratio was 3 : 1 ten years ago and will be 2 : 1 ten years from now. Is that consistent?
Yes, and it has to be. The ratio can only move towards 1 : 1 as time passes, so a later ratio closer to 1 : 1 is exactly what you expect. It would be the reverse — 2 : 1 earlier and 3 : 1 later — that is impossible for two real people.
Five years ago their ages were in the ratio 1 : 2. Can you write Vijay = x and Radha = 2x?
Only if you then remember that those are their ages five years ago, not today. Present ages would be (x + 5) and (2x + 5). Both setups are correct; mixing them — writing 1 : 2 against present ages — is the single most common mistake in the chapter.
Why do you insist on brackets when you shift an age in time?
Because 3x + 5 is not 8x. A variable and a plain number cannot be added into one term. When a multiplier is applied on top of a shift it matters even more: twice the age five years hence is 2(x + 5) = 2x + 10, not 2x + 5.
Is one ratio at one moment ever enough to find two ages?
No. A ratio is one equation in two unknowns, so it fixes the shape of the answer and not its size — 2 : 3 fits 10 and 15 just as well as 200 and 300. You need a second independent fact: another ratio at a different moment, a sum, a difference, or one actual age.
What if the two ratios I am given are the same ratio at two different moments?
Then they still are not sufficient, and the algebra tells you so — the two equations become the same line. It also makes physical sense: a ratio that has not moved in ten years can only belong to two people who are the same age, in which case it is 1 : 1 and there is nothing to find.
What is the fastest route when both ratios have the same difference in units?
The unit-shift shortcut. One unit of the ratio equals the elapsed years divided by the shift in units. For 2 : 3 today and 3 : 4 in five years, Vijay's share went from 2 units to 3 units over 5 years, so 1 unit = 5 and he is 10. When the two differences are unequal, scale one ratio up until they match — 3 : 1 has a gap of 2 units, so pair it with 4 : 2 rather than 2 : 1.
A student gets a present age of 42.5 years. Is that a mistake?
Not by itself. Nothing in the chapter forces whole-number ages, and standard exam questions do produce halves — a 15-year gap split across a 3 : 1 ratio gives units of 7.5. Treat a fraction as a signal to re-check, not as proof of error. A negative age is proof of error.
When would you actually use any of this outside an exam?
Directly, almost never — nobody solves for their cousin's age with simultaneous equations. What transfers is the habit: separating the quantity that is invariant from the quantity that moves. That is the same move as tracking a total instead of an average, or a difference instead of a percentage, and those show up constantly.

08 Practice problems

Six that test the reading, not the algebra

Every one of these is solvable with the bracket method and a gap check. Two of them have answers that are not whole numbers, on purpose — do not talk yourself out of a correct answer.

One person, two time zones

Easy
Aman's age after 15 years will be 4 times his age 15 years back. Find his present age.
Follow-up
There is only one person, so there is no gap and no ratio to drift — but both time shifts still need brackets, and the two of them are 30 years apart, not 15.
Show the hint
Write his present age as x, then form the equation x + 15 = 4(x − 15).

A gap and a past ratio

Easy
Nisha is 15 years elder to Romi. Five years ago Nisha was 3 times as old as Romi. Find Nisha's present age.
Follow-up
The 15-year gap was also 15 years five years ago, so it can be spent directly on the past ratio. The answer is not a whole number, and that is not a mistake.
Show the hint
In the past ratio 3 : 1 the difference is 2 units, and those 2 units are the 15-year gap.

Both facts away from today

Medium
Five years ago the ages of X and Y were in the ratio 6 : 5. Ten years hence the ratio will be 8 : 7. Find Y's present age.
Follow-up
Neither fact mentions the present, so you have to travel to today at the end rather than the start. Both ratios have a 1-unit difference, so the shortcut applies across a 15-year jump.
Show the hint
Y's share moves from 5 units to 7 units, and those 2 units cover the whole 15 years.

When the unit differences disagree

Medium
The present ages of two brothers are in the ratio 1 : 2, and five years ago the ratio was 1 : 3. Find the ratio of their ages five years from now.
Follow-up
1 : 2 has a difference of 1 unit and 1 : 3 has a difference of 2 units, so the shortcut does not apply until you rescale one of them. The answer is a ratio, not an age, so do not stop at the ages.
Show the hint
Double the present ratio to 2 : 4 so both facts describe a gap of 2 units, then compare.

A gap in years, an answer in ratio

Medium
The difference between the present ages of M and N is 14 years, and seven years ago their ages were in the ratio 5 : 7. Find the ratio of their ages fourteen years from now.
Follow-up
The gap is handed to you in years and the ratio in units, so one line converts between them. Then the question asks for a ratio at a third moment, which is where the drift becomes the point rather than a footnote.
Show the hint
In 5 : 7 the difference is 2 units and that equals 14 years, so one unit is 7.

The time itself is unknown

Hard
k years ago the ages of A and B were in the ratio 3 : 4, and k years from now they will be in the ratio 5 : 6. The sum of their present ages is 72 years. Find k.
Follow-up
Here the elapsed time is the unknown and the ratios are given, which inverts the usual question. The unit-shift relation still holds, and applying it with a symbolic jump of 2k years is the whole solution — it also shows why that shortcut is a real identity and not a coincidence about nice numbers.
Show the hint
Both ratios have a 1-unit difference and the shift is 2 units over 2k years, so one unit is k itself. Now express both present ages in units of k and use the sum.