Problems on Ages Tricks: Unit Method

📑 On this page

⚡ The short answer

In every problem on ages, the gap between two people's ages never changes: everyone ages one year per year. So sum and difference give each age directly, two ratios compare once their differences match (the unit method), and 'k times as old' means the younger is gap ÷ (k − 1). In the exam, test the options.

🎬 The 30-second story

The birthday that broke Meher's brain

“I'm 16 today and Papa's 48. He is exactly three times my age!”

🎂

Meher

Cake, then maths.

“Enjoy it. In 16 years I'll be only twice your age.”

⏳

Kabir

Kabir, cutting the cake.

“So she's catching up? One day you'll be the same age?”

🤯

Aarav

Aarav panics for Kabir.

“Never. He's 32 years older today and he'll be 32 years older at 80.”

📏

Zoya

Zoya ends the debate.

Tara's maths: The gap (32 years) is frozen forever. The ratio slides: 3 : 1 today, 2 : 1 in 16 years, 3 : 2 in 48, towards 1 : 1 but never touching it.

Why is the age gap the only number that never changes?

Your birthday comes once a year. So does your dad's. In 16 years you both add exactly 16, so the gap is fixed the day the younger one is born. Ratios, sums and averages move. The difference does not, and that frozen number is your handle on every ages question.

📊 The data

Kabir and Meher over time: the gap stays 32

KabirMeher

Meher is born · no ratio yet

32 yrs

0 yrs

Today · 3 : 1

48 yrs

16 yrs

In 16 years · 2 : 1

64 yrs

32 yrs

In 32 years · 5 : 3

80 yrs

48 yrs

In 48 years · 3 : 2

96 yrs

64 yrs

The gap is always 32; the ratio slides 3 → 2 → 1.67 → 1.5.

Why does the ratio shrink? The same 16 years is a huge slice of Meher's 16 and a small slice of Kabir's 48, so the ratio is pulled towards 1 : 1. Go back in time and it stretches: eight years ago, 40 : 8 = 5 : 1.

📌 Three facts, every question

1. Everyone ages by the same t years.
2. The difference never changes.
3. The sum of n ages changes by n × t: a family of 4 gains 12 years in 3 years, not 3.

How do you set up any problems-on-ages question?

Pick one unknown: the younger person's age now. Write everyone else in terms of it, then shift every row by the years mentioned. One unknown, one equation.

🧮 The formula

Ages in one line

🎂

Age now

⏳

± t years

the same t for everyone

📅

Age then

Person	5 years ago	Now	In 10 years
Son	x − 5	x	x + 10
Father	4x − 5	4x	4x + 10
Gap	3x	3x	3x

→ scroll to see more

A father is 4 times his son's age now; 5 years ago he was 7 times as old. Read one column: 4x − 5 = 7(x − 5), so 3x = 30 and x = 10. Son 10, father 40. Gap check: 40 − 10 = 30 ✓.

What is the ratio-unit method for ages?

Two ratios at two times? Skip the algebra. 4 : 3 means 4 units and 3 units. Both people gain the same years, so both gain the same number of units. Once the unit differences match, the jump in units is the time that passed.

✍️ Worked example Warm-up ⏱ 30 sec

Sum and difference given

Zoya and her brother's ages add up to 34; she is 6 years older. Find both ages and their ratio 5 years ago.

1
Elder = (sum + difference) ÷ 2 = (34 + 6) ÷ 2 = 20.
2
Younger = (sum − difference) ÷ 2 = (34 − 6) ÷ 2 = 14.
3
Five years ago both were 5 younger: 15 and 9.
4
15 : 9 simplifies to 5 : 3.

Answer: Zoya 20, brother 14; ratio 5 years ago 5 : 3

⚡ The exam-hall shortcut

Sum and difference given? No equation: halve (sum + difference).

✍️ Worked example CLAT level ⏱ 40 sec

Ratio now, ratio later

Aarav and Rohan's ages are in the ratio 4 : 3. In 6 years it will be 5 : 4. Find their ages.

1
Unit differences: 4 − 3 = 1 and 5 − 4 = 1. Already equal.
2
Each person moves up 1 unit (4 → 5, 3 → 4) in 6 years, so 1 unit = 6 years.
3
Aarav = 4 × 6 = 24. Rohan = 3 × 6 = 18.
4
Check: in 6 years, 30 : 24 = 5 : 4 ✓.

Answer: Aarav 24, Rohan 18

⚡ The slow route

(4k + 6) ÷ (3k + 6) = 5/4 gives 16k + 24 = 15k + 30, so k = 6. Same answer, more ink.

✍️ Worked example Tricky ⏱ 75 sec

Two ratios, ten years apart

Five years ago, Kabir's uncle and the uncle's son were aged in the ratio 7 : 3. Five years from now it will be 2 : 1. Find their ages now.

1
The ratios are 10 years apart.
2
Differences: 7 − 3 = 4 units, but 2 − 1 = 1. The gap is fixed, so scale 2 : 1 to 8 : 4.
3
7 : 3 → 8 : 4: each gained 1 unit in 10 years, so 1 unit = 10 years.
4
Then 70 and 30; now 75 and 35.
5
Check: in 5 years, 80 : 40 = 2 : 1 ✓.

Answer: Uncle 75, son 35

⚡ The exam-hall shortcut

Everyone skips matching the differences. Skip it and units 'fall' from 7 to 2, impossible for people getting older.

What's the fastest way to crack ages questions in the exam?

Ages are whole numbers and the options are four tidy integers, so put each option into the sentence and see which survives. Otherwise, the tree picks your tool.

🧭 Decision flow

Which ages method?

❓ Sum and difference given?

Yes →

Elder = (sum + difference) ÷ 2

No ↓

❓ Two ratios at two times?

Yes →

Unit method: match differences, 1 unit = years ÷ units gained

No ↓

❓ Options are four numbers?

Yes →

Plug in the options; non-whole ages die first

No →

x = younger's age now, one equation

Enough data?

💡 The k-times shortcut

When the elder is k times the younger, the younger is gap ÷ (k − 1). Kabir and Meher (gap 32): three times when she is 32 ÷ 2 = 16, twice when she is 32. 'Twice as old' means the younger's age equals the gap.

✍️ Worked example CLAT level ⏱ 25 sec

Let the options do the work

Five years ago, Rohan was 3 times as old as his niece. Five years from now, he will be twice as old as her. How old is Rohan now? (a) 30 (b) 35 (c) 40 (d) 25

1
Five years ago Rohan's age was 3 × a whole number.
2
Options minus 5: 25, 30, 35, 20. Only 30 divides by 3, so only (b) 35 works.
3
Confirm: niece 10 then, 15 now, 20 later; Rohan 40 = 2 × 20 ✓.

Answer: (b) 35 years

⚡ Using the k-times shortcut

Gap g: the niece was g ÷ 2 then and g later, so g ÷ 2 + 10 = g, g = 20. Niece 15, Rohan 35.

🎛️ Try it yourself — change any number

Age finder: sum and gap to ratio

Kabir + Meher now (years) Kabir − Meher (years) Years ahead (negative = ago)

Elder now

Younger now

Ratio then (elder : younger)

2 : 1

Elder = (sum + difference) ÷ 2; the difference never changes

Kabir (48) and Meher (16): 2 : 1 in 16 years. Try 32 (5 : 3) or −8 (5 : 1).

How do ages hide inside a CLAT caselet?

Since 2020 CLAT Quant is all caselets, with no one-line ages question. But a family caselet can test the same 'who was how old when' logic. Find the one person whose present age is given and hang everyone else off that anchor using the fixed gaps.

📊 The data

The Mehra family today (years)

Usha · 45 + 25

70 yrs

Mr Mehra · 16 + 29

45 yrs

Mrs Mehra · 45 − 3

42 yrs

Ira · 12 + 4

16 yrs

Dev · the anchor

12 yrs

Total 185 years across five people, so the average age is 37.

🧪 CLAT-style practice (not a past question) Ages chain in a caselet

The Mehra family has five members living together. Grandmother Usha was 25 when her son, Mr Mehra, was born. Mr Mehra was 29 when his daughter Ira was born, and his wife is 3 years younger than him. Ira was 4 years old when her brother Dev was born. Dev turns 12 today.

What is the average age of the Mehra family today?

A30.6 years

B37 years

C38.2 years

DCannot be determined

Reveal answer + reasoning

Answer: B

Anchor on Dev = 12. Ira = 16. Mr Mehra = 16 + 29 = 45. Mrs Mehra = 42. Usha = 45 + 25 = 70. Sum 185 ÷ 5 = 37. About 60 seconds.

Why the others fail

A — Makes Ira 4 years younger than Dev (8).
C — Makes Mrs Mehra 3 years older (48) instead of younger.
D — Dev's age anchors the chain, so it can be.

⚠️ Where 'cannot be determined' comes from

Delete 'Dev turns 12 today' and every gap survives, yet no age can be fixed. Gaps give relative ages only; you need one absolute fact (an age, a sum, or two ratios). One ratio alone is not enough.

Which traps cost the most marks in ages questions?

'Ago' for one person only. 'Five years ago the father was 3 times the son' is F − 5 = 3(S − 5), not F − 5 = 3S.
Sum shifted once. Four people, five years: the total drops by 20, not 5.
'When A was born, B was k'. That just means B − A = k.
A fractional age. Answers come out whole. Re-read the numbers, then the setup.
Noise. A mother's age may be there only to slow you. Which gaps link the two people you need?

How has CLAT asked problems on ages?

Both real ages questions are from CLAT 2019, the last paper with stand-alone arithmetic. There is no official 2019 key, so these are our solutions.

📝 Asked in CLAT 2019 · Q48 Age now vs age then

Beena got married 8 years ago. Today, her age is 1¼ times her age at the time of marriage. If her daughter's age is 1/10 times her age, then her daughter's age is:

A3 years

B4 years

C5 years

D2 years

Reveal answer + reasoning

Answer: B

Our solution. Our copy of the paper's text prints '11/4'; it is the mixed number 1¼ (as 11/4, Beena would be 88/7 ≈ 12.6). Let her age now be x: x = 5/4 × (x − 8), so 4x = 5x − 40 and x = 40. Daughter = 40 ÷ 10 = 4 years.
Options route (30 seconds): Beena is 10 × each option: 30, 40, 50 or 20. Only 1¼ × (40 − 8) = 40 fits.

Why the others fail

A — Beena 30, but 1¼ × 22 = 27.5.
C — Beena 50, but 1¼ × 42 = 52.5.
D — Beena 20, but 1¼ × 12 = 15.

📝 Asked in CLAT 2019 · Q51 Birth-order timeline

Kiran's brother is 5 years older to her. Her father was 30 years old when Kiran's sister was born, while her mother was 28 years old when Kiran was born. If Kiran's sister was 2 years old when her brother was born, what was the age of their father when Kiran's brother was born?

A32

B34

C37

D30

Reveal answer + reasoning

Answer: A

Our solution. Birth order: sister, then brother 2 years later, then Kiran 5 years after him. The father was 30 at the sister's birth, so at the brother's he was 30 + 2 = 32. The mother's age and Kiran's gap are noise. 30 seconds.

Why the others fail

B — Counts the 2-year gap twice.
C — 30 + 2 + 5 adds Kiran's gap, but Kiran came after the brother.
D — The father's age at the sister's birth.

💥 Myth vs fact

Ages myths that cost marks

🙅 Myth

If Papa is 3 times my age now, he'll always be 3 times my age.

Tap to bust it →

✅ Fact

Only the difference is constant. The ratio slides towards 1 : 1: 3 : 1, then 2 : 1, then 3 : 2.

🙅 Myth

Live long enough and the younger one catches up.

Tap to bust it →

✅ Fact

Never. Both add a year each year, so the gap is permanent.

🙅 Myth

A family's total age goes up by 1 every year.

Tap to bust it →

✅ Fact

It rises by 1 for each member: 5 per year for five people. The average rises by exactly 1.

Can you solve these five ages questions?

🎯 Quick quiz

Five ages questions, one frozen gap

0 / 5

Rohan is 7 years older than Zoya.

Q1. How much older than Zoya will Rohan be in 9 years?

Aarav and his cousin's ages are 5 : 2 now and will be 3 : 2 in 8 years.

Q2. How old is Aarav now?

Four fest-committee members have a total age of 68.

Q3. What was their total age 5 years ago?

Mr Mehra is 45 and Ira is 16.

Q4. In how many years will Mr Mehra be exactly twice Ira's age?

Zoya and Aarav's ages are in the ratio 6 : 5.

Q5. How old is Zoya?

🎯 Problems on ages in a nutshell

The difference between two ages never changes; ratios, sums and averages do.
Sum and difference given: elder = (sum + difference) ÷ 2.
Two ratios: match the differences, then 1 unit = years passed ÷ units gained.
'k times as old': the younger is gap ÷ (k − 1) at that moment.
Only gaps and no anchor: cannot be determined.

📎 Formula card

Sum S, difference D: elder = (S + D) ÷ 2; younger = (S − D) ÷ 2.
t years ago, A was k times B: A − t = k(B − t).
Ratio a : b now and c : d after t years, with a − b = c − d: 1 unit = t ÷ (c − a).
If a − b ≠ c − d, scale first (2 : 1 → 8 : 4 to match 7 : 3).
Elder = k × younger when younger = D ÷ (k − 1).
Sum of n ages changes by n × t; their average by exactly t.

Frequently asked questions

What is the trick for problems on ages?

The age difference between two people never changes, because both get older at the same rate. So sum and difference give each age directly, two ratios compare once their differences match, and 'k times as old' means the younger is the gap divided by k − 1. Then test the options: ages come out whole.

How do you solve age problems with ratios?

Treat each ratio as units. If ages are 4 : 3 now and 5 : 4 in 6 years, both differences are 1 unit, so each person gained 1 unit in 6 years. One unit is 6 years, and the ages are 24 and 18. If the differences don't match, scale one ratio until they do.

Are problems on ages asked in CLAT?

CLAT 2019 asked two stand-alone ages questions, Q48 and Q51. Since 2020 Quant has been built on data caselets, so a one-line ages question is unlikely. A family caselet can still test the same 'who was how old when' logic, which the constant-gap method handles in under a minute.

Does the ratio of two people's ages change over time?

Yes. Adding the same years to both ages pulls the ratio towards 1 : 1. A father of 48 and a daughter of 16 are 3 : 1; in 16 years they are 2 : 1, and in 48 years 3 : 2. It never reaches 1 : 1, because the 32-year gap never closes.