📑 On this page
⚡ The short answer
In a mixture question, find the part that doesn't change and track it in litres. Adding water? The acid is fixed. Adding acid? The water is. For replacement with water, each round keeps the same fraction: Pure left = T × (1 − x/T)n. The trap: subtracting the same amount every round.
🎬 The 30-second story
The cold-coffee cover-up
1
“I'll take 2 litres for the team and top it up with water. Nobody will know.”
☕
Aarav
College fest. A 20-litre cold-coffee dispenser.
2
“Did it three times. 6 litres gone, so it's 70% coffee. Fine.”
🔁
Aarav
Three bottles later.
3
“Not 6 litres of coffee. Your later bottles were already part water.”
🧮
Zoya
4
“Wait. So it's actually stronger than 70%?”
🥤
Meher
The judges are walking over.
Tara's maths: Each refill leaves 18 of 20 litres untouched, so 9/10 of the coffee survives every round. After three: 20 × (9/10)3 = 14.58 L of coffee, i.e. 72.9%. Aarav took 5.42 L of coffee, not 6.
What is concentration, and what actually changes in a mixture?
A mixture is two things sharing a container: milk and water, acid and water, ethanol and petrol. Concentration is one part's share of the whole: acid ÷ total × 100. Every question here is a percentage question in a lab coat. The skill is noticing what each action does: pour some out and both parts leave together; add water or boil it and only the water moves.
🟦 Picture the percent
Iced-tea syrup: what pouring and diluting do
Full jug: 30%
30 parts syrup in 100
One glass: 30%
a sample has the same strength
Doubled with water: 15%
same syrup, twice the total
📌 Pouring out never changes the strength
A glass from a stirred jug is a mini copy of the jug: same strength, and so is what's left. Concentration changes only when you add or remove one part on its own.
How do you solve mixture questions by tracking quantities?
Convert every percentage or ratio into litres of each part, apply the action, then convert back. This ledger never fails.
| You do this | What changes | What stays fixed | So track |
|---|---|---|---|
| Add water | Water, total | The solute | The solute |
| Add pure solute | Solute, total | The water | The water |
| Boil / evaporate | Water, total | The solute | The solute |
| Pour some out | Both, in ratio | The % | The % |
| Draw out, top up with water | Solute falls each round | The total | Fraction kept |
→ scroll to see more
✍️ Worked example
Warm-up
⏱ 30 sec
Kabir's chai base
Kabir's stall has 45 litres of chai base with milk and water in the ratio 7 : 2. Customers say it's too thick. How much water must he add to make the ratio 7 : 3?
- 1Turn the ratio into litres: 9 parts = 45 L, so 1 part = 5 L. Milk = 35 L, water = 10 L.
- 2Water leaves the milk alone: still 35 L = 7 parts, so a part is still 5 L. 7 : 3 needs 15 L of water.
- 3Check: 35 : 15 = 7 : 3.
Answer: 5 litres of water.
⚡ The exam-hall shortcut
Milk's 7 parts never move, so water's 2 parts become 3: one extra part = 45 ÷ 9 = 5 L.
✍️ Worked example
CLAT level
⏱ 35 sec
Boiling down the fest syrup
Meher has 40 kg of sugar syrup that is 15% sugar. The recipe needs 24% syrup, so she boils off water. How many kg of water must evaporate?
- 1Boiling removes only water, so sugar is frozen: 15% of 40 = 6 kg.
- 2At 24%, those 6 kg must be 24% of the new total: 6 ÷ 0.24 = 25 kg.
- 3Water boiled off = 40 − 25 = 15 kg.
- 4Check: 6 ÷ 25 = 24%.
Answer: 15 kg of water.
⚡ C₁ × Q₁ = C₂ × Q₂
Strength × quantity is constant when one part is fixed: 15 × 40 = 24 × Q, so Q = 600 ÷ 24 = 25 kg.
What happens when you remove some and replace it with water?
This is where Aarav went wrong. Only the first draw is pure coffee; every later draw is diluted, so it removes less coffee. Think in fractions instead: drawing 2 L from 20 L leaves 18/20 of everything, and the water top-up adds no coffee. So each round multiplies the coffee by 9/10, a successive percentage decrease, just like depreciation.
📊 The data
Coffee left in the 20-litre dispenser
Actual coffee leftAarav's 'minus 2 L' guess
Start
20 L
20 L
After refill 1 · same: the first draw is pure
18 L
18 L
After refill 2 · 18 × 0.9
16.2 L
16 L
After refill 3 · 16.2 × 0.9
14.58 L
14 L
The real amount left is always more than the linear guess.
🧮 The formula
The replacement formula
🧪
Pure stuff at the start
e.g. 20 L
×
🔁
(1 − x/T)n
x drawn, T total, n rounds
=
✅
Pure stuff left
20 × 0.93 = 14.58 L
Works for percentages too: 100% × 0.93 = 72.9%.
✍️ Worked example
Tricky
⏱ 60 sec
Working the formula backwards
A can holds 80 litres of pure milk. Some quantity is drawn out and replaced with water, twice. The milk : water ratio is now 16 : 9. How many litres were drawn each time?
- 1Milk left = 16 of 25 parts, so (1 − x/80)2 = 16/25.
- 2Square root: 1 − x/80 = 4/5, so x/80 = 1/5.
- 3x = 80 ÷ 5 = 16 L. Check: 80 × 0.8 × 0.8 = 51.2 L milk, 28.8 L water; 51.2 : 28.8 = 16 : 9.
Answer: 16 litres each time.
⚡ The exam-hall shortcut
16 : 9 adds to 25: a perfect square over a perfect square signals two rounds of 4/5 each.
🎛️ Try it yourself — change any number
The dispenser simulator
Final strength
72.9%
Pure liquid left
14.58 L
100% × (1 − 2/20)3 = 72.9%
It opens on the story: 72.9%, 14.58 L. Set the times to 10: still 34.87%, not 0%. Take 10 L (half) three times: 12.5%.
When does the replacement formula break?
⚠️ Only for a top-up with none of the thing you track
(1 − x/T)n assumes the top-up adds zero of the tracked part and that x and T never change. Different draws? Multiply the separate fractions: 4 L then 5 L from 20 L leaves 20 × 16/20 × 15/20. A top-up that is itself a mixture needs the tweak below.
✍️ Worked example
Tricky
⏱ 75 sec
Topping up with a weaker juice
Kabir's 30-litre juice drum is 60% fruit pulp. Twice, he draws out 10 litres and replaces it with a cheaper 30%-pulp drink. What is the pulp concentration now?
- 1Round 1: 20 L left at 60% = 12 L pulp. The 10 L top-up adds 3 L pulp. 15 ÷ 30 = 50%.
- 2Round 2: 20 L left at 50% = 10 L pulp, plus 3 L again. 13 ÷ 30 = 43.33%.
- 3The plain formula gives 60% × (2/3)2 = 26.67%: far too low, because the top-up isn't water.
Answer: About 43.33% pulp.
⚡ The gap version
Shrink the gap above the top-up's strength: 30 + (60 − 30) × (2/3)2 = 30 + 13.33 = 43.33%.
What is the fastest way to crack mixture questions in the exam?
- ✓Name the frozen part first: 'acid fixed', 'water fixed'. Half the question is done.
- ✓Dilution and evaporation: C₁ × Q₁ = C₂ × Q₂.
- ✓Know the 0.9 ladder: drawing a tenth leaves 90%, 81%, 72.9%, 65.61%. A fifth leaves 80%, 64%, 51.2%. A quarter leaves 75%, 56.25%, 42.19%.
- ✓Kill options with the linear guess: after two or more rounds, the real amount left is above T − n × x.
- ✓Squares and cubes give it away: 16 : 9 or 64 : 61 mean (4/5)2 or (4/5)3 is left.
How do mixture questions hide in a CLAT caselet?
CLAT won't say 'a vessel contains milk and water'. It describes a fuel depot, a dairy or a lab, gives a blend % and a target, and asks how much to add. The skill is spotting the frozen part.
✍️ Worked example
CLAT level
⏱ 50 sec
The fuel depot's new blend
A fuel depot holds 50,000 litres of fuel that is 10% ethanol, 90% petrol. To sell a 20%-ethanol blend, it adds pure ethanol. How many litres?
- 1Adding ethanol leaves the petrol fixed: 90% of 50,000 = 45,000 L.
- 2In the new blend, petrol is 80%: 45,000 ÷ 0.8 = 56,250 L in total.
- 3Ethanol added = 56,250 − 50,000 = 6,250 L.
- 4Check: ethanol = 5,000 + 6,250 = 11,250; 11,250 ÷ 56,250 = 20%.
Answer: 6,250 litres of ethanol.
⚡ The exam-hall shortcut
The tempting 5,000 L gives only 10,000 ÷ 55,000 ≈ 18.2%, because adding ethanol also raises the total.
📊 The data
The depot tank, before and after
PetrolEthanol
Before: 50,000 L · 10% ethanol
45,000 L
5,000 L
After: 56,250 L · 20% ethanol
45,000 L
11,250 L
The grey bar never moves: that's your frozen part.
Which mixture traps cost the most marks?
| Trap | What it looks like | The fix |
|---|---|---|
| Wrong frozen part | Tracking acid when acid is being added | Track whatever you didn't touch |
| Linear replacement | 2 L drawn 3 times = 6 L lost | Multiply by (1 − x/T) each round |
| Old total as the base | Water ÷ 100 L when the tank is 120 L | Divide by the final total |
| Adding ratios | 5 : 3 plus 3 : 1 = 8 : 4 | Convert each to litres first |
→ scroll to see more
How has CLAT asked mixture and replacement questions?
One stand-alone question in 2018, solved by the ledger. The 2020–25 papers haven't set one, but a blend caselet fits their format exactly, so a CLAT-style question follows.
📝 Asked in CLAT 2018
· Q142
Draw out, top up with two parts
A stand-alone question in the 2018 paper.
In a 120 litre of solution of Acid and water, acid is 75%. A person takes out 20 litres of this solution and added 16.2 litres of acid and 3.8 litres of water in the remaining solution. What is the percentage of water in the final solution?
A22
B24
C25
D28
Reveal answer + reasoning
Answer: B
Ledger. Start: acid 75% of 120 = 90 L, water 30 L.Take out 20 L at the same 3 : 1 split: 15 L acid, 5 L water. Left: 75 L acid, 25 L water.
Add 16.2 L acid and 3.8 L water: 91.2 L acid, 28.8 L water, total back to 120 L.
Water % = 28.8 ÷ 120 × 100 = 24%. Check: 91.2 + 28.8 = 120. Time: 40 seconds.
Why the others fail
- A — (25 + 19) ÷ 2: a plain average of the leftover's 25% and the top-up's 19%. The leftover is five times bigger.
- C — The water share after the removal alone. The top-up changes it.
- D — 28.8 L is litres, not %. Divide by the final 120 L.
🧪 CLAT-style practice (not a past question)
Repeated replacement in a caselet
Milk at a village collection centre is stored in a 100-litre steel can. On Monday morning the can was full of pure milk. A dishonest helper drew out 10 litres, sold it privately, and poured in 10 litres of water so the level looked unchanged. He did exactly the same on Tuesday.
What percentage of the can was water after Tuesday's top-up?
A20%
B19%
C81%
D10%
Reveal answer + reasoning
Answer: B
Each round keeps 9/10 of the milk: 100 × 0.9 × 0.9 = 81 L, so water = 19 L of 100 L: 19%. The second draw took only 9 L of milk. Time: 25 seconds.
Why the others fail
- A — The linear trap: 10 L + 10 L of water. The second draw removed 1 L of water too.
- C — That's the milk share, not the water.
- D — Only one round counted.
💥 Myth vs fact
Myths that cost marks
🙅 Myth
Drawing 10% and topping up three times removes 30% of the milk.
Tap to bust it →
✅ Fact
It removes 27.1%. Each round takes 10% of what's left: 0.93 = 0.729.
🙅 Myth
In every mixture question, track the acid (or milk).
Tap to bust it →
✅ Fact
Track whatever is untouched. Adding pure acid? The water is the fixed part.
🙅 Myth
Keep replacing with water and the pure stuff eventually hits zero.
Tap to bust it →
✅ Fact
It only gets smaller: 0.9n is never 0. Ten rounds still leave 34.87%.
🎯 Quick quiz
Five quick mixtures
0 / 5
Meher has 60 litres of salt solution that is 20% salt.
Q1. How much water must evaporate to make it 30% salt?
Salt fixed at 12 L. 12 ÷ 0.3 = 40 L total, so 20 L must go.A 50-litre mix has milk and water in the ratio 4 : 1.
Q2. How much pure milk must be added so that water is only 10% of the mix?
Adding milk fixes the water: 10 L. At 10%, total = 100 L, so add 50 L.A 50-litre drum of pure juice. 5 litres are drawn and replaced with water, three times.
Q3. How much pure juice is left?
50 × 0.93 = 36.45 L. A is the linear guess.A can holds 40 litres of pure milk. 8 litres are drawn and replaced with water, three times.
Q4. What is the final milk : water ratio?
(4/5)3 = 64/125 is milk, so milk : water = 64 : 61. C is milk : total; A stops at two rounds.Zoya mixes 30 litres of a 3 : 2 milk-water mix with 20 litres of a 1 : 1 mix.
Q5. What is the milk : water ratio of the result?
Litres first. 30 L at 3 : 2 = 18 milk + 12 water; 20 L at 1 : 1 = 10 + 10. Total 28 : 22 = 14 : 11. A adds the ratios; C is flipped.🎯 Mixtures in a nutshell
- Concentration = part ÷ whole. Every mixture question is a % question.
- Find the frozen part and track it in litres.
- Replacement with water: multiply by (1 − x/T) every round, never subtract.
- Divide by the final total, not the old one.
📎 Formula card
- Add water / evaporate: C₁ × Q₁ = C₂ × Q₂ (solute fixed).
- Add pure solute: (100 − C₁) × Q₁ = (100 − C₂) × Q₂ (water fixed).
- Replacement with water: pure left = T × (1 − x/T)n.
- Ratio after n rounds: pure : water = (T − x)n : [Tn − (T − x)n].
- Top-up at c%: final % = c + (start − c) × (1 − x/T)n.
- Unequal draws: multiply (1 − x₁/T) × (1 − x₂/T) × …
Frequently asked questions
What is the formula for removal and replacement in mixtures?
If a container holds T litres and x litres are drawn out and replaced with water n times, the pure liquid left is T × (1 − x/T)^n. Each round keeps the same fraction, so you multiply, never subtract. For 20 litres with 2 litres replaced three times, 20 × 0.9 × 0.9 × 0.9 = 14.58 litres remain.
How do I find how much water to add to dilute a solution?
Adding water doesn't change the amount of solute, so fix it first. Divide it by the target concentration to get the new total, then subtract the old total. Twenty-five litres of 80% juice holds 20 L of juice; at 50% the total must be 40 L, so add 15 L of water.
Why isn't the loss the same in every replacement round?
After the first round you draw out a mixture, not the pure liquid. The second draw already contains some water, so it removes less of the pure stuff. The amount left falls by a fixed percentage each round, like depreciation, and always stays a little above the subtract-each-time guess.
Are mixture and replacement questions still asked in CLAT?
CLAT 2018 had a stand-alone question on a 120-litre acid solution. Since 2020 the section has been built on caselets, and blends and compositions are exactly what those passages describe. The ledger method, tracking litres of each part, answers them quickly.
Keep going
← Previous topic
Alligation: the Cross Method
Two prices, one target: the cross gives the mixing ratio in seconds
Next topic →
Average Speed: Why It Is Not the Average of Speeds
Why 60 there and 40 back is 48, not 50, and the 2xy/(x+y) shortcut
📚
Part of the chapter
Averages, Mixtures & Alligation for CLAT Quantitative Techniques
See how this topic fits the whole of Averages, Mixtures & Alligation →
Connected topics in other chapters
Alligation: the Cross Method
When the question asks for the mixing ratio to hit a target, the cross is quicker than a ledger.
Successive Percentage Change
Repeated replacement is a successive % decrease in disguise.
Population Growth and Depreciation
The same (1 − r)ⁿ curve, applied to machines and towns instead of milk.
Weighted Average
The concentration of two mixtures poured together is a weighted average by volume.
Dividing in a Given Ratio
Turning a milk : water ratio into litres is the first step of every ledger.
Every topic in Averages, Mixtures & Alligation
- 1.Average: Formula, Meaning and the Balance Idea
- 2.When Someone Joins or Leaves: How the Average Shifts
- 3.Weighted Average: Why You Can't Just Average the Averages
- 4.Alligation: the Cross Method
- 5.Mixtures, Concentration and Replacement (you're here)
- 6.Average Speed: Why It Is Not the Average of Speeds