Practice Questions
Concept Explanation
In replacement, a fixed quantity is removed from a vessel and an equal quantity of another liquid (often water) is added. The concentration of the original liquid decreases.
Key Formulas
- After one replacement of a litres from V litres of pure liquid with water: pure left = V − a = V(1 − a/V)
- Fraction of original left after one draw = (V − a)/V
- After replacement, new concentration = old concentration × (V − a)/V
Example
100 L pure milk. Remove 20 L, add 20 L water. Pure milk left = 80 L. Concentration = 80%.
Common Trap
Replacement is not the same as simple dilution. You remove a mixture (if not pure initially), not just the solute.
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Q1 Easy80 litres of pure milk has 16 litres removed and replaced by water once. How much milk remains?ExplanationAfter one replacement from pure milk, milk left = V − a = 80 − 16 = 64 litres (equivalently 80 × (1 − 16/80) = 64). Hence the answer is 64 litres.
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Q2 EasyAfter replacing 20 litres from 100 litres of pure spirit with water, the concentration of spirit is:ExplanationSpirit left = 100 − 20 = 80 litres in 100 litres total, so concentration = 80%. Hence the answer is 80%.
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Q3 EasyFraction of original liquid left after replacing 1/4 of a vessel with water once is:ExplanationFraction left after one replacement of fraction f is (1 − f). Here f = 1/4, so fraction left = 3/4. Hence the answer is 3/4.
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Q4 MediumA vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?ExplanationSyrup fraction initially = 5/8. Let fraction x of the mixture be replaced with water. Syrup left = (5/8)(1 − x). For half water and half syrup, syrup fraction = 1/2. So (5/8)(1 − x) = 1/2 ⇒ 5(1 − x) = 4 ⇒ 1 − x = 4/5 ⇒ x = 1/5. Hence 1/5 of the mixture must be drawn off and replaced with water.
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Q5 MediumA container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?ExplanationAfter n replacements of a litres from V litres of pure liquid, quantity left = V(1 − a/V)^n. Here V = 40, a = 4, n = 3. Milk left = 40(1 − 4/40)^3 = 40(9/10)^3 = 40 × 729/1000 = 29160/1000 = 29.16 litres. Hence the answer is 29.16 litres.
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Q6 MediumA vessel contains 100 litres of milk. From this vessel, 10 litres of milk was taken out and replaced by water. This process was repeated further two times. The quantity of milk left in the vessel is:ExplanationFormula: milk left = V(1 − a/V)^n. Here V = 100, a = 10, n = 3. Milk left = 100(1 − 10/100)^3 = 100(0.9)^3 = 100 × 0.729 = 72.9 litres. Hence the answer is 72.9 litres.
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Q7 MediumFrom 50 litres of juice, 10 litres are drawn off and replaced with water, then the same is done once more. Juice left is:ExplanationJuice left = 50(1 − 10/50)^2 = 50(4/5)^2 = 50 × 16/25 = 32 litres. Hence the answer is 32 litres.
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Q8 MediumA tank has 60 litres of oil. 12 litres are replaced by petrol, and this is repeated once. Oil remaining is:ExplanationOil left = 60(1 − 12/60)^2 = 60(4/5)^2 = 60 × 16/25 = 38.4 litres. Hence the answer is 38.4 litres.
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Q9 Medium25 litres pure acid; 5 litres drawn and replaced with water twice. Acid left is:ExplanationAcid left = 25(1 − 5/25)^2 = 25(4/5)^2 = 25 × 16/25 = 16 litres. Hence the answer is 16 litres.
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Q10 HardA can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the can initially?ExplanationLet initial quantity = 12x litres, so A = 7x and B = 5x. In 9 litres drawn, A removed = (7/12) × 9 = 21/4 litres and B removed = 15/4 litres. After filling with B: A left = 7x − 21/4 and B left = 5x − 15/4 + 9 = 5x + 21/4. Given (7x − 21/4) : (5x + 21/4) = 7 : 9. So 9(7x − 21/4) = 7(5x + 21/4) ⇒ 63x − 189/4 = 35x + 147/4 ⇒ 28x = 336/4 = 84 ⇒ x = 3. Initial A = 7 × 3 = 21 litres. Hence the answer is 21 litres.
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Q11 HardFrom a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He repeated this process thrice. The ratio of wine to water is 343:169. Capacity of container:ExplanationAfter 3 replacements, wine/capacity = (1 − 15/C)^3. Wine:water = 343:169 ⇒ wine/total = 343/(343 + 169) = 343/512 = (7/8)^3. So 1 − 15/C = 7/8 ⇒ 15/C = 1/8 ⇒ C = 120 litres. Hence the capacity is 120 litres.
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Q12 Hard8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16:65. How much wine did the cask hold originally?ExplanationThe operation is performed 4 times in all. Wine left / capacity = (1 − 8/C)^4. Wine:water = 16:65 ⇒ wine/total = 16/(16 + 65) = 16/81 = (2/3)^4. So 1 − 8/C = 2/3 ⇒ 8/C = 1/3 ⇒ C = 24 litres. Hence the cask originally held 24 litres.
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Q13 Hard90 litres of milk; 9 litres replaced by water thrice. Milk left is:ExplanationMilk left = 90(1 − 9/90)^3 = 90(0.9)^3 = 90 × 0.729 = 65.61 litres. Hence the answer is 65.61 litres.
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Q14 HardMilk left after replacing 5 litres thrice from a 50-litre pure milk vessel is:ExplanationMilk left = 50(1 − 5/50)^3 = 50(0.9)^3 = 50 × 0.729 = 36.45 litres. Hence the answer is 36.45 litres.
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DS1 Easy
How much milk remains in the container?
Statement IA 40-litre container of pure milk has 4 litres replaced by water, and this process is done thrice in all.
Statement IIMilk left equals 40 × (0.9)^3 litres.
ExplanationStatement I alone: milk = 40(1 − 4/40)^3 = 29.16 litres. Statement II alone states the same computation. Either alone is sufficient. Answer: D. -
DS2 Easy
What fraction of the mixture must be replaced with water to make water and syrup equal.
Statement IInitially water:syrup = 3:5 in the vessel.
Statement IIThe target composition is half water and half syrup.
ExplanationStatement I alone lacks the target. Statement II alone lacks the initial composition. Together: (5/8)(1 − x) = 1/2 ⇒ x = 1/5. Both are needed. Answer: C.
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DS3 Medium
An HR analytics quiz asks for the capacity of a wine container. What is the capacity?
Statement I15 litres of wine were stolen and replaced with water thrice, and finally wine:water = 343:169.
Statement IIOnly the final wine:water ratio is 343:169.
ExplanationStatement I alone: (1 − 15/C)^3 = 343/512 = (7/8)^3 ⇒ C = 120 litres. Statement II alone is insufficient. Answer: A. -
DS4 Hard
What was the initial quantity of liquid A in the can?
Statement IA and B were in the ratio 7:5; 9 litres were drawn and replaced with B, after which A:B = 7:9.
Statement IIOnly the final ratio A:B is 7:9.
ExplanationStatement I alone is the classic problem and yields initial A = 21 litres. Statement II alone lacks the initial ratio and draw quantity. Answer: A.