Practice Questions
Concept Explanation
When a fixed quantity is repeatedly removed from a vessel and replaced with impurity (e.g. water), the pure quantity decreases exponentially.
Key Formulas
- After n replacements of x out of V (starting pure): pure left = V(1 − x/V)n
- Concentration after n operations = (1 − x/V)n × 100%
- After one step: pure = V − x
Example
64 L pure milk. Remove 8 L, add 8 L water, repeat 3 times. Pure = 64 × (7/8)³ = 42.875 L.
Common Trap
Apply the power n only after the same operation is repeated n times with the same x and V.
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Q1 EasyVessel has 60 litres milk. 6 litres replaced by water. Milk left:ExplanationOne replacement from pure milk: milk left = 60 − 6 = 54 L (or 60 × (1 − 6/60) = 54 L). Hence the answer is 54 L.
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Q2 Easy50 L pure oil: 5 L replaced by water once. Concentration of oil becomes:ExplanationOil left = 50 × (1 − 5/50) = 45 L in 50 L ⇒ 90%. Hence the answer is 90%.
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Q3 EasyAfter n replacements of x from V starting pure, pure left equals:ExplanationEach replacement multiplies remaining pure fraction by (1 − x/V). After n times: pure = V(1 − x/V)^n. Hence the answer is V(1 − x/V)^n.
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Q4 Easy200 L pure: 40 L replaced by water once. Fraction of pure left is:ExplanationPure left = 200 × (1 − 40/200) = 160 L. Fraction = 160/200 = 4/5. Hence the answer is 4/5.
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Q5 Medium40 litres milk. 4 litres replaced by water thrice. Milk left:ExplanationFormula: milk left = V(1 − x/V)^n. Here V = 40, x = 4, n = 3. Milk left = 40 × (1 − 4/40)^3 = 40 × (0.9)^3 = 40 × 0.729 = 29.16 L. Hence the answer is 29.16 L.
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Q6 Medium100 L pure milk: 10 L replaced by water, twice. Milk left is:ExplanationMilk left = 100 × (1 − 10/100)^2 = 100 × (0.9)^2 = 81 L. Hence the answer is 81 L.
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Q7 MediumA cask has 125 L pure wine. 25 L is replaced by water twice. Wine left is:ExplanationWine left = 125 × (1 − 25/125)^2 = 125 × (0.8)^2 = 125 × 0.64 = 80 L. Hence the answer is 80 L.
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Q8 Medium90 L pure milk: 9 L replaced by water, twice. Milk left is:ExplanationMilk left = 90 × (1 − 9/90)^2 = 90 × (0.9)^2 = 90 × 0.81 = 72.9 L. Hence the answer is 72.9 L.
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Q9 Medium1000 L pure wine: 100 L replaced by water, 3 times. Wine left is:ExplanationWine left = 1000 × (1 − 100/1000)^3 = 1000 × (0.9)^3 = 729 L. Hence the answer is 729 L.
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Q10 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 water is 16:81. How much wine did the cask hold originally?ExplanationLet original capacity be C litres. After 4 identical replacements of 8 L: wine left = C(1 − 8/C)^4. Given wine : water = 16 : 81 ⇒ fraction of wine left = 16/81 (classic Indiabix form). So (1 − 8/C)^4 = 16/81 = (2/3)^4 ⇒ 1 − 8/C = 2/3 ⇒ 8/C = 1/3 ⇒ C = 24. Hence the answer is 24 L.
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Q11 HardA vessel is filled with milk and water. After drawing out 20 litres and adding water, milk:water = 7:9. Drawing out 20 litres again and adding water, milk:water becomes 7:11. Capacity of vessel:ExplanationLet capacity = V. After first draw-and-replace with water, milk = (7/16)V. After second: milk = (7/16)V × (V − 20)/V = 7(V − 20)/16. Also milk = (7/18)V. So 7(V − 20)/16 = 7V/18 ⇒ (V − 20)/16 = V/18 ⇒ 18(V − 20) = 16V ⇒ 18V − 360 = 16V ⇒ 2V = 360 ⇒ V = 180. Hence the answer is 180 L.
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Q12 Hard80 L pure juice: 8 L replaced by water, 3 times. Juice left is:ExplanationJuice left = 80 × (1 − 8/80)^3 = 80 × (0.9)^3 = 80 × 0.729 = 58.32 L. Hence the answer is 58.32 L.
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Q13 HardFrom 64 L pure milk, 8 L is replaced by water three times. Milk left is:ExplanationMilk left = 64 × (1 − 8/64)^3 = 64 × (7/8)^3 = 64 × 343/512 = 21952/512 = 42.875 L. Hence the answer is 42.875 L.
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Q14 Hard27 L pure: 3 L replaced by water, 3 times. Pure left is:ExplanationPure left = 27 × (1 − 3/27)^3 = 27 × (8/9)^3 = 27 × 512/729 = 512/27 = 18.96 L (approx.). Hence the answer is 18.96 L.
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DS1 Easy
How much milk is left after one replacement?
Statement IVessel had 60 L pure milk.
Statement II6 L was replaced by water.
ExplanationStatement I alone gives initial volume only. Statement II alone gives the replaced volume only. Together: milk left = 60 − 6 = 54 L. Hence both are needed. -
DS2 Medium
How much milk remains after three replacements?
Statement IVessel had 40 L pure milk; each time 4 L is replaced by water.
Statement IIMilk left = 40 × (0.9)^3.
ExplanationI alone: 40 × (0.9)^3 = 29.16 L. II alone gives the same. Either sufficient.
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DS3 Hard
What was the original capacity of the wine cask?
Statement I8 L is replaced by water four times and wine left : water = 16 : 81.
Statement II(1 − 8/C)^4 = 16/81.
ExplanationI alone leads to (1 − 8/C)^4 = 16/81 ⇒ C = 24. II alone is the same equation. Either alone is sufficient. -
DS4 Hard
What is the capacity of the vessel?
Statement IAfter drawing 20 L and adding water, milk : water = 7 : 9.
Statement IIAfter repeating once more, milk : water = 7 : 11.
ExplanationBoth ratios are needed to form 7(V − 20)/16 = 7V/18 and get V = 180. Neither alone determines V. Hence both together.