A 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 refilled with B, the ratio of A to B becomes 7 : 9. How many litres of liquid A were in the can initially?
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A18 litres
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B21 litres
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C24 litres
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D28 litres
Answer
Correct Answer: 21 litres
Explanation
Problem restatementFrom a can containing a well-mixed liquid A and B in the ratio 7 : 5, 9 L are drawn out and replaced by pure B. The new ratio becomes 7 : 9. Find the original litres of A.
Given data
- Initial ratio A : B = 7 : 5 (i.e., A = (7/12) of total, B = (5/12) of total)
- Drawn off and replaced volume = 9 L (drawn from the mixture, refilled with pure B)
- Final ratio A : B = 7 : 9
Concept/ApproachWhen a well-mixed quantity is withdrawn, each component reduces proportionally. Let the can's capacity be T litres. After removing 9 L, the remaining amounts of A and B are each multiplied by (T − 9)/T. Then 9 L of pure B is added. Set up the final ratio and solve for T, then compute initial A = (7/12)T.
Step-by-step calculationInitial A = (7/12)T, Initial B = (5/12)TAfter removing 9 L (proportionally):Arem = (7/12)T − (7/12)×9 = (7/12)(T − 9)Brem = (5/12)T − (5/12)×9 = (5/12)(T − 9)After adding 9 L of pure B: Bnew = (5/12)(T − 9) + 9Final ratio given: Arem : Bnew = 7 : 9⇒ 9 × (7/12)(T − 9) = 7 × (5/12)(T − 9) + 9Multiply by 12: 63(T − 9) = 35(T − 9) + 75628(T − 9) = 756 ⇒ T − 9 = 27 ⇒ T = 36Initial A = (7/12) × 36 = 21 litres
Verification/AlternativeWith T = 36: After removal, remaining total = 27 L.Arem = (7/12)×27 = 15.75 L; Brem = (5/12)×27 = 11.25 L.After adding 9 L of B: Bnew = 11.25 + 9 = 20.25 L.Ratio = 15.75 : 20.25 = (divide by 2.25) = 7 : 9 — matches perfectly.
Common pitfalls
- Forgetting that the 9 L removed is a mixture (so both A and B reduce proportionally).
- Adding 9 L to both A and B (only B increases because the refill is pure B).
- Setting up the ratio as totals rather than component-wise after each operation.
Final Answer21 litres