Mixture in a 90-L container: A vessel initially contains two liquids, A and B, with unknown amounts. From the vessel, 60% of liquid A and 30% of liquid B are withdrawn. After this withdrawal the container is 40% empty (that is, 60% full). Determine the initial quantities of liquid A and liquid B in liters.
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AA = 30 L, B = 60 L
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BA = 36 L, B = 54 L
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CA = 50 L, B = 40 L
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DA = 45 L, B = 45 L
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EA = 60 L, B = 30 L
Answer
Correct Answer: A = 30 L, B = 60 L
Explanation
Introduction / Context: This problem combines percentage removal with total volume tracking. The key is to express what remains of each liquid after withdrawing fixed percentages and then use the final fullness (60% of capacity) to solve for the initial split of A and B.
Given Data / Assumptions:
- Total capacity = 90 L.
- 60% of liquid A withdrawn; 30% of liquid B withdrawn.
- After withdrawal, the vessel is 60% full ⇒ remaining volume = 54 L.
- Let initial A = x L, initial B = 90 − x L.
Concept / Approach: After withdrawal, remaining A = 40% of x (i.e., 0.4x). Remaining B = 70% of (90 − x) (i.e., 0.7*(90 − x)). Their sum must equal 54 L because the container is 60% full.
Step-by-Step Solution:
Remaining A = 0.4x.Remaining B = 0.7(90 − x) = 63 − 0.7x.0.4x + (63 − 0.7x) = 54.−0.3x = −9 ⇒ x = 30.Therefore A = 30 L and B = 90 − 30 = 60 L.Verification / Alternative check: Withdrawn A = 60% of 30 = 18 L; remaining A = 12 L. Withdrawn B = 30% of 60 = 18 L; remaining B = 42 L. Total remaining = 12 + 42 = 54 L = 60% of 90 L, consistent.
Why Other Options Are Wrong: Other pairs do not satisfy the condition 0.4x + 0.7(90 − x) = 54; they either overfill or underfill relative to the 60% final level.
Common Pitfalls: Subtracting percentages from the total instead of per-component; forgetting that 40% empty means 60% full (54 L).
Final Answer: A = 30 L, B = 60 L