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A mixture of 200 litres contains acid and water in the ratio 3:2. If 40 litres of the mixture is replaced with water, what will be the percentage of acid in the final mixture?

1. 42%
2. 45%
3. 48%
4. 50%

Correct answer: 48%

Solution

Initially, acid = $\frac{3}{5}\times 200=120$ L. In 40 L of mixture, acid removed = $\frac{3}{5}\times 40=24$ L, so remaining acid = 96 L. Final volume is still 200 L, so percentage of acid = $\frac{96}{200}\times 100=48\%$.

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