Discussion
How much pure alcohol must be added to 400 ml of a 15% solution to make its strength 32%?
Answer: A
Correct answer: 100 ml
The 400 ml contains \( 400 \times 0.15 = 60 \) ml of alcohol and 340 ml of water. The water is unchanged by adding alcohol, so anchor on it. If the alcohol is to be 32%, the water must be 68% of the new total: \[ 340 = 0.68 \times \text{new total} \;\Rightarrow\; \text{new total} = \frac{340}{0.68} = 500 \text{ ml}. \] The alcohol to add is therefore \[ 500 - 400 = 100 \text{ ml}. \] Checking: 500 ml containing \( 60 + 100 = 160 \) ml of alcohol is \( 160/500 = 32\% \). The alternative is alligation between the 15% solution and pure alcohol at 100%, with a mean of 32: \[ \frac{\text{solution}}{\text{alcohol}} = \frac{100 - 32}{32 - 15} = \frac{68}{17} = \frac{4}{1}, \] so for 400 ml of solution, 100 ml of alcohol - the same answer by a shorter route.
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