Liquid P is \(1\frac{3}{7}\) times as heavy as water, and water is \(1\frac{2}{5}\) times as heavy as liquid Q, volume for volume. How many litres of P must be added to 7 litres of Q so that the mixture weighs as much as an equal volume of water?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4\frac{2}{3}\) litres
Take the weight of 1 litre of water as 1 unit. The mixture must weigh 1 unit per litre, so the extra weight of P must make up for the lighter weight of Q.
1 litre of P weighs \(1\frac{3}{7}=\frac{10}{7}\) units
Water is \(\frac{7}{5}\) times as heavy as Q, so 1 litre of Q weighs \(\frac{5}{7}\) unit
Each litre of P is \(\frac{10}{7}-1=\frac{3}{7}\) unit heavier than water
Each litre of Q is \(1-\frac{5}{7}=\frac{2}{7}\) unit lighter than water, so 7 litres of Q are \(7\times\frac{2}{7}=2\) units short
Litres of P needed \(\displaystyle =2\div\frac{3}{7}=\frac{14}{3}=4\frac{2}{3}\)
\(4\frac{2}{3}\) litres of P must be added.