Question: A liquid P is $ 1\frac{3}{7} $ times as heavy as water and water is $ 1\frac{2}{5} $ times as heavy as another liquid Q. The amount of liquid P that must be added to 7 L of the liquid Q, so that the mixture may weight as much as an equal volume of water will be
Options:
A) $ 7L $
B) $ 5\frac{1}{6}L $
C) $ 5L $
D) $ 4\frac{2}{3}L $
Show Answer
Answer:
Correct Answer: D
Solution:
- Let x L of liquid P be mixed to 7 L of Q.
According to the question,
$ x\times \frac{10}{7}+\frac{5}{7}\times 7=x+7 $
$ \Rightarrow $ $ 10x+35=7x+49 $
$ \Rightarrow $ $ 3x=14 $
$ \Rightarrow $ $ x=\frac{14}{3}=4\frac{2}{3}L $