| 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 |
A) \[7\,\,L\]
B) \[5\frac{1}{6}\,\,L\]
C) \[5\,\,L\]
D) \[4\frac{2}{3}\,\,L\]
Correct Answer: D
Solution :
| Let x L of liquid P be mixed to 7 L of Q. |
| According to the question, |
| \[x\times \frac{10}{7}+\frac{5}{7}\times 7=x+7\] |
| \[\Rightarrow \]\[10x+35=7x+49\]\[\Rightarrow \]\[3x=14\] |
| \[\Rightarrow \]\[x=\frac{14}{3}=4\frac{2}{3}L\] |
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