A) 120 h
B) 115 h
C) 109 h
D) 112 h
Correct Answer: D
Solution :
Work done by the two pipes in 1 h |
\[=\left( \frac{1}{14}+\frac{1}{16} \right)=\frac{15}{112}\] |
\[\therefore \]Time taken by these pipes to fill the tank\[=\frac{112}{5}\]h = 7 h 28 min. |
Due to leakage, time taken = 7 h 28 min + 38 min = 8 h |
\[\therefore \]Work done by (two pipes + leak) in 1 h \[=\frac{1}{8}\] |
Work done by the leak in 1 h \[=\left( \frac{15}{112}-\frac{1}{8} \right)=\frac{1}{112}\] |
\[\therefore \] Leak will empty the full cistern in 112 h. |
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