A) 90
B) 85
C) 115
D) 112
Correct Answer: D
Solution :
Work done by two pipes in 1 h \[=\left( \frac{1}{4}+\frac{1}{16} \right)=\frac{15}{112}\] Total time taken by two pipes and leak in 1 h \[=\left( \frac{112}{15}+\frac{32}{60} \right)\]h = 8 h Work done by the leak in 1 h \[=\left( \frac{15}{112}-\frac{1}{8} \right)=\frac{1}{112}\] Hence, the leak will empty the full cistern in 112 hYou need to login to perform this action.
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