Two pipes can fill a tank in 10 h and 16 h, respectively. A third pipe can empty the tank in 40 h. If all the three pipes are opened and function simultaneously then in how much time the tank will be full? |
A) \[6\frac{1}{11}h\]
B) \[9\frac{9}{11}h\]
C) \[8\frac{9}{11}h\]
D) \[7\frac{3}{11}h\]
E) \[6\frac{5}{11}h\]
Correct Answer: D
Solution :
Part of the tank filled in 1 min \[=\frac{1}{10}+\frac{1}{16}-\frac{1}{40}\] |
\[=\frac{8+5-2}{80}=\frac{11}{80}\] |
Thus, required time for the tank to be filled |
\[=\frac{80}{11}=7\frac{3}{11}h\] |
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