Three taps A, B and C can fill a tank in 12, 15 and 20 h, respectively. If A is open all the time and B and C are open for one hour each alternately, then the tank will be filled in |
A) 6 h
B) 7 h
C) 5 h
D) None of these
Correct Answer: B
Solution :
Filling done by pipe A and Bin 1 h\[=\frac{1}{12}+\frac{1}{15}=\frac{3}{20}\] |
Filling done by pipe A and C in 1 h \[=\frac{1}{12}+\frac{1}{20}=\frac{2}{15}\] |
Filling done in \[2\,\,h=\frac{3}{20}+\frac{2}{15}=\frac{17}{60}\] |
Filling done in \[6\,\,h=\frac{17}{60}\times 3=\frac{51}{60}\] |
Remaining filling \[=1-\frac{51}{60}=\frac{3}{20}\] |
Now, in 7 h filling done by A and B |
So, \[\text{time}=\frac{\frac{3}{20}}{\frac{3}{20}}=1\,\,h\] |
\[\therefore \]Total time \[=7\,\,h\] |
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