| 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, C are open for one hour each alternately, in what time will the tank be full? |
A) 9 h
B) 7 h
C) 8 h
D) 10 h
E) 11 h
Correct Answer: B
Solution :
| Part of tank filled in 1 h by\[A=\frac{1}{2}\] |
| Part of tank filial in 1 h by \[B=\frac{1}{15}\] |
| Part of tank filled in 1 h by \[C=\frac{1}{20}\] |
| Part of tank filled in 1 h by A and B |
| \[=\frac{1}{12}+\frac{1}{15}=\frac{9}{60}=\frac{3}{20}\] |
| Part of tank filled in 1 h by A and C |
| \[C=\frac{1}{12}+\frac{1}{20}+\frac{2}{15}\] |
| Tank filled in 1st \[2\,h=\frac{3}{20}+\frac{2}{15}=\frac{17}{60}\] |
| Tank filled In \[6\,h=\frac{17}{60}\times 3=\frac{51}{60}\] |
| Remaining \[=1-\frac{51}{60}=\frac{3}{20}\] |
| Now, in the 7th hour, tank filled by taps A and B. |
| So, \[\text{time}=\frac{3/20}{3/20}=1\,h\] |
| \[\therefore \] Total time to fill the tank \[=6+1=7\,h\] |
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