Question: 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?
Options:
A) 9 h
B) 7 h
C) 8 h
D) 10 h
E) 11 h
Show Answer
Answer:
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, $ time=\frac{3/20}{3/20}=1,h $
$ \therefore $ Total time to fill the tank $ =6+1=7,h $