Deloitte Pipes and Cisterns Questions and Answers Quiz-1

Net part of the tank filled in 1 hour = (1/3+1/4-1/6) = 5/12
Therefore, the tank will be full in = 12/5 hours
=2 2/5 hours

1/20 +1/30
50/600 = 1/12= 12 minutes

Time by A = 10 min
Time by B= 20 min
Total time =(20 x 10)/(20+10)
= 200/30
= 20/3 min

First, it was 3/5 full i.e. (60%)
2 buckets back in it make it 1/2 (50%)
10% means 2 buckets
To empty 100% it require = (100/10)×2
=20 buckets

Pipe can fill in 1 hour = 1/20 parts of tank

Pipe can empty in 1 hour = 1/15 parts of tank

So, in [(20 x 15)/(20-15)] =(20×15)/5 = 60 hrs the tank will be completely full.

 

Suppose the slower pipe alone can fill the tank in $x$ minutes.
Then the faster pipe can fill the tank in $\frac{\mathrm{x/}}{4}$ minutes.

Part filled by the slower pipe in 1 minute $=\frac{\mathrm{1/}}{x}$
Part filled by the faster pipe in 1 minute $=\frac{\mathrm{4/}}{x}$
Part filled by both the pipes in 1 minute $=\frac{\mathrm{1/}}{x}+\frac{\mathrm{4/}}{x}$

Given that both the pipes together can fill the tank in 36 minutes.
=> Part filled by both the pipes in 1 minute $=\frac{\mathrm{1/}}{36}$

$\frac{\mathrm{1/}}{x}+\frac{\mathrm{4/}}{x}=\frac{\mathrm{1/}}{36}\frac{\mathrm{=>5/}}{x}=\frac{\mathrm{1/}}{36}\mathrm{=>x}=5\times 36=180$

i.e.,the slower pipe alone can fill the tank in 180 minutes.

20*2*1= x*2*4
X= 5

Pipes A and B can fill the tank in 6 and 5 hours respectively. Therefore,
part filled by pipe A in 1 hour =1/6
part filled by pipe B in 1 hour = 1/5
Pipe C can empty the tank in 18 hours. Therefore,
part emptied by pipe C in 1 hour =1/18
Net part filled by Pipes A,B,C together in 1 hour
1/5+1/6-1/18=28/90=14/45 hrs.