Model 1 Basic Pipes & Cisterns problems Practice Questions Answers Test with Solutions & More Shortcuts
pipes & cisterns PRACTICE TEST [3 - EXERCISES]
Model 1 Basic Pipes & Cisterns problems
Model 2 Filling tank by parts or fractions
Model 3 Opening both taps and leaks
Question : 6 [SSC CPO S.I.2004]
Two pipes can fill a tank in 15 hours and 20 hours respectively, while the third can empty it in 30 hours. If all the pipes are opened simultaneously, the empty tank will be filled in
a) 15$1/2$ hours
b) 10 hours
c) 12 hours
d) 15 hours
Answer »Answer: (c)
Using Rule 2,
Part of tank filled in 1 hour when all three pipes are opened simultaneously
= $1/15 + 1/20 - 1/30$
= ${4 + 3 - 2}/60 = 5/60 = 1/12$
Hence, the tank will be filled in 12 hours.
Question : 7 [SSC MTS 2013]
Two pipes, P and Q, together can fill a cistern in 20 minutes and P alone can in 30 minutes. Then Q alone can fill the cistern in
a) 51 minutes
b) 62 minutes
c) 60 minutes
d) 61 minutes
Answer »Answer: (c)
Using Rule 7,
Part of the cistern filled by pipe Q in 1 minute
= $1/20 - 1/30 = {3 - 2}/60 = 1/60$
Required time = 60 minutes
Question : 8 [SSC CPO S.I.2008]
Three pipes P, Q and R can separately fill a cistern in 4,8 and 12 hours respectively. Another pipe S can empty the completely filled cistern in10 hours. Which of the following arrangements will fill the empty cistern in less time than others ?
a) P, Q and S are open.
b) Q alone is open.
c) P and S are open.
d) P, R and S are open.
Answer »Answer: (a)
Using Rule 2 and 7,
Part of the cistern filled in 1 hour
when pipes P and S are open
= $1/4 - 1/10 = {5 - 2}/20 = 3/20$
Hence, the cistern will be filled in $20/3$ hours ≈ 6.6 hours
Part of the cistern filled in 1 hour
when pipes P, R and S are open
= $1/4 + 1/12 - 1/10$
= ${15 + 5 - 6}/60 = 14/60 = 7/30$
Hence, the cistern will be filled in $30/7$ hours ≈ 4.3 hours
Part of the cistern filled in I hour
when pipes P, Q and S are open
= $1/4 + 1/8 - 1/10$
= ${10 + 5 - 4}/40 = 11/40$
Hence, the cistern will be filled in $40/11$ hours ≈ 3.6 hours
Cistern can be filled faster when P, Q & S are open
Question : 9
12 pumps working 6 hours a day can empty a completely filled reservoir in 15 days. How many such pumps working 9 hours a day will empty the same reservoir in 12 days ?
a) 12
b) 15
c) 9
d) 10
Answer »Answer: (d)
Hours/day | Days | Pumps |
6 | 15 | 12 |
↑ | ↑ | ↓ |
9 | 12 | x |
Let x be number of pumps
9 : 6 : : 12 : x = 12 : 15 : : 12 : x
9 × 12 × x = 6 × 12 × 15
$x = {6 × 12 × 15}/{9 × 12}$ = 10
Question : 10 [SSC CHSL 2010]
Three taps A,B and C together can fill an empty cistern in 10 minutes. The tap A alone can fill it in 30 minutes and the tap B alone in 40 minutes. How long will the tap C alone take to fill it ?
a) 40 minutes
b) 16 minutes
c) 24 minutes
d) 32 minutes
Answer »Answer: (c)
Using Rule 1 and 7,
Part of the cistern filled by taps
A, B and C in 1 minute = $1/10$
Part of the cistern filled by taps
A and B in 1 minute
= $1/30 + 1/40 = {4 + 3}/120 = 7/120$
Part of the cistern filled by tap C in 1 minute
= $1/10 - 7/120 = {12 - 7}/120 = 5/120 = 1/24$
Tap C will fill the cistern in 24 minutes.
IMPORTANT quantitative aptitude EXERCISES
Model 1 Basic Pipes & Cisterns problems Shortcuts »
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