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 : 21 [SSC FCI Assistant Grade-III 2013]
Two pipes A and B can fill a cistern in 3 hours and 5 hours respectively. Pipe C can empty in 2 hours. If all the three pipes are open, in how many hours the cistern will be full?
a) 30 hours
b) can’t be filled
c) 10 hours
d) 15 hours
Answer »Answer: (a)
Using Rule 2,If x, y, z, ........... all taps are opened together then, the time required to fill/empty the tank will be:$1/x ± 1/y ± 1/z ± ... = 1/T$Where T, is the required timeNote: Positive result shows that the tank is filling and Negative result shows that the tank is getting empty.
Part of cistern filled by three pipes in an hour
= $1/3 + 1/5 - 1/2 = {10 + 6 - 15}/30 = 1/30$
Hence, the cistern will be filled in 30 hours.
Question : 22 [SSC CGL Tier-II 2012]
A tank can be filled by pipe A in 2 hours and pipe B in 6 hours. At 10 A.M. pipe A was opened. At what time will the tank be filled if pipe B is opened at 11 A.M.?
a) 12 P.M.
b) 12.45 A.M.
c) 5 P.M
d) 11.45 A.M.
Answer »Answer: (d)
Part of the tank filled in 1 hour by pipe A = $1/2$
Part of the tank filled by both pipes in1 hour
= $1/2 + 1/6 = {3 + 1}/6 = 2/3$
So, Time taken to fill $2/3$ part = 60 minutes
Time taken to fill $1/2$ part
= ${60 × 3}/2 × 1/2$ = 45 minutes
The tank will be filled at 11:45 A.M.
Question : 23 [SSC CGL Prelim 2000]
A tap can empty a tank in one hour. A second tap can empty it in 30 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?
a) 45 minutes
b) 20 minutes
c) 30 minutes
d) 40 minutes
Answer »Answer: (b)
1 hour = 60 minutes.
Rate of emptying the tank by the two taps are $1/60$ and $1/30$ of the tank per minute respectively.
Rate of emptying the tank when both operate simultaneously
= $1/60 + 1/30 = {1 + 2}/60 = 3/60 = 1/20$
of the tank per minute.
Time taken by the two taps together to empty the tank = 20 minutes
Using Rule 6,Two taps 'A; and 'B' can empty a tank in 'x' hours and 'y' hours respectively. If both the taps are opened together, then time taken to empty the tank will beRequired time = $({xy}/{x + y})$ hrs
Here, x = 60, y = 30
Required time = $({xy}/{x + y})$ minutes
= $({60 × 30}/{60 + 30})$ minutes = 20 minutes.
Question : 24 [SSC CAPFs SI 2014]
A water tank can be filled by a tap in 30 minutes and another tap can fill it in 60 minutes. If both the taps are kept open for 5 minutes and then the first tap is closed, how long will it take for the tank to be full ?
a) 45 minutes
b) 20 minutes
c) 25 minutes
d) 30 minutes
Answer »Answer: (a)
Using Rule 2,
Part of the tank filled by both taps in 5 minutes
= $5(1/30 + 1/60)$
= $5({2 + 1}/60) = 5 × 3/60 = 1/4$
Remaining part = $1 - 1/4 = 3/4$ that is filled by second tap.
Time taken = $3/4 × 60$ = 45 minutes
Question : 25 [SSC CGL Prelim 2007]
A pipe can fill a tank in ‘x’ hours and another pipe can empty it in‘y’ (y > x) hours. If both the pipes are open, in how many hours will the tank be filled ?
a) ${xy}/{y - x}$ hours
b) (x - y) hours
c) (y - x) hours
d) ${xy}/{x - y}$ hours
Answer »Answer: (a)
Part of the tank filled in 1 hour = $1/x$
Part of the tank emptied in 1 hour = $1/y$
Part of the tank filled in 1 hour
when both are opened
= $1/x - 1/y = {y - x}/{xy}$
Tank will be filled in ${xy}/{y - x}$ hours.
IMPORTANT quantitative aptitude EXERCISES
Model 1 Basic Pipes & Cisterns problems Shortcuts »
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