Model 3 Opening both taps and leaks 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 : 11 [SSC CISF ASI 2010]
A tap can fill a cistern in 40 minutes and a second tap can empty the filled cistern in 60 minutes. By mistake without closing the second tap, the first tap was opened. In how many minutes will the empty cistern be filled ?
a) 120
b) 108
c) 72
d) 84
Answer »Answer: (a)
Using Rule 7,
Tricky Approach
Part of the cistern filled in 1 minute by both the taps
= $1/40 - 1/60 = {3 - 2}/120 = 1/120$
Empty cistern will be filled in 120 minutes.
Question : 12
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternatively, the tank will be full in :
a) 7$1/2$ hours
b) 7 hours
c) 6 hours
d) 6$1/2$ hours
Answer »Answer: (b)
Using Rule 1,Two taps 'A' and 'B' can fill a tank in 'x' hours and 'y' hours respectively. If both the taps are opened together, then how much time it will take to fill the tank?Required time = $({xy}/{x + y})$ hrs
Part filled by A and B in 1 hour
= $1/12 + 1/15 = {5 + 4}/60 = 3/20 +$...(i)
Part filled by A and C in the next 1 hour
= $1/12 + 1/20 = {5 + 3}/60 = 2/15$
Part filled in 2 hours
= $3/20 + 2/15 = {9 + 8}/60 = 17/20$
Part filled in 6 hours = $51/60$
Remaining part
= $1 - 51/60 = 9/60 = 3/20$
This part will be filled by (A+B) in 1 hour. [By (i)]
Total time taken = 7 hours
Question : 13 [SSC CGL Prelim 2007]
A tap takes 36 hours extra to fill a tank due to a leakage equivalent to half of its inflow. The inflow can fill the tank in how many hours ?
a) 18 hrs
b) 30 hrs
c) 36 hrs
d) 24 hrs
Answer »Answer: (a)
Using Rule 7,
Let the inflow fill the tank in x hours.
$1/x - 1/{2x} = 1/36$
[leakage being half of inflow]
= ${2 - 1}/{2x} = 1/36$
2x = 36
$x = 36/2$ = 18 hours
Question : 14 [SSC CGL Prelim 2004]
Two pipes A and B can fill a cistern in 37$1/2$ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled just in half an hour, if the pipe B is turned off after :
a) 9 minutes
b) 5 minutes
c) 15 minutes
d) 10 minutes
Answer »Answer: (a)
Pipe A fills the tank in $75/2$ minutes.
Part of the tank filled by A in 30 minutes
= $2/75 × 30 = 4/5$
Remaining part
= $1- 4/5 = 1/5$
Now, 1 part is filled by pipe B in 45 minutes
$1/5$ part is filled in
= 45 × $1/5$ = 9 minutes
Hence, the pipe B should be turned off after 9 minutes.
Question : 15 [SSC CGL Tier-II 2015]
A tank has two pipes. The first pipe can fill it in 4 hours and the second can empty it in 16 hours. If two pipes be opened together at a time, then the tank will be filled in
a) 5$1/3$ hours
b) 6 hours
c) 5$1/2$ hours
d) 10 hours
Answer »Answer: (a)
Using Rule 7,
Part of tank filled by both the pipes in 1 hour
= $1/4 - 1/16 = {4 - 1}/16 = 3/16$
Required time = $16/3 = 5{1}/3$ hours
IMPORTANT quantitative aptitude EXERCISES
Model 3 Opening both taps and leaks Shortcuts »
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