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 : 16 [SSC Constable (GD) 2013]
Three taps A, B, C can fill an overhead tank in 4, 6 and 12 hours respectively. How long would the three taps take to fill the tank if all of them are opened together ?
a) 5 hrs.
b) 2 hrs.
c) 4 hrs.
d) 3 hrs.
Answer »Answer: (b)
Using Rule 2,
Part of the tank filled by all three taps in an hour
= $1/4 + 1/6 + 1/12 = {6 + 4 + 2}/24 = 1 2$
Hence, the tank will be filled in 2 hours.
Question : 17
Two pipes A and B can separately fill a tank in 2 hours and 3 hours respectively. If both the pipes are opened simultaneously in the empty tank, then the tank will be filled in
a) 1 hour 20 minutes
b) 1 hour 12 minutes
c) 2 hours 30 minutes
d) 1 hour 15 minutes
Answer »Answer: (b)
Using Rule 1,
Part of tank filled by pipes A and B in 1 hour
= $1/2 + 1/3 = {3 + 2}/6 = 5/6$ parts
Required time = $6/5$ hours
= 1 hour $1/5$ × 60
= 1 hour 12 minutes
Question : 18 [SSC CGL Tier-II 2015]
Pipe A can fill an empty tank in 6 hours and pipe B in 8 hours. If both the pipes are opened and after 2 hours pipe. A is closed, how much time B will take to fill the remaining tank?
a) 3$1/3$ hours
b) 7$1/2$ hours
c) 2$2/5$ hours
d) 2$1/3$ hours
Answer »Answer: (a)
Using Rule 1,
Part of tank filled by pipes A and B in 2 hours
= $2(1/6 + 1/8)$
=$2({4 + 3}/24) = 7/12$
Remaining part = $1 - 7/12 = 5/12$
This part is filled by pipe B.
Required time = $5/12$ × 8
= $10/3$ hours = 3$1/3$ hours
Question : 19 [SSC CPO S.I.2003]
Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. If both pipes are opened together, the time taken to fill the tank is :
a) 15 minutes
b) 50 minutes
c) 12 minutes
d) 25 minutes
Answer »Answer: (c)
Part of the tank filled by both pipes in one minute
= $1/20 + 1/30$
Required time = $1/{1/20 + 1/30}$
= ${20 × 30}/50$ = 12 minutes
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
Here, x = 20, y = 30
Required time = $({xy}/{x + y})$ minutes
= $({20 × 30}/{20 + 30})$ minutes = 12 minutes.
Question : 20
One tap can fill a water tank in 40 minutes and another tap can make the filled tank empty in 60 minutes. If both the taps are open, in how many hours will the empty tank be filled ?
a) 3.5 hours
b) 2 hours
c) 2.5 hours
d) 3 hours
Answer »Answer: (b)
Using Rule 7,
Part of the tank filled when both taps are opened together
= $1/40 - 1/60 = {3 - 2}/120 = 1/120$
Hence, the tank will be filled in 120 minutes 2 hours.
IMPORTANT quantitative aptitude EXERCISES
Model 1 Basic Pipes & Cisterns problems Shortcuts »
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