Model 1 Basic Pipes & Cisterns problems Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 3 EXERCISES
The following question based on pipes & cisterns topic of quantitative aptitude
(a) 9 hours
(b) 6 hours
(c) 7 hours
(d) 8 hours
The correct answers to the above question in:
Answer: (b)
Using Rule 1,
Part of the cistern filled by both pipes in 1 hour
= $1/10 + 1/15 = {3 + 2}/30 = 1/6$
The cistern will be filled in 6 hours.
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Read more basic pipes and cisterns problems Based Quantitative Aptitude Questions and Answers
Question : 1
A cistern is provided with two pipes A and B. A can fill it in 20 minutes and B can empty it in 30 minutes. If A and B be kept open alternately for one minute each, how soon will the cistern be filled ?
a) 120 minutes
b) 121 minutes
c) 110 minutes
d) 115 minutes
Answer »Answer: (d)
Using Rule 7,
Part of the tank filled in first two minutes
= $1/20 - 1/30 = {3 - 2}/60 = 1/60$
Part of tank filled in 114 minutes
= $57/60 = 19/20$
Remaining part of cistern will be filled in 115th minute
Question : 2
A cistern can be filled with water by a pipe in 5 hours and it can be emptied by a second pipe in 4 hours. If both the pipes are opened when the cistern is full, the time in which it will be emptied is :
a) 20$1/2$ hours
b) 9 hours
c) 18 hours
d) 20 hours
Answer »Answer: (d)
According to the question
Cistern filled in 1 hour = $1/5$ part
Cistern emptied in 1 hour = $1/4$ part
When the both pipes are opened, simultaneously ;
Cistern emptied in 1 hour
= $1/4 - 1/5 = {5 - 4}/20 = 1/20$ part
The time in which it will be emptied = 20 hours.
Using Rule 7,A tap 'A' can fill a tank in 'x' hours and 'B' can empty the tank in 'y' hours. Then (a) time taken to fill the tankwhen both are opened = $({xy}/{x - y})$ : x > yb) time taken to empty the tankwhen both are opened = $({xy}/{y- x})$ : y > x
Here, x = 5, y = 4
Required time = $({xy}/{x - y})$ hrs
= ${5 × 4}/{5 - 4}$ hrs = 20 hrs.
Question : 3
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 : 4
Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 50 minutes. In how much time the third pipe alone can empty the cistern?
a) 90 minutes
b) 110 minutes
c) 100 minutes
d) 120 minutes
Answer »Answer: (c)
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.
Let the third pipe empty the cistern in x minutes.
Part of cistern filled in 1 minute when all three pipes are opened simultaneously
= $1/60 + 1/75 - 1/x$
According to the question,
$1/60 + 1/75 - 1/x =1/50$
$1/x = 1/60 + 1/75 - 1/50$
= ${5 + 4 - 6}/300 = 3/300 ⇒ 1/x = 3/300$
$x = 300/3$ = 100 minutes
Question : 5
A pipe can fill a tank in x hours and another can empty it in y hours. They can together fill it in (y > x)
a) ${xy}/{x - y}$ hours
b) ${xy}/{y - x}$ hours
c) x - y hours
d) y - x hours
Answer »Answer: (b)
When both pipes are opened simultaneously,
part of the tank filled in 1 hour
= $1/x - 1/y = {y - x}/{xy}$
Required time = ${xy}/{y - x}$ hours
GET pipes & cisterns PRACTICE TEST EXERCISES
Model 1 Basic Pipes & Cisterns problems
Model 2 Filling tank by parts or fractions
Model 3 Opening both taps and leaks
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