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) 90 minutes
(b) 110 minutes
(c) 100 minutes
(d) 120 minutes
The correct answers to the above question in:
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
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Read more basic pipes and cisterns problems Based Quantitative Aptitude Questions and Answers
Question : 1
Two pipes can fill a cistern separately in 10 hours and 15 hours. They can together fill the cistern in
a) 9 hours
b) 6 hours
c) 7 hours
d) 8 hours
Answer »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.
Question : 2
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 : 3
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 : 4
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
pipes & cisterns Shortcuts and Techniques with Examples
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