Practice Pipes and cisterns - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) Two pipes X and Y can fill a cistern in 24 minutes and 32 minutes respectively. If both the pipes are opened together, then after how much time (in minutes) should Y be closed so that the tank is full in 18 minutes ?
(a)
(b)
(c)
(d)
If pipe y be closed after x minutes, then
$18/24 + x/32$ = 1
$x/32 = 1 - 18/24 = 1- 3/4 = 1/4$
$x = 32/4$ = 8 minutes
Using Rule 8,Two taps A and B can fill a tank in x hours and y hours respectively. If both the pipes are opened together, then the time after which pipe B should be closed so that the tank is full in t hoursRequired time = $[y(1 –{t/x})]$ hours
x = 24, y = 32, t = 18
Required time = $[y(1 –{t/x})]$ minutes
= $[32(1 - {18/24})]$ minutes
= $[32(1 - {3/4})] = 32 × 1/4$ = 8 minutes
Q-2) 12 pumps working 6 hours a day can empty a completely filled reservoir in 15 days. How many such pumps working 9 hours a day will empty the same reservoir in 12 days ?
(a)
(b)
(c)
(d)
| Hours/day | Days | Pumps |
| 6 | 15 | 12 |
| ↑ | ↑ | ↓ |
| 9 | 12 | x |
Let x be number of pumps
9 : 6 : : 12 : x = 12 : 15 : : 12 : x
9 × 12 × x = 6 × 12 × 15
$x = {6 × 12 × 15}/{9 × 12}$ = 10
Q-3) A tap can fill a cistern in 8 hours and another tap can empty it in 16 hours. If both the taps are open, the time (in hours) taken to fill the tank will be :
(a)
(b)
(c)
(d)
Part of the cistern filled in 1 hour = $1/8$
Part of the cistern emptied in 1 hour = $1/16$
When both the taps are opened simultaneously, part of cistern filled in 1 hour
= $1/8 - 1/16 = {2 - 1}/16 = 1/16$
Hence, the cistern will be filled in 16 hours.
Using Rule 7,
Here, x = 8, y = 16
Required time = ${xy}/{y- x}$
= ${8 × 16}/{16 - 8}$ = 16 hours
Q-4) Three pipes A, B and C can fill a tank in 6 hours. After working together for 2 hours, C is closed and A and B fill the tank in 8 hours. The time (in hours) in which the tank can be filled by pipe C alone is
(a)
(b)
(c)
(d)
Part of the tank filled by
(A + B + C) in 1 hour = $1/6$
Part of tank filled by these in 2 hours
= $2/6 = 1/3$
Remaining part = $1 - 1/3 = 2/3$
Time taken by A and B in filling
$2/3$ rd part = 8 hours
Time taken by A and B in filling the whole tank
= ${8 × 3}/2$ = 12 hours
Part of tank filled by C in an hour
= $1/6 - 1/12 = 1/12$
Hence, required time = 12 hours
Q-5) Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. The water pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes C is opened. Total time (in minutes) in which the tank will be filled up is :
(a)
(b)
(c)
(d)
Using Rule 1 and 2,
Part of tank filled by pipes A and B in 1 minute
= $1/30 + 1/45 = {3 + 2}/90 = 1/18$ part
Part of tank filled in 12 minutes
= $12/18 = 2/3$ part
Remaining part
= $1 - 2/3 = 1/3$ part
When pipe C is opened,
Part of tank filled by all three pipes
= $1/30 + 1/45 - 1/36$
= ${6 + 4 - 5}/180 = 5/180 = 1/36$
Time taken in filling $1/3$ part
= $1/3$ × 36 = 12 minutes
Total time = 12 + 12 = 24 miuntes
Q-6) 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)
(b)
(c)
(d)
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
Q-7) Two pipes can fill a cistern in 3 hours and 4 hours respectively and a waste pipe can empty it in 2 hours. If all the three pipes are kept open, then the cistern will be filled in :
(a)
(b)
(c)
(d)
Using Rule 2,
Part of the cistern filled in 1 hour
= $1/3 + 1/4 - 1/2$
[Cistern filled by 1st pipe + Cistern filled by 2nd pipe - Cistern emptied by 3rd pipe]
= ${4 + 3 - 6}/12 = 1/12$
Hence, the cistern will be filled in 12 hours.
Q-8) 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)
(b)
(c)
(d)
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.
Q-9) There are two pumps to fill a tank with water. First pump can fill the empty tank in 8 hours, while the second in 10 hours. If both the pumps are opened at the same time and kept open for 4 hours, the part of tank that will be filled up is :
(a)
(b)
(c)
(d)
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 of the tank filled in an hour by both pumps
= $1/8 + 1/10 = {5 + 4}/40 = 9/40$
Part of the tank filled in 4 hours
= ${4 × 9}/40 = 9/10$
Q-10) Pipes P and Q can fill a tank in 10 and 12 hours respectively and C can empty it in 6 hours. If all the three are opened at 7 a.m., at what time will one-fourth of the tank be filled ?
(a)
(b)
(c)
(d)
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 tank filled in 1 hour when all three pipes are opened
= $1/10 + 1/12 - 1/6= {6 + 5 - 10}/60 = 1/60$
The tank will be filled in 60 hours.
One fourth of the tank will be filled in 15 hours $[1/4 × 60]$
i.e. the tank will be filled at 10 p.m.
Q-11) Two pipes A and B can fill a tank in 36 minutes and 45 minutes respectively. Another pipe C can empty the tank in 30 minutes. First A and B are opened. After 7 minutes, C is also opened. The tank is filled up in
(a)
(b)
(c)
(d)
Part of the tank filled by pipes A and B in 1 minute
= $1/36 + 1/45 = {5 + 4}/180$
= $9/180 = 1/20$
Part of the tank filled by these pipes in 7 minutes
= $7/20$
Remaining unfilled part
= $1 - 7/20 = {20 - 7}/20 = 13/20$
When all three pipes are opened.
= $1/20 - 1/30 = {3 - 2}/60 = 1/60$
Time taken in filling $13/20$ part
= $13/20$ × 60 = 39 minutes
Required time = 39 + 7 = 46 minutes
Q-12) 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)
(b)
(c)
(d)
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.
Q-13) 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)
(b)
(c)
(d)
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.
Q-14) 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)
(b)
(c)
(d)
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
Q-15) $3/4$ part of a tank is full of water. When 30 litres of water is taken out, the tank becomes empty. The capacity of the tank is
(a)
(b)
(c)
(d)
Let the capacity of the tank be x litres.
According to the question,
${3x}/4 = 30$
3x = 30 × 4
$x = {30 × 4}/3$ = 40 litres
Q-16) 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)
(b)
(c)
(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.
Q-17) 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)
(b)
(c)
(d)
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.
Q-18) 20 buckets can fill a tank when the capacity of each bucket is 12 liters. If the capacity of each bucket is 10 liters, find the number of buckets required to fill the tank.
(a)
(b)
(c)
(d)
Capacity of each bucket = 12 liters
20 buckets can fill the tank. So, capacity of tank = 20 * 12= 240 liters
New capacity of bucket = 10 liters
So, 10 liters can be poured into the tank by one bucket
240 liters will be poured by $1/10$ x 240 = 24 buckets
Q-19) Three pipes A, B and C can fill a tank in 6 hours, 9 hours and 12 hours respectively. B and C are opened for half an hour, then A is also opened. The time taken by the three pipes together to fill the remaining part of the tank is
(a)
(b)
(c)
(d)
Using Rule 1 and 2,
Part of the tank filled by B and C in half an hour
= $1/2(1/9 + 1/12)$
= $1/2({4 + 3}/36) = 7/72$
= Remaining part
= $1 - 7/72 = {72 - 7}/72 = 65/72$
Part of tank filled by three pipes in an hour
= $1/6 + 1/9 + 1/12$
= ${6 + 4 + 3}/36 = 13/36$
Time to fill remaining part
= $65/72 × 36/13 = 5/2 = 2{1}/2$ hours
Q-20) 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)
(b)
(c)
(d)
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