Practice Basic pipes and cisterns problems - 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) If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?
(a)
(b)
(c)
(d)
Using Rule 1,
If the slower pipe fills the tank in x hours, then
$1/x + 1/{x - 10} = 1/12$
${x -10 + x}/{x(x - 10)} = 1/12$
$x^2 - 10x = 24x - 120$
$x^2 - 34x + 120$ = 0
$x^2 - 30x - 4x + 120$ = 0
$x (x - 30) - 4 (x - 30)$ = 0
$(x - 4) (x - 30)$ = 0
$x$ = 30 because $x ≠ 4$
Required time = 30 - 10 = 20 hours
Q-5) 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)
(b)
(c)
(d)
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.
Q-6) Two pipes can fill a cistern separately in 10 hours and 15 hours. They can together fill the cistern in
(a)
(b)
(c)
(d)
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.
Q-7) A tank can be filled by pipe A in 2 hours and pipe B in 6 hours. At 10 A.M. pipe A was opened. At what time will the tank be filled if pipe B is opened at 11 A.M.?
(a)
(b)
(c)
(d)
Part of the tank filled in 1 hour by pipe A = $1/2$
Part of the tank filled by both pipes in1 hour
= $1/2 + 1/6 = {3 + 1}/6 = 2/3$
So, Time taken to fill $2/3$ part = 60 minutes
Time taken to fill $1/2$ part
= ${60 × 3}/2 × 1/2$ = 45 minutes
The tank will be filled at 11:45 A.M.
Q-8) 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-9) 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-10) 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)
(b)
(c)
(d)
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.