Home ≫ Arithmetic Aptitute ≫ Pipes and Cisterns ≫ Page 5
41. One fill pipe A is 4 times slower than the second fill pipe B and takes 30 minutes more time than the fill pipe B. When will the cistern be full if both fill pipes are opened together? Also find, how much time will A and B separately take to fill the cistern?
  A.  8 minutes; A fills in 40 min; B fills in 10 min
  B.  5 minutes; A fills in 30 min; B fills in 20 min
  C.  10 minutes; A fills in 45 min; B fills in 10 min
  D.  6 minutes; A fills in 40 min; B fills in 8 min
     
   
View Answer

Shortcut:
If one filling pipe X is 'z' times slower and takes 'a' minutes more time than the other filling pipe Y, then the time, they will take to fill a cistern, if both the pipes are opened together, is
za / z2 − 1
minutes.
X will fill the cistern in
za / z − 1
minutes and
Y will fill the cistern in
a / z − 1

Note: Here, X is the slower filling pipe and Y is the faster one.
Here, z = 4, a = 30
Using these values in the shortcut, we get:
Required time =
4 x 30 / (4)2 − 1

=
4 x 30 / 16 − 1

=
4 x 30 / 15
= 4 x 2 = 8
Hence, the cistern will get full in 8 minutes.
A will fill the cistern =
za / z − 1

=
4 x 30 / 4 − 1
=
4 x 30 / 3
= 4 x 10 = 40
Hence, A will the cistern in 40 minutes.
Y will fill the cistern =
30 / 4 − 1

=
30 / 3
= 10
Hence, B will the cistern in 10 minutes.

42. 4 pipes are fitted to a water tank. Some of these are filling pipes and the other emptying pipes. Each filling pipe can fill the tank in 8 hours and each waste pipe can empty the tank in 4 hours. On opening all the pipes, and empty tank is filled in 8 hours. Find the number of filling pipes and waste pipes.
  A.  Filling pipes: 2; Waste pipe: 1
  B.  Filling pipes: 3; Waste pipe: 1
  C.  Filling pipes: 1; Waste pipe: 4
  D.  Filling pipes: 3; Waste pipe: 2
     
   
View Answer

Shortcut:
N pipes are fitted to a water tank. Some of these are filling pipes and the other emptying pipes. Each filling pipe can fill the tank in 'a' hours and each waste pipe can empty te tank in 'b' hours. On opening all the pipes, an empty tank is filled in 'Z' hours. Then the number of filling pipes is
b + NZ / a + b
x
a / Z
and the number of waste pipes is
NZ − a / a + b
x
b / Z

Here, N = 4, a = 8, b = 4, Z = 8
Using these values in the shortcut, we get:
Number of filling pipes =
4 + 4 x 8 / 8 + 4
x
8 / 8

=
4 + 32 / 12
x 1
=
36 / 12
= 3
Hence, the number of filling pipes are 3.
Number of waste pipes =
4 x 8 − 8 / 8 + 4
x
4 / 8

=
32 − 8 / 12
x
1 / 2

=
24 / 12
x
1 / 2

= 2 x
1 / 2
= 1
Hence, the number of waste pipe is 1.

43. One fill pipe A takes 1
2 / 7
minutes more to fill the cistern than two fill pipes A and B opened together to fill it. Second fill pipe B takes 28 minutes more to fill cistern than two fill pipes A and B opened together to fill it. When will the cistern be full if both pipes are opened simultaneously?
  A.  5 minutes
  B.  3 minutes
  C.  6 minutes
  D.  8 minutes
     
   
View Answer

Shortcut:
Two filling pipes X and Y opened together can fill a cistern in 't' minutes. If the first filling pipe X alone takes 'a' minutes more or less than 't' and the second fill pipe Y alone takes 'b' minutes more or less than 't' is given by t = √(ab) minutes.
Here, a =
9 / 7
, b = 28
Using these values in the shortcut, we get:
Required time =
√(
9 / 7
x 28
)

= √(9 x 4) = √(36) = 6
Hence, the cistern will get full in 6 minutes.

44. Two fill pipes A and B can fill a cistern in 4 and 6 minutes respectively. Both fill pipes are opened together, but 1 minutes before the cistern is full, one pipe A is closed. How much time will the cistern take to fill?
  A.  6 minutes
  B.  9 minutes
  C.  3 minutes
  D.  5 minutes
     
   
View Answer

Shortcut:
If first fill pipe can fill a cistern in m1 minutes alone, second fill pipe can fill the same alone in m2 minutes, and similarly, first empty pipe can empty the full cistern alone in a1 minutes, second empty pipe can empty the full cistern alone in a2 minutes, then the
alone filling time for first fill pipe = m1 minutes
alone filling time for second fill pipe = m2 minutes
alone emptying time for first empty pipe = a1 minutes
alone emptying time for second empty pipe = a2 minutes.
Now, if first fill pipe and second fill pipe are opened for t1 minutes and t2 minutes respectively. First empty pipe and second empty pipe are opened for t3 minutes and t4 minutes respetively, then
Step I: To find the amount of work (filling or emptying) done by each pipe (fill/empty), we use the following formula:
Amount of work(filling or emptying) done =
Number of minutes opened / Alone tme (empty or fill)

Note : Put (-ve) sign for 'emptying work'

Step II: Add the amount of work done by each pipe and equate it to the part of cistern filled.
(i)
t1 / m1
+
t2 / m2
−
t3 / a1
−
t4 / a2
= 1 (If cistern is filled completely)
(ii)
t1 / m1
+
t2 / m2
−
t3 / a1
−
t4 / a2
=
1 / 2
(If cistern is half filled)
(iii)
t1 / m1
+
t2 / m2
−
t3 / a1
−
t4 / a2
= 0 (If full cistern is emptied)
Step III: Find the unknown value.
Let the cistern be filled in y minutes.
∴ Pipe B is opened for y minutes and pipe A is opened for (y − 4) minutes.
Using these values in the shortcut, we get:
y − 1 / 4
+
y / 6
= 1
3(y − 1) + 2y / 12
= 1
3y − 3 + 2y / 12
= 1
5y − 3 / 12
= 1
5y − 3 = 12
5y = 12 + 3
5y = 15
y =
15 / 5
= 3
Hence, the cistern will get full in 3 minutes.

45. Two fill taps A and B can separately fill a cistern in 25 and 20 minutes respectively. They started to fill a cistern together but fill tap A is turned off after few minutes and fill tap B fills the rest part of cistern in 2 minutes. After how many minutes, was tap A turned off?
  A.  10 minutes
  B.  18 minutes
  C.  8 minutes
  D.  15 minutes
     
   
View Answer

Shortcut:
If first fill pipe can fill a cistern in m1 minutes alone, second fill pipe can fill the same alone in m2 minutes, and similarly, first empty pipe can empty the full cistern alone in a1 minutes, second empty pipe can empty the full cistern alone in a2 minutes, then the
alone filling time for first fill pipe = m1 minutes
alone filling time for second fill pipe = m2 minutes
alone emptying time for first empty pipe = a1 minutes
alone emptying time for second empty pipe = a2 minutes.
Now, if first fill pipe and second fill pipe are opened for t1 minutes and t2 minutes respectively. First empty pipe and second empty pipe are opened for t3 minutes and t4 minutes respetively, then
Step I: To find the amount of work (filling or emptying) done by each pipe (fill/empty), we use the following formula:
Amount of work(filling or emptying) done =
Number of minutes opened / Alone tme (empty or fill)

Note : Put (-ve) sign for 'emptying work'

Step II: Add the amount of work done by each pipe and equate it to the part of cistern filled.
(i)
t1 / m1
+
t2 / m2
−
t3 / a1
−
t4 / a2
= 1 (If cistern is filled completely)
(ii)
t1 / m1
+
t2 / m2
−
t3 / a1
−
t4 / a2
=
1 / 2
(If cistern is half filled)
(iii)
t1 / m1
+
t2 / m2
−
t3 / a1
−
t4 / a2
= 0 (If full cistern is emptied)
Step III: Find the unknown value.
Let the tap A be turned off after y minutes.
∴ Tap B is opened for y + 2 minutes.
Using these values in the shortcut, we get:
y / 25
+
y + 2 / 20
= 1
4y + 5(y + 2) / 100
= 1
4y + 5y + 10 / 100
= 1
4y + 5y + 10 = 100
9y = 100 − 10
9y = 90
y =
90 / 9
= 10
Hence, tap A was turned off after 10 minutes.

46. Three pipes A, B and C are attached to a cistern. A can fill it in 5 minutes and B in 20 minutes. C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding however, that the waste pipe had been left open, he closes it and the cistern now gets filled in 4 minutes. In how much time the pipe C, if opened alone, empty the full cistern?
  A.  3 minutes
  B.  7 minutes
  C.  6 minutes
  D.  4 minutes
     
   
View Answer

Shortcut:
Three pipes X, Y and Z are attached to a cistern. X can fill it in 'a' minutes and Y in 'b' minutes. Z is a waste pipe for emptying it. After opening both the pipes X and Y, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the wste pipe had been left open, he closes it and the cistern now gets filled in 't' minutes. The time in which the pipe Z, if opened alone, empty the full cistern in
(
ab / a + b
)
2 x
1 / t
minutes
Here, a = 5, b = 20, t = 4
Using these values in the shortcut, we get:
Required time =
(
5 x 20 / 5 + 20
)
2 x
1 / 4

=
(
100 / 25
)
2 x
1 / 4

= (4)2 x
1 / 4

= 16 x
1 / 4
= 4
Hence, pipe C can emtpy the full cistern in 4 minutes.

47. A pipe P can fill a tank in 15 hours. Due to development of a whole in the bottom of the tank
1 / 4
th of the water filled by the pipe P leaks out. Find the time when the tank will be full?
  A.  20 hours
  B.  23 hours
  C.  15 hours
  D.  18 hours
     
   
View Answer

Shortcut:
If a pipe X fills a cistern in 'a' hours and suddently a leak develops through which every hour 'm' part of the water filled the pipe X leaks out, then the time in which tank will be full is
a / 1 − m
hours.
Here, a = 15, m = 1/4
Using these values in the shortcut, we get:
Required time =
15 / 1 − 1/4

=
15 / (4 − 1)/4

=
15 x 4 / 3
= 5 x 4 = 20
Hence, the tank will be full in 20 hours.

48. Three pipes P, Q and R can fill a tank in 6 minutes, 8 minutes and 12 minutes respectively. The pipe R is closed 6 minutes before the tank is filled. In what time will the tank be full?
  A.  4 minutes
  B.  8 minutes
  C.  6 minutes
  D.  5 minutes
     
   
View Answer

Shortcut:
If pipes X, Y and Z can fill a cistern in a, b and c hours respectively. If pipe Z is closed 't' hours before the cistern is full, then the tank will be filled in
ab(t + c) / ab + bc + ac
hours.
Here, a = 6, b = 8, c = 12, t = 6
Using these values in the shortcut, we get:
Required time =
6 x 8(6 + 12) / 6 x 8 + 8 x 12 + 6 x 12

=
6 x 8(18) / 48 + 96 + 72

=
6 x 144 / 216
= 4
Hence, the tank will be full in 4 minutes.

49. One filling pipe P is 10 times faster than second filling pipe Q. If Q can fill a cistern in 22 minutes, then find the time when the cistern will be full if both fill pipes are opened together.
  A.  5 minutes
  B.  2 minutes
  C.  6 minutes
  D.  3 minutes
     
   
View Answer

Shortcut:
One filling pipe X is 'z' times faster than the other filling pipe Y. If Y can fill a cistern in 'a' hours, then the time in which the cistern will be full, if both the filling pipes are opened together, is
a / z + 1
hours.
Note: Value of the slower filling pipe is given.
Here, a = 22, z = 10
Using these values in the shortcut, we get:
Required time =
22 / 10 + 1
=
22 / 11
= 2
Hence, the cistern will get full in 2 minutes.

50. There are 6 filling pipes each capable of filling a cistern alone in 16 minutes and 4 emptying pipes each capalbe of emptying a cistern alone in 20 minutes. All pipes are opened together and as a result, tank fills 14 litres of water per minute. Find the capacity of the tank.
  A.  75 litres
  B.  85 litres
  C.  70 litres
  D.  80 litres
     
   
View Answer

Shortcut:
To find out the capacity (C) of the cistern in litres, if 'a' number of filling pipes, each capable of filling a cistern alone in 'y' minutes, and 'b' number of emptying pipes, each capable of emptying a cistern alone in 'z' minutes, are opened together and as a result 'q' is the rate at which the tank fills per minute, the following formula is used, C =
qyz / az − by
litres.
Here, a = 6, y = 16, b = 4, z = 20, q = 14
Using these values in the shortcut, we get:
Required capacity =
14 x 16 x 20 / 6 x 20 − 4 x 16

=
14 x 16 x 20 / 120 − 64

=
14 x 16 x 20 / 56
= 4 x 20 = 80
Hence, the capacity of the tank is 80 litres.

Copyright © 2020-2022. All rights reserved. Designed, Developed and content provided by Anjula Graphics & Web Desigining .