# Model 1 Basic Pipes & Cisterns problems Practice Questions Answers Test with Solutions & More Shortcuts

#### pipes & cisterns PRACTICE TEST [3 - EXERCISES]

Question : 1 [SSC CGL Tier-I 2015]

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

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

Question : 2 [SSC CGL Prelim 2003]

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

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 time
Note: 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 : 3 [SSC DEO 2009]

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

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 : 4 [SSC CGL Tier-I 2014]

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

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 : 5 [SSC CGL Prelim 2002]

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

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 tank
when both are opened = \$({xy}/{x - y})\$ : x > y
b) time taken to empty the tank
when 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.

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