# CAT Questions on Pipes and Cisterns Set-3 PDF

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## CAT Questions on Pipes and Cisterns Set-3 PDF

Download important Pipes and Cisterns Set-3 Questions for CAT PDF based on previously asked questions in CAT exam. Practice Pipes and Cisterns Set-3 Questions PDF for CAT exam.

Question 1: A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in 8 hours while an outlet pipe can empty the full tank in 12 hours. If all pipes are left open simultaneously, it takes 6 hours to fill the empty tank. What is the relationship between M and N?

a) M : N = 1 : 1

b) M : N = 2 : 1

c) M : N = 2 : 3

d) M : N = 3 : 2

e) None of the above

Question 2: Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?

a) Pipe A was open for 19 hours.

b) Pipe A was open for 19 hours 30 minutes.

c) Pipe B was open for 19 hours 30 minutes.

d) Pipe C was open for 19 hours 50 minutes.

e) The situation is not possible.

Question 3: A cylindrical overhead tank is filled by two pumps – P1 and P2. P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours. There is a pipe P3 which can empty the tank in 8 hours. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open P3 when the tank is half filled so that tank is exactly filled up by the time he is back. Due to technical fault P3 opens when tank is one third filled. If the supervisor comes back as per the plan what percent of the tank is still empty?

a) 25% tank

b) 12% tank

c) 10% tank

d) None of the above

Question 4: Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately in 5 hours, 10 hours and 15 hours respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at 3/4th of their normal rates for the first hour after which they supplied water at the normal rate. Pipe C supplied water at 2/3rd of its normal rate for first 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled.

a) 1.05 Hours

b) 2.05 Hours

c) 3.05 Hours

d) None of these

Question 5: A tank is connected with both inlet pipes and outlet pipes. Individually, an inlet pipe can fill the tank in 7 hours and an outlet pipe can empty it in 5 hours. If all the pipes are kept open, it takes exactly 7 hours for a completely filled-in tank to empty. If the total number of pipes connected to the tank is 11, how many of these are inlet pipes?

a) 2

b) 4

c) 5

d) 6

Question 6: Four two-way pipes A, B, C and D can either fill an empty tank or drain the full tank in 4, 10, 12 and 20 minutes respectively. All four pipes were opened simultaneously when the tank is empty. Under which of the following conditions the tank would be half filled after 30 minutes?

a) Pipe A filled and pipes B, C and D drained

b) Pipe A drained and pipes B, C and D filled

c) Pipes A and D drained and pipes B and C filled

d) Pipes A and D filled and pipes B and C drained

e) None of the above

Question 7: A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in 8 hours. If the outlet pipe is open then the inlet pipe fills the empty tank in 10 hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full?

a) 20

b) 30

c) 40

d) 45

M inlet pipe can fill $(\frac{M}{8})^{th}$ part of the tank in 1 hour.

Similarly, N outlet pipes can empty $(\frac{N}{12})^{th}$ part of the tank in 1 hour.

=> $\frac{M}{8} – \frac{N}{12} = \frac{1}{6}$

=> $6 M – 4 N = 8$

=> $M = \frac{4 + 2 N}{3}$

If, N = 1, => M = 2 => $M : N = 2 : 1$

If, N = 4, => M = 4 => $M : N = 1 : 1$

Thus, Ans – (E)

Height of cone comes down to 50%, => it becomes $\frac{1}{2}$

=> Volume would become $\frac{1}{8}$ as radius will also become half by similar triangles.

Let the capacity of cone = 24 litres

Volume of water run-off = $24 – \frac{1}{8} \times 24 = 21$ litres

Volume of water left in the cone = $\frac{1}{8} \times 24 = 3$ litres

Pipe A’s efficiency = $\frac{24}{8} = 3$ litres/hr

Pipe B’s efficiency = $\frac{24}{12} = 2$ litres/hr

Pipe C’s efficiency = $\frac{24}{-4} = -6$ litres/hr

All will run 19 hours simultaneously (going by the options)

=> Net effect = $(3 + 2 – 6) \times 19 = -19$ litres

This means that after 19 hours, 19 litres of water has been removed, we need to remove 2 more litres as per the requirement. Thus,  C will definitely run for another hour.

If we run A and C together for the 20th hour, net effect = $(3 – 6) \times 1 = -3$ litres

Run B for 30 minutes => $2 \times \frac{1}{2} = 1$ litres

$\therefore$ Volume of water removed = $-19 -3 + 1 = -21$ litres

Thus, Pipe B was open for 19 hours 30 minutes.

P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours therefore when Both the pumps are opened simultaneously they can fill $\frac{1}{8}$ + $\frac{1}{12}$ = $\frac{5}{24}$ this much of tank in an hour.Therefore Time required by P1 and P2 to fill full tank is $\frac{24}{}$ Hence they can fill half the tank in $\frac{12}{5}$ = 2.4 hours. After all the 3 pipes are opened the tank will be filled at the rate of $\frac{1}{8}$ + $\frac{1}{12}$ – $\frac{1}{8}$ = $\frac{1}{12}$ therefore for filling the remaining tank we will need 6 more hours. If supervisor was planning to come exactly when the tank is fully filled then he must come after 8.4 hours

Time required by P1 and P2 to fill 1/3rd of the tank is $\frac{24}{5*3}$ = $\frac{8}{5}$ = 1.6 hours therefore supervisor will come after = 8.4 – 1.6 = 6.8 hours

In 6.8 hours all the three pipe will fill $\frac{6.8}{12}$ = 0.56 of the capacity of tank

therefore remaining tank to be filled when the supervisor comes = 1- 0.56 – 0.33 $\approx$ 0.1 i.e. 10%

Therefore option ‘C’ is our answer

Let the capacity of the tank be 60 litres.
Capacity of the first pipe = 12 l/hr
Capacity of the second pipe = 6 l/hr
Capacity of the third pipe = 4 l/hr
In 2 hrs, first pipe fills (9 + 12) l = 21 l
In 2 hrs, second pipe fills (4.5 + 6) = 10.5 l
In 2 hrs, third pipe fills (16/3) l
In 2 hrs, tank filled = (21 + 10.5 + 5.33) l = 36.83 l
Tank left to be filled = (60 – 36.83) l = 23.17 l
Time required = (23.17/22) hr = 1.05 hrs
Total time = 3.05 hrs
Hence, option C is the correct answer.

Let the number of inlet pipes be x, then the number of outlet pipes will be 11-x.

The rate of emptying the tank is more than filling the tank if all the pipes are kept opened.

$\frac{11-x}{5}$ – $\frac{x}{7}$ = $\frac{1}{7}$

Solving, 12x =72,Hence x = 6

Therefore, the number of inlet pipes is 6.

Let us assume the volume of the tank to be 60 litres.
A can fill or empty 60/4 = 15 litres in a minute.
B can fill or empty 60/10 = 6 litres in a minute.
C can fill or empty 60/12 = 5 litres in a minute.
D can fill or empty 60/20 = 3 litres in a minute.

We have to find the combination for which the tank will be half-full in 30 minutes (i.e., completely filled in 1 hour).
Therefore, the combination must result in a net input of 60/60 = 1 litre per minute.

Let us evaluate the options.
Option B:
Pipe A drained and pipes B, C and D filled

The net result will be -15 + 6 + 5 + 3 = -1 litre/minute. We can eliminate option B.

Option C:
Pipes A and D drained and pipes B and C filled

The net result will be -15-3+6+5 = -7 litres/minute. We can eliminate option C as well.

Option D:
Pipes A and D filled and pipes B and C drained
The net result will be 15+3-6-5 = 7 litres/minute. We can eliminate option D as well.

Option A:
Pipe A filled and pipes B, C and D drained

The net result will be 15-6-5-3 = 1 litre/minute.
Therefore, option A is the right answer.

Then, $\frac{1}{8} – \frac{1}{x} = \frac{1}{10}$
=> $x = 40$