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# Pipes & Cistern Questions

Home > Quantitative Aptitude > Pipes & Cistern > General Questions
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A pump can be used either to fill or to empty a tank. The capacity of the tank is 3600 m. The emptying capacity of the pump is 10 m 3/min higher than its filling capacity. What is the emptying capacity of the pump if the pump needs 12 more minutes to fill the tank than to empty it?

A50 m

B60 m

C30 m

DNone of these

Explanation:

Let â€˜fâ€™ m^3/min be the filling capacity of the pump. Therefore, the emptying capacity of the
pump will be = (f + 10 ) m^3/ min.
The time taken to fill the tank will be = 3600/f minutes
And the time taken to empty the tank will be = 3600/(f+10).
We know that it takes 12 more minutes to fill the tank than to empty it
=> 3600/f - 3600/(f+10) = 12
=> 36000 = 12 (f^2 + 10 f) => 3000
=> f^2 + 10f - 3000 = 0
Solving for positive value of â€˜fâ€™ we get, f = 50.
So, the emptying capacity of the pump = 50 + 10 = 60 m

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Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical Exputions P, Q and R respectively. What is the proportion of Expution R in the liquid in the tank after 3 minutes?

A7/11

B3/11

C5/11

D6/11

Explanation:

Part filled by A in 1 min. = 1/30
Part filled by B in 1 min. = 1/20
and Part filled by C in 1 min. = 1/10
So, Part filled by (A + B + C) in 3 minutes = 3(1/30 + 1/20 + 1/10) = 11/20.
Part filled by C in 3 minutes = 3/10
Hence, the Required ratio = 3/10 * 20/11 = 6/11.

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There are two water tanks A and B, A is much smaller than B. While water fills at the rate of 1 liter every hour in A, it gets filled up like, 10, 20, 40, 80, 160 in tank B. (At the end of first hour, B has 10 liters, second hour it has 20 liters and so on). If tank B is 1/32 filled of the 21 hours, what is total duration of hours required to fill it completely?

A26

B25

C5

D27

Explanation:

for every hour water in tank in B is doubled,
Let the duration to fill the tank B is x hours.
x/32 part of water in tank of B is filled in 21 hours,
Next hour it is doubled so,
2*(x/32) part i.e (x/16) part is filled in 22 hours,
Similarly (x/8)th part in 23 hours,(x/4)th part is filled in 24 hours,
(x/2)th part is filled in 25 hours, (x)th part is filled in 26 hours

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If a pipe can fill the tank within 6hrs.But due to leak it takes 30 min more. Now the tank is full then how much time will it take to empty the tank through the leak?

A78

B56

C66

D59

Explanation:

Pipe A takes 6 hours to fill the tank. Therefore, it fills 1/6th of the tank in an hour or 16.66% of the tank in an hour.

When there is a leak it takes 6 hours 30 minutes for the tank to fill. i.e 13/2 hrs hours to fill the tank or 2/13th or 15.38% of the tank gets filled in 1 hr.

On account of the leak, (16.66âˆ’15.38)%=1.28% of the water gets wasted every hour. Therefore, the leak will take 100/1.28 i.e 78.125 hrs to drain a full tank.

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An empty tank be filled with an inlet pipe A in 42 minutes. After 12 minutes an outlet pipe B is opened which can empty the tank in 30 minutes. After 6 minutes another inlet pipe C opened into the same tank, which can fill the tank in 35 minutes and the tank is filled. Find the time taken to fill the tank?

A45.5 mins

B50.5 mins

C58.5 mins

Dnone of these

Explanation:

For first 12 mins tank is 2/7 filled
For first 18 mins tank is 8/35 filled. i.e pipe A is opened for 18 mins and pipe B is opened for 6 mins. (3/7)-(1/5)=8/35. that means after 18 mins tank is 8/35 filled.
so, after 18 mins all the 3 pipes work simultaneously.
Now calculate the time required to fill the tank by 3 pipes,
(1/42)+(1/35)-(1/30)=2/105. It means 105/2 mins.
But the tank is already 8/35 filled.
The time taken to fill 27/35 of the tank=(105/2)*(27/35)= 40.5 mins.
The time taken to fill the tank= (18+40.5) mins = 58.5 mins.

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It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in 10 minutes?

A4 times the draining rate

B3 times the draining rate

C2.5 times the draining rate

D2 times the draining rate

Explanation:

The equation to use is RT = Q
Q is equal to 1/2 the tank.
T is equal to 30 minutes
R is derived from the equation as follows:
RT = Q
R*30 = 1/2
R = 1/60
the tank is being drained at the rate of 1/60 of the tank per minute.
at that rate, the remaining 1/2 of the tank will be drained in 30 minutes.

the rate of filling and the rate of draining are opposing forces so you need to subtract one from the other to get the rate of filling.

you want to fill 1/2 the tank while at the same time you are draining the tank.
the combined formula would be as follows:
(RF - RD) * 10 = 1/2
you know RD is 1/60 because we just solved for that.
(RF - 1/60) * 10 = 1/2
simplify this to get:
10*RF - 10/60 = 1/2
add 10/60 to both sides of this equation to get:
10*RF = 1/2 + 10/60 which becomes:
10*RF = 30/60 + 10/60 which becomes:
10*RF = 40/60.
divide both sides of this equation by 10 to get:
RF = 4/60.

4/60 is 4 times 1/60.

fill rate is 4 times the drain rate.

you start at 1/2 a tank which is the same as 3/6 of a tank.
in 10 minutes you will have drained 10*1/60 = 10/60 = 1/6 more of the tank.
in the same 10 minutes you will have filled 10*4/60 = 40/60 = 4/6 of the tank.
sum of fill and drain is equal to 3/6 - 1/6 + 4/6 which is equal to 6/6 which is equal to 1.

at the end of the 10 minutes, the tank is full.

you could have solved this another way as well.
you still had to find the drain rate which is equal to 1/60 of the tank per minute.

in 10 minutes, you will have drained 1/6 more of the tank.
the tank is now 1/2 - 1/6 = 3/6 - 1/6 = 2/6 full.

in order to fill the tank in the same 10 minutes, you have to fill 4/6 of the tank.

the same formula is used.
RT=Q
10R = 4/6
R = 4/60.

the fill rate is 4 times the drain rate.

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Two pipes can fill a tank in 10 hours and 12 hours respectively while a third, pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?

A8.5 hrs.

B6 hrs.

C8 hrs.

D7.5 hrs.

Explanation:

Given, All the pipes are operate simultaneously
=> Tank filled in 1 hrs = 1/10+1/12-1/20 = 16/120 = 2/15
So, time taken to fill the tank = 15/2 = 7.5 hrs.

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The water from one outlet, flowing at a constant rate, can fill the swimming pool in 9 hours. The water from second outlet, flowing at a constant rate can fill up the same pool in approximately in 5 hours. If both the outlets are used at the same time, approximately what is the number of hours required to fill the pool?

A3.21 hrs.

B3.34 hrs.

C2.32 hrs.

DNone of these

Explanation:

Lets assume tank capacity = x Liters.
Given:
First pipe fills the tank in 9 hours.
So first pipe capacity = x/9 Liters/ Hour.
Second pipe fills the tank in 5 hours.
So its capacity is x/ 5 Liters/Hour

Condition: If both pipes are opened together
number of hrs. required to fill a tank of capacity x liters = x/(x/5+x/9) = x/(14x/45) = 45/14 = 3.21 hrs.

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Two pipes A and B can fill a tank in 4 hours and 5 hours respectively. If they are turned up alternatively for one hour each,then time taken to fill the tank is:

A2 hrs 15 mins

B4 hrs 24 mins

C5 hrs

D3 hrs

Explanation:

L.C.M of 4hr and 5hr will be 20units. (tank size)
Given. pipe A and B fill a tank in 4 hrs and 5 hrs res.
So, A can fill 5u in 1hr and B can fill 4u in 1hr.
(5u+4u)+(5u+4u)=18u in 4hr (alternatively opend)
Remaining 2u will be filled by pipe A
5u will take=60mins
Hence 1u =60/5min=12mins
Hence 2u will take 12*2=24mins.
Total time required = 4 hrs. + 24 min.

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The number of times a bucket of capacity 4 litres to be used to fill up a tank is less than the number of times another bucket of capacity 3 litres used for the same purpose by 4. What is the capacity of the tank?

A360 litres

B256 litres

C48 litres

D525 litres

Explanation:

Let the capacity of the tank be x
no.of times a bucket of capacity 4 liters :x/4
for 3 liters: x/3
x/3-x/4=4
x=48

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