# Aptitude::Pipes & Cistern

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*(20)*

Example 1 / 10

Time taken by pipe A alone to fill the tank= 36 hours

So, part filled by pipe A in 1 hour= (1/36)

Time taken by pipe B alone to fill the tank= 40 hours

So, part filled by pipe B in 1 hour= (1/40)

Now, if both the pipes are opened, then part filled by both in 1 hour= (1/36 + 1/40) = (19/360)

Therefore, time taken by A and B together to fill the tank= (360/19) hours = 18 hours 57 min.

So, part filled by pipe A in 1 hour= (1/36)

Time taken by pipe B alone to fill the tank= 40 hours

So, part filled by pipe B in 1 hour= (1/40)

Now, if both the pipes are opened, then part filled by both in 1 hour= (1/36 + 1/40) = (19/360)

Therefore, time taken by A and B together to fill the tank= (360/19) hours = 18 hours 57 min.

**Ans.**
Simply we have to calculate the part filled by a particular tap in unit time. This unit time can be in hour, minutes and day. After getting do LCM for both the taps and then calculate the total time.

NA

NA

Example 2 / 10

Let us name the three pipes as pipe A, B and C respectively.

Time taken by pipe A alone to fill the tank= 10 hours

So, part filled by pipe A in 1 hour= (1/10)

Time taken by pipe B alone to fill the tank= 15 hours

So, part filled by pipe B in 1 hour= (1/15)

Time taken by pipe C alone to empty the tank= 25 hours

So, part emptied by pipe C in 1 hour= (1/25)

Now, if all the three pipes operate at the same time, part filled by A, B and C in 1 hour = (1/10 + 1/15 - 1/25) = (19/150)

Therefore, Time taken to fill the tank if all three pipes operate = (150/19) hours 7 hrs. 53 min.

Time taken by pipe A alone to fill the tank= 10 hours

So, part filled by pipe A in 1 hour= (1/10)

Time taken by pipe B alone to fill the tank= 15 hours

So, part filled by pipe B in 1 hour= (1/15)

Time taken by pipe C alone to empty the tank= 25 hours

So, part emptied by pipe C in 1 hour= (1/25)

Now, if all the three pipes operate at the same time, part filled by A, B and C in 1 hour = (1/10 + 1/15 - 1/25) = (19/150)

Therefore, Time taken to fill the tank if all three pipes operate = (150/19) hours 7 hrs. 53 min.

**Ans.**
We have done negative because the third tap is going to decrease the content of the tank. Rest as type 1

NA

NA

Example 3 / 10

Let the reservoir is filled by first pipe in x hours.

Then, second pipe will fill it in (x+10) hours

Then 1/x +1/(x+10) = 1/12 ----- (eq.)

By Solving above equation, we will get x=20

So, the second pipe will take 20+10=30 hrs to fill the reservoir.

Then, second pipe will fill it in (x+10) hours

Then 1/x +1/(x+10) = 1/12 ----- (eq.)

By Solving above equation, we will get x=20

So, the second pipe will take 20+10=30 hrs to fill the reservoir.

**Ans.**
It is similar to type 1 only one difference is there that here we know the combined work of both the pipes and one of the pipe. We only need to calculate the work done by another pipe.

NA

NA

Example 4 / 10

Let us name the two taps which are filling the cistern as A and B, and the waste pipe which is emptying the cistern as C.

Time taken by A alone to fill the tank= 12 hours

So, part filled by A in 1 hour = (1/12)

Time taken by B alone to fill the tank = 16 hours

So, part filled by B in 1 hour = (1/16)

Time taken to fill the cistern when all three operates = 20 hours

Part filled in 1 hour when all three operate = (1/20)

Now, if all the three operate at the same time then we can write,

(1/A)+(1/B)-(1/C) = (1/20)

Or

(1/12)+(1/16)-(1/C) = (1/20)

By, Solving the above equation we get,

C = (240/23) hours

Therefore, the waste pipe C takes (240/23) hours to empty the cistern alone.

Time taken by A alone to fill the tank= 12 hours

So, part filled by A in 1 hour = (1/12)

Time taken by B alone to fill the tank = 16 hours

So, part filled by B in 1 hour = (1/16)

Time taken to fill the cistern when all three operates = 20 hours

Part filled in 1 hour when all three operate = (1/20)

Now, if all the three operate at the same time then we can write,

(1/A)+(1/B)-(1/C) = (1/20)

Or

(1/12)+(1/16)-(1/C) = (1/20)

By, Solving the above equation we get,

C = (240/23) hours

Therefore, the waste pipe C takes (240/23) hours to empty the cistern alone.

**Ans.**
We just have to calculate the work done by the waste pipe in unit time. We will get negative value because it is decreasing the contents of the cistern hence negative.

NA

Work done by waste pipe in 1 minute = 1/20 - (1/12+1/16) = -23/240

Hence the waste pipe will empty the full cistern = 240/23 min. = 10.43 minutes

Hence the waste pipe will empty the full cistern = 240/23 min. = 10.43 minutes

**Ans.**Example 5 / 10

Let the time taken by the leak to empty the tank be 'L' hours.

Time taken by the electric pump to fill the tank= 4 hours

So, part filled by the electric pump in 1 hour = (1/4)

Time taken by the leak to empty the empty the tank = L hours

So, part emptied in 1 hour = (1/L)

Therefore,

(1/4) - (1/L) = 1/4.5

Solving the above equation we get,

L= 36 hours

Hence, time taken by the leak to empty the tank = 36 hours

Time taken by the electric pump to fill the tank= 4 hours

So, part filled by the electric pump in 1 hour = (1/4)

Time taken by the leak to empty the empty the tank = L hours

So, part emptied in 1 hour = (1/L)

Therefore,

(1/4) - (1/L) = 1/4.5

Solving the above equation we get,

L= 36 hours

Hence, time taken by the leak to empty the tank = 36 hours

**(Ans)**
The catch of the question is the extra time which is taken is that amount of work done by the leak.

NA

Work done by the leak in in the tank in 1 hour = [1/4 - 1/4.5] = 1/36

Hence, the leak in tank will empty the tank completely = 36 hrs.

Hence, the leak in tank will empty the tank completely = 36 hrs.

**(Ans)**Example 6 / 10

Work done by the 2 pipes in 1 hour = 1/10+1/16 = 13/80

∴ Time taken by these pipes to fill the tank = 80/13 =6 hr 2minutes

Due to leakage, time taken = 6 hr 2 mins +28 mins= 6 hr 30 minutes

∴ Work done by (2 pipes + leak) in 1 hr =1/6.5

Work done by the leak in 1 hour = 13/80 - 1/6.5 = 9/1040

∴ Leak will empty the pipe in 115.55 hrs

∴ Time taken by these pipes to fill the tank = 80/13 =6 hr 2minutes

Due to leakage, time taken = 6 hr 2 mins +28 mins= 6 hr 30 minutes

∴ Work done by (2 pipes + leak) in 1 hr =1/6.5

Work done by the leak in 1 hour = 13/80 - 1/6.5 = 9/1040

∴ Leak will empty the pipe in 115.55 hrs

**(Ans.)**
NA

NA

NA

Example 7 / 10

Time taken by pipe A to fill the tank alone = 40 mins

So, part filled by pipe A in 1 hour = (1/40)

Similarly, time taken by pipe B to fill the tank alone = 45 mins

So, part filled by pipe B in 1 hour = (1/45)

And, time taken by pipe C to empty the tank alone = 50 mins

So, part emptied by pipe C in 1 hour = (1/50)

Now, part filled by pipe A and B together = (1/40) + (1/45) = 17/360

So, part filled by both in 7 minutes = 7 x (17/360) = 119/360

Remaining part = 1-(119/360) = 241/360

Now, part filled in 1 min when A, B and C are operating at the same time = (1/40) + (1/45) - (1/50) = 49/1800

But, time taken to fill (49/1800)

Or, time taken to fill 1 part of the tank = (1/(49/1800)) min = (1800/49) mins

Therefore, time taken to fill the remaining (241/360)

=(1205/49) mins = 24.59 mins

Therefore, total time taken = (24.59 + 7) mins = 31.59 mins

So, part filled by pipe A in 1 hour = (1/40)

Similarly, time taken by pipe B to fill the tank alone = 45 mins

So, part filled by pipe B in 1 hour = (1/45)

And, time taken by pipe C to empty the tank alone = 50 mins

So, part emptied by pipe C in 1 hour = (1/50)

Now, part filled by pipe A and B together = (1/40) + (1/45) = 17/360

So, part filled by both in 7 minutes = 7 x (17/360) = 119/360

Remaining part = 1-(119/360) = 241/360

Now, part filled in 1 min when A, B and C are operating at the same time = (1/40) + (1/45) - (1/50) = 49/1800

But, time taken to fill (49/1800)

^{th}part of the tank = 1 minOr, time taken to fill 1 part of the tank = (1/(49/1800)) min = (1800/49) mins

Therefore, time taken to fill the remaining (241/360)

^{th}part of the tank = (1800/49) x (241/360)=(1205/49) mins = 24.59 mins

Therefore, total time taken = (24.59 + 7) mins = 31.59 mins

**Ans.**
First approach with the amount of work done by the 2 pipes in 7 minutes and after that all the three pipes will work.

NA

NA

Example 8 / 10

Let us assume that pipe B be closed after 'x' mins.

Part filled by A and B together in 1 min = (1/20) + (1/30) = 1/12

So, part filled by A and B in 'x' mins = x/12

Now, part filled by A alone in 1 min = 1/20

So, part filled by A alone in (18-x) mins = (18-x)/20

According to the question:

(x/12) + (18-x)/20 = 1 [Since, total work done is equal to 1]

By, solving the above equation we get:

=> x = 3 mins

Therefore, pipe B should be closed after 3 mins.

Part filled by A and B together in 1 min = (1/20) + (1/30) = 1/12

So, part filled by A and B in 'x' mins = x/12

Now, part filled by A alone in 1 min = 1/20

So, part filled by A alone in (18-x) mins = (18-x)/20

According to the question:

(x/12) + (18-x)/20 = 1 [Since, total work done is equal to 1]

By, solving the above equation we get:

=> x = 3 mins

Therefore, pipe B should be closed after 3 mins.

**Ans.**
NA

NA

NA

Example 9 / 10

Tank can be emptied by leak in 1 hour = 1/10

or

Time required to emptied 1/10 part of tank = 1 hour

Given full tank can be emptied by leak = 15 hours

or

Time need to emptied 1/15 part of tank by leak = 1 hours

so (1/10 - 1/15) part can be filled by tap = 1 hours

=> 1/30 part can be filled by tap = 1 hours or

Time needed to fill the cistern by tap = 30 hours = 1800 minutes

So, Capacity of cistern = 8 * 1800 =

or

Time required to emptied 1/10 part of tank = 1 hour

Given full tank can be emptied by leak = 15 hours

or

Time need to emptied 1/15 part of tank by leak = 1 hours

so (1/10 - 1/15) part can be filled by tap = 1 hours

=> 1/30 part can be filled by tap = 1 hours or

Time needed to fill the cistern by tap = 30 hours = 1800 minutes

So, Capacity of cistern = 8 * 1800 =

**14400 liters (Ans.)**
NA

NA

NA

Example 10 / 10

Time taken by A, B and C to fill the tank together = 6 hours

So, part filled by A, B and C in 1 hour = 1/6

part filled by A, B and C in 2 hours = 2 x (1/6) = 1/3

Remaining part = 1 - (1/3) = 2/3

Now, part of the tank filled by A and B in 7 hours = 2/3

part of the tank filled by A and B In 1 hour = (2/3) / 7 = 2/21

[Part filled by (A+B+C) in 1 hour] - [Part filled by (A+B) in 1 hour] = C's 1 hour work

C's 1-hour work = (1/6) - (2/21) = 1/14

Therefore, time taken by C alone to fill the tank = 14 hours

So, part filled by A, B and C in 1 hour = 1/6

**Or**part filled by A, B and C in 2 hours = 2 x (1/6) = 1/3

Remaining part = 1 - (1/3) = 2/3

Now, part of the tank filled by A and B in 7 hours = 2/3

**Or**part of the tank filled by A and B In 1 hour = (2/3) / 7 = 2/21

[Part filled by (A+B+C) in 1 hour] - [Part filled by (A+B) in 1 hour] = C's 1 hour work

**Or**C's 1-hour work = (1/6) - (2/21) = 1/14

Therefore, time taken by C alone to fill the tank = 14 hours

**Ans.**
NA

NA

NA