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# Aptitude - Pipes & Cisterns Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to **Pipes & Cisterns**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

Q 1 - A tap can fill storage in 8 hours and another tap can discharge it in 16 hours. In the event that both the taps are open, the time taken to fill the tank will be:

### Answer : C

### Explanation

Net part filled in 1 hr = (1/8 ? 1/16)= 1/16 Total time taken to fill the tank = 16 hrs.

Q 2 - A channels can fill a tank in x hours and another funnel can exhaust it in y (y>x) hours. In the event that both the funnels are open, in how long will the tank be filled?

### Answer : D

### Explanation

Work done by filling pipe in 1 hr = 1/x Work done by emptying pipe in 1 hr = 1/y Net filling work done by both in 1 hr = (1/x- 1/y) = (y-x)/xy ∴The tank will be filled in xy/(y-x) hrs.

Q 3 - A funnel can discharge a tank in 40 minutes. A second pipe with distance across twice as much as that of the first is likewise joined with the tank to purge it. The two together can exhaust the tank in:

### Answer : B

### Explanation

A pipe with double diameter will take half time. So, the second pipe can empty the full tank in 20 min. Part emptied by both in 1 min. (1/40+ 1/20) = 3/40 Time taken to empty the full tank = 40/3 min.

Q 4 - Two channels can fill a tank in 15 hours and 12 hours separately and a third pipe can purge it in 4 hours. In the event that the channels are opened all together at 8 am, 9 am and 11am separately, the tank will be exhausted at

### Answer : D

### Explanation

Let the tank be emptied in x hrs after 8 am. Work done by A in x hrs, by B in (x-1) hrs and C in (x-3) hrs = 0 ⇒x/15+ (x-1)/12- (x-3)/4 = 0 ⇒ 4x+5(x-1) - 15(x-3) = 0 ⇒6x= 40 ⇒x= 20/3 hrs. ⇒x= 6 hrs. 40 min after 8 am Hence the tank will be emptied at 14 hrs 40 min, i.e., 2:40 pm

Q 5 - A tank is fitted with two taps A and B. A can fill the tank totally in 45 minutes and B can purge the full tank in 60 minutes. On the off chance that both the taps are opened on the other hand for 1 minute, then in what amount of time the unfilled tank will be filled totally?

### Answer : D

### Explanation

Work done by A in 1st minutes and B 2nd minute= (1/45- 1/60)= 1/180 Part filled in 2 min = 1/180 Part filled in 358 min = (1/360*358) = 358/360 = 179/180 Remaining part = (1-179/180) = 1/180 1/45 part is filled by A in (45*1/180) min= 1/4 min. Total time taken to fill it = 358 1/4 min = 5 hrs.58 min 15 sec.

Q 6 - A substantial tanker can be filled by two pipes A and B in an hour and 40 minutes separately. How long will it take to fill the tanker from unfilled state if B is utilized for a fraction of the time and A and B fill it together for the other half?

### Answer : D

### Explanation

Let the total time taken be x minute. Then, (1/40*x/2) + (1/60+ 1/40) x/2= 1 ⇒ x/80 + x/48 = 1 ⇒3x+ 5x= 240 ⇒8x= 240 ⇒x= 30 Hence, the required time is 30 minutes.

Q 7 - A break in the base of a tank can purge the full tank in 6 hours. A channel funnel fills water at the rate of 4 liters for each moment. At the point when the tank is full, the channel is opened and because of the break, the tank is void in 8 hours. The limit of the tank is

### Answer : B

### Explanation

Part filled in 1 hour = (1/6-1/8)= 1/24 So, the inlet can fill the tank in 24 hours. Capacity of the tank= volume of water that flows in 24 hrs = (4*60*24) ltr. = 5760 liters.

Q 8 - A cistern has 3 pipes A, B and C. A and B can ill it in 3 hours and 4 hours respectively while C can empty the completely filled cistern in 1 hour. If the pipes are opened in order at 3, 4 and 5 pm respectively, at what time will the cistern be empty?

### Answer : A

### Explanation

In 2 hours Pipe A will fill = 2/3 tank In 1 hour Pipe B will fill = 1/4 tank Part of tank filled till 5 PM = (2/3) + (1/4) = 11/12 Remaining part = 1 - (11/12) = 1/12 Net part emptied when A, b and C are opened = (1/3) +( 1/4) - 1 = (4+3-12)/12 = -5/12 ∴ 5/12 part is emptied in 1 hour. ∴ 11/12 is emptied in (12/5)*(11/12) = 11/5 = 2 hr 12 min ∴ Tank will be emptied at 7:12 PM.

Q 9 - Two pipes A and B can fill a water tank in 20 and 24 min respectively. A third pipe C can empty at the rate of 3 gallons per minute. If A, B and C opened together fill the tank in 15 min, the capacity of the tank (in gallons) is:

### Answer : C

### Explanation

Let the capacity of the tank = x gallons Quantity of the water filled in the tank in 1 min when all the pipes A, B and C are opened simultaneously= x/20 + x/24 - 3 According to question, x/20 + x/24 - 3 = x/15 or, x/20 + x/24 - x/15 = 3 or, (6x + 5x - 8x)/120 = 3 or, 3x/120 = 3 or, x = 120 gallons

Q 10 - Taps A, B and C are connected to a water tank and the rate of flow of water is 42 ltr/hr, 56 ltr/hr and 48 ltr/hr respectively. A and ill fill the tank while tap C empties the tank. If the three taps are opened simultaneously, the tank gets filled up completely in 16 hours. What is the capacity of the tank?

### Answer : A

### Explanation

Net amount of water filled in 1 hr when all the pipes are opened simultaneously = 42 + 56 ? 48 = 50 ltr Tank gets completely filled in 16 hrs. ∴ Capacity of tank = 16*50 = 800 liters.