Two pipes P1 and P2 can fill a cistern in 6 minutes and 9 minutes

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Two pipes running together can fill a tank in 11 1/9 minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

Q.1.Pipe M and N running together can fill a cistern in 6 minutes. If M takes 5 minutes less than N to fill the cistern, then the time in which N alone can fill the cistern will be

Sol : Option A
Explanation: Let pipe M fills the cistern in x minutes. Therefore, pipe N will fill the cistern in (x+5) minutes. Now, 1/x + 1/(x+5) = 1/6 → x = 10

Thus, the pipe M can fill in 10 minutes, so N can fill in 10+5 =15 minutes.

Q.2. A cistern normally takes 10 hours to be filled by a tap but because of one open outlet pipe, it takes 5 hours more. In how many hours will the outlet pipe will empty a full cistern?

Sol : Option C
Explanation: As cistern is filled in 10 hours, therefore in 1 hour, filled part → 1/10th Now, due to outlet pipe, filled part in 1 hour = 1/15th Part of the cistern emptied, due to leakage in 1 hour = 1/10 - 1/15 = 1/30th

Therefore the leak will empty the full cistern in 30 hrs.

Q.3. Two pipes can fill a tank in 12 and 20 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, 30 minutes extra are taken for the cistern to be filled up. If the cistern is full, in what time would the leak empty it?

Sol : Option A
Explanation: Cistern filled by both pipes in one hour = 1/12+1/20=2/15th Therefore both pipes filled the cistern in 15/2hrs. Now, due to leakage both pipes filled the cistern in 15/2+30/60=8hrs. Therefore Due to leakage, filled part in one hour = 1/8 Therefore part of cistern emptied, due to leakage in one hour = 2/15-1/8= 1/120th

∴ In 120 hrs, the leak would empty the cistern.

Q.4.Two pipes P and Q can fill a cistern in 36 and 48 minutes respectively. Both pipes are opened together, after how many minutes should Q be turned off,so that the cistern be fill in 24 minutes?

Sol : Option B
Explanation: P can fill the cistern in 36 minutes,so in 1 min, P can fill the cistern = 1/36th part In 24 min, P can fill the cistern = 24/36 = 2/3rd. Remaining part = 1- 2/3 = 1/3rd As Q can fill full cistern in 48 minutes,so it will fill

1/3rd part in 16 minutes.

Q.5.Two pipes A and B can fill a tank in 20 and 16 hours respectively. Pipe B alone is kept open for 1/4 of time and both pipes are kept open for remaining time. In how many hours, the tank will be full?

Sol : Option C
Explanation: Let the required time be x hours, then [image]
⇒ x/16+3x/80 = 1⇒ x= 11=10 hours.

Q.6.Two taps M and N can separately fill a cistern in 30 and 20 minutes respectively. They started to fill a cistern together but tap A is turned off after few minutes and tap B fills the rest part of cistern in 5 minutes. After how many minutes was tap M turned-off?

Sol : Option A
Explanation: Let M was turned off after x min. Then, cistern filled by M in x min + cistern
filled by N in (x+5) min = 1 ⇒ x/30 + (x+5)/20 = 1 ⇒ 5x+15=60 ⇒ x = 9 min.

Q7.Three fill pipes A, B and C can fill separately a cistern in 12, 16 and 20 minutes respectively. A was opened first. After 2 minute, B was opened and after 2 minutes from the start of B, C was also opened. Find the time when the cistern will be full after opening of C?

Sol : Option A
Explanation: Let cistern will be full in x min. Then part filled by A in x min + part filled by B in (x-2) min + part filled by C in (x-4) min = 1

⇒ x/12 + (x-2)/16 + (x-4)/20 = 1 ⇒ 47x - 78 = 240⇒ x = 162/47 = 321/47 min

Q8.A cistern filled in 20 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is thrice as fast as A. How much time will pipe A alone take to fill the tank?

Sol : Option A
Explanation: Suppose pipe A alone takes x hours to fill the tank. Then pipes B and C will take x/3 and x/6 hours respectively to fill the tank.

Therefore 1/x + 3/x + 6/x= 1/20 ⇒ 10/x = 1/20 ⇒ x = 200hrs

Q9.Three taps P,Q and R can fill a tank in 10,20 and 30hours respectively. If P is open all the time and Q and R are open for one hour each alternately, then the tank will be full in :

Sol : Option C
Explanation: (P+Q)’s 1 hour’s work = (1/10+1/20) = 3/20 (A+C)’s 1 hour’s work = (1/10+1/30) = 2/15 Part filled in 2 hrs = (3/20+2/15) = 17/60 Part filled in 6 hrs = (3×17/60) = 17/20 Remaining Part = (1-17/20) = 3/20 Now, it is the turn of P and Q and 3/20 part is filled by P and Q in 1 hour.

Therefore, Total time taken to fill the tank = (6+1) hrs = 7 hrs

Q10.A tank has a leak which would empty it in 10 hours,a tap is turned on which admits 4 litre a minute into the tank and now it emptied in 12 hours.The capacity of the tank is:

Sol : Option B
Explanation: Let speed of the bike be x km/hr. Let speed of the electric car be y km/hr ∴ 200/x + 600/y = 10 ∴ 300/x + 500/y = 11 Part filled in 1 hour = (1/10-1/12) = 1/60 Time taken to fill the tank = 60 hours

Water filled in 60 hours =4*60*60=1440 litres

Two pipes P1 and P2 can separately fill a cistern in 30 and 20 minutes respectively and a waste pipe P3 can carry off 11 litres per minute. If all the pipes are opened when the cistern is full,it is emptied in 42 hours. How many litres does the cistern hold?

    131.4 litre

    132 litre

    130 litre

    135 litre

Explanation:

Part filled by P1 in 1 minute =1/30 Suppose P3 can empty the cistern in x minutes. Then, Part emptied by P3 in 1 minute =1/x Net part emptied by all pipes together in 1 min =$\dfrac{1}{42×60}=\dfrac{1}{2520}$ $\dfrac{1}{x}−\dfrac{1}{30}−\dfrac{1}{20}=\dfrac{1}{2520}$ ⇒$\dfrac{1}{x}=\dfrac{211}{2520}$ ⇒x=$\dfrac{2520}{211}$ i.e., P3 alone can empty the cistern in $\dfrac{2520}{211}$ minutes.

Capacity of the cistern =$\dfrac{2520}{211}$×11≈131.4 litre

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