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total work done combined p,q,r
10 * (1/20+1/25-1/30) = 10*17/300 ; 17/30
so now left with 13/30 work
which would be completed in ;
x * ( 1/20+1/25) = 13/30
x= 13*50/ 45*3
x= 130/27
total time taken ; 130/27+ 10 ; 400/27 hours
OPTION C



Bunuel
Two pipes P and Q can fill a tank in 20 hrs and 25 hrs respectively while a third pipe R can empty the tank in 30 hrs. All the pipes are opened together for 10 hrs and then pipe R is closed.In what time the tank will be filled?

A. 400/21 hrs
B. 400/23 hrs
C. 400/27 hrs
D. 200/23 hrs
E. 200/27 hrs


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lacktutor
for the 2nd part ; you have taken combined rate of P & R ; where as it should be P & Q ; ( 1/20+1/25) ; :)


lacktutor
\(10( \frac{1}{20}+ \frac{1}{25} —\frac{1}{30}) = 10( \frac{17}{300}) = \frac{17}{30}\) ( Amount of work done within 10 hours)

\(1 —\frac{17}{30} = \frac{13}{30}\) (remaining amount of work)

—>\(x*(\frac{1}{20} + \frac{1}{30}) = \frac{13}{30}\)

—> \(x= (\frac{13}{30})(\frac{100}{9}) = \frac{130}{27}\) (amount of time to finish the remaining amount of work with both pipes P and Q)

In total, \(10 + \frac{130}{27} = \frac{400}{27}\) hours

Answer (C)

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lacktutor
for the 2nd part ; you have taken combined rate of P & R ; where as it should be P & Q ; ( 1/20+1/25) ; :)


lacktutor
\(10( \frac{1}{20}+ \frac{1}{25} —\frac{1}{30}) = 10( \frac{17}{300}) = \frac{17}{30}\) ( Amount of work done within 10 hours)

\(1 —\frac{17}{30} = \frac{13}{30}\) (remaining amount of work)

—>\(x*(\frac{1}{20} + \frac{1}{30}) = \frac{13}{30}\)

—> \(x= (\frac{13}{30})(\frac{100}{9}) = \frac{130}{27}\) (amount of time to finish the remaining amount of work with both pipes P and Q)

In total, \(10 + \frac{130}{27} = \frac{400}{27}\) hours

Answer (C)

Posted from my mobile device

Thank you, Archit3110
Corrected it.
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Bunuel
Two pipes P and Q can fill a tank in 20 hrs and 25 hrs respectively while a third pipe R can empty the tank in 30 hrs. All the pipes are opened together for 10 hrs and then pipe R is closed.In what time the tank will be filled?

A. 400/21 hrs
B. 400/23 hrs
C. 400/27 hrs
D. 200/23 hrs
E. 200/27 hrs


In 10 hours the amount of work done (i.e., the fraction of the tank that is filled) is:

10 x (1/20 + 1/25 - 1/30)

10 x (15/300 + 12/300 - 10/300)

10 x 17/300 = 17/30

Thus, we still have 1 - 17/30 = 13/30 of the tank unfilled. Since pipe R will then be closed, the combined rate of pipes P and Q is 1/20 + 1/25 = 5/100 + 4/100 = 9/100. Thus, the amount of time it will take to pipes P and Q to fill the remaining 13/30 of the tank is:

(13/30)/(9/100) = 13/30 x 100/9 = 13/3 x 10/9 = 130/27 hours

Therefore, the total amount of time to completely fill the tank is 10 + 130/27 = 270/27 + 130/27 = 400/27 hours.

Answer: C
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Bunuel
Two pipes P and Q can fill a tank in 20 hrs and 25 hrs respectively while a third pipe R can empty the tank in 30 hrs. All the pipes are opened together for 10 hrs and then pipe R is closed.In what time the tank will be filled?

A. 400/21 hrs
B. 400/23 hrs
C. 400/27 hrs
D. 200/23 hrs
E. 200/27 hrs


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Bunuel isnt the question a bit ambiguous ? if the question were "what is the total time required for tank to be filled" would be better cause "In what time the tank will be filled?" could mean as well time required to fill tank after R is closed :grin:
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Let's use LCM Method to solve this time & work related question.

Assume that total volume of the tank = LCM(20,25,30) = 300 L

Per hour Work calculation:

P can do 300/20 = 15 L/hr
Q can do 300/25 = 12 L/hr
R can do 300/30 = 10 L/hr (Negative work )

They all work together for 10 h.
Amount of water filled by P,Q and R in 10 hrs = (15 + 12 - 10 ) *10 = 170 L
Now pipe R is closed , and remaining 130 L will be filled by P and Q.

Time taken by P and Q to fill the remaining water = 130/(15+12) = 130/27 hrs

Total time is 130/27+10 = 400/27 hours.

Option C is the answer.

Thanks,
Clifin J Francis
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