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Pipe P can drain the liquid from a tank in 3/4 the time that
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Updated on: 02 Jul 2014, 05:34
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Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank? A. 9/29 B. 8/23 C. 3/8 D. 17/29 E. 3/4
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Originally posted by XxxyyY on 17 Jun 2006, 18:32.
Last edited by Bunuel on 02 Jul 2014, 05:34, edited 2 times in total.
Edited the question and added the OA.




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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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17 Jun 2006, 19:35
A  9/29
Plugging numbers
P6hrs
Q8hrs
R9hrs
Together in one hr they can drain = 1/6+1/8+1/9 = 29/72 tank
thus together they can drain tank in 72/29 hrs
in 1 hr q drains 1/8 tank
in 72/29 hrs  72/29*(1/8) = 9/29




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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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17 Jun 2006, 19:45
You got it, Thanks. Apreciate your explanation.
Cheers
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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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02 Jul 2014, 11:17
X & Y wrote: Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank?
A. 9/29 B. 8/23 C. 3/8 D. 17/29 E. 3/4 Suppose Q can drain in 1 hr. So, rQ = 1/1 = 1 So, rP = 1/[(3/4)rQ] = 4/3 Also, rP = rR/(2/3) => 4/3 = rR/(2/3) => rR = 8/9 Let H is the time it takes to drain by running all 3 pipes simultaneously So combined rate = rC = 1/H = 1 + 4/3 + 8/9 = 29/9 = 1/(9/29) Thus running simultaneously, Pipe Q will drain 9/29 of the liquid. Thus answer = A.



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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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06 Jul 2014, 16:53
Bunuel wrote: gautamodi wrote: is the OA A or D???
Even i got it as A=9/29 It's A. Edited. Thank you. Hi Bunuel, I got the answer by following steps below: R : P : Q 2 : 3 4 : 3 combined ratios can be obtained as: 8:12:9 So Q's part is 9/29 of total Is this the correct way of solving?



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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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07 Jul 2014, 04:03



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Pipe P can drain the liquid from a tank in 3/4 the time that
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27 Apr 2015, 14:37
sandeep1756 wrote: Bunuel wrote: gautamodi wrote: Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank?
A. 9/29 B. 8/23 C. 3/8 D. 17/29 E. 3/4
is the OA A or D???
Even i got it as A=9/29 It's A. Edited. Thank you. Hi Bunuel, I got the answer by following steps below: R : P : Q 2 : 3 4 : 3combined ratios can be obtained as: 8:12:9 So Q's part is 9/29 of total Is this the correct way of solving? I'm doubting if the ratios are correct. trying to figure out how the ratios are seted up, the problem says " P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it." or P=(3/4)xQ => P:Q = 3:4 not 4:3 as stated above in red. same thing for the other ratio P=(2/3)xR => P:R = 2:3 now if we even out P to be 6 ( for 2 and 3) we get P:Q:R = 6:8:9 and Q part will be 8/(8+6+9) = 8/23 which is not correct answer. Am I missing something or what is wrong here, what do you think Bunuel and sandeep1756



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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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29 Apr 2015, 03:11
Quote: I'm doubting if the ratios are correct. trying to figure out how the ratios are seted up, the problem says " P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it." or P=(3/4)xQ => P:Q = 3:4 not 4:3 as stated above in red. same thing for the other ratio P=(2/3)xR => P:R = 2:3 now if we even out P to be 6 ( for 2 and 3) we get P:Q:R = 6:8:9 and Q part will be 8/(8+6+9) = 8/23 which is not correct answer. Am I missing something or what is wrong here, what do you think Bunuel and sandeep1756Hi kzivrev, The question gives us the ratio of time taken by Pipes P, Q and R to do a particular work i.e. draining the tank. We know from the time work equation that Work = Rate * Time. So rate is inversely proportional to the time taken. So if the time taken are in the ratio of 2:1 the rates will be in the ratio of 1:2. For example if John can make 6 sandwiches in an hour and Mike can make 6 sandwiches in half an hour the ratio of time taken by John and Mike is 2:1 and hence the ratio of their rates would be 1:2 i.e Mike can make sandwiches twice as fast as John.Coming back to the question, if all the 3 pipes work simultaneously, they would be working for the same amount of time.Hence to find the proportion of work done by pipe Q we need to find the ratios of rates of pipes P, Q and R. Since the ratio of time taken by pipes P & Q to do a particular work is 3:4, the ratio of their rates would be 4:3. On the same lines, if the ratio of time taken by pipes P & R to do a particular work is 2:3, the ratio of their rates would be 3:2. Hence if we assume the rate of pipe P as \(x\), we can write the rate of pipe Q as \(\frac{3x}{4}\) and rate of pipe R as \(\frac{2x}{3}\). The proportion of work done by pipe Q can be calculated as rate of pipe Q divided by the total rates of pipe P, Q and R. The sum of the rates of pipes P, Q and R is \(\frac{29x}{12}\). Hence the proportion of work done by pipe Q would be \(\frac{3x * 12}{4 * 29x} = \frac{9}{29}\). Hope its clear Regards Harsh
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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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14 Feb 2018, 10:00
X & Y wrote: Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank?
A. 9/29 B. 8/23 C. 3/8 D. 17/29 E. 3/4 We can let p = the time it takes pipe P to drain the liquid from the tank. Thus, the rate of pipe P = 1/p. Since pipe P can drain the liquid from the tank in 3/4 the time that it takes pipe Q to drain it, the rate of pipe Q is: (3/4)(1/p) = 3/(4p) Since pipe P can drain the liquid from the tank in 2/3 the time that it takes pipe R to do it, the rate of pipe R is: (2/3)(1/p) = 2/(3p) Thus, the portion of liquid pipe Q drains is: [3/(4p)] / [1/p + 3/(4p) + 2/(3p)] Multiplying the numerator and denominator by 12p, we have: 9/(12 + 9 + 8) 9/29 Answer: A
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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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14 Feb 2018, 22:33
X & Y wrote: Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank?
A. 9/29 B. 8/23 C. 3/8 D. 17/29 E. 3/4 relative times & rates are: P: 6/8 & 96/72 Q: 8/8 & 72/72 R: 9/8 & 64/72 combined rate=232/72 portion drained by Q=(72/72)/(232/72)=9/29 A



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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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22 Feb 2018, 10:50
P  rate of P, Q  rate of Q, R  rate of R P = 4/3Q (as P takes 3/4 th time of Q) P = 3/2R (as P takes 2/3 th time of R) => R = 8/9Q P : Q : R = 4/3 : 1 : 8/9 = 12 : 9 : 8 Q's portion = 9/(12+9+8) = 9/29



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Re: Pipe P can drain the liquid from a tank in 3/4 the time that
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06 Apr 2018, 10:53
hellosanthosh2k2 wrote: P  rate of P, Q  rate of Q, R  rate of R P = 4/3Q (as P takes 3/4 th time of Q) P = 3/2R (as P takes 2/3 th time of R) => R = 8/9Q P : Q : R = 4/3 : 1 : 8/9 = 12 : 9 : 8 Q's portion = 9/(12+9+8) = 9/29 I was able to solve through until I got to the highlighted portion, is someone able to explain how to go from R= 8/9Q to this three part ratio? Math is hard




Re: Pipe P can drain the liquid from a tank in 3/4 the time that &nbs
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