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Three pipes A, B and C fill a tank in 4,6, and 8 minutes res [#permalink]
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29 Apr 2010, 11:03
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Three pipes A, B and C fill a tank in 4,6, and 8 minutes respectively. Pipe A is opened and after a minute, pipe B is opened and after another minute pipe C is opened. When will the tank be full?
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Re: Pipes and cisterns  2 [#permalink]
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29 Apr 2010, 11:30
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aiming4mba wrote: Three pipes A, B and C fill a tank in 4,6, and 8 minutes respectively. Pipe A is opened and after a minute, pipe B is opened and after another minute pipe C is opened. When will the tank be full? After 1 min \(\frac{1}{4}\) of the tank will be filled (as only pipe A works); After 2 mins \(\frac{1}{4}+(\frac{1}{4}+\frac{1}{6})=\frac{2}{3}\) of the tank will be filled and \(\frac{1}{3}\) will be empty; Starting from the 3rd minute the combined rate of three pipes would be \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}=\frac{13}{24}\) job/min; To fill remaining \(\frac{1}{3}\) of the tank they'll need \(time=\frac{job}{rate}=\frac{1}{3}*\frac{24}{13}=\frac{8}{13}\) min. So total time \(2+\frac{8}{13}=2\frac{8}{13}\) min.
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Re: Pipes and cisterns  2 [#permalink]
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31 Jan 2014, 22:10
Bunuel wrote: aiming4mba wrote: Three pipes A, B and C fill a tank in 4,6, and 8 minutes respectively. Pipe A is opened and after a minute, pipe B is opened and after another minute pipe C is opened. When will the tank be full? After 1 min \(\frac{1}{4}\) of the tank will be filled (as only pipe A works); After 2 mins \(\frac{1}{4}+(\frac{1}{4}+\frac{1}{6})=\frac{2}{3}\) of the tank will be filled and \(\frac{1}{3}\) will be empty; Starting from the 3rd minute the combined rate of three pipes would be \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}=\frac{13}{24}\) job/min; To fill remaining \(\frac{1}{3}\) of the tank they'll need \(time=\frac{job}{rate}=\frac{1}{3}*\frac{24}{13}=\frac{8}{13}\) min. So total time \(2+\frac{8}{13}=2\frac{8}{13}\) min. Hi Bunuel Kindly explain again with detailed formation of equation... I have gone becoz too lazy to understand



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Re: Pipes and cisterns  2 [#permalink]
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01 Feb 2014, 04:23
prasannajeet wrote: Bunuel wrote: aiming4mba wrote: Three pipes A, B and C fill a tank in 4,6, and 8 minutes respectively. Pipe A is opened and after a minute, pipe B is opened and after another minute pipe C is opened. When will the tank be full? After 1 min \(\frac{1}{4}\) of the tank will be filled (as only pipe A works); After 2 mins \(\frac{1}{4}+(\frac{1}{4}+\frac{1}{6})=\frac{2}{3}\) of the tank will be filled and \(\frac{1}{3}\) will be empty; Starting from the 3rd minute the combined rate of three pipes would be \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}=\frac{13}{24}\) job/min; To fill remaining \(\frac{1}{3}\) of the tank they'll need \(time=\frac{job}{rate}=\frac{1}{3}*\frac{24}{13}=\frac{8}{13}\) min. So total time \(2+\frac{8}{13}=2\frac{8}{13}\) min. Hi Bunuel Kindly explain again with detailed formation of equation... I have gone becoz too lazy to understand The rate of A 1/4 tank/minute. The rate of B 1/6 tank/minute. The rate of C 1/8 tank/minute. Combined rate of all the three pipes: \(\frac{1}{4}+\frac{1}{4}+\frac{1}{8}=\frac{13}{24}\) tank/minute. After 1 min \(\frac{1}{4}\) of the tank will be filled (as only pipe A works); After 2 mins \(\frac{1}{4}+(\frac{1}{4}+\frac{1}{6})=\frac{2}{3}\) of the tank will be filled and \(\frac{1}{3}\) will be empty (A works for 2 minutes, while B works for 1 minute); To fill remaining \(\frac{1}{3}\) of the tank they'll need \(time=\frac{(job)}{(combined \ rate)}=\frac{(\frac{1}{3})}{(\frac{13}{24})}=\frac{1}{3}*\frac{24}{13}=\frac{8}{13}\) min. Hope it's clear.
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Re: Three pipes A, B and C fill a tank in 4,6, and 8 minutes res [#permalink]
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01 Feb 2014, 09:29
aiming4mba wrote: Three pipes A, B and C fill a tank in 4,6, and 8 minutes respectively. Pipe A is opened and after a minute, pipe B is opened and after another minute pipe C is opened. When will the tank be full? LCM of 4,6,8 = 24 Suppose the capacity of the tank is 24 gallons. A fills = 24/4 = 6 gallons per minute B fills = 24/6 = 4 gallons per minute C fills = 24/8 = 3 gallons per minute A + B + C = 6 + 4 + 3 =13 gallons per minute Pipe A is on for 2 minutes = 6 x 2 = 12 gallons Pipe B is on for 1 minutes = 4 gallons After end of two minutes, I have 12 + 4 = 16 gallons full in tank Remaining capacity = 24  16 = 8 gallons A + B + C fill 13 gallon in 1 minute A + B + C will fill 1 gallon in 1/13 minute A + B + C will fill 8 gallon in 8/13 minute = Answer
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Re: Three pipes A, B and C fill a tank in 4,6, and 8 minutes res [#permalink]
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03 Feb 2014, 07:57
Yes Bunuel ...100% clear



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Re: Three pipes A, B and C fill a tank in 4,6, and 8 minutes res [#permalink]
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18 Mar 2014, 04:11
Let the tank is full in T minutes then Pipe A (R1=1/4) has worked for T minutes Pipe B (R2=1/6)has worked for T1 minutes Pipe C (R3=1/8)has worked for T2 minutes Then, R1*T1 + R1*T2 + R3*T3 = Total Work T/4 + (T1)/6 + (T2)/9 = 1 T= 34/13
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Re: Three pipes A, B and C fill a tank in 4,6, and 8 minutes res [#permalink]
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Re: Three pipes A, B and C fill a tank in 4,6, and 8 minutes res [#permalink]
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02 May 2016, 07:53
aiming4mba wrote: Three pipes A, B and C fill a tank in 4,6, and 8 minutes respectively. Pipe A is opened and after a minute, pipe B is opened and after another minute pipe C is opened. When will the tank be full? 24 total work 6+6+4 in 2 mins 2mins 8/13 sec (Ans)
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Re: Three pipes A, B and C fill a tank in 4,6, and 8 minutes res [#permalink]
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