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# Three pipes A, B and C fill a tank in 4,6, and 8 minutes res

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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, 12:03
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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, 12:30
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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, 23:10
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, 05:23
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, 10:29
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, 08:57
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, 05:11
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 T-1 minutes
Pipe C (R3=1/8)has worked for T-2 minutes

Then,

R1*T1 + R1*T2 + R3*T3 = Total Work

T/4 + (T-1)/6 + (T-2)/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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02 May 2016, 08:53
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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04 Sep 2017, 20:59
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Re: Three pipes A, B and C fill a tank in 4,6, and 8 minutes res   [#permalink] 04 Sep 2017, 20:59
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