Pipes and Cisterns: Filling with pipes A and B

Shovon

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Two pipes A and B can fill a tank in 12 and 16 minutes respectively. Both pipes are opened together but 4 minutes before the tank is full, pipe A is closed. In how many minutes will the tank be full?
Options:
1. 9 mins 8 secs
2. 10 mins 9 secs
3. 11 mins 19 secs
4. 11 mins 29 secs
 
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lev888

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From posting guidelines:
"Show some of your work or explain where you're stuck. Don't worry that your work might be wrong; learning is a process of making mistakes. We want to check your efforts, to see where you're at. If you show no work at all, some tutors may assume that you need links to lessons or just the first step. Other tutors might think you're looking for somebody to do your homework and might not help at all. Even if you're asking about only the very end of the solution process, please include your intermediate steps. Errors may have occurred earlier than you realize; for example, correcting part (b) may clear up your confusion on part (d). "
 

Dr.Peterson

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One reason we ask to show work is that for many problems, including this one, there are several techniques that can be used. We need to see which you are learning, as well as where you are having trouble.

Probably you are expected to write an equation. Please at least show your work on that.

If you have no idea what equation to write, one way to get yourself started is to try checking one of the proposed answers. Working through the problem with a specific number can prime your mind to write an equation, putting a variable in place of the number you used.
 

Shovon

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I tried solving it using the LCM method ....the answer is not matching with any of the given....the question came in RRB NTPC 2016.
 

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lev888

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You know that pipe B alone is open for 4 min. But in your solution it's more than 9 min. Ideas?
 

Dr.Peterson

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The trouble with your work is that you supposed that pipe A was closed 4 minutes before they would have filled the tank if they were both left open!

In fact, though, you have B continuing for 9 1/3 minutes after A is closed, not 4 minutes.

I would tend to switch to an algebraic method for this reason: you can only know when to close A after solving the problem!

But you know what you have been taught; do you know a way to get around this? It can be done; you might, for example, reverse your work by finding how many units B alone fills in those last 4 minutes, and then finding how long it takes to have already filled the rest.
 

Shovon

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You know that pipe B alone is open for 4 min. But in your solution it's more than 9 min. Ideas?
Pipe B could have been open for just 4 mins had A & B worked together, but since A closed 4 mins before completion ....B had to work alone for 5(1/3) mins extra (total 9(1/3) mins) to fill the remaining .....may be I am wrong ....kindly provide a solution as per your knowledge!
 

MarkFL

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Like you, I would observe that in 48 minutes, the two pipes working together can complete 7 tasks. Now, if pipe A is closed for 4 minutes, then pipe B must complete 1/4 of the task alone. So, working together it takes both pipes the following number of minutes to complete the remaining 3/4:

\(\displaystyle \frac{3}{4}\cdot\frac{48}{7}=\frac{36}{7}\)

Thus the total time in minutes is:

\(\displaystyle T=\frac{36}{7}+4=\frac{64}{7}\)

This is about 9 minutes 8.571428571428571 seconds.
 

Subhotosh Khan

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A & B together fills (1/12+ 1/16 =) 7/48 part of the tank per minute.

When the pipe A was shutoff (4*7/48 =) 7/12 part of the tank was yet to be filled.

B alone will take [(7/12)/(1/16) =112/12 = 28/3 =] 9 minutes and 20 seconds to fill the tank.
 

Dr.Peterson

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Here's a basic algebraic approach:

If x is the number of minutes until it is filled, pipe A fills 1/12 tank per minute for x-4 minutes, and pipe B fills 1/16 tank per minute for x minutes.

Thus, (x-4)/12 + x/16 = 1. Multiplying by 48, 4(x-4) + 3x = 48, and the solution is x = 64/7 = 9 min 8.57 sec.

In solving, you are in effect adding the 16 "units" pipe A would have filled in the last 4 minutes to the 48 units in the tank, and dividing by the 7 units filled by A and B together each minute. This suggests an alternate non-algebraic method, based on the fact that closing pipe A for 4 minutes adds 1/3 of a tank that has to have been filled by both.
 
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Dr.Peterson

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A & B together fills (1/12+ 1/16 =) 7/48 part of the tank per minute.

When the pipe A was shutoff (4*7/48 =) 7/12 part of the tank was yet to be filled.

B alone will take [(7/12)/(1/16) =112/12 = 28/3 =] 9 minutes and 20 seconds to fill the tank.
This is essentially what the OP did, with the same error. But a small change makes it work (and it becomes more or less what I suggested initially).

When pipe A is shut off, the amount remaining to be filled is the amount that B alone (not A and B) can fill in 4 minutes, namely 4*1/16 = 1/4 of the tank, not 7/12. To fill the tank to that point (3/4 full), both together need (3/4 tank)(48/7 min/tank) = 36/7 min; add on the 4 minutes and you get 64/7.

Most of this looks just like MarkFL's, just explained differently.
 

Shovon

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This is essentially what the OP did, with the same error. But a small change makes it work (and it becomes more or less what I suggested initially).

When pipe A is shut off, the amount remaining to be filled is the amount that B alone (not A and B) can fill in 4 minutes, namely 4*1/16 = 1/4 of the tank, not 7/12. To fill the tank to that point (3/4 full), both together need (3/4 tank)(48/7 min/tank) = 36/7 min; add on the 4 minutes and you get 64/7.

Most of this looks just like MarkFL's, just explained differently.
Thank you sir, your algebraic approach is less prone to making errors....and easily understandable ...Thank you so much...!
 
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