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Math questions for you smart people.

T

thereds

Diamond Member
A gas cylinder has two vents, V1 and V2, for discharge. The cylinder empties in 9 minutes when only vent V1 is open. The cylinder empties in 4 minutes when both vent V1 and vent V2 are open. About how many minutes will the cylinder take to empty if only vent V2 is open?


A large water reservoir has two outlet pipes, a small one and a large one. It takes 8 hours to empty the reservoir when only the small outlet pipe is open. It takes 5 hours to empty the reservoir when only the large outlet pipe is open. About what fraction of the reservoir will be emptied in one hour if both the pipes are open?
 
Does the gas discharge from each vent at a constant rate? It the rate of discharge of one vent affected by the open/closed state of the other vent?
 
Oh well...here's the answer

V1 alone empties the cylinder in 9 minutes
this means in 1 min V1 alone will empty = 1/9 of the cylinder

Let, V2 alone empties the cylinder in x minutes
this means in 1 min V2 alone will empty = 1/x of the cylinder

When both are working together, the cylinder is emptied in 4 minutes
this means in 1 minute both (V1 and V2) will empty 1/4 th of the cylinder

so we can set up the equation
work done by V1 alone in 1 minute + work done by V2 alone in 1 minute =
work done by V1 and V2 together in 1 minute.

1/9 + 1/x = 1/4

x = 7.2 minutes
 
uhhh, the reds,

1. why did you ask that question if you intended to answer it?

2. I'm not sure without thinking for 1 more minute than I want to, but I believe your equation is wrong.
Reason: Since cylinder A empties in 9 minutes, you say it's emptying at a rate of 1/9 per minute. That assumption is incorrect. As the cylinder empties, the rate at which it is emptying will decrease. Thus, during the first minute, perhaps 1/8 or even 1/7 of the cylinder empties. In the final minute, maybe only 1/30th of the cylinder will empty.

However, I the context of a typical algebra I text book, that's an assumption they expect you to make.

However, your second problem is one that I would expect my calculus class to be able to tackle after almost 1 year of calculus. It leads to a differential equation. Note: the deeper the water, the faster the water will come out. Simple experiment to prove it: put a hole through the side of a paper (or styrofoam) cup near the bottom. The deeper the water, the farther horizontal it sprays because it's coming out faster. Thus, you cannot assume a constant rate.
 
btw, I like the 2nd problem... it's extremely challenging for a calc student - it may make a great take home project for extra credit. Thanks.
 
Originally posted by: DrPizza
btw, I like the 2nd problem... it's extremely challenging for a calc student - it may make a great take home project for extra credit. Thanks.
Click to expand...

Actually, once you see for the answer to that, you're gonna hit your head 🙂
 
Do you have to take into account the change in pressure inside the cylinder as the gas is discharging because that would affect the rate of discharge.
 
actually, thereds, I know what to expect as your answer to that, and quite simply, that answer is wrong.

However, since you're taking (I'm guessing) Algebra I, they expect you to make some assumptions which are fundamentally flawed. One of the problems I've had in teaching physics students (although we don't do problems at this level of difficulty in my class - a greater level of difficulty than you think, if done correctly) is that physics students typically come to class with a lot of previously acquired knowledge about how the real world works. Unfortunately, much of that knowledge was incorrectly passed on to them.
 
Originally posted by: duke
Do you have to take into account the change in pressure inside the cylinder as the gas is discharging because that would affect the rate of discharge.
Click to expand...

Yes, you do. But, it's not done for this simple solution.
 
The pressure change will go like this:

dP/dt = - r *P

where P=pressure, t=time, r=rate constant for the valve. Integrate and you get:

r = ln (P/Po) / t

where Po=final pressure of the tank (the pressure of the surrounding gas). Thus the rate constant for valve #1 is:

r1 = ln (P/Po)/ 9 min

The rate constant for both valves is:

rboth = r1 + r2 = ln (P/Po) / 4 min

Thus the rate constant for the second valve can be easilly found by subtraction:

r2 = ln (P/Po)/ 4 min - ln (P/Po)/ 9 min = ln (P/Po) / (1/4 min - 1/9 min)

Now how long will it take to empty with just the second valve open?

t2 = ln (P/Po)/ r2 = 1/(1/4 min - 1/9 min) = 7.2 minutes.

Notice how in the final answer, all the pressures cancel out, and you thus don't need to consider it.
 
Originally posted by: DrPizza
actually, thereds, I know what to expect as your answer to that, and quite simply, that answer is wrong.

However, since you're taking (I'm guessing) Algebra I, they expect you to make some assumptions which are fundamentally flawed. One of the problems I've had in teaching physics students (although we don't do problems at this level of difficulty in my class - a greater level of difficulty than you think, if done correctly) is that physics students typically come to class with a lot of previously acquired knowledge about how the real world works. Unfortunately, much of that knowledge was incorrectly passed on to them.
Click to expand...

Dude, these are GMAT questions, you throw out real work knowledge here.

I'm a masters student by the way 🙂
 
Damn it! I hate it when I'm wrong. Thanks, Dullard.... I had a scary feeling that the pressures cancelled out when the gas cylinder discharged completely. However, since in the 2nd problem, he's not finding the time for the reservoir to empty completely, I don't think everything will cancel out.


edit: for the 2nd problem that I'm more familiar with, you'd have to use Torricelli's law (I think; again, I don't think everything cancels out) The rate of change of the volume (the flow rate) will be proportional to the square root of the depth of the reservoir at any point in time. So, the shape of the reservoir will also be a factor (although, I think that one may cancel out)
 
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