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# A cylindrical tank with radius 3 meters is filled with a solution. The

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A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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Updated on: 03 Feb 2016, 07:46
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A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6+(5/9)
B. 6+(8/9)
C. 7+(1/3)
D. 7+(5/9)
E. 8

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Originally posted by bhandariavi on 31 Jan 2011, 21:30.
Last edited by chetan2u on 03 Feb 2016, 07:46, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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01 Feb 2011, 01:01
1
hmm, without having a proper answer I would leave the equation out and say its 7/3 * d high
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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01 Feb 2011, 05:39
V (4) = 56π
V (6) = 60π

56π = (π/3) * (9) * (h - d)

60π = (π/3) * (9) * (h - d/3)

Solving further, h = 20.66

I agree, this is definitely a time consuming question.

Cheers,
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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27 Jan 2015, 02:50
I think there is an easier way than this. But i am not getting the answer. Experts pls help

in 4 minutes D part was empty of the tank.
in 6 min 1/3 part is empty.

So in 2 minutes 2/3rd of the cylinder is full.

So in another 1 min 1/3 of the cylinder will be filled. So the cylinder is filled in 7 min.

There fore V = (84-49+24 ) Pai
V= 59Pai
Volume of a cylinder = 2pai r (r+h)
But on equating am getting anw as 41/6.
Pls help in identifying the issue
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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12 Aug 2015, 02:35
bhandariavi wrote:
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6*(5/9)
B. 6*(8/9)
C. 7*(1/3)
D. 7*(5/9)
E. 8

Hi bunuel,

Could you please provide your comments on this. I am getting 56 pi at 4 minutes and 60 pi at 6 minutes. But how to proceed from here please clarify.

Thanks

Thanks.
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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12 Aug 2015, 06:36
For V = (-t^2 + 12t + 24)π we have 56pi at 4 minutes and d remaining to fill the cilinder. At 6 minutes we have 60pi, wich means that 4pi = (2/3)d, therefore d = 6. With this we have the total volume as 62pi.

The volume of the cilinder: v = pi*(R^2)*h = pi*(3^2)*h

Now that we found the volume (62), we can find the height: h = 62pi/9pi

What have I missed? Can someone help? Bunuel?
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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13 Aug 2015, 06:24
Mascarfi wrote:
For V = (-t^2 + 12t + 24)π we have 56pi at 4 minutes and d remaining to fill the cilinder. At 6 minutes we have 60pi, wich means that 4pi = (2/3)d, therefore d = 6. With this we have the total volume as 62pi.

The volume of the cilinder: v = pi*(R^2)*h = pi*(3^2)*h

Now that we found the volume (62), we can find the height: h = 62pi/9pi

What have I missed? Can someone help? Bunuel?

Got the same answer... Don't know what's wrong...
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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14 Aug 2015, 06:23
bhandariavi wrote:
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6*(5/9)
B. 6*(8/9)
C. 7*(1/3)
D. 7*(5/9)
E. 8

at 4 minutes, V = 88pi+9pid
ar 6 minutes V = 132pi+9pid/3
Equating d = 7 and 1/3
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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03 Feb 2016, 02:15
Hi, the answer can be found by:- first find out the height at the volume of tank after 4 minutes (using the volume formula), which will work out to 56/9 ('h1').
At this point we know that a further height d is remaining. So the total height of the tank will be height after 4 minutes + the value of 'd'
in order to find 'd' , we can use the information given regarding the volume of tank after 6 minutes. The height works to 60/9 ('h2').
We know that 2/3 d was filled up these 2 minutes. (i.e. h2 - h1 = 2/3 d). Hence, d= 2/3 ( 60/9 - 56/9 = 2/3 d. Therefore by solving for d, we get d= 2/3). Now, the height of the tank will be 56/9+2/3 = 62/9, which can be written as 6*(8/9)
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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03 Feb 2016, 07:27
One very important thing to notice in this question is that the function for the volume is NON-LINEAR!

That means that just because a certain volume was filled in 4 minutes, doesn't mean that half that volume will be filled in 2 minutes. We can't make that assumption.

The other major thing to notice is that we are given the formula for the VOLUME, not the height. Since we're dealing with a cylinder, $$V=\pi r^2h$$ so $$h=\frac{V}{\pi r^2}$$

We are told the radius is 3, so we can find the height at t=4 and t=6:
Height at t=4: 56/9
Height at t=6: 60/9
The difference in the height at these two times is 4/9, and we are told that this is equal to $$\frac{2}{3}d$$

$$\frac{4}{9}=\frac{2}{3}d$$

$$d=\frac{4*3}{9*2} = \frac{2}{3}$$

So the total height of the cylinder is the height at 4 minutes + d (or the height at 6 minutes + d/3)

$$h=\frac{56}{9} + \frac{2}{3} = \frac{62}{9} = 6 \frac{8}{9}$$

I believe the answer choices were not written with the correct formatting. The answer is B, but it should not be written as 6*(8/9). It should have been written as either 6+(8/9) or $$6 \frac{8}{9}$$

Cheers
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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03 Feb 2016, 07:54
bhandariavi wrote:
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6+(5/9)
B. 6+(8/9)
C. 7+(1/3)
D. 7+(5/9)
E. 8

HI,
Many of us have got the answer, but the Q is flawed in its equation of volume..
the equation is that of a parabola and parabola has a maximum or minimum value at a certain point..
here that point is at t=6min..
this means the volume will start decreasing beyond 6 minutes..
try at 7 minutes, which is the time required to fill the tank..
it is -7^2+12*7+24..= 59..
whereas at 6 it is 60..
SO IN ONE MINUT FROM 6 TO 7 MINUTE, OUR VOLUME HAS ACTUALLY REDUCED FROM 60 TO 59...

V = (-t^2 + 12t + 24)π...
V= {t(12-t) + 24}π..
IF YOU SEE HERE THE VOLUME WILL REDUCE TO 24 AT T=12..
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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03 Feb 2016, 08:25
chetan2u wrote:
bhandariavi wrote:
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6+(5/9)
B. 6+(8/9)
C. 7+(1/3)
D. 7+(5/9)
E. 8

HI,
Many of us have got the answer, but the Q is flawed in its equation of volume..
the equation is that of a parabola and parabola has a maximum or minimum value at a certain point..
here that point is at t=6min..
this means the volume will start decreasing beyond 6 minutes..
try at 7 minutes, which is the time required to fill the tank..
it is -7^2+12*7+24..= 59..
whereas at 6 it is 60..
SO IN ONE MINUT FROM 6 TO 7 MINUTE, OUR VOLUME HAS ACTUALLY REDUCED FROM 60 TO 59...

V = (-t^2 + 12t + 24)π...
V= {t(12-t) + 24}π..
IF YOU SEE HERE THE VOLUME WILL REDUCE TO 24 AT T=12..

You're right that the volume will start to decrease after t=6, but that doesn't matter. We are never asked to find the time at which the cylinder is full. We are given enough information to determine the height of the cylinder, so the actual volume of liquid in the cylinder at times other than t=4 and t=6 is irrelevant.

The only problem with the question was the form of the answer choices, but you fixed that.

Cheers
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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03 Feb 2016, 08:46
davedekoos wrote:
chetan2u wrote:
bhandariavi wrote:
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6+(5/9)
B. 6+(8/9)
C. 7+(1/3)
D. 7+(5/9)
E. 8

HI,
Many of us have got the answer, but the Q is flawed in its equation of volume..
the equation is that of a parabola and parabola has a maximum or minimum value at a certain point..
here that point is at t=6min..
this means the volume will start decreasing beyond 6 minutes..
try at 7 minutes, which is the time required to fill the tank..
it is -7^2+12*7+24..= 59..
whereas at 6 it is 60..
SO IN ONE MINUT FROM 6 TO 7 MINUTE, OUR VOLUME HAS ACTUALLY REDUCED FROM 60 TO 59...

V = (-t^2 + 12t + 24)π...
V= {t(12-t) + 24}π..
IF YOU SEE HERE THE VOLUME WILL REDUCE TO 24 AT T=12..

You're right that the volume will start to decrease after t=6, but that doesn't matter. We are never asked to find the time at which the cylinder is full. We are given enough information to determine the height of the cylinder, so the actual volume of liquid in the cylinder at times other than t=4 and t=6 is irrelevant.

The only problem with the question was the form of the answer choices, but you fixed that.

Cheers

Hi,
it does make a difference...
1)one may get the right answer taking that..
at 4 mins volume- 56 pi..
at 6 mins- 2/3 d is done and it becomes 60 so with enitre d it will become 62 pi..
ans is B
that was one way..

2)but another way some one can find is, first checking the time it takes to fill the tank..
from 4 to 6 minutes, that is, in 2 mins it fills up 2/3 * d, so in 3 minutes, it will fill 'd' mtrs..
it took 4 minutes to fill up total -d..
so total time =4+3=7..
check for volume at 7 minutes..
-7^2+12*7+24= 59 pi..
why should some one not work on that because finally one should believe the equation one is given...
now one can say that it cannot be 59, since it has to be>60..
but that will not stand as there is no restrictions on the equation..

SO it is a wrong /flawed Q, which can get you to two answers, and GMAT will never give you such equations..
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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03 Feb 2016, 09:52
Hi Chetan2u,

The error in your second method is assuming that the tank is being filled at a constant rate.

Chetan2u wrote:
in 2 mins it fills up 2/3 * d, so in 3 minutes, it will fill 'd' mtrs..
it took 4 minutes to fill up total -d..
so total time =4+3=7..

The rate at which the tank is being filled is not linear (not constant), so you cannot take the two points given and draw a straight line (shown in orange below) to represent the volume of liquid in the tank. The function for the volume of the liquid in the tank is clearly stated, and it is a parabola with a vertex at t=6, not a straight line.

See below for the illustration
Attachment:

Cylinder volume.png [ 14.09 KiB | Viewed 3702 times ]

Similarly, based on the volumes at t=4 and t=6 (56$$\pi$$, and 60$$\pi$$ respectively), you might be tempted to think that the volume at t=2 minutes would be 52$$\pi$$, but it's not, it's 44$$\pi$$

Again, we are not asked to find the time when the tank is filled completely (because that time doesn't exist, the tank is never completely full), we are asked to use the information given to determine the height of the tank, which we can do.

If you assume that the rate of change of the volume in the tank is constant (orange line), then you are ignoring part of the information given to you in the question, namely that $$V=(-t^2+12t+24)\pi$$

And we all know that ignoring the information given to you is a very, VERY bad idea on the GMAT.

Cheers
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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03 Feb 2016, 18:14
Hi davedekoos,

There is no error in any of the method because I am already writing that teh equatin is that of a parabola and the water is not filled up constantly..
that is where the error/flaw lies..

can you have an equation given for volume and it starts giving you a lesser volume with increased time of filling
The reply that we do not have to bother about time does not have any merit since Q is based on timings only..
V = (-t^2 + 12t + 24)π m3...
it is clearly given volume is based on time, and there are no restrictions given on "till when".

Just because one can get an answer to a flawed Q in one method, it does not become a correct answer.

Its good that you know the filling of tank is not constant.
If you read my reply, I have even explained how parabola functions, so I am pretty aware of parabola graph. ANYWAY thanks for drawing it.

Since you seem to be advocating correctness of tyhe Q, can I ask you in how much time will the tank get filled up..
Mind you its not a MAGICAL tank, which cannot get filled up..

I am surprised with the Volume given clearly as V = (-t^2 + 12t + 24)π m3..., you feel there is no importance of the time..

Quote:
And we all know that ignoring the information given to you is a very, VERY bad idea on the GMAT.

It is even worse if take any flawed Q as correct and do not put logic behind it.
Its best to avoid such Q because actual GMAT will never give you Qs like this

I hope you have got the essence of what I am trying to say..
Cheers!
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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04 Feb 2016, 07:36
Hi Chetan2u,

I'm glad we're having this discussion, and I hope others benefit from it.

However, I must continue to point out that there is no flaw in the question, only in your assumptions. You rightly point out that the equation for the volume of liquid in the tank follows a parabola, as I have drawn in case anyone was unsure, but then you proceed to approach the problem as if the filling of the tank was linear (your method #2). Keep in mind that we are not being asked to find the time at which the tank will be filled, we are asked to find the height of the tank. It is a subtle but very important difference.

I mentioned in my post that there is no time at which the tank will be filled up. That does not mean it is a magical tank, it means that liquid is being added or removed from the tank such that the volume of liquid in the tank follows the equation given. What is wrong with that? If we were to continue to pour liquid into the tank of course the tank would fill up, but we are not continuing to pour liquid into the tank. The question doesn't explicitly explain the reason for why the volume of liquid in the tank follows a parabola, it just says that the tank is being filled such that the volume =$$(-t^2+12t+24)\pi$$. If you're getting hung up on the word "filled" in the question, I agree that the wording could have been improved slightly, as in:

Quote:
"A cylindrical tank with has a radius of 3 meters is filled with a solution. The volume at t minutes is given by..."

But it is not so unclear as to render the question "flawed". With the equation given for the volume, it becomes very clear.

Imagine a pump connected to the bottom of the tank that can add or remove liquid, or a natural condensation-evaporation cycle, or a pack of squirrels arriving one by one to have a party and drink from the tank with festive straws. There are countless practical explanations for why the volume of liquid in the tank would follow the curve of a parabola, but again, that is irrelevant. We don't care why the volume of liquid in the tank follows a parabola, we just have to know that it does. At t=0 the volume of liquid is 24$$\pi$$. The pump is turned on and it pumps liquid into the tank such that the volume increases according to $$V=(-t^2+12t+24)\pi$$. At t=6 the volume of liquid is 60$$\pi$$. At that time the pump is reversed and begins to remove liquid from the tank, still following the curve of the parabola. At $$t=6+2\sqrt{15}$$ the tank will be empty. Terrific, but none of that is important to the question being asked.

There is nothing wrong or flawed about the situation described above. The question asks us to determine the height of the tank and provides the information needed to find the answer. There is only one answer. There can only be one answer. If you use a different "technique" and come up with a different answer, it either means that you made a calculation error or that the technique is wrong.

davedekoos wrote:
We are given enough information to determine the height of the cylinder, so the actual volume of liquid in the cylinder at times other than t=4 and t=6 is irrelevant.

Do we care what the volume is at t=1, t=10? No, we don't. Can we calculate it if we want to? Yes, because we have been given the equation for the volume of liquid in the tank at any time. But to answer the question, we don't need to know the volume of liquid at any times other than t=4 and t=6.

The lesson here: It a good idea to try different approaches to a problem. BUT IF, while trying two different approaches, you get two different answers, the conclusion should not be that the question is flawed, it should be that one (or both) of your approaches is flawed. If this happens, take the opportunity to review and scrutinize the approaches you have used and discover where you went wrong. While practicing for the GMAT I highly encourage all students to approach every question from multiple angles to gain a deeper understanding of the question and how best to approach similar questions that may be encountered.

Cheers
_________________

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Math Expert
Joined: 02 Aug 2009
Posts: 7037
Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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04 Feb 2016, 08:10
davedekoos wrote:
Hi Chetan2u,

I'm glad we're having this discussion, and I hope others benefit from it.

However, I must continue to point out that there is no flaw in the question, only in your assumptions. You rightly point out that the equation for the volume of liquid in the tank follows a parabola, as I have drawn in case anyone was unsure, but then you proceed to approach the problem as if the filling of the tank was linear (your method #2). Keep in mind that we are not being asked to find the time at which the tank will be filled, we are asked to find the height of the tank. It is a subtle but very important difference.

I mentioned in my post that there is no time at which the tank will be filled up. That does not mean it is a magical tank, it means that liquid is being added or removed from the tank such that the volume of liquid in the tank follows the equation given. What is wrong with that? If we were to continue to pour liquid into the tank of course the tank would fill up, but we are not continuing to pour liquid into the tank. The question doesn't explicitly explain the reason for why the volume of liquid in the tank follows a parabola, it just says that the tank is being filled such that the volume =$$(-t^2+12t+24)\pi$$. If you're getting hung up on the word "filled" in the question, I agree that the wording could have been improved slightly, as in:

Quote:
"A cylindrical tank with has a radius of 3 meters is filled with a solution. The volume at t minutes is given by..."

But it is not so unclear as to render the question "flawed". With the equation given for the volume, it becomes very clear.

Imagine a pump connected to the bottom of the tank that can add or remove liquid, or a natural condensation-evaporation cycle, or a pack of squirrels arriving one by one to have a party and drink from the tank with festive straws. There are countless practical explanations for why the volume of liquid in the tank would follow the curve of a parabola, but again, that is irrelevant. We don't care why the volume of liquid in the tank follows a parabola, we just have to know that it does. At t=0 the volume of liquid is 24$$\pi$$. The pump is turned on and it pumps liquid into the tank such that the volume increases according to $$V=(-t^2+12t+24)\pi$$. At t=6 the volume of liquid is 60$$\pi$$. At that time the pump is reversed and begins to remove liquid from the tank, still following the curve of the parabola. At $$t=6+2\sqrt{15}$$ the tank will be empty. Terrific, but none of that is important to the question being asked.

There is nothing wrong or flawed about the situation described above. The question asks us to determine the height of the tank and provides the information needed to find the answer. There is only one answer. There can only be one answer. If you use a different "technique" and come up with a different answer, it either means that you made a calculation error or that the technique is wrong.

davedekoos wrote:
We are given enough information to determine the height of the cylinder, so the actual volume of liquid in the cylinder at times other than t=4 and t=6 is irrelevant.

Do we care what the volume is at t=1, t=10? No, we don't. Can we calculate it if we want to? Yes, because we have been given the equation for the volume of liquid in the tank at any time. But to answer the question, we don't need to know the volume of liquid at any times other than t=4 and t=6.

The lesson here: It a good idea to try different approaches to a problem. BUT IF, while trying two different approaches, you get two different answers, the conclusion should not be that the question is flawed, it should be that one (or both) of your approaches is flawed. If this happens, take the opportunity to review and scrutinize the approaches you have used and discover where you went wrong. While practicing for the GMAT I highly encourage all students to approach every question from multiple angles to gain a deeper understanding of the question and how best to approach similar questions that may be encountered.

Cheers

Hi,
1) firstly, I am aware of the two ways that it could have been done and that is the reason why I mentioned the two.
2) The reasoning for second method, by which the volume comes down, is not necessarily linear.. it can be there for a curve . Of course I am aware it is a parabola, and my second method started off by saying that the equation is of parabola. So I was very well aware when the solution was being given.
3) ASSUMPTIONS :- this is exactly what you are doing where in the a pump connected to the bottom of the tank that can add or remove liquid, or a natural condensation-evaporation cycle, or a pack of squirrels arriving one by one to have a party and drink from the tank with festive straws. too jump in to make it a strong assumptions. I am yet to see any GMAT Quant leaving things to assumption.
4) I am not, for a second, assuming anything. I am just saying the volume of the solution cannot have the equation of parabola. The Actual GMAT will clearly mention the assumptions you have taken above and then mention that it is equation of parabola or may be equation of waves, where the squirrels come back after regular interval to party.
5) We can justify our stand but it is not our stand that matters. I will be more than willing to accept your point of view if you can provide even a single Q following similar lines or even anything close to it. If there is none, we can say that this Q is a great effort by someone to test people's figment of imagination.

I always encouage all students to believe blindly only in OG and actual GMAT Qs and not to waste too much time in methods/reasoning in Q, which do not follow actual GMAT.

Cheers!
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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04 Feb 2016, 11:00
We seem to agree on almost everything actually, but what I still don't understand is why you think that the volume of liquid in the tank cannot follow the equation of a parabola. You have not explained any reasoning for this. It clearly CAN follow a parabola, because that is the information given, and there is nothing physically or mathematically preventing it. We don't need an explanation of WHY it follows a parabola. It doesn't matter why. The possible scenarios I described were just to show that the volume of liquid in the cylinder could in fact reasonably follow a parabola and have a physical explanation. But an explanation is not necessary to solve the problem.

No assumptions are necessary for this problem, and I make no assumptions; I only use the information given in the problem to solve it. Again, my scenarios were only to demonstrate that the volume following a parabola is a reasonable physical possibility (and an attempt at levity), and because you seem to think that the volume COULD NOT follow a parabola. It can and does, and there is nothing wrong with that. Can you explain what assumptions you think a student would have to make to solve this problem? Or explain what information you think the problem is missing?

I wasn't very clear when I said that your error in your second approach was because the volume in the tank is not linear. I know you understand very well what is happening with the volume in the tank, but I'm not sure why you would propose that using the height of the liquid at t=7 is a valid approach, when it clearly is not. The height of the liquid in the tank at t=7 is not the height of the tank. Of the two approaches you used to solve the problem, the first one was correct, the second one was not. Using an incorrect technique to find an incorrect answer does not invalidate the problem.

Another main issue seems to be that you think this is not an official GMAT question. Maybe it is, and maybe it isn't. I didn't write it, and I don't have all the OG questions at my fingertips. But I see nothing in the question that would suggest that it is not a good practice question. Other than the slight rewording I suggested in a previous post, I think the question is very clear, and tests a student's ability to evaluate an expression and then use the information to figure out the answer to the question. No red flags there. The question could be reworded many different ways, in fact, the formula for the volume could have been almost anything as well but the essence of what it is testing is valid and appropriate for the GMAT.

Regardless of the formula for the volume of liquid in the tank (parabola, hyperbola, 5th order equation, spiral, hedgehog, straight line, etc.) the approach to solving the problem remains the same:
1. Evaluate the volume at t=4 and t=6
2. Determine the height of the liquid at t=4 and t=6 based on the area of the base of the tank
3. Equate the difference in height of the liquid between t=4 and t=6 to $$\frac{2}{3}d$$
4. Solve for d
5. Add d to the height of the liquid at t=4 to solve for the height of the tank
6. Done

To me that's a good GMAT question regardless of whether it is from the OG or not.

I'm happy to continue to discuss this, but if it's only a matter of your opinion that this question is not worthy of the GMAT, then that is up to you. Changing your mind is not why I'm here, I just want to help students who are studying to have the best tools to be able to get a great score on the GMAT.

Cheers
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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17 Jul 2016, 06:19
1
bhandariavi wrote:
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6+(5/9)
B. 6+(8/9)
C. 7+(1/3)
D. 7+(5/9)
E. 8

Wasted almost 20 minutes because the expression V = (-t^2 + 12t + 24) should have been written as V = (-(t)^2 + 12t + 24)
t=4 and 6 in the expression -t^2 will an incorrect value for solving the question
t=4 and 6 in the expression -(t)^2 will give the correct value

God knows where the person who posted the question is now ??
It's highly unacceptable and callous on his part to waste other peoples time.
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Re: A cylindrical tank with radius 3 meters is filled with a solution. The  [#permalink]

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02 Mar 2018, 09:39
LogicGuru1 wrote:
bhandariavi wrote:
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)π m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3*d meters of space remaining. What is the height of the tank?

A. 6+(5/9)
B. 6+(8/9)
C. 7+(1/3)
D. 7+(5/9)
E. 8

Wasted almost 20 minutes because the expression V = (-t^2 + 12t + 24) should have been written as V = (-(t)^2 + 12t + 24)
t=4 and 6 in the expression -t^2 will an incorrect value for solving the question
t=4 and 6 in the expression -(t)^2 will give the correct value

God knows where the person who posted the question is now ??
It's highly unacceptable and callous on his part to waste other peoples time.

Dear LogicGuru1,

I know it's a late response, but nevertheless would like to address this as it may be helpful for others.

-t^2 implies -(t^2). It is a convention.

Regards,
Louis
Re: A cylindrical tank with radius 3 meters is filled with a solution. The &nbs [#permalink] 02 Mar 2018, 09:39
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