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DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
12 Feb 2009, 23:00

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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a swimming pool full of water with vertical walls and a flat bottom that holds 56 cubic meters of water?

1. The depth of the pool is 2 meters 2. The pool is circular with a radius of 3 meters

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
01 Jun 2010, 08:29

2

This post received KUDOS

Expert's post

misterchipper wrote:

I agree that the answer should be D.

However, what I find interesting is that the 2 answers are not exactly the same.

1) Depth=2, Volume=56, ergo surface area=28

2) Surface Area= 9Pi==28.274. And thus our depth is different

If the answer is D, shouldn't both answers be exactly equal.

Am I missing something?

Above point is correct. +1 for spotting this.

In DS two statements never contradict, nor they are giving different answers to the question: statement 1 can not give the numerical value different from the numerical value calculated using statement 2.

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
07 Jun 2013, 04:34

1

This post received KUDOS

Expert's post

Rocky423 wrote:

Its not required for the answers to be same. The question is, what is sufficient to solve the problem. In this case it can be solved by either of the option.

That's not correct.

In DS two statements never contradict, nor they are giving different answers to the question: statement 1 can not give the numerical value different from the numerical value calculated using statement 2.

Re: DS - Water evaporates at the rate of two liters [#permalink]
12 Feb 2009, 23:59

alpha_plus_gamma wrote:

xALIx wrote:

Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a swimming pool full of water with vertical walls and a flat bottom that holds 56 cubic meters of water?

1. The depth of the pool is 2 meters 2. The pool is circular with a radius of 3 meters

B.

2. is suff to calculate surface area.

D

We need to calculate surface area, since the walls are vertical, the surface area is same at base and at the water level.

Statement 1 : Area = volume / height --> Suff Statement 2 : Area = pi * 3^2 --> Suff

Re: DS - Water evaporates at the rate of two liters [#permalink]
15 Feb 2009, 19:28

xALIx wrote:

durgesh79 wrote:

D

We need to calculate surface area, since the walls are vertical, the surface area is same at base and at the water level.

Statement 1 : Area = volume / height --> Suff Statement 2 : Area = pi * 3^2 --> Suff

OA is D. Can you explain the rest of the calculation, I don't get how we can solve the problem. Thanks

Question : How long will it take to evaporate 30 liters of water,

Water evaporates @ 2 liters per hrs per square meter of surface, We have the volume of water (56 Cubic meter). so if we can find the surface area of the pool, we can answer the question.

The question becomes : what is the surface area of the pool ?.

No remember, it a DS question, so we dont need to find the actual area in term of square meter, we just have see if we have sufficient data or not

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
01 Jun 2010, 07:54

misterchipper wrote:

I agree that the answer should be D.

However, what I find interesting is that the 2 answers are not exactly the same.

1) Depth=2, Volume=56, ergo surface area=28

2) Surface Area= 9Pi==28.274. And thus our depth is different

If the answer is D, shouldn't both answers be exactly equal.

Am I missing something?

In case of DS question we should just check if we can get the ans, that is if we have all the suffient data to solve the problem . We are not required to get the actual answer. _________________

_________________ If you like my post, consider giving me a kudos. THANKS!

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
03 Apr 2011, 02:23

xALIx wrote:

Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a swimming pool full of water with vertical walls and a flat bottom that holds 56 cubic meters of water?

1. The depth of the pool is 2 meters 2. The pool is circular with a radius of 3 meters

This question has minor discrepancies. This reminded me of Units and Dimensions from Physics.

Rate of evaporation = Volume of water evaporated/(Time*Surface Area occupied by the water evaporated)

Time = Volume of water evaporated/(Rate of evaporation*Total Surface Area occupied by the water evaporated)

Volume of water evaporated=30 liters Rate of evaporation = 2 l/(h*m^2) Total Surface Area = ? Time = (30/(2*Area)) hours = (15/Area) hours

Q: What is the total surface area of the water occupied by the water evaporated?

1. The depth of the pool is 2 meters We don't know the shape of the pool and at least the information about the shape of the pool is required to find the area. We may require few more information but certainly this is insufficient. The shape of the pool may be cylinder, rectangular box, triangular prism, pentagonal prism, cube. Note: the shape can't be conical as the walls are vertical and a cone is slanting. Not Sufficient.

2. The pool is circular with a radius of 3 meters. We now know that the pool is cylindrical in shape and the radius of the cylinder=3m

The volume of water to be evaporated = 30l \(1 liter=10^{-3} m^3\) \(30 liters=30*10^{-3} m^3 = 0.03 m^3\)

Volume of a cylinder \(= \pi*r^2*h\) \(0.03 = \pi*3^2*h\)

\(h = \frac{0.03}{9*\pi} = \frac{0.01}{3*\pi} m\)

Now, we have the height of the water evaporated and also the radius of the water evaporated. We can find the total surface area.

I don't think that the two statements contradict each other. But in my opinion, the answer is not "D". According to our finding, the swimming pool is a cylinder. Pool's radius=3m and height=2m.

Stem says; volume of the pool=56 cubic meter \(\pi*r^2*h=\pi*3^2*2=\pi*18 \approx 56\). Yes, I still don't like this approximate value though. _________________

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
03 Apr 2011, 03:29

Hey fluke Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a swimming pool full of water with vertical walls and a flat bottom that holds 56 cubic meters of water? ---> the volume of water in pool is known. Actually you are trying to find this volume laterally. But this is given to you. Now even if we dont know the geometry of the pool, but we know how much water it can hold. And from S1 we also know the height of the pool. Knowing the one dimension and knowing that pool is vertical, I can write

surface area of base * height = volume. Hence S1 is very sufficient information since we want the surface area of the base. I may be wrong, pls correct

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
03 Apr 2011, 04:15

gmat1220 wrote:

Hey fluke Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a swimming pool full of water with vertical walls and a flat bottom that holds 56 cubic meters of water? ---> the volume of water in pool is known. Actually you are trying to find this volume laterally. But this is given to you. Now even if we dont know the geometry of the pool, but we know how much water it can hold. And from S1 we also know the height of the pool. Knowing the one dimension and knowing that pool is vertical, I can write

surface area of base * height = volume. Hence S1 is very sufficient information since we want the surface area of the base. I may be wrong, pls correct

My interpretation of the problem is different from yours and many others, not necessarily correct. Surface area of the base can be found. But, I think we need to consider Total surface area of the water evaporated and not the surface area of the base. If we had to find the area of the base alone to get the answer, both statements alone would have sufficed.

Case I: The base of the rectangular pool l=14 w=2 height=2 Volume=56

56 -> 2m 1 -> 2/56 0.03 -> (2/56)*0.03=0.001 height=0.001m for 30 liters of water. Total surface area of the water evaporated = 2(lw+wh+lh) = 2(14*2+2*0.001+0.001*14)=2(28+0.002+0.014)=31.2*2=56.03 m^2

CaseII:

The base of the rectangular pool l=7 w=4 height=2 Volume=56

56 -> 2m 1 -> 2/56 0.03 -> (2/56)*0.03=0.001 height=0.001m for 30 liters of water. Total surface area of the water evaporated = 2(lw+wh+lh) = 2(7*4+4*0.001+0.001*7)=2(28+0.004+0.007)==56.022 m^2

Thus, we are getting different total surface area for these two scenarios. It will differ even if we considered the shape as a prism or cylinder. _________________

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
03 Jun 2011, 06:21

adinit wrote:

From stmt 1, how do we know the pool is in cylindrical shape?

Really, we don't need worry about the shape of the pool, whether its circular/ oval/ rectangular on top.

Volume of the pool is given as \(58 m^3\)

Another key is, vertical walls and flat bottom. So, it rules out cone/ bowl shaped pool with irregular depth. Thus, surface area is even allover.

Surface area \(X m^2\) of the pool could be found, if we know 1. either height (or) length & breadth in case of rectangular pool 2. either height (or) radius in case of circular pool

which is given as statement 1 & 2. Though the answer from both statement are not matching exactly, but still it's close enough.

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
16 Jul 2011, 14:13

It has to be D.

Statement 1 gives us enough information to calculate surface area since we already know volume (this works irrespective of shape - being circle, square, or whatever).

Statement 2 tells us its a circle and gives us enough info to calculate surface area.

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
07 Aug 2011, 12:23

The pool's depth is 2m.

That does not imply that water itself is 2m deep in the pool. What if the water is 1.5m deep? Neither does the question imply that the swimming pool is filled to the brim.

I can see why D would be right (if the above reasoning were included in the question). But I picked B as I am not convinced yet. _________________

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
07 Jun 2013, 04:28

Its not required for the answers to be same. The question is, what is sufficient to solve the problem. In this case it can be solved by either of the option.

Re: DS - Water evaporates at the rate of two liters (m07q26) [#permalink]
07 Jun 2013, 04:31

Expert's post

xALIx wrote:

Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a swimming pool full of water with vertical walls and a flat bottom that holds 56 cubic meters of water?

1. The depth of the pool is 2 meters 2. The pool is circular with a radius of 3 meters

A right circular cylinder of 72 cubic meters is completely filled with water. If water evaporates from the cylinder at a constant rate of two liters per hour per one square meter of surface, how long will it take for 30 liters of water to evaporate?

To find the time needed for 30 liters of water to evaporate we need to find the surface area of the top of the cylinder: \(\frac{area}{2}\) will be the amount of water that evaporates each hour, thus \(time=\frac{30}{(\frac{area}{2})}\).

On the other hand since \(volume=\pi{r^2}h=72\) then \(area=\pi{r^2}=\frac{72}{h}\). So, basically all we need is ether the area of the surface or the height of the cylinder.

(1) The height of the cylinder is 2 meters. Sufficient. (2) The radius of the base of the cylinder is \(\frac{6}{\sqrt{\pi}}\) meters --> \(area=\pi{r^2}=36\). Sufficient.