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Volume of a rectangular room

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Volume of a rectangular room

by Sm1520 » Wed Sep 29, 2010 1:49 am
What is the volume, in cubic feet, of a rectangular room that has length x feet, width y feet, and height z feet?

1) x=1200/yz
2) The area of the ceiling of this room is 120 square feet and the area of each of two facing walls of this room is 100 square ft.

[spoiler] OA: A

[/spoiler] This might be a very simple question, but I am confused over why Statement 2 is insufficient? How I see it is: Since we are given the ceiling of the room as 120 sq feet, x could be either 12 or 10. Similarly, Y could be 10 or 12. For the walls, XZ=100 or YZ=100. Either ways, the volume seems to be 1200. Am I missing something? Do we consider other value of x and y (eg. 30,4) for this problem?

X---Y---Z : Volume
10 12 10 : 1200
12 10 10 : 1200

Thanks in advance!

SM
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by mgm » Wed Sep 29, 2010 4:14 am
XY * YZ * XZ = 120 * 100 * 100 ==> X^2 * Y^2 * Z^2 = 120 * 100 * 100 ==> XYZ = 100 * Root 120 ..

I am not sure why the OA is A :(
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by tlt2372 » Wed Sep 29, 2010 7:49 am
I am pretty sure that the answer lies in how statement 2 is worded. Youre right we have the area of the ceiling (or floor- however you want to look at it) but it says the AREA of the facing wall is 100 sf. So the hieght dimension could be 10 ( if the wall was a 10x10) or 1 ( if the wall was 100x1). Without a definite z (height) dimension you cant solve.

Hope this helps!
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by shovan85 » Wed Sep 29, 2010 9:22 am
A is Correct I believe.

In B: Besides ceiling and floor we have two more sets of facing walls.
ceiling is 120, facing wall is 100 => say x= 10, y = 12, z= 10 so volume = 1200
one more possibility is there x=20, y=6, z=5 so volume = 600

Hope my explanation helps :)
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by Sm1520 » Wed Sep 29, 2010 10:24 am
Ahh ok... so there ARE more possibilities for the height n walls other than the 12 and 10
i c it now
thanks shovan, thanx tlt! :)
makes more sense now !

SM
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by rehandossani » Wed Sep 29, 2010 11:52 am
As far as I see it, A should be the answer. For example xy= area of ceiling and xz or yz = area of wall. In both the cases we have 3 unknown variables and two equations. So B is insuff.
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by alivapriyada » Thu Sep 30, 2010 3:26 am
Sm1520 wrote:What is the volume, in cubic feet, of a rectangular room that has length x feet, width y feet, and height z feet?

1) x=1200/yz
2) The area of the ceiling of this room is 120 square feet and the area of each of two facing walls of this room is 100 square ft.

[spoiler] OA: A

[/spoiler] This might be a very simple question, but I am confused over why Statement 2 is insufficient? How I see it is: Since we are given the ceiling of the room as 120 sq feet, x could be either 12 or 10. Similarly, Y could be 10 or 12. For the walls, XZ=100 or YZ=100. Either ways, the volume seems to be 1200. Am I missing something? Do we consider other value of x and y (eg. 30,4) for this problem?

X---Y---Z : Volume
10 12 10 : 1200
12 10 10 : 1200

Thanks in advance!

SM
this is a very important and frequent question in GMAT.
while considering B you are missing the other aspect

X-----Y-----Z :Volume
12 10 10:1200
12 10 10:1200
120 1 100:12000(you missed it)
so B is not sufficient.
hope this helps!!
Regards
Aliva
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by nitinthesailor » Tue Jun 14, 2011 9:36 am
Lets consider ceiling xy=120 and the areas of teh facing walls yz=100 and xz=100.

Given: yz=xz
Hence y=x
x2=120 (using xy=120)
x=Root 120
Volume xyz=(Root 120) X 100
Hence , D
Guys please let me know if I am wroong here.
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by goalevan » Sun Jul 10, 2011 3:07 pm
Remember that facing walls in a rectangular room have the same dimensions.

Given: zy = 120 and zx = 100 (the ceiling and only one wall), we cannot solve for xyz. We can solve for z(x+y), z(x-y), x/y, or y/x, but not for x, y, z, or xyz.
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