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
Volume of a rectangular room
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- tlt2372
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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!
Hope this helps!
- shovan85
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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
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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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.
- alivapriyada
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this is a very important and frequent question in GMAT.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
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
- nitinthesailor
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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.
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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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.
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.