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605-655 Level|   Geometry|      
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Bunuel
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What is the volume, in cubic feet, of a rectangular room that has leng 10 Jul 2018, 21:21
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A
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59% (01:40) correct 41% (01:37) wrong based on 139 sessions

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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 = 1,200/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 feet
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A
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What is the volume, in cubic feet, of a rectangular room that has leng 10 Jul 2018, 21:36
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Bunuel
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 = 1,200/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 feet
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Question mean: x*y*z=?

1.
x=1200/yz=> xyz=1200
sufficient

2.
xy=120
yz=100
x=12; y=10; z=10=>xyz=1200
x=60; y=2; z= 50=>xyz= 6000
insufficient.
hence: A
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What is the volume, in cubic feet, of a rectangular room that has leng 10 Jul 2018, 23:30
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Bunuel
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 = 1,200/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 feet
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Question stem:- Volume=x*y*z=?

St1:- xyz=12000
Suffcient.

St2:- The area of the ceiling of this room is 120 square feet and the area of each oftwo facing walls of this room is 100 square feet

Ceiling area=length*width=xy=120
Area of each opposite facing walls=length*height or width*height=100
So, xz=100 (what is y?) or yz=100 (what is x?)
So, xyz can't be determined.
Insufficient.

Ans. (A)
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What is the volume, in cubic feet, of a rectangular room that has leng 11 Jul 2018, 07:00
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Answer : A

1) cross multiply, we can get XYZ which is the volume LxBxH
2) not enough. We have XY = 120 , YZ = 100. Too many combinations for XY and YZ.
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What is the volume, in cubic feet, of a rectangular room that has leng 29 May 2021, 08:38
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Hello from the GMAT Club BumpBot!

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