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17 Oct 2012, 19:21
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
52% (01:55) correct
48%
(02:19)
wrong
based on 379
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A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper?
(1 yard = 3 Ft)
A. 8
B. 9
C. 18
D. 27
E. 36
(1 yard = 3 Ft)
A. 8
B. 9
C. 18
D. 27
E. 36
17 Oct 2012, 20:59
Kudos
Bookmarks
aeros232
Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).
The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet.
Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)).
So, it must be true that \(6a^2\leq{63}\).
If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube.
If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet.
Answer: D.
If we won't limit the value of a side to integers only, then: \(6a^2={63}\) --> \(a=\sqrt{10.5}\) --> \(volume=a^3=(\sqrt{10.5})^3\).
Hope it's clear.
General Discussion
