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30 Nov 2017, 22:58
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
82% (00:49) correct
18%
(01:10)
wrong
based on 90
sessions
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What is the maximum number of cubes, each 3 centimeters on an edge, that can be packed into a rectangular box with inside dimensions as shown above?
(A) 360
(B) 120
(C) 90
(D) 40
(E) 20
Attachment:
2017-12-01_0947_002.png [ 5.29 KiB | Viewed 7642 times ]
30 Nov 2017, 23:23
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D
Along the side of length 12 only four cubes can be placed. Similarly along the side of 15 only 5 cubes can be placed. Hence 4*5=20 cubes can be placed at bottom.
Also one more layer of cubes can be placed along the side of length 6. Hence 20*2=40 cubes only can be placed in the given arrangement.
Sent from my iPhone using GMAT Club Forum
Along the side of length 12 only four cubes can be placed. Similarly along the side of 15 only 5 cubes can be placed. Hence 4*5=20 cubes can be placed at bottom.
Also one more layer of cubes can be placed along the side of length 6. Hence 20*2=40 cubes only can be placed in the given arrangement.
Sent from my iPhone using GMAT Club Forum
02 Dec 2017, 10:34
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Bunuel
Box's total volume / each cube's volume = number of cubes
Box's total volume: L* W* H
15 * 12 * 6 = 1,080
Shortcut: Double and halve factors (combine last two: 12*6 = 72)
15 * 72
30 * 36
60 * 18
120 * 9 = 1,080
Each cube's volume: 3 * 3 * 3 = 27
\(\frac{1,080}{27} = 40\) cubes
OR
Box side 15 = 3 * 5
Box side 12 = 3 * 4
Box side 6 = 3 * 2
Volume of each cube: (3*3*3) = 27
\(\frac{BoxVolume}{OneCubeVolume}=\frac{3*5*3*4*3*2}{3*3*3}=5*4*3=40\)
Answer D
