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What is the maximum number of cubes, each 3 centimeters on an edge, th

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What is the maximum number of cubes, each 3 centimeters on an edge, th  [#permalink]

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New post 30 Nov 2017, 21:58
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Difficulty:

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Question Stats:

86% (00:47) correct 14% (01:12) wrong based on 31 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
2017-12-01_0947_002.png [ 5.29 KiB | Viewed 665 times ]

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Re: What is the maximum number of cubes, each 3 centimeters on an edge, th  [#permalink]

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New post 30 Nov 2017, 22:23
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.


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What is the maximum number of cubes, each 3 centimeters on an edge, th  [#permalink]

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New post 02 Dec 2017, 09:34
Bunuel wrote:
Image
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

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
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What is the maximum number of cubes, each 3 centimeters on an edge, th &nbs [#permalink] 02 Dec 2017, 09:34
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