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

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Math Expert
Joined: 02 Sep 2009
Posts: 42559

Kudos [?]: 135309 [0], given: 12686

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

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30 Nov 2017, 22:58
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Question Stats:

94% (00:26) correct 6% (00:00) wrong based on 18 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

[Reveal] Spoiler:
Attachment:

2017-12-01_0947_002.png [ 5.29 KiB | Viewed 249 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 135309 [0], given: 12686

Manager
Joined: 17 Oct 2016
Posts: 148

Kudos [?]: 38 [0], given: 89

Location: India
Concentration: Operations, Strategy
GPA: 3.7
WE: Design (Real Estate)
Re: What is the maximum number of cubes, each 3 centimeters on an edge, th [#permalink]

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30 Nov 2017, 23:23
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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Help with kudos if u found the post useful. Thanks

Kudos [?]: 38 [0], given: 89

VP
Joined: 22 May 2016
Posts: 1108

Kudos [?]: 397 [0], given: 640

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

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02 Dec 2017, 10:34
Bunuel wrote:

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

[Reveal] Spoiler:
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

$$\frac{BoxVolume}{OneCubeVolume}=\frac{3*5*3*4*3*2}{3*3*3}=5*4*3=40$$

Kudos [?]: 397 [0], given: 640

What is the maximum number of cubes, each 3 centimeters on an edge, th   [#permalink] 02 Dec 2017, 10:34
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