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Retired Moderator
Joined: 16 Apr 2020
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WE:Education (Education)
A big cube is formed by rearranging the 160 colored and 56 non-colored
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22 Aug 2021, 22:43
22 Aug 2021, 22:43
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A big cube is formed by rearranging the 160 colored and 56 non-colored similar cubes in such a way that the exposure of the colored cubes to the outside is minimum.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Percentage of exposed area that is colored |
\(25.90\)% |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: A big cube is formed by rearranging the 160 colored and 56 non-colored
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02 Sep 2021, 00:59
02 Sep 2021, 00:59
1
Explanation:
Let edge length of each small cube be \(1 unit\)
Total number of small cubes = \(160 + 56 = 216\)
Volume of big cube = \(216unit^3\)
Side of big cube = \(216^{\frac{1}{3}} = 6 unit\)
Since the exposure of the colored cubes to the outside has to be minimum, we should keep maximum number of colored cubes inside the big cube.
i.e. let us place \(4^3 = 64\) small colored cubes inside the big cube.
Now, exposed cubes = \(216 - 64 = 152\) cubes, out of which \(96\) are colored ones.
T.S.A of big cube = \(6a^2 = 6(6^2) = 216 unit^2\)
Col. A: \(\frac{96}{216}(100) = 44.44%\)
Col. B: \(25.90%\)
Hence, option A
Let edge length of each small cube be \(1 unit\)
Total number of small cubes = \(160 + 56 = 216\)
Volume of big cube = \(216unit^3\)
Side of big cube = \(216^{\frac{1}{3}} = 6 unit\)
Since the exposure of the colored cubes to the outside has to be minimum, we should keep maximum number of colored cubes inside the big cube.
i.e. let us place \(4^3 = 64\) small colored cubes inside the big cube.
Now, exposed cubes = \(216 - 64 = 152\) cubes, out of which \(96\) are colored ones.
T.S.A of big cube = \(6a^2 = 6(6^2) = 216 unit^2\)
Col. A: \(\frac{96}{216}(100) = 44.44%\)
Col. B: \(25.90%\)
Hence, option A
Re: A big cube is formed by rearranging the 160 colored and 56 non-colored
[#permalink]
18 Oct 2023, 10:08
18 Oct 2023, 10:08
KarunMendiratta wrote:
Explanation:
Let edge length of each small cube be \(1 unit\)
Total number of small cubes = \(160 + 56 = 216\)
Volume of big cube = \(216unit^3\)
Side of big cube = \(216^{\frac{1}{3}} = 6 unit\)
Since the exposure of the colored cubes to the outside has to be minimum, we should keep maximum number of colored cubes inside the big cube.
i.e. let us place \(4^3 = 64\) small colored cubes inside the big cube.
Now, exposed cubes = \(216 - 64 = 152\) cubes, out of which \(96\) are colored ones.
T.S.A of big cube = \(6a^2 = 6(6^2) = 216 unit^2\)
Col. A: \(\frac{96}{216}(100) = 44.44%\)
Col. B: \(25.90%\)
Hence, option A
Let edge length of each small cube be \(1 unit\)
Total number of small cubes = \(160 + 56 = 216\)
Volume of big cube = \(216unit^3\)
Side of big cube = \(216^{\frac{1}{3}} = 6 unit\)
Since the exposure of the colored cubes to the outside has to be minimum, we should keep maximum number of colored cubes inside the big cube.
i.e. let us place \(4^3 = 64\) small colored cubes inside the big cube.
Now, exposed cubes = \(216 - 64 = 152\) cubes, out of which \(96\) are colored ones.
T.S.A of big cube = \(6a^2 = 6(6^2) = 216 unit^2\)
Col. A: \(\frac{96}{216}(100) = 44.44%\)
Col. B: \(25.90%\)
Hence, option A
How do you know that we should take 4 as the max? Why not 5?
