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19 Mar 2013, 23:17
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A cube is made up of equal smaller cubes. Two of the sides of the larger cube are called A and B. What is the total number of smaller cubes?
(1) When n smaller cubes are painted on A, n is 1/9 of the total number of smaller cubes.
(2) When m smaller cubes are painted on B, m is 1/3 of the total number of smaller cubes
(1) When n smaller cubes are painted on A, n is 1/9 of the total number of smaller cubes.
(2) When m smaller cubes are painted on B, m is 1/3 of the total number of smaller cubes
tizen
Joined: 26 Apr 2015
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23 May 2015, 14:51
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I solved this problem differently.
We are trying to solve for \(x{^3}\), the number of smaller cubes that compose the larger cube.
1) On Surface A we paint n cubes where n = \(\frac{1}{9}x{^3}\). We also know that on any surface there \(x{^2}\) cubes so the maximum number of cubes we can paint is \(x{^2}\). This means n = \(\frac{1}{9}x{^3}\le{x{^2}}\) => \(x\le{9}\). Since n is an integer we know that \(x{^3}\)must be a multiple of 9. Therefore x=3 OR x=6 OR x = 9. Insufficient.
2) Using the same math as before we get \(x\le{3}\) with \(x{^3}\) a multiple of 3 to meet the requirement that n is an integer. The only options we have for x are 1, 2, 3. Only x = 3 gives a multiple of 3. Sufficient.
We are trying to solve for \(x{^3}\), the number of smaller cubes that compose the larger cube.
1) On Surface A we paint n cubes where n = \(\frac{1}{9}x{^3}\). We also know that on any surface there \(x{^2}\) cubes so the maximum number of cubes we can paint is \(x{^2}\). This means n = \(\frac{1}{9}x{^3}\le{x{^2}}\) => \(x\le{9}\). Since n is an integer we know that \(x{^3}\)must be a multiple of 9. Therefore x=3 OR x=6 OR x = 9. Insufficient.
2) Using the same math as before we get \(x\le{3}\) with \(x{^3}\) a multiple of 3 to meet the requirement that n is an integer. The only options we have for x are 1, 2, 3. Only x = 3 gives a multiple of 3. Sufficient.
Updated on: 25 Sep 2013, 16:56
Originally posted by SVaidyaraman on 21 Mar 2013, 06:09.
Last edited by SVaidyaraman on 25 Sep 2013, 16:56, edited 1 time in total.
Last edited by SVaidyaraman on 25 Sep 2013, 16:56, edited 1 time in total.
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SravnaTestPrep
Fact 1: When a cube is made of equal smaller cubes, the number of smaller cubes is \(r^3\) where r is natural number.
Fact 2: Out of the six sides of the bigger cube, we are considering two sides and have named them A and B resp.
Statement 1: when on side A, n smaller cubes are painted we have that value of n equal to 1/9 of the total number of smaller cubes. So n is divisible by 9. Or the number of smaller cubes is a multiple of 9. Statement is not sufficient to answer the question because for example both 27 smaller cubes and 729 smaller cubes satisfy the conditions that the number of smaller cubes is a multiple of 9 and satisfies fact 1.
Statement 2: In this case the number of smaller cubes is a multiple of 3. The number of smaller cubes cannot be greater than \(3^3\) based on this statement alone.
Assume first the number of smaller cubes which is a multiple of 3 and satisfies fact 1, is 27. In this case each side has 9 smaller cubes exposed. If the number of smaller cubes painted, which is m, is 9.we have m =1/3 of the number of smaller cubes.
Assume next the the number of smaller cubes which is a multiple of 3 and satisfies fact 1, is 216. In this case each side has 36 smaller cubes exposed. So you could paint a maximum of 36 smaller cubes. Thus maximum value of m is 36/216 = 1/6 which cannot be equal to 1/3.
For higher multiples of 3, the value of m would keep on decreasing and would never satisfy the conditions mentioned.
Thus statement 2 alone is sufficient to answer this question.
Therefore answer is choice B.
