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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Bunuel
Official Solution:

If a cube with a side length of 4 cm is cut into smaller cubes, each with a side length of 1 cm, what is the percentage increase in the total surface area of the resulting smaller cubes?

A. 4%
B. 166%
C. 266%
D. 300%
E. 400%


A cube has 6 faces.

The surface area of a cube with a side length of 4 cm is \(6*4^2 = 6*16\) square centimeters.

Given that the volume of the larger cube is \(4^3 = 64\) cubic centimeters and the volume of each smaller cube is \(1^3 = 1\) cubic centimeter, the larger cube can be divided into \(\frac{64}{1} = 64\) smaller cubes. Each of these smaller cubes has a surface area of \(6*1^2 = 6\) square centimeters, resulting in a total surface area of \(6* 64\) square centimeters for all 64 smaller cubes.

The total surface area of the smaller cubes, \(6*64\), is 4 times greater than the surface area of the larger cube, \(6*16\), which corresponds to a 300% increase.

Alternatively: To calculate the percentage increase, use the formula for percent change: \(\text{Percent} = \frac{\text{Change{\text{Original *100\).

The percentage increase can be calculated as follows: \(Percent = \frac{\text{Change{\text{Original *100 = \frac{6*64 - 6*6}{6*16}*100 = 300\%\).


Answer: D



Hi bunnel, shoudn't the larger cube be divided into 256 smaller cubes with side of 1cm each - since right now we have one cube with side 4cm each. If we have 64 smaller cubes, won't there be a part of larger cube left?
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Bunuel
Official Solution:

If a cube with a side length of 4 cm is cut into smaller cubes, each with a side length of 1 cm, what is the percentage increase in the total surface area of the resulting smaller cubes?

A. 4%
B. 166%
C. 266%
D. 300%
E. 400%


A cube has 6 faces.

The surface area of a cube with a side length of 4 cm is \(6*4^2 = 6*16\) square centimeters.

Given that the volume of the larger cube is \(4^3 = 64\) cubic centimeters and the volume of each smaller cube is \(1^3 = 1\) cubic centimeter, the larger cube can be divided into \(\frac{64}{1} = 64\) smaller cubes. Each of these smaller cubes has a surface area of \(6*1^2 = 6\) square centimeters, resulting in a total surface area of \(6* 64\) square centimeters for all 64 smaller cubes.

The total surface area of the smaller cubes, \(6*64\), is 4 times greater than the surface area of the larger cube, \(6*16\), which corresponds to a 300% increase.

Alternatively: To calculate the percentage increase, use the formula for percent change: \(\text{Percent} = \frac{\text{Change{\text{Original *100\).

The percentage increase can be calculated as follows: \(Percent = \frac{\text{Change{\text{Original *100 = \frac{6*64 - 6*6}{6*16}*100 = 300\%\).


Answer: D



Hi bunnel, shoudn't the larger cube be divided into 256 smaller cubes with side of 1cm each - since right now we have one cube with side 4cm each. If we have 64 smaller cubes, won't there be a part of larger cube left?

We have a cube with a side length of 4 cm; hence its volume is 4^3 = 64 cm^3. This cube, when cut into smaller cubes with a side length of 1 cm, so when cut into smaller cubes of 1 cm^3, will give 64/1 = 64 small cubes."
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation.
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I think this is a poor-quality question and I agree with explanation. Geometry is no longer included in the focus. If required the question shall include the formula
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I think this is a poor-quality question and I agree with explanation. Geometry is no longer included in the focus. If required the question shall include the formula
­The area of a square, rectangle, the volume of a cube or a rectangular solid, and the Pythagorean theorem are not considered by the GMAT as specific geometry knowledge and can still be tested on the exam. There are several questions involving this in the GMAT Prep Focus mocks. Thus, the question above is not about geometry; it's rather on percents.
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Bunuel
juansecalderon
I think this is a poor-quality question and I agree with explanation. Geometry is no longer included in the focus. If required the question shall include the formula
­The area of a square, rectangle, the volume of a cube or a rectangular solid, and the Pythagorean theorem are not considered by the GMAT as specific geometry knowledge and can still be tested on the exam. There are several questions involving this in the GMAT Prep Focus mocks. Thus, the question above is not about geometry; it's rather on percents.
I agree. Even if you forget the formula for surface area or worst just assume the cube to be a square and take the surface area to be \(a^2\), you still get the same answer since any additional constants in the surface area get cancelled when you calculate the percentage increase. While attempting this question I forgot the formula and took it to be 4\(a^2\) instead of 6\(a^2\) and still got the correct answer.
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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This is a great question that’s helpful for learning. This is the geometry question,and in Gmat focus geometry is obsolete.
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hritik04
This is a great question that’s helpful for learning. This is the geometry question,and in Gmat focus geometry is obsolete.

­The area of a square, rectangle, the volume of a cube or a rectangular solid, and the Pythagorean theorem are not considered by the GMAT as specific geometry knowledge and can still be tested on the exam. There are several questions involving this in the GMAT Prep Focus mocks. Thus, the question above is not about geometry; it's rather on precents.
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