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16 Sep 2014, 00:15
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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. 4%
B. 166%
C. 266%
D. 300%
E. 400%
16 Sep 2014, 00:15
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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: \(Percent = \frac{\text{Change} }{\text{Original} } *100 = \frac{6*64 - 6*16}{6*16}*100 = 300\%\).
Note: 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.
Answer: D
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: \(Percent = \frac{\text{Change} }{\text{Original} } *100 = \frac{6*64 - 6*16}{6*16}*100 = 300\%\).
Note: 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.
Answer: D
19 Feb 2016, 12:42
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Bunuel, is it just luck that it worked out by ratios?
The way I got it is that since the ratio is 4:1 then 4 is 300% greater than 1.
If we had changed the larger cube for a ratio of 5:1 it would have been 400% greater.
If we had changed the smaller cubes for a ratio of 4:2 the surface area would have been 100% greater.
The way I got it is that since the ratio is 4:1 then 4 is 300% greater than 1.
If we had changed the larger cube for a ratio of 5:1 it would have been 400% greater.
If we had changed the smaller cubes for a ratio of 4:2 the surface area would have been 100% greater.
