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M01-11

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M01-11 [#permalink]

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

A. 4%
B. 166%
C. 266%
D. 300%
E. 400%
[Reveal] Spoiler: OA

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Official Solution:

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

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


A cube has 6 faces.

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

Now, since the volume of the big cube is \(4^3=64\) \(cm^3\) and the volume of the smaller cubes is \(1^3=1\) \(cm^3\), then when the big cube is cut into the smaller cubes we'll get \(\frac{64}{1}=64\) little cubes. Each of those little cubes will have the surface area equal to \(6*1^2=6\) \(cm^2\), so total surface are of those 64 little cubes will be \(6*64\) \(cm^2\).

\(6*64\) is 4 times more than \(6*16\) which corresponds to 300% increase.

Or: general formula for percent increase or decrease, (percent change): \(\text{Percent} = \frac{\text{Change}}{\text{Original}}*100\)

So the percent increase will be: \(Percent=\frac{\text{Change}}{\text{Original}}*100=\frac{6*64-6*16}{6*16}*100=300%\).


Answer: D
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New post 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.

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Re: M01-11 [#permalink]

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New post 25 Jul 2016, 12:25
how did you ascertain that there would be 64 smaller cubes?

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New post 25 Jul 2016, 12:33
devbond wrote:
how did you ascertain that there would be 64 smaller cubes?


Since the volume of the big cube is \(4^3=64\) \(cm^3\) and the volume of the smaller cubes is \(1^3=1\) \(cm^3\), then when the big cube is cut into the smaller cubes we'll get \(\frac{64}{1}=64\) little cubes.
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Re: M01-11 [#permalink]

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New post 12 Sep 2016, 06:04
Hi,

How did you got below :

Each of those little cubes will have the surface area equal to 6∗1^2= 6 cm2

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New post 12 Sep 2016, 06:14

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New post 29 Sep 2016, 04:58
I did everything correctly and fell for (E) which is a trap answer... Gotta always remember to read the question correctly.

The new total surface area is 400% of the original, which means 400%-100% = 300% increase. Dang it.

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New post 17 Sep 2017, 07:32
What confirms that we must calculate the volume of original cube and divide with the volume of smaller cube to know the number of smaller cubes.
can not we cut the original cube in dimension (length, width and height ) to know the number of cubes resulting from original cube.

am I missing something or have I got the concept all wrong ?

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New post 17 Sep 2017, 08:41
jyotipes21@gmail.com wrote:
What confirms that we must calculate the volume of original cube and divide with the volume of smaller cube to know the number of smaller cubes.
can not we cut the original cube in dimension (length, width and height ) to know the number of cubes resulting from original cube.

am I missing something or have I got the concept all wrong ?


It's not clear what are you trying to do. Can you please elaborate?
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Kudos [?]: 129048 [0], given: 12187

Re: M01-11   [#permalink] 17 Sep 2017, 08:41
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