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# M01-11

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Math Expert
Joined: 02 Sep 2009
Posts: 55801

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16 Sep 2014, 00:15
2
21
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Difficulty:

55% (hard)

Question Stats:

61% (01:48) correct 39% (02:22) wrong based on 220 sessions

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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%

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Math Expert
Joined: 02 Sep 2009
Posts: 55801

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16 Sep 2014, 00:15
2
2
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: $$Percent=\frac{\text{Change}}{\text{Original}}*100=\frac{6*64-6*16}{6*16}*100=300%$$.

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Joined: 05 Jul 2015
Posts: 97
GMAT 1: 600 Q33 V40
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19 Feb 2016, 12:42
3
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.
Intern
Joined: 26 May 2014
Posts: 39
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

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25 Jul 2016, 12:25
how did you ascertain that there would be 64 smaller cubes?
Math Expert
Joined: 02 Sep 2009
Posts: 55801

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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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Joined: 28 Nov 2015
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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
Math Expert
Joined: 02 Sep 2009
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12 Sep 2016, 06:14
karandedhia wrote:
Hi,

How did you got below :

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

A cube has 6 faces, the area of each is 1^2 = 1 cm^2, so the total surface area is 6∗1^2= 6 cm^2.
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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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Joined: 20 Sep 2016
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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 ?
Math Expert
Joined: 02 Sep 2009
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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?
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Joined: 09 Aug 2016
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01 Oct 2018, 03:03
Hi Bunuel,

I calculated the number of cubes through surface area.
Surface area of original cube = 96
Surface area of each smaller cube =6
Total cubes = 96/6 = 16 cubes.

Although I understood how you solved the question, what is the error in my approach?
Intern
Joined: 05 Aug 2018
Posts: 18
GMAT 1: 650 Q46 V34

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08 Jan 2019, 07:29
Bunuel wrote:
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: $$Percent=\frac{\text{Change}}{\text{Original}}*100=\frac{6*64-6*16}{6*16}*100=300%$$.

Hey Bunuel,

The formula you used for percentage increase does not correspond to 300%

Percent Increase = Change/Original * 100% = 256-96/96 * 100% = 190/96 * 100% = 166.67%.

Math Expert
Joined: 02 Sep 2009
Posts: 55801

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08 Jan 2019, 07:32
sakshamjinsi wrote:
Bunuel wrote:
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: $$Percent=\frac{\text{Change}}{\text{Original}}*100=\frac{6*64-6*16}{6*16}*100=300%$$.

Hey Bunuel,

The formula you used for percentage increase does not correspond to 300%

Percent Increase = Change/Original * 100% = 256-96/96 * 100% = 190/96 * 100% = 166.67%.

6*64 = 384, not 256
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09 Jan 2019, 01:02
Bunuel wrote:
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%

SA of original cube : 6s^2 = 6* 16
and vol 4^3 = 64
the cube of 64 cm^3 is being cut into 1 cm small length pieces or say 64 pieces

the new SA would be 6*1^3 = 6 * 64
% change = (6*64-6*16)/6*16 = 300% IMO D
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Joined: 21 Jul 2018
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01 Mar 2019, 08:25
DJ1986 wrote:
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.

Hi Bunuel, chetan2u,

Any comments on above question. Re-posting it as I am also wondering the same.
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28 Apr 2019, 18:43
I solved this incorrectly because I was using the surface area to determine the number of cubes that would form after the big cube is broken up.

Now I realise you use volume.
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Re: M01-11   [#permalink] 28 Apr 2019, 18:43
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