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Confusing Percentages

by manish_shr » Tue Jan 20, 2009 10:40 pm
A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface are of each of the new solids is not painted red?

a. 15%
b. 16(2/3)%
c. 20%
d. 25%
e. 33(1/3)%

i always end up getting 25% but thats not the OA. Can anyone plz explain this one?
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Re: Confusing Percentages

by piyush_nitt » Wed Jan 21, 2009 3:12 am
manish_shr wrote:A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface are of each of the new solids is not painted red?

a. 15%
b. 16(2/3)%
c. 20%
d. 25%
e. 33(1/3)%

i always end up getting 25% but thats not the OA. Can anyone plz explain this one?
IMO E

side of a cube = a (say)
surface area of each cube = 3a^2(6a^2/2)

on partition , 5 sides of the new rectangular block has to be painted.

therefore area of remaining side is a^2

hence %age = a^2/3a^2 = 1/3 = 33.33
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Re: Confusing Percentages

by piyush_nitt » Wed Jan 21, 2009 3:13 am
manish_shr wrote:A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface are of each of the new solids is not painted red?

a. 15%
b. 16(2/3)%
c. 20%
d. 25%
e. 33(1/3)%

i always end up getting 25% but thats not the OA. Can anyone plz explain this one?
IMO E

side of a cube = a (say)
surface area of each cube = 3a^2(6a^2/2)

on partiation , only 1 side is unpainted

therefore area of remaining side is a^2

hence %age = a^2/3a^2 = 1/3 = 33.33
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Re: Confusing Percentages

by sanju09 » Wed Jan 21, 2009 5:14 am
piyush_nitt wrote:
manish_shr wrote:A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface are of each of the new solids is not painted red?

a. 15%
b. 16(2/3)%
c. 20%
d. 25%
e. 33(1/3)%

i always end up getting 25% but thats not the OA. Can anyone plz explain this one?
IMO E

side of a cube = a (say)
surface area of each cube = 3a^2(6a^2/2)

on partition , 5 sides of the new rectangular block has to be painted.

therefore area of remaining side is a^2

hence %age = a^2/3a^2 = 1/3 = 33.33
:lol:

No dear! You are at error if you think that when a cube is cut into equal halves in the fashion as given in the question, the surface area of each smaller part becomes half of the big one; this in fact happens to its volume. See what happens:

If 'a' was each side of the original cube, then the new rectangular solid so formed would have its dimensions as 'a by a by a/2'; and only one square face (i.e. a * a = a ^ 2) will remain unpainted out of the total surface area, which is 2 [a * a + a * a/2 + a * a/2] = 4 a ^ 2, leaving the unpainted part's percentage = [(a ^ 2)/(4 a ^ 2)] * 100 = 25%.

So my answer is 'd'.


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by sjd00d » Wed Jan 21, 2009 9:49 pm
D. 25% in my opinion. Let the sides be 1 before cut, after the cut 1 side not painted but 1/2+1/2+1/2+1/2+1 painted, 1/4 = 25%
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Re: Confusing Percentages

by aroon7 » Wed Jan 21, 2009 11:00 pm
manish_shr wrote:A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface are of each of the new solids is not painted red?

a. 15%
b. 16(2/3)%
c. 20%
d. 25%
e. 33(1/3)%

i always end up getting 25% but thats not the OA. Can anyone plz explain this one?
What is the OA?
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by sanju09 » Thu Jan 22, 2009 1:12 am
sjd00d wrote:D. 25% in my opinion. Let the sides be 1 before cut, after the cut 1 side not painted but 1/2+1/2+1/2+1/2+1 painted, 1/4 = 25%
:oops: This is really very pathetic elucidation to the problem.
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Re: Confusing Percentages

by sanju09 » Thu Jan 22, 2009 1:16 am
aroon7 wrote:
manish_shr wrote:A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface are of each of the new solids is not painted red?

a. 15%
b. 16(2/3)%
c. 20%
d. 25%
e. 33(1/3)%

i always end up getting 25% but thats not the OA. Can anyone plz explain this one?
What is the OA?
:

? I do not understand the meanings of IMO and OA. Please brief me somebody.
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by DanaJ » Thu Jan 22, 2009 1:46 am
OA = original answer

IMO = in my opinion
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by Ian Stewart » Thu Jan 22, 2009 2:58 am
sanju09 wrote:
sjd00d wrote:D. 25% in my opinion. Let the sides be 1 before cut, after the cut 1 side not painted but 1/2+1/2+1/2+1/2+1 painted, 1/4 = 25%
:oops: This is really very pathetic elucidation to the problem.
There is nothing 'pathetic' about sjd00d's solution; indeed, it's probably the fastest path to an answer here. In any case, I don't know why you would say such a thing - it certainly isn't helpful to anybody.
sanju09 wrote:
? I do not understand the meanings of IMO and OA. Please brief me somebody.
IMO- 'In my opinion'
OA- 'Original answer'

Manish - what makes you say that the OA is not D? We begin with six red faces, and then cut the cube to expose two white faces equal in size to the red faces, so 2/(2+6) = 25% of the surface of the resulting shapes will be white.
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by sanju09 » Thu Jan 22, 2009 3:42 am
Ian Stewart wrote:
sanju09 wrote:
sjd00d wrote:D. 25% in my opinion. Let the sides be 1 before cut, after the cut 1 side not painted but 1/2+1/2+1/2+1/2+1 painted, 1/4 = 25%
:oops: This is really very pathetic elucidation to the problem.
There is nothing 'pathetic' about sjd00d's solution; indeed, it's probably the fastest path to an answer here. In any case, I don't know why you would say such a thing - it certainly isn't helpful to anybody.
sanju09 wrote:
? I do not understand the meanings of IMO and OA. Please brief me somebody.
IMO- 'In my opinion'
OA- 'Original answer'

Manish - what makes you say that the OA is not D? We begin with six red faces, and then cut the cube to expose two white faces equal in size to the red faces, so 2/(2+6) = 25% of the surface of the resulting shapes will be white.
I am extremely sorry if it hurt anybody, as myself being a GMAT instructor, I really never ever try to demoralize students with my deeds or words. Although in this case in particular, I agree with you sir, and would take care of it in future. And yes, this was really the fastest way to go to the answer. Thank you..
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by krisraam » Thu Jan 22, 2009 1:04 pm
2 parts of the 6 parts is not painted. so 33 1/3 %.
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by aroon7 » Thu Jan 22, 2009 7:28 pm
krisraam wrote:2 parts of the 6 parts is not painted. so 33 1/3 %.
the question is "What percent of the surface are of each of the new solids is not painted red?"
so we should consider only one of the pieces...
am i right?
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by techwiz » Thu Jan 22, 2009 10:06 pm
After cutting, the total area = 4a*a

Area of the unpainted side = a*a

Hence, 25%
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by krisraam » Fri Jan 23, 2009 5:38 am
Oh Yeah . I missed it. The answer is 25%. 1 part of 4 parts is not painted.
I was calulating as 2D for some reason.
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