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Very tough Math problem!!

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Very tough Math problem!!

by santosh_surathkal » Mon Feb 22, 2010 9:08 am
Topic: Square countertop
Sat Oct 25, 2008 8:26 am
Elapsed Time:
00:00
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A square countertop has a square tile inlay in the center, leaving an untiled strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39, which of the following could be the width, in inches, of the strip?

I. 1
II. 3
III. 4

A) I only

B) II only

C) I and II only

D) I and III only

E) I, II and III

Could you please help ?

I feel the ans could only be multiple of 1.5 ... and there is no similar option
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by arzanr » Mon Feb 22, 2010 10:49 am
You answered the question yourself. 3 is a multiple of 1.5. :) The answer should be B IMO.
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by sanju09 » Tue Feb 23, 2010 3:13 am
santosh_surathkal wrote:Topic: Square countertop
Sat Oct 25, 2008 8:26 am
Elapsed Time:
00:00
Why a timer is critical to improving your score
A square countertop has a square tile inlay in the center, leaving an untiled strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39, which of the following could be the width, in inches, of the strip?

I. 1
II. 3
III. 4

A) I only

B) II only

C) I and II only

D) I and III only

E) I, II and III

Could you please help ?

I feel the ans could only be multiple of 1.5 ... and there is no similar option
If x is the side of the square countertop and y that of the square tile inlay in the center, then ½ (x - y) is the uniform width around the tile. Now, the tiled area is y^2 and the un-tiled area is x^2 - y^2, such that y^2/ (x^2 - y^2) = 25/39, here x^2 - y^2 is a certain multiple of 39, or ½ (x - y) could be 3/2 or 13/2 or a multiple thereof. We can see 3 in the choices as would be integer of 3/2, but no would be integer of 13/2 is there in the choices, hence [spoiler]II[/spoiler] only.


[spoiler]B[/spoiler]
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by lunarpower » Tue Feb 23, 2010 5:02 am
hey people.

the way this problem is written, it can be solved in a few seconds, and the answer is (e).

reasoning:
all we have is a RATIO. we don't have either of the following:
* ANY concrete lengths (e.g., sides of the inner square, sides of the outer square, etc.)
* ANY restrictions on the dimensions (e.g., they have to be whole numbers, etc.)

absent either of these conditions, the countertop can be absolutely any size whatsoever (as long as the inlay is kept proportional to the 25:39 ratio).

i can see the intent of the question writer(s), but they were sloppy - they forgot to include restriction(s).

what's the source of the question?
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by sanju09 » Tue Feb 23, 2010 5:11 am
In the absense of restrictions, I agree with lunarpower.

[spoiler]me2E[/spoiler]
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by kstv » Tue Feb 23, 2010 7:38 am
The ratio of tiled to untiled is 25 : 39. So tiled area is to the whole area is 25 : 64. Therefore ratio of the sides of respective squares is 5 : 8 . So the width of the strip could be 3.
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by Ian Stewart » Tue Feb 23, 2010 10:55 am
lunarpower wrote: i can see the intent of the question writer(s), but they were sloppy - they forgot to include restriction(s).

what's the source of the question?
I don't think they were sloppy at all - it's a question from the Official Guide! I think the entire purpose of the question is to test precisely what Ron points out above, that is, whether you can see that the setup is only based on ratios, and that any width is possible. Of course, one could do a lot of algebra here, but there's no reason to.
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by rahul.s » Tue Feb 23, 2010 10:49 pm
This is a problem from the OG, and it's been discussed earlier.

https://www.beatthegmat.com/square-count ... 21427.html
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by lunarpower » Tue Feb 23, 2010 10:54 pm
Ian Stewart wrote:
lunarpower wrote: i can see the intent of the question writer(s), but they were sloppy - they forgot to include restriction(s).

what's the source of the question?
I don't think they were sloppy at all - it's a question from the Official Guide! I think the entire purpose of the question is to test precisely what Ron points out above, that is, whether you can see that the setup is only based on ratios, and that any width is possible. Of course, one could do a lot of algebra here, but there's no reason to.
hmm.

that's interesting.

in retrospect, this does seem like the sort of stunt that the official guide would pull - but i wouldn't expect to see it in the problem solving section!

this sort of thinking is already the crux of the ENTIRE data sufficiency section - i.e., "is this enough stuff to narrow / solve the problem?" - so i'm surprised to see it bleeding into the PS problems.
very surprised, actually.

i wonder if there's some DS version of this problem floating around out there (in gmatprep, or in the real exam), and this is an alternate version.
they have done that in CR - they write the same passage, but have two or more questions based on the passage - so i could see that. this just doesn't seem like a problem that was originally written as a PS problem.

the same goes for the rare PS problems that have as an answer choice "the quantity cannot be determined". again, it's weird to see such things bleeding into the PS section, since they are already the entire raison d'être of data sufficiency.

weird.
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by ganeshraj » Thu Feb 25, 2010 3:06 am
Hi Ron,

Can you pls clarify this.

Can the length of a side be 13.3333333333.....................

shouldn't it end somewhere.how can the length be an unending number.

If i solve for the side of the tiled square (the inner one ), i end up with the a solution where the side =13.333333.......... and this just goes on.

this is how i solved it. can you pls say where iam going wrong :

say the side of the inner tiled square = d
the width of the untiled strip = w

then,

the area of inner tiled square = d^2

the area of the outer untiled strip = 4 (w^2)+ 4(w*d).

the area of the outer untiled square can be considered as 4 rectangles (w*d) and 4 squares (w*w).


Now, the ratio of the area of the inner square to the ratio of the area of the outer strip = d^2 : 4(w^2)+ 4(w*d).

In the question this ratio is given as 25:39.


therefore, 25:39 = d^2 : 4(w^2)+ 4(w*d).

=> 39 (d^2) = 25 [4(w^2)+ 4(w*d)].
=> 39 (d^2) =100 w (w+d).

Substituting w=4 the width.we get,

39 (d^2) - 400 d-1600 =0

Solving this quadratic equation,

we get 2 solutions for d = -3.076923076923077 and 13.3333333333.................

Both this values can not be the length of a side , offcourse that is just what i believe.

therefore width cannot be 4,III cannot be the solution.

Can you pls say whether my reasoning is correct and clarify incase if it is not.


many thanks.
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by lunarpower » Thu Feb 25, 2010 3:18 am
a couple of things.

1 * did you read the earlier posts in this thread?
it's not possible to solve this problem for a single length - the desired square can have quite literally any side length at all.

2 *
ganeshraj wrote:Hi Ron,

Can you pls clarify this.

Can the length of a side be 13.3333333333.....................

shouldn't it end somewhere.how can the length be an unending number.
sure, that can be the length of a side. written as a mixed number, that's 13 1/3 inches. (are you familiar with converting back and forth between fractions and repeating decimals?)

also -
ANY positive number can be the length of a side. this includes not only repeating decimals (like the one above), but also non-repeating decimals (i.e., irrational numbers).
for instance, if you have a 45-45-90 or 30-60-90 triangle, you MUST have at least one side length that is an irrational number - a decimal that neither repeats nor terminates.
for instance, if you have a 30-60-90 triangle whose short leg is 1, then the hypotenuse is 2, but the long leg is √3, which is 1.732050807...


3 *
this is how i solved it. can you pls say where iam going wrong :
right here:
we get 2 solutions for d = -3.076923076923077 and 13.3333333333.................

Both this values can not be the length of a side , offcourse that is just what i believe.
you're right about the negative value - you can't have a negative side length - but, again, ANY positive number can be a side length.
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