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A square wooden plaque has a square brass inlay in the
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15 Jan 2010, 03:00
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The Official Guide For GMAT® Quantitative Review, 2ND EditionA square wooden plaque has a square brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches, of the wooden 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 Problem Solving Question: 175 Category: Geometry Area Page: 85 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: hard problem OG Quant 2nd edition
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15 Jan 2010, 05:14
hrish88 wrote: so doest it mean any width <= 39/4 is possible. this is the 2nd last problem in OG.so i thought it would be difficult. No I mean ANY width is possible. Let the the side of small square be \(x\) and the big square \(y\). Given: \(\frac{x^2}{y^2x^2}=\frac{25}{39}\) > \(\frac{x^2}{y^2}=\frac{25}{64}\) > \(\frac{x}{y}=\frac{5}{8}\). We are asked which value of \(\frac{yx}{2}\) is possible. \(\frac{y\frac{5}{8}y}{2}=\frac{3}{16}y=?\). Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all. Hope it's clear.
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Re: hard problem OG Quant 2nd edition
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Re: hard problem OG Quant 2nd edition
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24 Mar 2010, 05:10
hrish88 wrote: A square wooden plaque has a square brass inlay in the center ,leaving a wooden strip of uniform width around the brass square.if the ratio of the brass area to the wooden area is 25 to 39,which of the following could be the width ,in inches ,of the wooden strip.
I. 1 II. 3 III. 4
A.I only B.II only C.III only D.I and III only e.I,II and III Area of brass square/ area of wooden strip = 25 /39 lets say length of the wooden plaque= l and length of the square brass = x then x^2 / (l^2  x^2) = 25/39 =>39x^2 = 25l^2  25x^2 =>64x^2 = 25l^2 =>8x = 5l Width of wooden strip should be lx =>x = 5l/8 so l x = l = 5l/8 = 3l/8 Now 3l/8 could be any value depending on the value of l so answer is E.
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Re: hard problem OG Quant 2nd edition
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15 Jan 2010, 04:03
so doest it mean any width <= 39/4 is possible. this is the 2nd last problem in OG.so i thought it would be difficult.



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Re: hard problem OG Quant 2nd edition
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15 Jan 2010, 08:49
let the width of the wooden part = w let the width of the brass part = b given brass area/wooden area = 25/39 area of brass part = b^2 area of wooden part = (b+2w)^2  b^2 Simplifying 64b^2=25(b+2w)^2 8b=5b+10w b=10w/3 b has to be an integer or a terminating decimal. We can't have a width of 10/3 in real life (note the question doesn't ask for approximate width.) hence w has to be a multiple of 3. Answer is B.
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A square wooden plaque has a square brass inlay in the
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15 Jan 2010, 09:44
atish wrote: let the width of the wooden part = w let the width of the brass part = b given brass area/wooden area = 25/39 area of brass part = b^2 area of wooden part = (b+2w)^2  b^2 Simplifying 64b^2=25(b+2w)^2 8b=5b+10w b=10w/3
b has to be an integer or a terminating decimal. We can't have a width of 10/3 in real life (note the question doesn't ask for approximate width.) hence w has to be a multiple of 3. Answer is B. Are you saying that in real life everything has integer or terminating decimal length? Why cannot we have repeated decimal or even irrational number as width of something? Take the square with side 1, diagonal would be \(\sqrt{2}\), it's not an integer or terminating decimal. Also it's possible to divide the line segment into three equal parts, Google it and you find that it's quite easy.
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Re: hard problem OG Quant 2nd edition
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15 Jan 2010, 10:25
Quote: Are you saying that in real life everything has the integer or terminating decimal length? Why can not we have repeated decimal or even irrational number as width of something?
Take the square with side 1, diagonal would be \(\sqrt{2}\), it's not an integer or terminating decimal.
Also it's possible to divide the line segment into three equal parts, google it and you find that it's quite easy. That takes the question to a whole new dimension, I do understand what you are saying though. If the width of something is 10/3 that means you can never (accurately) measure it. The width of the brass square can never be measured practically, it can be only measured mathematically. If such a square was to be made, the creator would have to take a square of 10/10 dimension, divide it into 9 exactly equal parts and use one of them, he/she could never make just the square as it would be impossible to measure 10/3 inches. I do get the concept, but don't like the fact that a question can be based on it.
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Re: hard problem OG Quant 2nd edition
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15 Jan 2010, 12:09
Bunuel wrote: Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all.
Hope it's clear.
Wow such a simple concept i must have left my brain someplace else. Nice explanation.You rock man as always.



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Re: hard problem OG Quant 2nd edition
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07 Aug 2010, 16:32
Bunuel wrote: hrish88 wrote: so doest it mean any width <= 39/4 is possible. this is the 2nd last problem in OG.so i thought it would be difficult. No I mean ANY width is possible. Let the the side of small square be \(x\) and the big square \(y\). Given: \(\frac{x^2}{y^2x^2}=\frac{25}{39}\) > \(\frac{x^2}{y^2}=\frac{25}{64}\) > \(\frac{x}{y}=\frac{5}{8}\). We are asked which value of \(\frac{yx}{2}\) is possible. \(\frac{y\frac{5}{8}y}{2}=\frac{3}{16}y=?\). Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all. Hope it's clear. To generalize then, since the answer does not seem to depend on the fact that the ration is 25/39, can it be said that regardless of what the ratio is, the width of strip can be ANYTHING?
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Re: hard problem OG Quant 2nd edition
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08 Aug 2010, 00:32
mainhoon wrote: Bunuel wrote: hrish88 wrote: so doest it mean any width <= 39/4 is possible. this is the 2nd last problem in OG.so i thought it would be difficult. No I mean ANY width is possible. Let the the side of small square be \(x\) and the big square \(y\). Given: \(\frac{x^2}{y^2x^2}=\frac{25}{39}\) > \(\frac{x^2}{y^2}=\frac{25}{64}\) > \(\frac{x}{y}=\frac{5}{8}\). We are asked which value of \(\frac{yx}{2}\) is possible. \(\frac{y\frac{5}{8}y}{2}=\frac{3}{16}y=?\). Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all. Hope it's clear. To generalize then, since the answer does not seem to depend on the fact that the ration is 25/39, can it be said that regardless of what the ratio is, the width of strip can be ANYTHING? Yes, width can have any positive value: the larger the width is the larger the whole square would be.
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Re: hard problem OG Quant 2nd edition
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19 Sep 2010, 10:29
Hi Bunuel,
Thanx for the explanation. I didn't understand one part:
We are asked which value of \frac{yx}{2} is possible. \frac{y\frac{5}{8}y}{2}=\frac{3}{16}y=?
Please enlighten me.



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Re: hard problem OG Quant 2nd edition
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Re: hard problem OG Quant 2nd edition
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19 Sep 2010, 14:41
I actually created equation and substituted width value to conclude E. I should have simply thought straight like Bunuel and marked E in 15 sec.
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Re: hard problem OG Quant 2nd edition
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17 Jan 2011, 11:50
Bunuel wrote: hrish88 wrote: A square wooden plaque has a square brass inlay in the center ,leaving a wooden strip of uniform width around the brass square.if the ratio of the brass area to the wooden area is 25 to 39,which of the following could be the width ,in inches ,of the wooden strip.
I. 1 II. 3 III. 4
A.I only B.II only C.III only D.I and III only e.I,II and III Why would ANY width of the strip be impossible? Answer: E. That was my rationale exactly. I got the answer correct, but the explanation in the book made me feel like I did not grasp the underlying math. Thanks Bunuel.



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Re: A square wooden plaque has a square brass inlay in the
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01 Apr 2012, 06:48
I didn't understand the statement "We are asked which value of (yx)/2 is possible" . Can someone explain?
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Re: hard problem OG Quant 2nd edition
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01 Sep 2012, 06:48
Bunuel wrote: hrish88 wrote: A square wooden plaque has a square brass inlay in the center ,leaving a wooden strip of uniform width around the brass square.if the ratio of the brass area to the wooden area is 25 to 39,which of the following could be the width ,in inches ,of the wooden strip.
I. 1 II. 3 III. 4
A.I only B.II only C.III only D.I and III only e.I,II and III Why would ANY width of the strip be impossible? Answer: E. Hi Bunuel, this was the most appropriate reason for the answer; however, can there be a case where such a condition("any possible width of the strip") might fail, provided there is no restriction on dimensions to be integral or non integral?
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Re: A square wooden plaque has a square brass inlay in the
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18 Sep 2012, 10:50
Bunuel wrote: mymbadreamz wrote: Why is it not yx? Why do we calculate (yx)/2? Consider the diagram below: Attachment: Wooden strip.png As you can see the width of the wooden strip (the width of grey strip) is \(\frac{yx}{2}\). total length is 8x and the length of the countertop is 5x.so the one side length of the untiled area is w = 8x5x/2 =3x/2 .Since x could be any value so the answer is E..Am I right Bunuel ?




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