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A square wooden plaque has a square brass inlay in the

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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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New post 25 May 2017, 13:06
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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) 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


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L = length of wooden plaque
X - length of brass inlay

We are told that x^2/(L^2-x^2) = 25/39

39x^2 = 25L^2 - 25x^2
64x^2 = 25L^2
8x = 5L
x/L = 5/8

Now we know that the width of the frame is (L - x)/2

8x/5 = L

(8x/5 - x)/2 = (16x - 5x)/10*2

11x/20

or

x = 5L/8
(L - 5/8L)/2

= 3L / 16

Any values for x or L are possible because neither needs to be an integer.
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New post 12 Sep 2017, 11:06
Bunuel wrote:
mymbadreamz wrote:
Why is it not y-x? Why do we calculate (y-x)/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 \(\frac{y-x}{2}\).


I still don't understand the fact that why do width of wooden strip is Y-X/2 but not Y-X.
Please do elaborate little more.

Posted from my mobile device
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New post 12 Sep 2017, 11:09
rkantsah wrote:
Bunuel wrote:
mymbadreamz wrote:
Why is it not y-x? Why do we calculate (y-x)/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 \(\frac{y-x}{2}\).


I still don't understand the fact that why do width of wooden strip is Y-X/2 but not Y-X.
Please do elaborate little more.

Posted from my mobile device


Please go through the whole thread: https://gmatclub.com/forum/a-square-woo ... l#p1068331

Hope it helps.
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A square wooden plaque has a square brass inlay in the  [#permalink]

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New post Updated on: 16 Apr 2018, 12:34
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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) I and II only
(D) I and III only
(E) I, II , and III


Let's say we want to create a plaque with a square brass inlay in the center, and we want the brass to wood ratio to be 25:39

Let's begin a square wooden board with ANY dimensions.
Image


Now place a square brass inlay in the middle of the wooden board, and keep adjusting the size of the brass inlay until we have a brass to wood ratio that is 25:39
Image

At this point, if we shrink or expand the plaque . . .
Image
. . . the brass to wood ratio will remain at 25:39

So, as you can see, this plaque can be ANY size, which means the width of the wooden strip can have ANY measurement.

Answer:

Cheers,
Brent
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Originally posted by GMATPrepNow on 15 Sep 2017, 12:19.
Last edited by GMATPrepNow on 16 Apr 2018, 12:34, edited 1 time in total.
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A square wooden plaque has a square brass inlay in the  [#permalink]

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New post 27 Dec 2017, 21:51
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. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) 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.


I have difficulties this this part \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\)

How do you get it?
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New post 27 Dec 2017, 22:06
1
Erjan_S 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. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) 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.


I have difficulties this this part \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\)

How do you get it?


\(\frac{x^2}{y^2-x^2}=\frac{25}{39}\);

Cross multiply: \(39x^2=25y^2-25x^2\);

Re-arrange: \(64x^2=25y^2\);

\(\frac{x^2}{y^2}=\frac{25}{64}\).

Hope it's clear.
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New post 01 Mar 2018, 10:45
Hi Bunuel, I found the stem very misleading...I thought it meant the ratio of the brass square to the ratio of the wooden square..not just the exposed part of the wooden square! It says wooden area..not wooden area that is not covered by the brass square....I don't think the question is put down very clearly. In this case equation would become

\(x^2/y^2\)

and the problem would become a bit more complicated to solve....I found that the only answer possible would be h=1 in that case...




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. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) 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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New post 19 Mar 2018, 03:20
VeritasPrepKarishma, chetan2u your take on this problem, please? Having serious issues understanding this problem. :(
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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New post 05 May 2018, 04:19
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.


hahaha Bunuel this is your most shortest explanation :) great sense of humor! :)
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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New post 05 May 2018, 05:34
1
Erjan_S 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. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) 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.


I have difficulties this this part \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\)

How do you get it?


hello generis :) i am honored to tag you and ask you three questions :)

Q # 1: i reviewed whole thread and all explanations but cant understand why do we divide by 2 here \(\frac{y-x}{2}\) ?

Q # 2: why do we subtract \(x^2\)from \(y^2\) here in the denominator \([m]\frac{x^2}{y^2-x^2}\) ? we know that ratio is \(\frac{25}{39}\), why to subtract ? :?

Q # 3: my approach to to tho problem was simple i rounded 39 to 40 so i simply got ratio \(\frac{25}{40} = \frac{5}{8}\) do you think its correct approach ? :? :)

Many thanks and have an awesome weekend ! :)

P.S. :cool: by the way in your signature isnt there a SC issue ? :)

it says "In the depths of winter, I finally learned
that within me there lay an invincible summer
."

but shouldnt we say so "In the depths of winter, I finally learned
that within me there is a thick layer of an invincible hot summer mood
" :)
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A square wooden plaque has a square brass inlay in the  [#permalink]

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New post 05 May 2018, 14:55
1
1
dave13 wrote:
Erjan_S wrote:
Bunuel wrote:
No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) 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.
I have difficulties this part \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\)
How do you get it?
hello generis :) i am honored to tag you and ask you three questions :)
Q # 1: SEE BELOW Q # 2: SEE BELOW Q # 3: SEE BELOW

Many thanks and have an awesome weekend ! :)

P.S. :cool: by the way in your signature isnt there a SC issue ? :)

it says "In the depths of winter, I finally learned
that within me there lay an invincible summer
."

but shouldnt we say so "In the depths of winter, I finally learned
that within me there is a thick layer of an invincible hot summer mood
"
:)

:lol: :lol: :lol: OMG you are irrepressible.

Hi dave13 -
Your second question is hard to answer.
I cannot tell whether you are following the algebra.
Others have had a hard time. Maybe this write-up will help.

I am assigning capital letter variables in order to write as abbreviated an
answer as possible.
Let P = the whole PLAQUE (which is a "big square"). Area of \(P = y^2\)
Let B = the square BRASS inlay (which is a "little square). Area of \(B = x^2\)
Let W = the WOODEN part (which is a uniform strip, like a "frame"). Area of \(W = y^2-x^2\)
Quote:
Q # 1: i reviewed whole thread and all explanations but cant understand why do we divide by 2 here \(\frac{y-x}{2}\)?

Short answer: because there are TWO border widths between \(y\) and \(x\). TAKE A LOOK:
Attachment:
woodenborder.png
woodenborder.png [ 8.24 KiB | Viewed 278 times ]

We are solving for the width of the border.
Does (side y) - (side x) = width of border?
No. \((y - x)\) = TWO times the width of the border

Why? On EACH side of \(x\), there is one border width, \(w\):
\(y = w + x + w\)
\(y - x = 2w\)
\(w = \frac{y-x}{2}\)


Try drawing the diagram. Let's say \(x = 2, y = 4\), and width of wooden frame, W = 1 (write values in)
Does \((y - x)\) = width of wooden border? \((y-x) = 2. W = 1\). Not correct.
\((y - x) = 2\) * width of the wooden border.
Width of border therefore equals \(\frac{y-x}{2}=\frac{4-2}{2}=1\)
Quote:
Q # 2: why do we subtract \(x^2\)from \(y^2\) here in the denominator \(\frac{x^2}{y^2-x^2}\) ? we know that ratio is \(\frac{25}{39}\), why to subtract ? :?

Short answer: We subtract because
1) \(y^2 - x^2\) equals the area of the wooden part, which
2) is the bottom part of the ratio given by the prompt
(I think you may be slightly confused about area of "the wooden part." It's the frame, the thin strip.)

We need this ratio: \(\frac{Area_{B}}{Area_{W}}\), in which \(Area_{W} = y^2-x^2\),
to get another ratio: side length of B to side length of P.

\(\frac{InnerSquareArea}{WoodenFrameArea} = \frac{25}{39}\)

1) inner square area of P = \(x^2\)
2) wooden frame area = \((y^2 - x^2)\)
Think of the area of the wooden frame as a "shaded region."
You would find the shaded region's area by subtracting B's area \((x^2)\) from P's area \((y^2)\).

We go from \(\frac{area of B}{area of W}\) to \(\frac{Side of B(=x)}{Side of P(=y)}\) to border width= some y expression

Border width depends on length of \(y\). That demonstration has steps

1) Use the given ratio of \(\frac{Barea}{Warea}\) to find ratio between the areas of the squares
The given ratio is NOT between the areas of the two squares (that's what we need to find)

GIVEN ratio is \(\frac{B}{W} = \frac{25}{39}\)

Area of W = (Area of P - Area of B) (think of W as a shaded region)
Area of P = \(y^2\)
Area of B = \(x^2\)
Area of W = \(y^2 - x^2\)
Substitute that RHS for W in the original ratio.

\(\frac{B}{W}=\frac{x^2}{y^2-x^2}=\frac{25}{39}\)
To find the ratio of areas of squares (P and B), we need \(x^2\) over \(y^2\)
Eliminate \(x^2\) in the denominator. Cross multiply:

\(\frac{x^2}{y^2-x^2}=\frac{25}{39}\)
\(39x^2 = 25y^2 - 25x^2\)
\(64x^2=25y^2\)

DIVIDE both sides by \(y^2\) and by \(64\)
\(\frac{x^2}{y^2}=\frac{25}{64}\)

(2) Use areas of squares to find ratio of side lengths of squares
Find ratio of side x to side y by taking the square root of \(\frac{x^2}{y^2}=\frac{25}{64}\):
\(\frac{x}{y}=\frac{5}{8}\)

(3) Express width of the border in terms of \(y\) only
At #2 we still have a ratio. Still no counting number.
Width of the border depends on \(y.\)
But \(y\) can be any length. NO limit on values for y = no limit on border width
Bunuel 's next steps simply detail THAT there are no limits on the longer side.
He shows that every answer depends on \(y\)

Defining border width in terms of \(y\)
Recall that border width = \(\frac{y-x}{2}\)
Define everything in terms of \(y\). Use the side ratio to do so
\(\frac{x}{y}=\frac{5}{8}\)
Multiply both sides by \(y\)
\(x = \frac{5}{8}y\)
Substitute that RHS for \(x\)

Border width calculation: \(\frac{y - x}{2} =\frac{y -\frac{5}{8}y}{2}\)

\(\frac{y-\frac{5}{8}y}{2}=(\frac{\frac{8}{8}y-\frac{5}{8}y}{2})=(\frac{\frac{3}{8}y}{2})\)

\(\frac{\frac{3}{8}y}{2}=(\frac{3}{8}y*\frac{1}{2})=\frac{3}{16}y\)

Border width equals \(\frac{3}{16}\) of \(y\)

That's nice, but how long can \(y\) be? \(y\)'s length can be ANY real number.

The width of the border, dependent on \(y\)'s length,
can be ANY value that is \(\frac{3}{16}\) of an infinite set of possibilities.
Quote:
Q # 3: my approach to to tho problem was simple i rounded 39 to 40 so i simply got ratio \(\frac{25}{40} = \frac{5}{8}\) do you think its correct approach ? :? :)

Excellent idea to try to find shortcuts. :thumbup:
But . . . wrong ratio. You got lucky.
To find side length, given area of squares, you must take a square root. AND you must be solving for the right parts.
\(\frac{25}{39}\) is the ratio of \(\frac{AreaB}{AreaW}\)
\(\frac{5}{8}\) is ratio of \(\frac{SideB}{SideP}\)

Your progression seems to be: \(\frac{25}{39}= \frac{AreaB}{AreaP}=\frac{SideB}{SideW}\)
The ratio you derived is not correct: \(P\neq{W}\)
Tip: be clear about the parts of the ratios you set up. WHICH side to WHICH side, e.g.

I hope that helps. :-)

(whew!)
You get kudos for making me laugh about the SC issues in my signature. I am a reading fool (fanatic). Elegant prose uses metaphor. SC questions are hard but . . . most do not display elegant prose. BTW, the author was an astonishing human being. But did you REALLY go after my pick for translation of “Au milieu de l'hiver, j'apprenais enfin qu'il y avait en moi un été invincible”? Thou art brave. :cool:
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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New post 29 Jun 2018, 02:59
Hello Bunuel .
Could you be so kind to post similar, hard questions to practice?
Thanks a lot.
Re: A square wooden plaque has a square brass inlay in the &nbs [#permalink] 29 Jun 2018, 02:59

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