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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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Re: hard problem OG Quant 2nd edition  [#permalink]

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

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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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?

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Re: hard problem OG Quant 2nd edition  [#permalink]

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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 l-x
=>x = 5l/8
so l -x = l = 5l/8 = 3l/8

Now 3l/8 could be any value depending on the value of l
##### General Discussion
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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.

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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Why is it not y-x? Why do we calculate (y-x)/2?

Consider the diagram below:
Attachment: Wooden strip.png [ 2.75 KiB | Viewed 69270 times ]
As you can see the width of the wooden strip (the width of grey strip) is $$\frac{y-x}{2}$$.
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Re: hard problem OG Quant 2nd edition  [#permalink]

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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  [#permalink]

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

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.

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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This tells us that there is a particular ratio for the width of the wooden frame to the width of the brass inlay.

This means there is a f(x) = $$(\frac{A}{B})(x)$$. Any x is possible and will yield a corresponding f(x).
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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. 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 At this point, if we shrink or expand the plaque . . . . . . 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.

Cheers,
Brent
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Originally posted by GMATPrepNow on 15 Sep 2017, 11:19.
Last edited by GMATPrepNow on 16 Apr 2018, 11:34, edited 1 time in total.
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Re: hard problem OG Quant 2nd edition  [#permalink]

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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.
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Re: hard problem OG Quant 2nd edition  [#permalink]

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

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.

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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I didn't understand the statement "We are asked which value of (y-x)/2 is possible" . Can someone explain?

Question asks about the possible width, in inches, of the wooden strip.

Let the the side of small square be $$x$$ and the big square $$y$$, then the width of the wooden strip would be $$\frac{y-x}{2}$$, which means that we are asked to determine the possible values of this exact expression.

Hope it's clear.
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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

The only restriction on the value of b(side of the brass square) is that b<w. From the given problem, we know that$$\frac{b^2}{(w^2-b^2)}$$= 25/39--> 39/25 = $$(\frac{w}{b})^2-1$$-->
$$\frac{w}{b}$$= $$\sqrt{64/25}$$--> b = $$\frac{5}{8}*w$$. Thus, any value of w, would still give a value where b<w.
E.
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Bunuel 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 $$\frac{y-x}{2}$$.

OMG !!! It took me ages to understand of what you said.... Is not too obvious at all.

For those that they are on the same boat with me it boils down to y-x / 2 BECAUSE...

Say d is the length of the of the wooden strip.... where Bunuel has the DOUBLE arrows. Then you know that Y = x + d (the left) + d (the right) so you end up having:

y = x + 2d hence d = (y - x) / 2
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hanyhamdani wrote:
Hi, I did it another way, but in my solution none of the given value satisfies.

Take wooden (bigger area) =$$a^2$$
let x be the width.
so brass area = $$(a-2x)^2$$
so,

$$(25/39)=( (a-2x)^2 )/ (a^2)$$

1. x=1
$$(a-2) = 5$$
so, a=7
and $$a^2 = 49$$

giving ratio : $$25/49$$ which is not as given one.

trying all these, give different ratios.

what did i do wrong?

Just because $$(25/39)=( (a-2x)^2 )/ (a^2)$$, doesn't necessarily mean that $$25=(a-2x)^2$$.

That's a general truth about ratios. If x/y = 12/19, x isn't necessarily 12. For instance, x could be 24 and y could be 38.
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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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. 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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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. 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. 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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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. 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
"    OMG you are irrepressible.

Hi dave13 -
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
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 [ 8.24 KiB | Viewed 1299 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 ? 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. 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. _________________
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Intern  Joined: 18 Jul 2009
Posts: 26
Re: hard problem OG Quant 2nd edition  [#permalink]

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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.  Re: hard problem OG Quant 2nd edition   [#permalink] 15 Jan 2010, 04:03

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# A square wooden plaque has a square brass inlay in the  