A square wooden plaque has a square brass inlay in the center, leaving

Solution attached.

Edit1: After reading Bunuel 's one line solution, I feel really stupid, having done so much calculation

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.

Answer: E

Cheers,
Brent

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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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 "

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 \(W = y^2-x^2\)

Short answer: because there are TWO border widths between \(y\) and \(x\). TAKE A LOOK:

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 \(\frac{y-x}{2}=\frac{4-2}{2}=1\)

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.

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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My explanation would be just like Brent's - we're only given a ratio, so we can make the width anything we like just by making the diagram bigger or smaller.

If that's not convincing, imagine that it's possible the width is 1 inch. Then it must also be possible that the width is 2.54 units, because 1 inch is 2.54 centimeters. But if that's true, then the width can be anything at all, just by changing units. But I prefer the first way of thinking about the problem.

Genoa2000
We cross paths again
Ratios are so open that we deal with them in every aspect of real world - finance, food, chemistry and even flags of countries and logos of companies .

Basically just remember that whenever we have been given a ratio and asked about peculiar value then we can play with the values. This implies that there can be n number of solutions to the value we are looking out.
OR
if we are asked about any ratio we can again play with the values to obtain the ratio within the given condition. So, n number of possibilities exists.

This is what Ian probably meant.

Take a look at this question. https://gmatclub.com/forum/in-the-figur ... 21855.html
It's a great question to understand how open ratios are and how values can be assumed to reach a solution.



In the question we are dealing, if x is the side of brass square and y is the side of big square including the wooden strip, then \(\frac{x^2}{y^2 - x^2} = \frac{25}{39}\). Note y = x + 2w.
\(\frac{y^2}{x^2} - \frac{x^2}{x^2} = \frac{25}{39}\)
\(\frac{y^2}{x^2} = \frac{25}{39} + 1 = \frac{64}{25}\)
\(\frac{(x+2w)^2}{x^2} = \frac{64}{25}\)

Now if we put w = 1 or w = 3 or w = 4, we get x's values accordingly because the ratio is maintained. We need not to think about the solutions of quadratic equation because x > 0, y > 0 and w >0. Also, we have been asked only about values of width of wooden strip as 1, 3 and 4.

Hence every value is possible. That is why Bunuel said 'Why would ANY width of the strip be impossible?".
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I didn't see that earlier question until now. I meant something much more precise. Suppose the question didn't mention any unit at all (and there's no particular reason the question needs to ask about "inches" -- it could just as easily ask about "kilometers" at the end, and every correct solution posted so far would still be valid). If it turned out the width could be 1, then we could certainly make the problem into a word problem where the width was 1 yard. But if the width can be 1 yard, it can also be 3 feet, because 1 yard and 3 feet are the same thing. But then it could also be 36 inches, because 36 inches and 3 feet are the same thing. So if the width can be 1, it can also be 3, or 36, or anything else.

You can't normally use that logic in a Geometry problem, because ordinarily you're given a fixed length (or area, or other quantity), and if you change your units from, say, yards to feet, you'd be changing one of the quantities provided in the question. But here, we're not given any fixed quantities at all. So if any solution is possible, every solution is possible. And since from the answer choices, we can see that one solution is possible, there's no reason to do any work here. The answer must be I, II and III.

Hi Bunuel, This is how I approached it, would it be correct?

First, I determined the area of the brass inlay versus the area of the ENTIRE figure.

25x : 25x + 39x = 64x

Then,

This means the side of the brass inlay is 5x and 8x for the whole figure. Going through the choices:

i) 1

1 + 1 + 5x = 8x --> 2 = 3x --> x = 1.5

ii) 3

3 + 3 + 5x = 8x --> x = 2

iii) 4

4 + 4 + 5x = 8x --> x = 3/8

So the answer is E ...and I am guessing that non-integer values are acceptable.

So first, is my approach correct? Secondly, if it is, then why does this very similar question not accept non-integer values? https://gmatclub.com/forum/a-square-cou ... 76050.html
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It is certainly correct to solve a question like this by calling one length x, and expressing everything in terms of x, but the question just gives us a ratio. Ratios never tell you anything about your actual numbers, except when numbers must be integers (if the ratio of apples to oranges is 5 to 2, then the number of apples must be a multiple of 5, and the number of oranges must be a multiple of 2). Lengths don't need to be integers, so in these two questions, the lengths can be anything at all. If the length can be 1, we'd just scale the diagram to get a length of 3, say (just like blowing up a photograph).

You say "this very similar question [does] not accept non-integer values", but there isn't any issue with non-integer values in that question. I read your solution there, and you seem to have concluded that the three answers were acceptable because they were integers, but that's not why those three values work. They work because literally any positive values work, integers or not.

As has been noted in earlier posts, no math is needed here.
Just use common sense.
Why couldn't the width of the strip be ANY value?
We could choose a width for the strip -- ANY width we want -- then shrink or expand the inlay until the ratio of inlay area:strip area = 25:39.
Since any width is possible, the correct answer is E.


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Like most geometry questions, this isn't about whether you can do math; it's about whether you can just reason your way to the solution.

Once we have the drawing of one square inside the other, we can scale the entire thing as needed to fit any dimension imaginable (I've shown four examples, but you can imagine a million others of all sorts of size).

Answer choice E.

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?

let \(i\) be the internal square side, let \(w\) be the wood strip width

set up \(\frac{i^2}{4w^2+2iw}=\frac{25}{29}\)
reduce to a second degree equation \(39i^2-50iw-100w^2=0\)
look at the coefficient, all the choice for w are >0, then, for the cartesio rule (change in coefficient sign between two consecutive term ax^2,bx^1,cx^0), there must exist a positive solution.
Descartes' rule of signs

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

Thank you! That's a perfect solution and a very clear explanation
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An alternative approach:

If the side of the outer square is ‘a’ and the width between the two squares is ‘x’, we are given that
\(\frac{(a+2x)^2}{a^2}\) =\(\frac{7^2}{5^2}\)
This gives a = 5x.

Now, a can take any value as per x.

Bunuel
Why do you divide by 2 for (y−x)/2 but not for x^2/y^2−x^2? Don't you need to be consistent?

Let me remind you what these expressions represent:



(y−x)/2 is the width, in inches, of the wooden strip.
x^2/(y^2−x^2) is the ratio of the brass area to the wooden area.

So, dividing x^2/(y^2−x^2) by 2 does not make any sense.
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Hey Bunuel, I am stuck with this Q.
I tried another way, so I assumed the side of Brass Square as X and then the width as Y, making the outer Square side to be X+2y.
However, on solving this the Eqns become pretty ugly and does not get so easily solved.

How do I not fall into such traps? Any tips please. I have the exam next week.