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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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01 Dec 2012, 20:18
2
2
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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Re: hard problem OG Quant 2nd edition  [#permalink]

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31 May 2013, 07:31
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.

Again,
This squestion is ambiguous.
Why cant anyone not interpret it as:
b^2/w^2 = 25/39
where b- is the side length for brass inlay,
and W - length of wooden box
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Re: hard problem OG Quant 2nd edition  [#permalink]

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31 May 2013, 08:23
cumulonimbus 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.

Again,
This squestion is ambiguous.
Why cant anyone not interpret it as:
b^2/w^2 = 25/39
where b- is the side length for brass inlay,
and W - length of wooden box

Not ambiguous at all (notice that the question is from OG).

The question says that "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 ...", so it should be b^2/(w^2-b^2)=25/39.

Consider the diagram below:

Only grey area is wooden, the remaining (white area) is brass.

Hope it's clear.
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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01 Jun 2013, 00:31
1
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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Re: hard problem OG Quant 2nd edition  [#permalink]

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02 Oct 2013, 13:48
1
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.

the ratio is 5/8, Lets say the number be 5n and 8n

So what we need is difference divided by 2 i.e.

8n-5n/2 = 3n/2 = 1.5n

Now you said it can take any value, so my question is n is an integer or n is an infinite set of Natural number?

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

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03 Oct 2013, 00:24
honchos 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.

the ratio is 5/8, Lets say the number be 5n and 8n

So what we need is difference divided by 2 i.e.

8n-5n/2 = 3n/2 = 1.5n

Now you said it can take any value, so my question is n is an integer or n is an infinite set of Natural number?

We are not told that the sides are integers, thus n may or may not be an integer.
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Re: hard problem OG Quant 2nd edition  [#permalink]

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03 Oct 2013, 00:47
N is absolutely infinite that means, infinite solutions. Make sense. Thanks for the discussion.

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

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19 Mar 2014, 20:06
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

Since the question indicates only ratio of areas, therefore any length is possible. Hence E is the answer.
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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23 Jan 2015, 19:32
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: $$\frac{x^2}{y^2-x^2}=\frac{25}{39}$$ --> $$\frac{x^2}{y^2}=\frac{25}{64}$$ --> $$\frac{x}{y}=\frac{5}{8}$$.

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 think we can use slightly different approach.

There is a square inside a square with a uniform border of thickness. We know that the larger side lets say S = s +2t, where t is the thickness of the border.

We are given that s^2 / (s+2t)^2 -s^2 = 25/39

Then on solving we get
s^2 / (s^2 + 4t^2+ 4ts - s^2) = 25/39

therefore: s^2/ ( 4t +4ts) = 25/39
on solving we get
25 x 4 x t ( 1+s) = 39s^2.

therefore : 39 s^2 - 100t - 100ts = 0

this is a quadratic of the form ax^2 + bx + c= 0 & we have a = positive, b= -100t & c=-100t

SO here he discriminant Sqrt ( b^2 - 4ac) will always be positive as the sign of c is neg. & we will have two value of the solution. Hence any value of t will satisfy the equation as long as it is positive length thickness.
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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23 Jan 2015, 23:21
I made a mistake- I thought "brass has to be a multiple of 5 and wood has to be a multiple of 1.5 ( since 5x is the brass portion and 3x is the wooden portion in the 8x whole, that i calculated from 25+39=64's root, AND 3x is splitted into 2 as width is what is asked and not the wooden portion)"
I saw Bunuel's first answer and totally agreed with it (I was amazed by it!!!). But could someone tell me what I assumed wrong.
Thanks
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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30 May 2015, 00:40
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?

Alternate approach

Let the uniform width be = x

=> (5+2x)²/25 = 39/25

4x²+25+10x =39
(4x+14) (x-1) = 0
x= 1 or
x =-14/4 (length cannot be negative)

=> Since we were dealing with ratios this means x can be a any positive multiple (as length cannot be negative) of 1

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

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05 Jul 2015, 04:35
I read the question and interpreted "wooden area" as the whole plaque. Bad move...but you do get some funky equations out of it...

Ah, the benefit of hindsight...
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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29 May 2016, 02:46
Hello Bunuel

I was not able to understand the part where you mentioned Area of the brass inlay (small square with side x) /area of the wooden area( large square with side y) =25/64 (39+25). Why did we add these two ?

Ratio of the same has been clearly mentioned in the question as :- If the ratio of the brass area (small square) to the the wooden area(larger square) is 25 to 39 , then how come we added the two numbers (25+39) to depict the area of the larger square ?

Please explain , I know it is a silly question , but i am not able to get the facts straight .

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

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30 May 2016, 11:42
Ujjwalkathpalia wrote:
Hello Bunuel

I was not able to understand the part where you mentioned Area of the brass inlay (small square with side x) /area of the wooden area( large square with side y) =25/64 (39+25). Why did we add these two ?

Ratio of the same has been clearly mentioned in the question as :- If the ratio of the brass area (small square) to the the wooden area(larger square) is 25 to 39 , then how come we added the two numbers (25+39) to depict the area of the larger square ?

Please explain , I know it is a silly question , but i am not able to get the facts straight .

Thanks

Please click Quote under the post you are referring in order for me to know which post you are talking abut. Thank you.
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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11 Aug 2016, 19:54
Also, am I wrong in considering 25/39 as the ratio of the brass area to the whole square and not just the border? A plaque with brass inlay is simply a layer of brass above the wooden part. Should they not have mentioned clearly that 25/39 is the ratio of the brass area to the wooden border or something like exposed wooden area or something?

If this were the case, we would have a constraint that the inlay area should not be more than the area of the plaque itself. I started out with this equation and the whole arithmetic turned out very clumsy since none were nice numbers.

May be all three numbers are still possible and the answer remains unchanged, but I am just trying to understand how is this not considered an ambiguity in the question.
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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27 Oct 2016, 09:01
1
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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A square wooden plaque has a square brass inlay in the  [#permalink]

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16 Dec 2016, 11:08
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?
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Re: A square wooden plaque has a square brass inlay in the  [#permalink]

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16 Dec 2016, 11:20
1
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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A square wooden plaque has a square brass inlay in the  [#permalink]

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17 Dec 2016, 19:36
Solution attached.

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

5.jpg [ 212.62 KiB | Viewed 2376 times ]

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

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09 May 2017, 23:41
any measure is possible because
area (wood) / area (brass) = 25/39
=> width (wood) / width (brass) = 5/ root sqrt(39) which is smaller than 1
then if width wood increases, then width of brass increases accordingly such as wood always lies in center of brass
Re: A square wooden plaque has a square brass inlay in the &nbs [#permalink] 09 May 2017, 23:41

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