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Re: A square wooden plaque has a square brass inlay in the center, leaving [#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
so answer is E.
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Re: A square wooden plaque has a square brass inlay in the center, leaving [#permalink]
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
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Re: A square wooden plaque has a square brass inlay in the center, leaving [#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.
Answer is B.
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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.
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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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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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: A square wooden plaque has a square brass inlay in the center, leaving [#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. :-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.


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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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. :-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.


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: A square wooden plaque has a square brass inlay in the center, leaving [#permalink]
Hi Bunuel,

Thanx for the explanation.
I didn't understand one part:

We are asked which value of \frac{y-x}{2} is possible. \frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?

Please enlighten me.
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appy001 wrote:
Hi Bunuel,

Thanx for the explanation.
I didn't understand one part:

We are asked which value of \frac{y-x}{2} is possible. \frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?

Please enlighten me.


We have that \(\frac{x}{y}=\frac{5}{8}\) --> \(x=\frac{5}{8}y\)

The width of the wooden strip would be \(\frac{y-x}{2}\), substitute \(x\): \(\frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y\).

So the question is: which of the following could be the value of \(\frac{3}{16}y\)?
Answer: expression \(\frac{3}{16}y\) can take ANY value depending on \(y\).

Hope it's clear.
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Re: A square wooden plaque has a square brass inlay in the center, leaving [#permalink]
I didn't understand the statement "We are asked which value of (y-x)/2 is possible" . Can someone explain?
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mymbadreamz wrote:
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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mymbadreamz wrote:
Why is it not y-x? Why do we calculate (y-x)/2?


Consider the diagram below:
Attachment:
Wooden strip.png
Wooden strip.png [ 2.75 KiB | Viewed 141083 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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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: A square wooden plaque has a square brass inlay in the center, leaving [#permalink]
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.



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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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. :-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.



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 center, leaving [#permalink]
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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