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The ratio of the area of a square mirror to its frame is 16

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The ratio of the area of a square mirror to its frame is 16 [#permalink]

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New post 13 Apr 2013, 09:06
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The ratio of the area of a square mirror to its frame is 16 to 33. If the frame has a uniform width (c) around the mirror, which of the following could be the value, in inches, of c ?

I. 2
II. 3 1/2
III. 5

Choices
A I only
B III only
C I and II only
D I and III only
E I, II, and III
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Apr 2013, 14:08, edited 1 time in total.
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Re: Help:areas [#permalink]

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New post 13 Apr 2013, 10:08
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Area of picture = 16x Area of frame = 33x

Area of picture / (Area of picture + area of frame) = 16x / 49x = 4x/7x.

This results in c value to be a multiple of 1.5 = multiple of 3/2 and since c is not constrained to be only integer all the three choices are possible

For i) the multiple of c is 4/3
For ii) the multiple of c is 7/3
For iii) the multiple of c is 10/3

The key is c can be any real positive number... very tricky problem.. took me > 15 minutes. :-(

Answer is E

//kudos please, if this explanation is good
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Re: Help:areas [#permalink]

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New post 13 Apr 2013, 10:12
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Before gettin started let me say that you have always great questions!

\(\frac{M}{F}=\frac{16}{33}\) Refer to the image : area \(M = x^2\), area Frame= areaTot - areaMirr = \((x+2c)^2-x^2=F\)
\(\frac{M}{F}=\frac{x^2}{(x+2c)^2-x^2}=\frac{16}{33}\) turn everything upside down
\(\frac{(x+2c)^2-x^2}{x^2}=\frac{33}{16}\) is \(\frac{(x+2c)^2}{x^2}=\frac{33+16}{16}\)
we arrive at \(\frac{(x+2c)^2}{x^2}=\frac{49}{16}\)

At this point we can go on and make this equation "better looking" but to me we can stop here.
I don't see any value of c for wich this cannot be true (maybe I am wrong...) but given C (and keeping in mind that x and c can be any number and not only integers) we can find X to make this equation work.
So my answer to you is: any C "fits". E
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Re: Help:areas [#permalink]

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New post 13 Apr 2013, 10:31
nt2010 wrote:

Area of picture / (Area of picture + area of frame) = 16x / 49x = 4x/7x.



The question asks for area of frame, not frame and mirror together. So the demoninator should be ( total area - area of mirror)
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Re: Help:areas [#permalink]

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New post 15 Apr 2013, 09:26
hitman5532

The question is asking to find possible widths of the frame and not the area. Getting width from area of frame is time consuming exercise and the shortcut is to find overall area and the corresponding width to find the frame width.

Hope this is clear.

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Re: The ratio of the area of a square mirror to its frame is 16 [#permalink]

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New post 15 Apr 2013, 13:36
We get:

x^2/(4c^2+4cx) = 16/33

Cross multiply and you'll get:
33x^2-64cx-64c^2=0 which is a quadratic in x.

In order to arrive at a real value of x, the discriminant of this quadratic must be >=0, which gives us:

(64c)^2+8448>=0 ---(Equation I)

Clearly, Equation I holds true for all real values of c.

Hence, the answer is E.
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Re: The ratio of the area of a square mirror to its frame is 16 [#permalink]

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Re: The ratio of the area of a square mirror to its frame is 16 [#permalink]

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Re: The ratio of the area of a square mirror to its frame is 16 [#permalink]

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Re: The ratio of the area of a square mirror to its frame is 16   [#permalink] 20 Sep 2017, 08:43
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