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A square counter top has a square tile inlay in the center

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A square counter top has a square tile inlay in the center [#permalink] New post 22 May 2008, 16:27
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A
B
C
D
E

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57% (01:07) correct 43% (00:22) wrong based on 28 sessions
A square counter top has a square tile inlay in the center, leaving an untiled strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39,which of the following could be the width, in inches, of the strip?

I. 1
II. 3
III. 4

A I only
B II only
C I & II only
D I & III only
E. I, II & III

Open discussion of this question is here: a-square-wooden-plaque-has-a-square-brass-inlay-in-the-89215.html
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Last edited by Bunuel on 30 Jun 2012, 08:39, edited 1 time in total.
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Re: Ratio problem [#permalink] New post 22 May 2008, 17:06
B

let the whole counter top = x
the inlay = y

(y^2)/((x^2) - (y^2)) = 25/39

39*y^2 = 25*x^2 - 25*y^2

64*y^2 = 25*x^2

8y = 5x

y = 5x/8

therefore the 2 widths (1 on each side) add together is 3x/8 or 1 width = 3x/16

you can substitute any number for the width and get the x value and use the x value to get the y value.

Therefore any width is possible, because there is no constrain on the size of the actual counter top.
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Re: Ratio problem [#permalink] New post 23 May 2008, 06:29
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tekno9000 wrote:
Please solve and explain.

A square counter top has a square tile inlay in the center, leaving an untiled strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39,which of the following could be the width, in inches, of the strip?
I. 1
II. 3
III. 4

A I only
B II only
C I & II only
D I & III only
E. I,II & III


A sample value is Square of SIDE 5 and STRIP of thickness 3; total counter 8;
Now, the ratio of canter square area and strip area is 25/39.
Since, there is no constraint on the total size, this exact shape can shrink/expand maintaining its ratios and achieveing any strip thickness including 1/3/4 - Hence E.
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Re: Ratio problem [#permalink] New post 23 May 2008, 06:53
gmatnub wrote:
B

let the whole counter top = x
the inlay = y

(y^2)/((x^2) - (y^2)) = 25/39

39*y^2 = 25*x^2 - 25*y^2

64*y^2 = 25*x^2

8y = 5x

y = 5x/8

therefore the 2 widths (1 on each side) add together is 3x/8 or 1 width = 3x/16

you can substitute any number for the width and get the x value and use the x value to get the y value.

Therefore any width is possible, because there is no constrain on the size of the actual counter top.


I think it´s E.

Picking the number at this fórmula, we have:
I. 8y = 5(y+2)
8y=5y+10
y=10/3 - POSSIBLE

II. 8y = 5(y+6)
8y=5y+30
y=30/3=10 - POSSIBLE

III. 8y = 5(y+8)
8y=5y+40
y=40/3 - POSSIBLE
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Re: Ratio problem [#permalink] New post 07 Feb 2009, 02:33
neither am i..
please explain and elaborate

thank you.
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Re: Ratio problem [#permalink] New post 07 Feb 2009, 22:17
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LoyalWater wrote:
neither am i..
please explain and elaborate

thank you.



Area Ratio = 25/39

Let say
outer square area = 25+39 = 64 = 8*8
inner square area (tile area) = 25=5*5

strip length = 8-5= 3

So Outer Square side length: tile lenght: strip = 8:5:3
Question is asking is 1,3,4 strip lengths are possible..? answer is yes.


Outer Square side length: tile lenght: strip = 8:5:3 (here strip length 3)
Outer Square side length: tile lenght: strip = 8/3:5/3:1 (here strip length 1)
Outer Square side length: tile lenght: strip = 32/3:20/3:4 (here strip length 1)

Answer is E.

Here tricky part is ... Question gave the ratio not actual areas.
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Re: Ratio problem [#permalink] New post 30 Jun 2012, 08:11
will vote for : E (question is could be, so can consider approx near by values)


tiled square area = 25x = side = 5 sqrt x
total area = 64x = side = 8 sqrt x

now strip = 8 sqrt x - 5 sqrt x

when x = 1
strip = 3

when x = 2 (sqrt 2 = 1.4)
strip = (approx) 3 (1.4) = 4.2

when x = 3 (sqrt 3 = 1.7)
strip = (approx) 3 (1.7) = 5.1

when x = 1/9
strip = 3 (1/3) = 1
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Re: Ratio problem [#permalink] New post 30 Jun 2012, 08:37
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Re: Ratio problem   [#permalink] 30 Jun 2012, 08:37
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