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# A framed picture is shown above. The frame, shown shaded, is 6 inches

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A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink]

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28 May 2016, 04:44
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A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture?

(1) The area of the shaded region is 24 square inches.
(2) The frame is 8 inches tall.
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Aug 2009
Posts: 5777
Re: A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink]

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28 May 2016, 05:20
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nalinnair wrote:
Attachment:
Image.png

A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture?

(1) The area of the shaded region is 24 square inches.
(2) The frame is 8 inches tall.

hi,

let the length be x, so dimensions of bigger rectangle = 6 and x...
Also let the width around be y, so dimensions are 6-2y and x-2y....

For the Q asked, we require value of x and y....

lets see the statements-

(1) The area of the shaded region is 24 square inches.
Area of shaded region = $$6*x - (6-2y)*(x-2y) = 6x - 6x+12y+2xy-4y^2 = 12y+2xy-4y^2 = 24$$
We can't solve for two variables through one equation...
Insuff

(2) The frame is 8 inches tall.
we know the dimensions of bigger frame...
nothing about the inner picture
Insuff

Combined-
12y+2xy-4y^2 = 24.......... x=8
so $$12y+16y-4y^2 = 24......................4y^2-28y+24 = 0...................y^2-7y+6 = 0.........(y-1)(y-6) = 0..$$
either y=1 or y=6.......
But if y=6, sides will become -ive, 6-2y = 6-2*6 = -6, so y=1.....

sides are 6-1*2 = 4 and 8-2*1=6.......
Suff

C
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Joined: 09 Aug 2016
Posts: 68
A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink]

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28 Jan 2017, 10:21
For x = height of the shaded region, the area of the shaded region can be expressed as:

24 = 6x - (Area of viewable portion V) => 24 = 6x - V

we have two unknowns so (1) and (2) they give one of the two variables x, V

Combining both we get the right answer hence C
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Re: A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink]

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15 Dec 2017, 06:13
Any chance to resolve it without going into equations? If the smaller quadrilateral with the area of 24 is inside of the larger quadrilateral with the area of 48 and the border is uniform that these two quadrilaterals should have the same proportion? or something like that? the pair of 6 and 4 immediately comes to mind... Is it a good tactic here just to realize that using facts A and B must give us a solution and move to the next problem?
Intern
Joined: 11 Sep 2017
Posts: 20
Re: A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink]

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08 Jan 2018, 10:42
Erjan_S wrote:
Any chance to resolve it without going into equations? If the smaller quadrilateral with the area of 24 is inside of the larger quadrilateral with the area of 48 and the border is uniform that these two quadrilaterals should have the same proportion? or something like that? the pair of 6 and 4 immediately comes to mind... Is it a good tactic here just to realize that using facts A and B must give us a solution and move to the next problem?

Hi Erjan,

I found this explanation helpful- http://www.gmatquantum.com/gmat-quantit ... view-2017/

Check it out as it might help you too.

Aiena.
Re: A framed picture is shown above. The frame, shown shaded, is 6 inches   [#permalink] 08 Jan 2018, 10:42
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# A framed picture is shown above. The frame, shown shaded, is 6 inches

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