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

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New post 28 May 2016, 05:20
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3
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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A framed picture is shown above. The frame, shown shaded, is 6 inches  [#permalink]

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New post 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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New post 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?
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Re: A framed picture is shown above. The frame, shown shaded, is 6 inches  [#permalink]

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New post 08 Jan 2018, 10:42
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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.

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

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New post 02 Jun 2018, 06:48
Hi chetan2u niks18 amanvermagmat gmatbusters


Quote:
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.


Quote:
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....


Quote:

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


Do I need to perform the said step by opening the brackets? Can I halt by seeing: A quadratic equation with two unknowns , clearly insufficient .

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


We know x, but have no info about another variable. Insuff

Combining both statements, I have a quadratic equation (which opens the possibility of even a negative
value as a root of equation, but a side can not be negative) and I do have an unique value of x from St 2.
Clearly suff, Let me know if my approach is valid.

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

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New post 02 Jun 2018, 08:50
1
Quote:

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

Reply: Sometimes, the terms cancel out and equation can be solved. But yes careful examination can show that it is unsolvable. But under exam pressure sometimes, one tends to miss small aspects which can lead to mistake, So it is better to open the brackets. Believe me , it will not take more than 5 secs. One need to improve the calculation speed.


Quote:
Combining both statements, I have a quadratic equation (which opens the possibility of even a negative
value as a root of equation, but a side can not be negative) and I do have an unique value of x from St 2.
Clearly suff, Let me know if my approach is valid.[/color]

Reply: Why you think that every quadratic equation gives one positive and negative root. See, even in this questions, both roots are positive.
It is better to solve the equation, Cutting corners ( for sake of saving some time) for the question one knows could be proved detrimental.


adkikani wrote:
Hi


Quote:
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.


Quote:
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....


Quote:

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


Do I need to perform the said step by opening the brackets? Can I halt by seeing: A quadratic equation with two unknowns , clearly insufficient .

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


We know x, but have no info about another variable. Insuff

Combining both statements, I have a quadratic equation (which opens the possibility of even a negative
value as a root of equation, but a side can not be negative) and I do have an unique value of x from St 2.
Clearly suff, Let me know if my approach is valid.

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Re: A framed picture is shown above. The frame, shown shaded, is 6 inches &nbs [#permalink] 02 Jun 2018, 08:50
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