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A border of uniform width is placed around a rectangular

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A border of uniform width is placed around a rectangular [#permalink] New post 20 Dec 2012, 04:33
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A border of uniform width is placed around a rectangular photograph that measures 8 inches by 10 inches. If the area of the border is 144 square inches, what is the width of the border, in inches?

(A) 3
(B) 4
(C) 6
(D) 8
(E) 9
[Reveal] Spoiler: OA
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 20 Dec 2012, 04:42
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Walkabout wrote:
A border of uniform width is placed around a rectangular photograph that measures 8 inches by 10 inches. If the area of the border is 144 square inches, what is the width of the border, in inches?

(A) 3
(B) 4
(C) 6
(D) 8
(E) 9


Consider the diagram below:
Attachment:
photograph.png
photograph.png [ 5.15 KiB | Viewed 7355 times ]
The area of just the photograph is 8*10=80 square inches.

The area of the photograph with the border is (10+2x)(8+2x)=4x^2+36x+80 square inches.

The difference is 144 square inches, thus (4x^2+36x+80)-80=144 --> 4x^2+36x-144=0 --> x^2+9x-36=0 --> (x-3)(x+12)=0 --> x=3.

Answer: A.
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 25 May 2013, 04:42
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Let the width of the uniform border be X

Area of photo 80 (given)

Area of Photo + border = (10+2X)*(8+2X) (deduced)

Difference in areas, (10+2X)*(8+2X) - 80 = 144 (given)

(10+2X)*(8+2X) =224
(10+2X)*(8+2X) =16*14
(10+2X)*(8+2X) =(10+6)(8+6)

=> 2X= 6
=>X=3
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 18 Oct 2013, 12:33
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 18 Oct 2013, 12:24
Why does this not work....

(2x+10)(2x+8)=224

Divide above by 2 to get...

(x+5)(x+4)=112
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 11 Feb 2014, 13:54
why are we adding both sides by 2x
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 12 Feb 2014, 00:21
Expert's post
kedusei wrote:
why are we adding both sides by 2x


Check the diagram:
Attachment:
Untitled.png
Untitled.png [ 4.15 KiB | Viewed 3804 times ]
The length of the longer side is 10 + x + x = 10 + 2x and the length of the shorter side is 8 + x + x = 8 + 2x.

Hope it's clear.
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 25 Aug 2014, 21:19
Once we get the idea that the width (w) is to be added twice to 10 and 8 we can find by substitution
This how I did the problem
Area of frame = 80 Area of photograph = 144
Total Area = 224

Now we know the new area will be (8+2w)*(10+2w)
Substitute option A for w
(8+6)*(10+6) = 14*16 = 224

Answer A
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 26 Aug 2014, 21:29
Backsolving:
we are looking for area 144+80=224
C. 22*20=440, to much
B. 18*16=288, to much

must be A
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Re: A border of uniform width is placed around a rectangular [#permalink] New post 27 Aug 2014, 01:27
Refer diagram below:

Attachment:
photograph.png
photograph.png [ 3.89 KiB | Viewed 592 times ]

Area of multi-coloured shaded region is given = 144

Area of each square (Yellow) = x^2 ; No. of squares = 4

Area of rectangle (pink) = 10x ; No. of rectangle (pink) = 2

Area of rectangle (blue) = 8x ; No. of rectangle (pink) = 2

Setting up the equation

4x^2 + 20x + 16x = 144

x^2 + 9x - 36 = 0

x = 3

Answer = A
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Re: A border of uniform width is placed around a rectangular   [#permalink] 27 Aug 2014, 01:27
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