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Re: A border of uniform width is placed around a rectangular pho [#permalink]
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bellatrix wrote:
The quickest way to solve this problem is to plug in the options

area of photograph (inner rectangle) = 8*10 = 80
area of the border (outer - inner) = 144

area of border = area of outer rectangle - area of inner rectangle

plugging in option A as possible width of the border, the dimensions of the outer rectangle become, 8+3+3= 14 and 10+3+3=16 (equal width on either side)

area of outer rectangle = 14*16=224
area of border = 224-80=144 which satisfies the question

Answer is A.

I think algebra is equally if not more faster than plugging

dimensions of the photo are 8inch by 10inch , so area 80 sq in

let the border width be x

so outer area = ( 10 + 2x) * (8 + 2x)

given that area of border = 144 = outer area - inner area = ( 10 + 2x) * (8 + 2x) - 80

144= 80 + 36x + 4 x^2 -80

4x^2 + 36x - 144 = 0

x^2 + 9x -36 = 0

(x+12)(x-3)=0

x = -12 or 3

width cant be negetive so answer is 3 (option A)
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Re: A border of uniform width is placed around a rectangular pho [#permalink]
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This is how I did it

Side note, doing Quadratic formula on GRE calculator sucks, I don't recommend it lol
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Re: A border of uniform width is placed around a rectangular pho [#permalink]
Hello from the GRE Prep Club BumpBot!

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