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A border of uniform width is placed around a rectangular [#permalink]
20 Dec 2012, 04:33

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Difficulty:

85% (hard)

Question Stats:

58% (03:24) correct
42% (02:18) wrong based on 673 sessions

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?

Re: A border of uniform width is placed around a rectangular [#permalink]
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 [ 5.15 KiB | Viewed 17605 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 \((4x^2+36x+80)-80=144\) --> \(4x^2+36x-144=0\) --> \(x^2+9x-36=0\) --> \((x-3)(x+12)=0\) --> \(x=3\).

Re: A border of uniform width is placed around a rectangular [#permalink]
02 Jan 2015, 17:03

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Length of Picture L1 = 10 Width of Picture W1 = 8

Area of Picture = L1 X W1 = 80

Total Area including frame = L2 X W2 = 144

So L2 and W2 should be multiples/factors of 144.

Prime factorization of 144 = 2 X 7 X 11

Only multiples that make any sense are 14 and 11. (2 x 77 is impossible and so is 22 x 7; total length/width can't be less than length/width of the photograph).

So length (L2) should be 14 and Width (W2) should be 11.

Re: A border of uniform width is placed around a rectangular [#permalink]
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

Re: A border of uniform width is placed around a rectangular [#permalink]
24 Apr 2015, 11:40

I didn't find this question difficult, however I still managed to make a mistake. I basically thought that the outer size of the board matches with the edges of the picture. This would give the following equation: 144= 2(8x) +2x(10-2x). However , the inner sides of the board match with the picture. I think that question did not state it clear enough.

A border of uniform width is placed around a rectangular [#permalink]
18 May 2015, 02:22

There is actually quite a nifty little way of doing this. For those who struggle with the whole quadratic solution.

You can do the following:

Since a uniform width is placed around it. We need to find a number where A remains 2 bigger than B. (Since just as much is added to the length as to the width. A will always be 2 more than B)

Since area is 224. We need to find a number for A that is 2 greater than B. You might ask yourself well how am I supposed to do that?

If you can realize that 224 is 1 less than a perfect square. You can use the following: \((x+y)(x-y)\), which in this case is \((15+1)(15-1)\). So we got a = 16 and b = 14.

Now, since the inner border is 8. and we add 2x we got 8 + 2x = 14. x = 3.

This might seem like a weird approach, but if you can quickly realize the relation here, you can solve it in less than a minute

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