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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]  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
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Re: A border of uniform width is placed around a rectangular [#permalink]  20 Dec 2012, 04:42
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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

Consider the diagram below:
Attachment:

photograph.png [ 5.15 KiB | Viewed 15577 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 $$(4x^2+36x+80)-80=144$$ --> $$4x^2+36x-144=0$$ --> $$x^2+9x-36=0$$ --> $$(x-3)(x+12)=0$$ --> $$x=3$$.

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Re: A border of uniform width is placed around a rectangular [#permalink]  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]  18 Oct 2013, 12:33
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Expert's post
runningguy wrote:
Why does this not work....

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

Divide above by 2 to get...

(x+5)(x+4)=112

If you divide by 2 you get: (x+5)(2x+8)=112.
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A border of uniform width is placed around a rectangular [#permalink]  26 Aug 2014, 21:29
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Backsolving:
we are looking for area 144+80=224
C. 22*20=440, too much
B. 18*16=288, too much

must be A

Last edited by Temurkhon on 12 Mar 2015, 19:42, edited 1 time in total.
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Re: A border of uniform width is placed around a rectangular [#permalink]  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]  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]  12 Feb 2014, 00:21
Expert's post
kedusei wrote:
why are we adding both sides by 2x

Check the diagram:
Attachment:

Untitled.png [ 4.15 KiB | Viewed 12012 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]  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

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

Attachment:

photograph.png [ 3.89 KiB | Viewed 8788 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

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

So Width of border = (W2-W1) = 11-8 = 3

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