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# A rectangular photograph is surrounded by a border that is 1

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A rectangular photograph is surrounded by a border that is 1 [#permalink]

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09 Nov 2008, 19:19
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A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

A. 34
B. 36
C. 38
D. 40
E. 42
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Oct 2013, 02:33, edited 1 time in total.
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09 Nov 2008, 19:46
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D.

First, we know that M is the area of the picture AND the border, and that the border is 1 in wide. If you have a border around a picture that is 1" wide, it doesn't add just 1" to the width and height, but actually 2" to the width and 2" to the height because you have an extra inch on each side for the border.

So, if we want to let x = width of the photograph (no border) and y = height of the photograph (no border), then we know that M = (border width + photograph width) * (border + photograph height).

A border adds twice its width to the width of the photograph. A 4" picture will have a 1" border on the right and left, so it adds 2", for a new width of 6". (Hypothetical example). For a 2" wide border, the added dimensions will be twice the width of the border, or 4" (2" for each left and right side).

(x+4)(y+4) = M + 52

But we know that M is the area of the photograph plus a 1" border. This is represented alegbraically like:

(x+2)(y+2) = M

So substitute in:

(x+4)(y+4) = (x+2)(y+2) + 52, Now lets reduce as far as we can. First do FOIL.

xy + 4x + 4y + 16 = xy + 2x +2y +4 +52

The xy on each side cancels each other out.

4x + 4y + 16 = 2x + 2y + 56
subtract the following values from each side: 2x, 2y and 16

2x + 2y = 40.

We can reduce further if you want to, but it isn't necessary.

If you have a rectangle with height (x) and width (y) and you want to find the perimeter, then you can do x + x + y + y = perimter. Also known as 2x + 2y = Perimeter.

This shows us that the perimter should be 40.

petercao wrote:
A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

A. 34
B. 36
C. 38
D. 40
E. 42

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09 Nov 2008, 19:54
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petercao wrote:
A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

A. 34
B. 36
C. 38
D. 40
E. 42

width = x, height = y,
2x+2y=?

(x+4)(y+4)-(x+2)(y+2)=M+52-M
2x+2y=40

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17 Aug 2010, 20:28
Sion wrote:
petercao wrote:
A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

A. 34
B. 36
C. 38
D. 40
E. 42

width = x, height = y,
2x+2y=?

(x+4)(y+4)-(x+2)(y+2)=M+52-M
2x+2y=40

D

Agreed.

Outer rectangle area when width is two inches on either side - Outer rectangle area when the width is 1 inch
= M + 52 - M.

2 (l+b) = 40 units.
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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]

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29 Oct 2013, 02:27
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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]

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29 Oct 2013, 02:43
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petercao wrote:
A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

A. 34
B. 36
C. 38
D. 40
E. 42

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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]

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29 Oct 2014, 03:38
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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]

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25 Jan 2015, 03:08
I did not understand how 1 inch border is counted as 2. Can somebody please explain?
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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]

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25 Jan 2015, 03:12
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Salvetor wrote:
I did not understand how 1 inch border is counted as 2. Can somebody please explain?

1 inch from right and left sides of the photograph give 2 inches. Similarly 1 inch from top and bottom of the photograph give 2 inches.
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A rectangular photograph is surrounded by a border that is 1 [#permalink]

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25 Jan 2015, 05:40
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Here we go----

Let Length and Breadth of photograph be L and B respectively.
Perimeter is given by 2*(L+b) -----(1)

According to the question:

(L+2)(B+2) = m ----(2)

and

(L+4)(B+4) = m + 52 ---------> (L+4)(B+4) - 52 = m ------(3)

Equating (2) and (3)

(L+2)(B+2) = (L+4)(B+4) - 52

LB + 2L + 2B + 4 = LB + 4L + 4B + 16 -52

Simplify
2L + 2B = 40 ------> 2(L+B) = 40 (Check eq (1))

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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]

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25 Jan 2015, 23:57
x*y=M

(x+2)*(y+2)=M+52, because every 1 inch in side gives additional 2 inches in lenght

2x+2y+4=52

2(x+y)=48, it is perimeter with border

48-8=40, where 8 is total lenght of border sides

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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]

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26 Jul 2015, 09:12
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File comment: Here is a visual that might help. Notice how I'm plugging in lengths and widths to make the prompt true: usually we plug in constants, but it's a good example of how you can even plug in variables.

Screen Shot 2015-07-26 at 10.09.22 AM.png [ 167.72 KiB | Viewed 12439 times ]

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01 Aug 2016, 08:17
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