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

D
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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]
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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 126704 times ]

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

We are given that a rectangular photograph is surrounded by a border that is 1 inch wide on each side and that the total area of the photograph and border is M square inches. If we let L = the length of the photograph and W = the width of the photograph, since the border surrounds the length and width on two sides, the length of the photograph and border is L + 2 and the width of the photograph and border is W + 2.

Let’s represent this in a diagram:

We can now represent the area of the border and photograph:

area = length x width

M = (L + 2)(W + 2)

M = WL + 2W + 2L + 4

We are also given that if the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. Thus, the new length of the border and photograph would be L + 4 and the new width of the border and photograph would be W + 4.

We can again represent this in a diagram:

The new area of the border and photograph is:

M + 52 = (L + 4)(W + 4)

M + 52 = WL + 4W + 4L + 16

M = WL + 4W + 4L – 36

We have two equations for M. Let’s equate them and simplify:

WL + 2W + 2L + 4 = WL + 4W + 4L – 36

2W + 2L + 4 = 4W + 4L – 36

2W + 2L = 4W + 4L – 40

2W + 2L = 40

Since perimeter = 2L + 2W, the perimeter of the photograph is 40 inches.

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A rectangular photograph is surrounded by a border that is 1 [#permalink]
LB=M
(L+2)(B+2) = M+52
2(L+B) = 48
Perimeter of pic = 2x (L-2+B-2) = 2(L+B)-8 = 48-8 =40

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

You can get a quick solution geometrically...
The area of just outer boundary is M+52-M=52..
Draw a line (red) to divide this boundary into 4 parts..
1) 2 parts are rectangle with sides (X+4) and 1
So area = $$(X+4)*1=X+4$$... Area of both parts = $$2(X+4)=2x+8$$
2) 2 parts are rectangle with sides (y+2) and 1
So area = $$(y+2)*1=y+2$$......area of both parts =$$2(y+2)=2y+4$$

Total area =$$52=2x+8+2y+4=2x+2y=12.$$..
So $$2x+2y=52-12=40........2(X+y)=40$$
2(X+y) is nothing but the perimeter of photograph.
Hence Ans 40
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Re: A rectangular photograph is surrounded by a border that is 1 [#permalink]
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petercao
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

Please find attached the explanation for the question.
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WhatsApp Image 2020-07-13 at 1.50.13 PM.jpeg [ 147.25 KiB | Viewed 49338 times ]

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

We are given that a rectangular photograph is surrounded by a border that is 1 inch wide on each side and that the total area of the photograph and border is M square inches. If we let L = the length of the photograph and W = the width of the photograph, since the border surrounds the length and width on two sides, the length of the photograph and border is L + 2 and the width of the photograph and border is W + 2.

Let’s represent this in a diagram:

We can now represent the area of the border and photograph:

area = length x width

M = (L + 2)(W + 2)

M = WL + 2W + 2L + 4

We are also given that if the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. Thus, the new length of the border and photograph would be L + 4 and the new width of the border and photograph would be W + 4.

We can again represent this in a diagram:

The new area of the border and photograph is:

M + 52 = (L + 4)(W + 4)

M + 52 = WL + 4W + 4L + 16

M = WL + 4W + 4L – 36

We have two equations for M. Let’s equate them and simplify:

WL + 2W + 2L + 4 = WL + 4W + 4L – 36

2W + 2L + 4 = 4W + 4L – 36

2W + 2L = 4W + 4L – 40

2W + 2L = 40

Since perimeter = 2L + 2W, the perimeter of the photograph is 40 inches.

ScottTargetTestPrep

I am confused by the following assumption "If we let L = the length of the photograph and W = the width of the photograph, since the border surrounds the length and width on two sides, the length of the photograph and border is L + 2 and the width of the photograph and border is W + 2." I have seen frames that are thicker in the width than the length and vice versa. To solve this problem, though it seems you need to assume the border has an even thickness throughout.

Another user above mentioned this along the same lines of what I am confused on: "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."
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