A rectangular photograph is surrounded by a border that is 1

Question

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

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

jallenmorris · Accepted Answer

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.

Bunuel · Answer

Similar questions to practice about rectangles: a-square-wooden-plaque-has-a-square-brass-inlay-in-the-89215.html a-border-of-uniform-width-is-placed-around-a-rectangular-144434.html a-rectangular-yard-is-20-yards-wide-and-40-yards-long-it-is-8453.html the-figure-above-represents-a-square-plot-measuring-x-feet-o-160568.html Similar questions to practice about circles: marty-s-pizza-shop-guarantees-that-their-pizzas-all-have-a-148428.html a-circular-lawn-with-a-radius-of-5-meters-is-surrounded-by-a-154737.html the-figure-above-shows-a-circular-flower-bed-with-its-cente-144448.html the-figure-shows-the-top-side-of-a-circular-medallion-made-o-101845.html

Sion · Answer

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

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

D

ezhilkumarank · Answer

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.

PareshGmat · Answer

Please refer below:

Salvetor · Answer

I did not understand how 1 inch border is counted as 2. Can somebody please explain?

Bunuel · Answer

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.

DesiGmat · Answer

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))

Answer is D

Temurkhon · Answer

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

mcelroytutoring · Answer

Please see attached visual.

ScottTargetTestPrep · Answer

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:

https://i.imgur.com/tMAwubz.png

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:
 
 https://i.imgur.com/zVNBRku.png

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.
 
Answer: D

sahilvijay · Answer

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

Answer is D

chetan2u · Answer

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

hemantbafna · Answer

Please find attached the explanation for the question.

woohoo921 · Answer

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."

harryzhang · Answer

helpful to draw a quick visual here

a rectangle with width = x, height = y, surrounded by a border = 1, so now new width is x+1+1 = x+2 (*note there's one "on each side") and new height is y+2; this first rectangle's area can be calculated as:
A1 = (x+2)(y+2) = M

similarly, we have the same setup except the new rectangle has new width = x+2+2 = x+4 and new height = y+2+2 = y+4; this second rectangle's area can be calculated as:
A2 = (x+4)(y+4) = M+52

now it's just a system of equation that we can simplify by first multiplying it out:
A1 = xy + 2x + 2y + 4 = M
A2 = xy + 4x + 4y + 16 = M+52

see how they have similar terms, we can further simply but subtracting the M out:
A2 - A1 = (xy + 4x + 4y + 16) - (xy + 2x + 2y + 4) = (M+52) - M

for left side, we now have:
(xy + 4x + 4y + 16) - (xy + 2x + 2y + 4) = 2x + 2y + 12

for right side, we now have:
52

setting them equal to each and solve for 2x+2y (perimeter formula) = 52-12 = (D) 40

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

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