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rectangular photograph-Gmat Prep -OA incorrect??

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by khurram » Fri Jun 13, 2008 10:09 pm
Q30:
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



solution is below
Area of photograph alone = LW
Area of photograph + 1inch border = LW + (L+2)(W+2) (We add 2 to the length and width because the 1 inch border adds 1 inch to EACH SIDE of the picture width wise and length wise!)

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

2 inch border:
(We will be adding 4 to the length and width this time since the border adds 2 inches to EACH SIDE of the picture width wise and length wise)

M+52 = LW + (L+4)(W+4)
We can replace the ‘M’ above with the 1st equation, giving us:

LW + (L+2)(W+2) + 52 = LW + (L+4)(W+4) …..After canceling, we are left with 2L + 2W = 40. The equation for the perimeter of rectangle is 2L + 2W or 2(L+W)

OA is 40 I think it is 38
(l+2)(b+2)=m+52
b+l=21 so times two is total perimter minus 4 outside =38

what am i doing wrong
thanks
Khurram
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by babachal » Fri Jun 13, 2008 10:38 pm
hi khurram, your name means to be happy, any way, when I solved the problem, i got it 40;

if let's say if 1 inch is added, it will be added 2 to to each side of rectangle.
so, W+2, and, L+2,
in oreder not to confuse, W+2=x and, L+2=y
here, x*y=M
if 1 inch is added again, 2 inch will be added to both side, then
(x+2)(y+2)=M+52;
xy+2x+2y+4=M+52;
xy=M, cancel them,
2x+2y=48
x+y=48/2=24
now plug them with the sides of photograph, which is:
W+2+ L+2=24
W+L=20
perimeter is 2(W+L)=20*2=40
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by khurram » Fri Jun 13, 2008 10:44 pm
Thanks

Slowly starting to make sense.

Yes, my name means happy and I am anything but Happy. Gmat next week on 24. PS very few people know meaning of name.

Good luck

Khurram
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by babachal » Fri Jun 13, 2008 10:54 pm
it is private messgae, where are you from? it sounds turkish, or uzbek something like that?
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by khurram » Fri Jun 13, 2008 11:07 pm
karachi, pakistani
but in US for half my life.
that is why I knew you were not from there, my name is hardly used in the happy context anymore there but still is up north.
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Re: rectangular photograph-Gmat Prep -OA incorrect??

by Ian Stewart » Sat Jun 14, 2008 2:08 am
khurram wrote:
Area of photograph + 1inch border = LW + (L+2)(W+2) (We add 2 to the length and width because the 1 inch border adds 1 inch to EACH SIDE of the picture width wise and length wise!)

what am i doing wrong
It's in this step (and another place later in your solution) where there is an error. The area of the photograph+1 inch border is not LW + (L+2)(W+2). It's just (L+2)(W+2). L+2 and W+2 are the length and width of the photo and the border together, so multiplying them gives the total area; no need to add on LW again.
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by Scott@TargetTestPrep » Wed Dec 13, 2017 5:25 pm
khurram 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
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:

Image

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:

Image

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

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