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Rectangle

by crackgmat007 » Fri May 08, 2009 9:32 am
Q. 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

I took a guess for D and got it right. But how do I arrive at 40? pls help!
Last edited by crackgmat007 on Fri May 08, 2009 9:51 am, edited 1 time in total.
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by DeepakR » Fri May 08, 2009 9:40 am
Going forward kindly ask 1 question per thread:

#1. Let the length of photograph=l and breadth of photograph=b
Area of photograph + Area of border = M
i.e. lb + (l+2)(b+2)=M ----1

Similarly, lb + (l+4)(b+4)=M+52 --- 2

Subtracting eqn, 2 -1 gives,
2l+2b=Perimeter = 40 = D.)

-Deepak
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by DeepakR » Fri May 08, 2009 9:51 am
#2. Given that, 2l + 2b= 560 Hence l + b=280 --- (1)
Also its given that l^2 + b^2=200^2 --- (2)

Now we can rewrite eqn (2) as follows:
(l + b)^2 - 2lb = 200^2 and substitute eqn (1) there:

(280^2) - 2lb = 200^2 hence 2lb = 280^2 - 200^2 its like a^2-b^2 so we can rewrite it as (a+b)(a-b)

So 2lb = (280+200) * (280-200) = 480 * 80

so lb= 480*80/2= 19200 = A.)

-Deepak
Last edited by DeepakR on Fri May 08, 2009 10:12 am, edited 1 time in total.
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by crackgmat007 » Fri May 08, 2009 10:03 am
got it....tx

minor error in the solution =>So 2lb = (280+80)- (280-80): should this be 2lb = (280+200)- (280-200)?
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by DeepakR » Fri May 08, 2009 10:11 am
Ya that was a typo..thanks for correcting that..i will edit my previous post.

-Deepak
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by vitaly » Fri May 08, 2009 10:47 am
>Area of photograph + Area of border = M
>i.e. lb + (l+2)(b+2)=M ----1

I think it's just (l+2)(b+2), no?
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Re: Rectangle

by Uri » Fri May 08, 2009 11:21 am
crackgmat007 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?
Let, length of photograph=l1 and width of photograph=w1
With border 1 inch in each side, length l and width w.
With border 2 inch in each side, length l+2 and width w+2.
l1=l-2 and w1=w-2
We need to find 2(l1+w1)
Given, lw=M ………(1) and
(l+2)*(w+2)=M+52
Simplifying and substituting the value of lw=M in the above equation, we get, 2(l+w)=48
Now, 2(l1+w1)= 2{(l-2)+(w-2)}=2(l+w)-8=48-8=40
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