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by abhasjha » Mon Jul 21, 2014 11:46 pm
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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by dmv » Tue Jul 22, 2014 12:20 am
(l+2)(b+2)= M (l+4)(l+4)= M+52
There fore 2(l+b) = 40
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by GMATGuruNY » Tue Jul 22, 2014 3:13 am
abhasjha 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 can PLUG IN THE ANSWERS, which represent the perimeter of the photograph.
To make the math easier, let the photograph be a SQUARE.
When the correct answer choice is plugged in, the area with a 2-inch border will be 52 more square inches than the area with a 1-inch border.

Answer choice D: 40
Here, each side of the square photograph = 10.
A 1-inch border increases the length by a total of 2 inches (1 inch to the left and 1 inch to the right) and the height by a total of 2 inches (1 inch upwards and 1 inch downwards).
Area with a 1-inch border = (10+2)(10+2) = 144
A 2-inch border increases the length by a total of 4 inches (2 inches to the left and 2 inches to the right) and the height by a total of 4 inches (2 inches upwards and 2 inches downwards).
Area with a 2-inch border = (10+4)(10+4) = 196.
Difference between the areas = 196 - 144 = 52.
Success!

The correct answer is D.
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