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30 Sep 2006, 08:41
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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
A- 34
B- 36
C- 38
D- 40
E- 42
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30 Sep 2006, 09:08
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let the sides of border be x and y
so xy = M
after increasing the width,
then (x+2)(y+2) = M+52
=> M + 2x + 2y + 4 = M + 52
=> x + y = 24
now sides of photograph are (x-2) and (y-2) because of 1 inch border
so perimeter of photograph = 2[ x -2 + y - 2] = 2 [ 24-4] = 40
hence D
so xy = M
after increasing the width,
then (x+2)(y+2) = M+52
=> M + 2x + 2y + 4 = M + 52
=> x + y = 24
now sides of photograph are (x-2) and (y-2) because of 1 inch border
so perimeter of photograph = 2[ x -2 + y - 2] = 2 [ 24-4] = 40
hence D
30 Sep 2006, 14:17
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TOUGH GUY
If X and Y are the length and width of the photograph respectively (excluding the border) then
(X + 2)(Y + 2) = M ( since border is 1 inch on each side)
XY + 2X + 2Y + 4 = M ---> A
Also,
(X + 4) (Y + 4) = M + 52 (if border is 2 inch wide on each side)
XY + 4X + 4Y + 16 = 52 ---> B
Subtract A from B
2X + 2Y + 12 = 52
2X + 2Y = 40
2X + 2Y = Perimeter of photograph
So, D.
