A frame 2 inches wide is placed around a rectangular picture with dime

Hi All,

This question is an example of a 'punch out' question - we have to find the area of EVERYTHING, then 'punch out' the part that we don't want.

We're told that a frame 2 inches wide is placed around a rectangular picture with dimensions 8 inches by 12 inches. We're asked for the area of the frame, in square inches.

Area of a rectangle = (length)(width)

So the area of the PICTURE is... (8)(12) = 96

The frame 'adds' 2 inches to the top, bottom, left and right 'sides' of the picture, so the area of EVERYTHING is...

(8+2+2)(12+2+2) = (12)(16) = 196

When we 'punch out' the area of the picture, we'll be left with the area of the FRAME:

192 - 96 = 96

Final Answer:

GMAT assassins aren't born, they're made,
Rich

area of rectangular picture = 12 *8 = 96 sq inches
length of picture with frame = 12 + 4 = 16 inches
width of picture with frame = 8 +4 = 12 inches
area of whole picture with frame = 16 * 12 = 192 sq inches
area of frame = area of whole picture with frame - area of rectangular picture =192 - 96 =96
correct answer option B

total area of picture + frame = 12+2+2 * 8+2+2 = 16*12 = 192
and picture area = 12*8= 96
so area of fram = 192-96 = 96 IMO B

Area of photo=8*12=96
Area of photo+frame= (8+2+2)*(12+2+2)=12*16=192

Hence area of frame= 192-96=96

IMo B

96

12+2+2=16*2=32 ×2=64
8*2=16*2=32
Add 96

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