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Easy one but answer not getting matched ! Try once

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by imhimanshu » Tue Aug 25, 2009 5:55 am
A rectangular picture is to be framed. If the width of the frame is 1, the area of the framed picture would be m; is the width of the frame is 2, the area of the framed picture would be m+52. What is the perimeter of the un-framed picture?

Pls post your explanantions
[spoiler]ImO its 40 but OA = 20[/spoiler]
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by shanrizvi » Tue Aug 25, 2009 11:50 am
l = length of the picture
w = width of the picture

perimeter of the picture : 2(l+w) = 2l+2w

when frame-width=1, total area = (l+2)(w+2) = m
when frame-width=2, total area = (l+4)(l+4) = m + 52

Eq 1: lw+2l+2w+4=m
Eq 2: lw+4l+4w+16=m+52

Eq2 - Eq1: 2l+2w+12=52

2l+2w=40 (perimeter)
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by ershovici » Tue Aug 25, 2009 7:22 pm
since l+2*w+2 - l*w = 52
lw+2L+4+2w -Lw = 52
2L + 2w = 48

now we have to substract the first frame, substract 2 from every side and the answer is 44
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by rish » Tue Aug 25, 2009 7:34 pm
ershovici wrote:since l+2*w+2 - l*w = 52
lw+2L+4+2w -Lw = 52
2L + 2w = 48

now we have to substract the first frame, substract 2 from every side and the answer is 44
Please recheck. Here m is (l+2)*(w+2).

IMO its 40
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by ershovici » Wed Aug 26, 2009 6:10 am
rish wrote:
ershovici wrote:since l+2*w+2 - l*w = 52
lw+2L+4+2w -Lw = 52
2L + 2w = 48

now we have to substract the first frame, substract 2 from every side and the answer is 44
Please recheck. Here m is (l+2)*(w+2).

IMO its 40
So you right, I forgot that there are 4 sides.

:lol:
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