A tailor trims 4 feet from opposite edges of a square piece : PS Archive
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# A tailor trims 4 feet from opposite edges of a square piece

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A tailor trims 4 feet from opposite edges of a square piece [#permalink]

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11 Jan 2004, 10:07
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A tailor trims 4 feet from opposite edges of a square piece of cloth, and 5 feet from the other two edges. If 120 square feet of cloth remain, what was the length of a side of the original piece of cloth?
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11 Jan 2004, 10:19
he took off 8 and 10 inches--

we need two divisors of 120 that have a difference of two.

10 and 12.

Originally, it was 20 * 20.

I think...
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11 Jan 2004, 12:17
The fastest way is to pick numbers and see if (x-8)*(x-10) was equal to 120. Otherwise, algebraically, it is:
(x-8)*(x-10)=120
x^2 - 18x - 40 = 0
(x-20) * (x+2) = 0
x = 20
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Paul

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11 Jan 2004, 16:37
How come he took 8 and 10 feet from each side? He (the tailor I mean) took 4 and 5 feet from oposing sides. If the side of the square were a, shouldn't it look like the figure below? The gray part is the part the tailor cut.

I must be making some dumb mistake because if you approach the problem like this you don't get to an answer listed among the options.
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11 Jan 2004, 17:33
I think you misunderstood the problem. It says the tailor trims 4 feet from opposite edges. So you have to take off 4 off from each opposite edges. So it is for 5 which would turn out to be 10. Agree that the problem could have been better worded by saying 4 feet from each opposite edges.
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30 Jan 2004, 04:39
[quote="MartinMag"]How come he took 8 and 10 feet from each side? He (the tailor I mean) took 4 and 5 feet from oposing sides. If the side of the square were a, shouldn't it look like the figure below? The gray part is the part the tailor cut.

I must be making some dumb mistake because if you approach the problem like this you don't get to an answer listed among the options

Which figure are u talking about????
How can we see the figure...i am a new joinee don't know much and is unable to find!!! Please let me know???
You can mail me at mdfrahim@yahoo.com
Thanks,
Frahim[/url]
30 Jan 2004, 04:39
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