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Old 05-29-2005, 02:32 PM   #1
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sat math help

could anyone with the blue collegeboard book teach me an easy method of solving #16 on page 551? thanks.
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Old 05-29-2005, 02:44 PM   #2
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Is the answer E? I think it is
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Old 05-29-2005, 02:47 PM   #3
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OK so here goes nothing:
I am not the best at explaining this stuff

you see that they want a 12L by 10L rectangle, which would give you a total of 120L units

Then, if you look on your image above the question you need to figure out how one L by L unit would compare to a L by W unit.

you can see from the picture that 2L=3W, so a ratio of 1.5 (from W to L)
So, you can multiply 120 units by 1.5 and come up with 180.
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Old 05-29-2005, 02:50 PM   #4
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Actually better way of exxplaining (i told you i suck)

since it 12L by 10L and you know that 1L=1.5W, change the 10L to equal 15W

then multiply 12 by 15 and get 180.
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Old 05-29-2005, 02:53 PM   #5
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From the picture you can see that 2L = 3W (by looking at the two sides), so W = (2/3)L, which means that the top side is equal to (5/3)L. This means that the dimension of the rectangle given is (5/3)L x 2L. You're looking for the number of the rectangles in a rectangular region with dimension 12L x 10L, so determine what number you would need to multiply the dimensions of one region by [(5/3)L x 2L] to obtain the desired region. This number turns out to be 6:

(5/3)L x 6 = 10L
2L x 6 = 12L

Keep in mind that that's 6 lengthwise and 6 width-wise, for a total of 36. That means that you need 36 of the given "patterns," and in each pattern there are 5 rectangles "of dimension L by W" -- 36 x 5 = 180, so the answer is E.

EDIT: sr's explanation is definitely a lot faster...
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Old 05-29-2005, 02:57 PM   #6
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Zach the answer it says in the book is E on pg560
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Old 05-29-2005, 03:00 PM   #7
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Yeah, I saw my mistake and corrected it.
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Old 07-01-2005, 09:55 AM   #8
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Just rephrasing sr's solution (I hate fracrions):
from the picture
2L = 3W
*5 *5
10L = 15W.
Region 12L*10L = 12L*15W = (12*15) (L*W) = 180 (L*W).
180 rectangles L by W is answer E.
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Old 02-15-2006, 11:32 PM   #9
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Can anyone explain #15 on the same page? There has to be an fast/easy way to do it
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