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sat math problem from new BB second ed.

desi_chickdesi_chick 1588 replies104 threadsRegistered User Senior Member
edited May 2011 in SAT Preparation
here's a graphic that goes with it: [url]http://i28.****.com/iw82sk.jpg[/url]

The pattern shown above is composed of rectangles. This pattern is used repeatedly to completely cover a rectangular region 12L units long and 10L units wide. How many rectangles of dimension L by W are needed?

answer choices:
A) 30
B) 36
C) 100
D) 150
E) 180 << that's the answer.

i don't get how to arrive at this solution, thanks!
edited May 2011
12 replies
Post edited by desi_chick on
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Replies to: sat math problem from new BB second ed.

  • Bigb14Bigb14 1970 replies74 threadsRegistered User Senior Member
    2L = 3W

    therefore, each rectangle is L x 2/3L = 2/3 L^2

    Total area is 120L^2

    120L^2/ (2/3)L^2 = 180
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  • UrcUrc 206 replies12 threadsRegistered User Junior Member
    well, from the image, we see that 2L=3W.

    we can see the dimensions of the pattern are 2L x (L+W)= 2L x (5/3 L). luckily, if you use 6 of these as the length, and 6 as the width (creating a 6x6 array of this pattern), the dimensions would be (2L * 6) x (5/3 L * 6) = 12L x 10L. thus there would be 36 of the pattern shown, and since there are 5 L x W rectangles in each pattern, there are 36 * 5 =180 L x W rectangles needed.
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  • Bigb14Bigb14 1970 replies74 threadsRegistered User Senior Member
    ^That's unnecessarily complex :P
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  • pckellerpckeller 1059 replies20 threadsRegistered User Senior Member
    It is even easier if you make up concrete numbers. But you have to make it so 2 L's = 3 W's.

    Say you use L=6 and W=4. That makes the areal of the big region 12L x 10L = 72 x 60 = 4320.

    Then, since each little rectangle has area 4 x 6 = 24, then the number of them you need is 4320 / 24 = 180
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  • OrangeMathOrangeMath 1 replies0 threadsRegistered User New Member
    Bigb14: Your reply to Urc is incorrect. Urc VERIFIED that there was a solution. You did not. Urc's answer is better (proved) as a result. pckeller has the same problem. 180 means nothing without the Urc's verification.
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  • ibSleepyibSleepy 221 replies15 threadsRegistered User Junior Member
    I substituted.
    2L = 3w
    L = 3/2w
    Area of Rectangle = L x w = 3/2w x w = 3/2w^2
    Area of Rectangular Region = 12L x 10L = 18w x 15w = 270w^2
    (270w^2)/(1.5w^2) = 180.
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  • ShamoilShamoil 24 replies2 threadsRegistered User New Member
    My way is different L=1.5W

    so

    W+x(1.5W)=15W
    x(1.5W)=14W
    x=28/3

    so theres 12 rectangles going down vertically and then 18 going down horizontally in each line. So I just did (28/3)*18=168 and +12 = 180. Different method probably not as efficient as those mentioned above.
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  • ShamoilShamoil 24 replies2 threadsRegistered User New Member
    More easily understood though. Probably the hardest SAT question iv'e ever seen (my opinion)
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  • DrSteveDrSteve 1266 replies24 threadsRegistered User Senior Member
    In the expression 2L=3W, the simplest choices for L and W are L=3 and W=2 (because clearly 2*3=3*2).

    Now just apply the simple strategy "to see how many 2-dimensional objects fit inside another 2-dimensional object, divide the areas."

    So the answer is (12L*10L)/6 = (12*3*10*3)/6 = 180.
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  • DivineEDivineE 100 replies39 threadsRegistered User Junior Member
    @DrSteve‌ how did you get 6?
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  • DrSteveDrSteve 1266 replies24 threadsRegistered User Senior Member
    The area of each small rectangle is LW. I chose 3 for L and 2 for W. So the area of each small rectangle is 3*2=6.
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  • xiggixiggi 24571 replies872 threadsRegistered User Senior Member
    This thread offers a perfect example of the benefits of using this site. Some will view this problem as very basic and others as very hard. This thread shows 3-4 various ways to solve the problem through simple (and correct) approaches. This is not unusual for the SAT and students who learn to balance straight high school "techniques" with reasoning (or graphical) approaches will really help themselves.
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