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08-06-2009, 04:57 AM
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#1 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
Posts: 1,701
| Math help center
Give me any SAT Reasoning Test math question (or a question from a simulated practice test) and I'll post a solution. Specify detail (just the math work or explicit reasoning or w/e) if you want. I'll post ridiculously detailed solutions by default.
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08-06-2009, 09:23 AM
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#2 | | Member
Join Date: Nov 2008 Location: UF '14
Posts: 501
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Umm..
#5 pg. 472 of Blue Book
#6 pg. 472 of BB
#8 pg. 473 of BB
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08-06-2009, 10:55 AM
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#3 | | Senior Member
Join Date: Apr 2009 Location: Where the wind of freedom blows
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08-06-2009, 03:15 PM
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#4 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
Posts: 1,701
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Harambee:
Image 1:
#15:
Since OD bisects <AOF, you know that m<AOD = m<DOF, and since you know that OC bisects <AOE, that m<AOB + m<BOC = m<COD + m<DOE, and since OB bisects <AOD, you know that m<BOC + m<COD = x = 40
Now since m<BOC + m<COD = 40, and since we know that OD bisects <AOF, we know that
40 + m<BOC + m<COD = 30 + m<DOE. Substituting, it is easy to see that the expression boils down to 40 + 40 = 30 + m<DOE, which gives that m<DOE is 50.
Finally, we want to find <BOE, which we can do by adding m<BOC + m<COD + m<DOE, which from the information we have derived simplifies to 40 + 50 = 90
#17:
Since we remove every 4 inches, it is useful to find out how many triangles we will remove. This can be found by: 80/4 = 20. Then, notice that for every 1 inch you remove, you gain back 2 inches. So we gain back 20(2-1) = 20 inches, which would make our total length 100 .
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08-06-2009, 03:19 PM
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#5 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
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Image 2:
#20
You know that the total rope length will be y + 4x, so the insight needed is an expression of y in terms of x. Luckily, we have area which gives xy = 4000. Now, rearranging, we see that x = 4000/y, which we can substitute into our expression to yield 16000/y, choice B |
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08-06-2009, 03:21 PM
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#6 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
Posts: 1,701
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Image 3:
#7
Sum of the arc measures is 2pi or 360 degrees. It's divided into 5 equal segments. ABC gets two of these segments, which leaves 3 of these segments to AEC. Thus ratio of ABC to AEC is 2:3, choice B .
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08-06-2009, 05:27 PM
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#7 | | Senior Member
Join Date: Apr 2009 Location: Where the wind of freedom blows
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Thanks! Much appreciated.
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08-06-2009, 05:28 PM
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#8 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
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Anytime. I enjoy these problems, so I figured I'd help you all out |
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08-06-2009, 06:31 PM
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#9 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
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@Vertigo220h: please type up the questions. I've long since discarded my blue book.
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08-06-2009, 06:31 PM
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#10 | | Senior Member
Join Date: Mar 2008 Location: Out of my mind. Back in five minutes.
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what is 2 + 2? I seem to be baffled by this question!
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08-06-2009, 06:33 PM
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#11 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
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Well, it depends. In base >= 4, 2+2 = 4. In base 1, the question is illogical. In base 2, the question is illogical. In base 3, it is 11.
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08-06-2009, 06:33 PM
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#12 | | Senior Member
Join Date: Mar 2008 Location: Out of my mind. Back in five minutes.
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Makes perfect sense! Thank you sensei
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08-06-2009, 06:38 PM
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#13 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
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Indeed. 'tis my job to be familiar with tricky questions like the one you posed.
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08-06-2009, 07:49 PM
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#14 | | Junior Member
Join Date: Jan 2009
Posts: 33
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Tom and Bill agreed to race across a 50-foot pool and back again. They started together, but Tom finished 10 feet ahead of Bill. If their rates were constant, and Tom finished the race in 27 seconds, how long did Bill take to finish it?
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08-06-2009, 08:21 PM
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#15 | | Senior Member
Join Date: Sep 2008 Location: Salem High '09 -> UVM '12
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GRE:
Tom's rate: 100 ft/27 sec
Bill's rate: 90 ft/27 sec.
(27sec/90ft)(100ft) = 30 seconds.
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