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11-16-2012, 06:21 PM
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#1 | | New Member
Join Date: Oct 2011
Posts: 14
| Daily Math Questions
Hey guys, I'm prepping for the Dec SAT and I think a daily math questions would help me improve. Can anyone help me with these questions? If you nee help too, feel free to post.
1. Kyle's lock combination consists of 3 two-digit numbers. The combination satisfies the three conditions below.
-One number is odd
-One number is a multiple of 5
-One number is the day of the month of Kyle's birthday
If each number satisfies exactly one of the conditions, which of the following could be the combination of the lock?
A. 14-20-13
B. 14-25-13
C. 15-18-16
D. 20-15-20
E. 34-30-21
I'm confused because A, B, and D all fit the conditions...
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11-16-2012, 06:35 PM
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#2 | | Member
Join Date: Jun 2012
Posts: 340
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I'm not agreeing with how the question is given to us; but I'll give you the answer they're looking for.
The only thing you missed was the word "one." ONE number is odd. ONE number is a multiple of 5. And ONE number.. you get it.
A. 14-20-13
B. 14-25-13 - Eliminate because it has 2 odd numbers.
C. 15-18-16 - Eliminate because 15 can't satisfy two requirements.
D. 20-15-20 - Eliminate because it has two multiple of 5's.
E. 34-30-21 - Eliminate because 21 and 30 work, but 34 doesn't satisfy the birthday requirement.
A!
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11-16-2012, 06:42 PM
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#3 | | New Member
Join Date: Oct 2011
Posts: 14
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Omg I feel so stupid. Thank you so much!
Here's another one I need help with:
Let x be defined as [x]=x^2-x for all values of x. If [a]=a-2, what is the value of a?
A. 1
B. 1/2
C. 3/2
D. 6/5
E. 3
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11-16-2012, 06:43 PM
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#4 | | Senior Member
Join Date: Mar 2012 Location: Cambridge, MA
Posts: 1,913
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This is not an SAT question, it's just an SAT-level question I made up.
Q: If p and q are positive integers such that pq = 50, which of the following cannot equal (p^2)q?
A: 50
B: 100
C: 200
D: 250
E: 500
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11-16-2012, 06:46 PM
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#5 | | Senior Member
Join Date: Mar 2012 Location: Cambridge, MA
Posts: 1,913
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@maraudersmap, are you sure your question is correctly typed? Here's what I did:
[a] = a^2 - a = a - 2
a^2 - 2a + 2 = 0
a = (2 +- sqrt(4 - 8))/2, i.e. two complex solutions.
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11-16-2012, 06:57 PM
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#6 | | Junior Member
Join Date: Apr 2011
Posts: 43
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I'm pretty sure the question is: Let x be defined as [x]=x^2-x for all values of x. If [a]=[a-2], what is the value of a?
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11-16-2012, 06:59 PM
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#7 | | New Member
Join Date: Oct 2011
Posts: 14
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The question is typed correctly, [ ] is one of those weird symbols
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11-16-2012, 07:11 PM
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#8 | | Senior Member
Join Date: Mar 2012 Location: Cambridge, MA
Posts: 1,913
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[a] = a-2 and [a] = [a-2] mean completely different things...
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11-16-2012, 07:13 PM
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#9 | | New Member
Join Date: Oct 2011
Posts: 14
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sorry, you guys are right. It's [a-2]
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11-16-2012, 07:17 PM
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#10 | | Senior Member
Join Date: Mar 2012 Location: Cambridge, MA
Posts: 1,913
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Then it's just
a^2 - a = (a-2)^2 - (a-2)
a^2 - a = a^2 - 5a + 6 (upon expanding, simplifying). a^2 cancels, leaving
-a = -5a + 6 --> a = 3/2
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11-16-2012, 07:21 PM
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#11 | | New Member
Join Date: Oct 2011
Posts: 14
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Thank you very much! I'll post when I have more questions, I'm currently not at home haha [:
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11-17-2012, 04:33 AM
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#12 | | New Member
Join Date: Jul 2012 Location: Austin, Texas
Posts: 10
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rspence, is the answer to your question C. ?
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11-17-2012, 10:36 AM
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#13 | | Senior Member
Join Date: Mar 2012 Location: Cambridge, MA
Posts: 1,913
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^JackTC yep. All the other answer choices are in the form 50p, where p is a factor of 50.
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11-17-2012, 10:48 AM
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#14 | | Junior Member
Join Date: Sep 2012
Posts: 57
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Can you go over how you got that question exactly? The way I did it took too long
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11-17-2012, 10:54 AM
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#15 | | Senior Member
Join Date: Mar 2012 Location: Cambridge, MA
Posts: 1,913
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The solution involves thinking of (p^2)(q) as p(pq). Since pq = 50, this is equal to 50p. All the answer choices are multiples of 50, but note that p must be a factor of 50, so 50p must be (a factor of 50) times 50.
200 is 50*4, and 4 is not a factor of 50, so C is the answer.
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