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mikeclarke44
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 Posted: Tue Jun 12, 2007 5:33 pm Post subject: Gmat preview questions |
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Here are a couple of recent ones that I am unable to figure out.
If 2 of the 4 expressions, x+y, x+5y, x-y, and 5x -y are choosen at random, what is the probabilty that their product will be of the form x^2 - (by)^2, where b is an integer?
a). 1/2
b). 1/3
c). 1/4
d). 1/5
e). 1/6
The answer is (e)
y< x+z/2 ?
1). y-x < z-y
2). z-y> z-x/2 |
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jayhawk2001
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| Posted: Tue Jun 12, 2007 8:39 pm Post subject: Re: Gmat preview questions |
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| mikeclarke44 wrote: | Here are a couple of recent ones that I am unable to figure out.
If 2 of the 4 expressions, x+y, x+5y, x-y, and 5x -y are choosen at random, what is the probabilty that their product will be of the form x^2 - (by)^2, where b is an integer?
a). 1/2
b). 1/3
c). 1/4
d). 1/5
e). 1/6
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Number of ways you can get a pair of expressions from a set of 4 = 4C2
Number of pairs of the form x^2 - b.y^2 = 1 [only (x+y)*(x-y) ]
So, probability = 1/4C2 = 1/6
| wrote: |
y< x+z/2 ?
1). y-x < z-y
2). z-y> z-x/2 |
Is it (x+z)/2 or x + z/2. Can you please clarify. 1 is sufficient
if it is (x+z)/2 |
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mikeclarke44
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| Posted: Wed Jun 13, 2007 7:36 am Post subject: |
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It is (x+z)/2
In the first problem, I did come up with 4C2 = 6, but I thought since there were two combinations it would be 2/6. Close but no cigar!! |
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jayhawk2001
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Location: Silicon valley, California
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| Posted: Wed Jun 13, 2007 6:58 pm Post subject: |
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| mikeclarke44 wrote: | It is (x+z)/2
In the first problem, I did come up with 4C2 = 6, but I thought since there were two combinations it would be 2/6. Close but no cigar!! |
In that case, is the answer D?
Rearranging 1, we get 2y < x+z or y < (x+z)/2
Rearranging 2, we get 2z - 2y > z - x; 2y < x+z; y < (x+z)/2
Don't worry about not getting the right answer in the first try  .
Over time your mind will automatically start to look for patterns
such as this. |
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