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Gmat preview questions

by mikeclarke44 » Tue Jun 12, 2007 5:33 pm
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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Re: Gmat preview questions

by jayhawk2001 » Tue Jun 12, 2007 8:39 pm
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
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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by mikeclarke44 » Wed Jun 13, 2007 7:36 am
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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by jayhawk2001 » Wed Jun 13, 2007 6:58 pm
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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by ellexay » Thu Feb 12, 2009 9:07 am
So, for the second question, we have both statement (1) and (2) repeating the information given in the question.

For those questions, do we just look at the 2 statements' agreeing with the question as a confirmation of the question? Hence, D?

Or do we have to look at the fact that they're different ways of saying the same thing as in the question? So we're back to square one? Hence, E?

Thoughts???
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by jhhatdaddy » Sat Feb 14, 2009 5:16 am
For the first one...

x^2 - (by)^2

I always remember the result listed above can only come from the the form (x-y)(x+y)

All the integer values in front of the x's and y's have to be equal so that the middle terms can cancel out.

Looking at your four options that means only the two terms can be multiplied together successfully to achieve the desired result.

Each term has a 1/4 chance of being selected and the remaining necessary term has a 1/3 chance. Hence 1/12. Multiplied by two because either term could be selected first or second and there you go...

Its not pretty, but its a different way of looking at it...
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by missrochelle » Sun Aug 29, 2010 4:03 pm
can someone explain 4C2. I thought the answer would be 4*3.

4 possiblities from choice 1, 3 possibilities from choice 3.
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