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Pls help, GMATprep problem

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by sumatitandon » Fri Jan 09, 2009 2:53 am
If two of the four expressions (x+y),(x+5y),(x-y),(5x-y) are chosen at random, what is the probability 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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Re: Pls help, GMATprep problem

by piyush_nitt » Fri Jan 09, 2009 3:00 am
sumatitandon wrote:If two of the four expressions (x+y),(x+5y),(x-y),(5x-y) are chosen at random, what is the probability 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
IMO B

Total Number of ways : 4C2 = 6

expression will be in form X^2 -(by)^2 only when

(x-y)(x+y)

or

(x+5y)(x-y)

therefore

2/6 = 1/3
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by welcome » Fri Jan 09, 2009 3:01 am
Total ways of selecting 2 out of 4 = 4C2 = 6.

Only 1 ways will give X^2-(by)^2 that is (x^2-y^2). Remember (x^2-5y^2) is not required expressing as 5y^2 cannt be written as (by)^2, b being an integer(sqrt5 is not).

So Answer = 1/6 - E.
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by piyush_nitt » Fri Jan 09, 2009 3:12 am
welcome wrote:Total ways of selecting 2 out of 4 = 4C2 = 6.

Only 1 ways will give X^2-(by)^2 that is (x^2-y^2). Remember (x^2-5y^2) is not required expressing as 5y^2 cannt be written as (by)^2, b being an integer(sqrt5 is not).

So Answer = 1/6 - E.
Good Catch welcome. I overlooked that interger part.

Well done!!!
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