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by cans » Sat May 28, 2011 8:21 am
If two of the four expressions (x+y), (x+5y), (x-y), and (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

OA Later.
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by SoCan » Sat May 28, 2011 8:50 am
There are 6 ways to pick 2 items from a set of 4, so right off the bat you can eliminate C and D.

The only two expressions that will give the form given are (x+y)(x-y), so answer is E. You could spot this by recognizing that the 5s in the other expressions will result in a "bxy" in the quadratic equation, but if you want to check it to make sure, it shouldn't take you too long.

(x+y)(x-y) = x^2 - y^2
(x+y)(x+5y) = x^2 + 6xy + 5y^2
(x+y)(5x-y) = 5x^2 + 4xy - y^2
(x+5y)(x-y) = x^2 + 4xy - 5y^2
(x+5y)(5x-y) = 5x^2 + 24xy - 5y^2
(x-y)(5x-y) = 5x^2 - 6xy + y^2
Last edited by SoCan on Sat May 28, 2011 8:53 am, edited 1 time in total.
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by manpsingh87 » Sat May 28, 2011 8:51 am
cans wrote:If two of the four expressions (x+y), (x+5y), (x-y), and (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

OA Later.
out of 4 expressions any two can be selected in 4C2 ways= 6;

now out of six possible selections only when selection (x-4)*(x+y)= x^2-y^2; will result in desired result; here b=1; hence probability should be 1/6 hence E
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by sivaelectric » Sat May 28, 2011 8:52 am
IMO E
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by sandeep800 » Sat May 28, 2011 9:31 am
IMO Answer is E
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by Stuart@KaplanGMAT » Sat May 28, 2011 6:38 pm
Here's a link to another discussion of this question:

https://www.beatthegmat.com/gmat-expressions-t10416.html
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by cans » Sun May 29, 2011 10:44 pm
OA E
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