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by neelgandham » Fri Apr 13, 2012 1:24 pm
Solution:
(x + y)(x + 5y) = x^2 + 6xy + 5y^2: Not in the form x^2- (by)^2
(x + y)(5x - y) = 5x^2 + 4xy - y^2: Not in the form
(x + 5y)(x - y) = x^2 + 4xy - 5y^2: Not in the form
(x + 5y)(5x - y) = 5x^2 + 24xy - 5y^2: Not in the form
(x - y)(5x - y) = 5x^2 - 6xy + y^2: Not in the form
(x + y)(x - y) = x^2 - y^2: This is in the required form

Hence, the required probability is 1/6.

Solution by Anurag@Gurome(https://www.beatthegmat.com/probability- ... 96791.html)
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by Anurag@Gurome » Fri Apr 13, 2012 5:28 pm
bobdylan 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 of x^2-(by)^2, where b is an integer
a)1/3
b)1/2
c)1/4
d)1/5
e)1/6
(x + y)(x + 5y) = x^2 + 6xy + 5y^2: Not in the form x^2- (by)^2
(x + y)(5x - y) = 5x^2 + 4xy - y^2: Not in the form
(x + 5y)(x - y) = x^2 + 4xy - 5y^2: Not in the form
(x + 5y)(5x - y) = 5x^2 + 24xy - 5y^2: Not in the form
(x - y)(5x - y) = 5x^2 - 6xy + y^2: Not in the form
(x + y)(x - y) = x^2 - y^2: This is in the required form

Hence, the required probability is [spoiler]1/6[/spoiler].
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