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Hi i_have_no_cool_username(:D It indeed is a cool name)i_have_no_cool_username 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?
i_have_no_cool_username 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) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6
Answer is E. Why?
Request noted, Neelgandham! If I'm still posting questions after my test on Monday (highly unlikely), I will keep that request in mindneelgandham wrote:Hi i_have_no_cool_username(:D It indeed is a cool name)i_have_no_cool_username 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?
Welcome to the forum. Before I answer the question, I would like to request you to conceal the answers of the questions posted.
How ? Look here -> https://www.beatthegmat.com/new-spoilers-t5302.html. Now the solution...
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, their product will be of the form of x^2-(by)^2 only if the coefficient of x in both the expressions is 1 and if the coefficient of y in both the expressions is equal and is of different signs. x-y,x+y is only such pair.
Total number of pairs = 3+2+1 = 6
Total number of pairs that satisfy the condition = 1
Probability = 1/6
