PS

[lionheart]

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 interger?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

[Nagm]

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 interger?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

Only product of X+Y and X-Y satisfy this condition.
so E

[AC]

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 interger?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

Only product of X+Y and X-Y satisfy this condition.
so E

X+5Y and X-Y also works. In both cases you need to select X-Y. The probability of selecting (X-Y) first is 1/4. Probability of then selecting X+Y or X+5Y is then 2/3. Multiplied together, you get 2/12 or 1/6.

The other possibility is that you select X+5Y or X+Y first which has a probability of 2/4. Probability of then selecting X-Y is then 1/3 for a total probability of 2/12. Add the two possibilities together and you get 4/12 or 1/4. This could be right or completely wrong.

MGMAT Staff?

[AC]

Sorry! Maybe I should work on my fraction simplification. I meant 1/3.

[Nagm]

You are looking for a product that has no XY term.
If you multiply X+5Y and X-Y, you get XY term in the product. Correct me if you are wrong.
And if you have XY term, you can not write in the required format.

[RonPurewal]

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 interger?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

the form listed is the DIFFERENCE OF SQUARES formula, which can only be had by multiplying the sum and the difference of the same 2 quantities. the only expressions fitting the bill, as has been mentioned by other posters here, are x + y and x - y.

the probability that you'd pick one of these at random with your first choice is 1/2; the probability that you'd pick the remaining one at random with your second choice is 1/3. therefore (e) is the answer. (you can also figure this out with mgmat's anagram method: the total number of ways of picking two of those four expressions is 4! / (2!2!) = 6, and you only want one of those combinations, so that's 1/6.)

--

if they wanted to be real bastards, they could have included 5x - y and 5x + y. those are the two components of a difference of squares, but they _still_ don't satisfy the required pattern (because x isn't allowed to have a coefficient). luckily, they aren't that mean.