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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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thiagod wrote:
Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.

Exactly that pair and only that pair gives x^2 -(by)^2. Please re-read the whole discussion.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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thiagod wrote:
Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.

That is the pair. That's the only pair that yields the correct product.

And there are 4C2 ways to pick a pair. 4C2 = $$\frac{4!}{2!2!}$$ = 6.

So the there's only 1 pair that satisfies out of the 6 possible pairs that could be chosen. $$\frac{1}{6}$$

Hope this helps!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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x^2-(by)^2
=(x+by)(x-by)
only x+y and x-y can be the pair
All occasions are "selection of 2 out of 4 = 4C2= 6
Probability is 1/6
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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total expressions possible 4c2 ; 6
so x^2 -(by)^2 ; 1/6
IMO E

Bunuel 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

Four expressions can be paired in $$C^2_4=6$$ ways (choosing 2 out of 4);

$$x^2 - (by)^2=(x-by)(x+by)$$ and only one pair is making such expression: $$(x+1*y)(x-1*y)$$, hence probability is 1/6.

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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Made the silly mistake of including (x+5y) in the selection range.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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Actually it is selecting two( (x+y) and (x-y)) out of four
Therefore the probability is 2/4 = 1/2
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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1
charanvemasani wrote:
Actually it is selecting two( (x+y) and (x-y)) out of four
Therefore the probability is 2/4 = 1/2

Hi charanvemasani,

Unfortunately, that is NOT how 'probability math' works. We're given the terms (X+Y), (X+5Y), (X-Y) and (5X-Y). We're asked for the probability that multiplying any randomly chose pair will give a result that is written in the format: X^2 - (BY)^2.

Since there are only 4 terms, and we're MULTIPLYING, there are only 6 possible outcomes.

From the prompt, you should notice that the 'first part' of the result MUST be X^2....and that there should be NO 'middle term'....which limits what the first 'term' can be in each of the parentheses....

By brute-forcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(X-Y) = X^2 - Y^2
(X+Y)(5X-Y) = X^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

Only the second option is in the proper format, so we have one option out of six total options.

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_________________ Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at   [#permalink] 06 Jan 2020, 15:51

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