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### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Posted:
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Wed Aug 13, 2014 4:40 am
Quote:
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 xÂ² - (by)Â², where b is an integer?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6
Okay, first recognize that xÂ² - (by)Â² is a DIFFERENCE OF SQUARES.
Here are some examples of differences of squares:
xÂ² - 25yÂ²
4xÂ² - 9yÂ²
49mÂ² - 100kÂ²

In general, we can factor differences of squares as follows:
aÂ² - bÂ² = (a-b)(a+b)

So . . .
xÂ² - 25yÂ² = (x+5y)(x-5y)
4xÂ² - 9yÂ² = (2x+3y)(2x-3y)
49mÂ² - 100kÂ² = (7m+10k)(7m-10k)

-----------

From the 4 expressions (x+y, x+5y ,x-y and 5x-y), only one pair (x+y and x-y) will result in a difference of squares when multiplied.

So, the question now becomes:
If 2 expressions are randomly selected from the 4 expressions, what is the probability that x+y and x-y are both selected?

P(both selected) = [# of outcomes in which x+y and x-y are both selected]/[total # of outcomes]

As always, we'll begin with the denominator.

total # of outcomes
There are 4 expressions, and we must select 2 of them.
Since the order of the selected expressions does not matter, we can use combinations to answer this.
We can select 2 expressions from 4 expressions in 4C2 ways (= 6 ways)

If anyone is interested, we have a free video on calculating combinations (like 4C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

# of outcomes in which x+y and x-y are both selected
There is only 1 way to select both x+y and x-y

So, P(both selected) = 1/6 = E

Cheers,
Brent

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### GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
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GMAT Score:
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Wed Aug 13, 2014 2:00 am
Nijo wrote:
If 2 of the expressions x-y, x+y, x+5y and 5x-y are chosen at random, what is the probability that product will be in 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
FOIL every possible pair:

(x-y)(x+y) = xÂ² - yÂ²
(x-y)(x+5y) = xÂ² + 4xy - 5yÂ²
(x-y)(5x-y) = xÂ² - 6xy + yÂ²
(x+y)(x+5y) = xÂ² + 6xy + 5yÂ²
(x+y)(5x-y) = 5xÂ² + 4xy - yÂ²
(x+5y)(5x-y) = 5xÂ² + 24xy - 5yÂ²

Of the 6 results, only the first is in the required form:
xÂ² - yÂ² = xÂ² - (1y)Â², where b=1.

Thus:
P(the product will be in the required form) = 1/6.

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