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by rakaisraka » Wed Aug 19, 2015 3:18 pm
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 x^2 - (by)^2, where b is an integer.
1) 1/2
2) 1/3
3) 1/4
4) 1/5
5) 1/6

Can anyone explain please.
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by Brent@GMATPrepNow » Wed Aug 19, 2015 3:21 pm
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 x² - (by)², where b is an integer?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6
Important concept tested here
First recognize that x² - (by)² is a DIFFERENCE OF SQUARES.
Here are some examples of differences of squares:
x² - 25y²
4x² - 9y²
49m² - 100k²

In general, we can factor differences of squares as follows:
a² - b² = (a-b)(a+b)

So . . .
x² - 25y² = (x+5y)(x-5y)
4x² - 9y² = (2x+3y)(2x-3y)
49m² - 100k² = (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: https://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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by Brent@GMATPrepNow » Wed Aug 19, 2015 3:21 pm
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 x² - (by)², where b is an integer?

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

Once we recognize (from my last post) that we must find P(x+y and x-y are both selected), we can write:
P(x+y and x-y are both selected) = P(one of the two expressions is selected on 1st pick AND the other expression is selected on 2nd pick)
= P(one of the two expressions is selected on 1st pick) x P(the other expression is selected on 2nd pick)
= 2/4 x 1/3
= [spoiler]1/6[/spoiler]
= E

Cheers,
Brent
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by [email protected] » Wed Aug 19, 2015 3:35 pm
Hi rakaisraka,

This question is based heavily on algebra patterns. If you can spot the patterns involved, then you can save some time; even if you can't spot it though, a bit of 'brute force' math will still get you the solution.

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.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich
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