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Probability Problem

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by Brent@GMATPrepNow » Tue Jan 31, 2017 6:56 am
Here's the complete question...
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 » Tue Jan 31, 2017 12:54 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] » Tue Jan 31, 2017 3:25 pm
Hi aditiniyer,

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

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Rich
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by Matt@VeritasPrep » Wed Feb 01, 2017 6:30 pm
This one is asked pretty often, so don't feel bad for struggling with it!

x² - (by)² factors as

(x + by) * (x - by)

So this will only work if both parts of the pair have the same coefficient on y. Given our four possibilities, the only pair that works is (x + y) * (x - y), since they each have 1y. So only one pair out of (4 choose 2), or 6 total, works, giving us 1/6.
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by Jay@ManhattanReview » Thu Feb 16, 2017 5:02 am
aditiniyer wrote:If 2 of the 4 expressions x+y, x+5y,x-y & 5x-y are chosen at random, what is the prob that their product will be of the form x^2-by^2 where b is an integer.

Ans 1/6, I got 1/4 :(
Many nice solutions are provided by experts. Here is my take on this question.

We have to choose two expressions out of four: (x+y), (x+5y), (x-y) & (5x-y) such that the product of them is in the form of (x^2-by^2), where b is an integer.

We see that in the desired quadratic expression (x^2-by^2), there is no term of 'xy', thus, we must choose the TWO terms such that we avoid 'xy' terms when they are multiplied.

Let's analyses possible pairs of two terms in (x+y), (x+5y), (x-y) & (5x-y).

1. (x+y) and (x+5y): It will have a 'xy' term when multiplied, so ignore it.
2. (x+y) and (x-y) = x^ - y^2: This in the form of x^2-by^2. Here b = 1. We need this!
3. (x+y) and (5x-y): It will have a 'xy' term when multiplied, so ignore it.
4. (x+5y) and (x-y): It will have a 'xy' term when multiplied, so ignore it.
5. (x+5y) and (5x-y): It will have a 'xy' term when multiplied, so ignore it.
6. (x-y) and (5x-y): It will have a 'xy' term when multiplied, so ignore it.

Out of six outcomes, the desired one is only one outcome: #2.

Thus, the probability = [spoiler]1/6[/spoiler]

Download free ebook: Manhattan Review GMAT Quantitative Question Bank Guide

Hope this helps!

-Jay
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