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Register now and save up to $200 Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code ## Probability Query tagged by: Brent@GMATPrepNow This topic has 4 expert replies and 3 member replies Nijo Senior | Next Rank: 100 Posts Joined 12 Jun 2014 Posted: 45 messages Thanked: 1 times #### Probability Query Wed Aug 13, 2014 1:27 am Elapsed Time: 00:00 • Lap #[LAPCOUNT] ([LAPTIME]) 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 OA is E Thanks Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums! ### GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 13388 messages Followed by: 1782 members Thanked: 12909 times GMAT Score: 790 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) = x² - y² (x-y)(x+5y) = x² + 4xy - 5y² (x-y)(5x-y) = x² - 6xy + y² (x+y)(x+5y) = x² + 6xy + 5y² (x+y)(5x-y) = 5x² + 4xy - y² (x+5y)(5x-y) = 5x² + 24xy - 5y² Of the 6 results, only the first is in the required form: x² - y² = x² - (1y)², where b=1. Thus: P(the product will be in the required form) = 1/6. The correct answer is E. _________________ Mitch Hunt GMAT Private Tutor GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "Thank" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert Brent@GMATPrepNow GMAT Instructor Joined 08 Dec 2008 Posted: 10768 messages Followed by: 1213 members Thanked: 5156 times GMAT Score: 770 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 x² - (by)², where b is an integer? A) 1/2 B) 1/3 C) 1/4 D) 1/5 E) 1/6 Okay, 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: 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 _________________ Brent Hanneson – Founder of GMATPrepNow.com Use our video course along with Check out the online reviews of our course Come see all of our free resources Thanked by: amina.shaikh309 GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! ### Top Member GMATinsight Legendary Member Joined 10 May 2014 Posted: 998 messages Followed by: 21 members Thanked: 203 times Wed Aug 13, 2014 7:29 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 Thanks If we Choose to go by Options... There are 4 Expressions given x-y, x+y, x+5y and 5x-y The total ways of combination of two out of 4 expressions = 4C2 = 6 i.e. Denominator of the probability can be either 6 or any factor of 6 eliminating options C and D Now (5x-y) when multiplied by any other expression given will not give us any result of the form x² - (by)² therefore we have to check the possible out comes of the product of expressions x+y, x+5y and x-y Out of 3 combinations that we can make using them, we can easily identify that only one product of two of these expressions x+y and x-y will give you the desired result. Therefore Favorable outcome = 1 Probability = 1/6 Answer: Option E _________________ Prosper!!! Bhoopendra Singh & Sushma Jha "GMATinsight" Contact Us Testimonials To register for One-on-One FREE ONLINE DEMO Class Call/e-mail e-mail: info@GMATinsight.com Mobile: +91-9999687183 / +91-9891333772 Get in touch for SKYPE-Based Interactive Private Tutoring One-On-One Classes fee - US$40 per hour &
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shrivats Junior | Next Rank: 30 Posts
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Wed Aug 13, 2014 7:43 pm
Its simple. the coefficient of 'x' in the question is 1.

so (x+5y), (5x -y) cannot be multiplied with any of the other options,

so that leaves only one combination, (x+y) * (x-y). Since there are 4 items, there are 4C2 ways of choosing the outcomes. i.e 6 ways.

hope it makes sense.

jervizeloy Junior | Next Rank: 30 Posts
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Tue Aug 09, 2016 1:22 pm
Brent@GMATPrepNow wrote:
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 x² - (by)², where b is an integer?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6
Okay, 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: 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
Hi Brent, how about I solve it considering the probability of choosing algebraic expressions 1 and 2 ( x-y and x+y) out of the 4 expressions, so I would have that P=((1/4)+(1/4))*(1/3). Would it be correct?

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Tue Aug 09, 2016 7:45 pm
Hi All,

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.

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

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Matt@VeritasPrep GMAT Instructor
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Fri Aug 19, 2016 1:42 am
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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