• Award-winning private GMAT tutoring
Register now and save up to $200 Available with Beat the GMAT members only code • Free Trial & Practice Exam 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 • Reach higher with Artificial Intelligence. Guaranteed Now free for 30 days 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 • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE 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 • 5-Day Free Trial 5-day free, full-access trial TTP Quant 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 ## 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 Upvotes: 1 #### Probability Query Wed Aug 13, 2014 1:27 am 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 ### GMAT/MBA Expert Rich.C@EMPOWERgmat.com Elite Legendary Member Joined 23 Jun 2013 Posted: 9301 messages Followed by: 478 members Upvotes: 2867 GMAT Score: 800 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. Final Answer: E GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert Matt@VeritasPrep GMAT Instructor Joined 12 Sep 2012 Posted: 2637 messages Followed by: 114 members Upvotes: 625 Target GMAT Score: V51 GMAT Score: 780 Fri Aug 19, 2016 1:42 am This one is asked pretty often, so don't feel bad for struggling with it! xÂ² - (by)Â² 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. Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now! ### GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 14025 messages Followed by: 1812 members Upvotes: 13060 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) = 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. 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 "UPVOTE" 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: 11397 messages Followed by: 1229 members Upvotes: 5254 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 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 _________________ Brent Hanneson â€“ 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 GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### Top Member GMATinsight Legendary Member Joined 10 May 2014 Posted: 1044 messages Followed by: 23 members Upvotes: 205 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 xÂ² - (by)Â² 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 _________________ Bhoopendra Singh & Sushma Jha - Founder "GMATinsight" Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772 To register for One-on-One FREE ONLINE DEMO Class Call/e-mail One-On-One Private tutoring fee - US$40 per hour & for FULL COURSE (38 LIVE Sessions)-US\$1000

shrivats Junior | Next Rank: 30 Posts
Joined
11 Jun 2014
Posted:
23 messages
5
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
Joined
08 Aug 2016
Posted:
13 messages
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 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
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?

### Best Conversation Starters

1 lheiannie07 88 topics
2 Roland2rule 70 topics
3 ardz24 60 topics
4 LUANDATO 59 topics
5 M7MBA 50 topics
See More Top Beat The GMAT Members...

### Most Active Experts

1 Rich.C@EMPOWERgma...

EMPOWERgmat

138 posts
2 GMATGuruNY

The Princeton Review Teacher

131 posts
3 Brent@GMATPrepNow

GMAT Prep Now Teacher

129 posts
4 Jeff@TargetTestPrep

Target Test Prep

114 posts
5 Scott@TargetTestPrep

Target Test Prep

100 posts
See More Top Beat The GMAT Experts