Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Schools: Richard Ivey School of Business (University of Western Ontario)

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

15 Apr 2010, 04:19

12

This post received KUDOS

80

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

57% (00:42) correct
43% (00:41) wrong based on 1667 sessions

HideShow timer Statistics

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 of x^2 -(by)^2, where b is an integer?

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 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

Four expressions can be paired in \(C^2_4=6\) ways (choosing 2 out of 4);

\(x^2 - (by)^2=(x-by)(x+by)\) and only one pair is making such expression: \((x+1*y)(x-1*y)\), hence probability is 1/6.

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

21 Sep 2010, 22:54

3

This post received KUDOS

1

This post was BOOKMARKED

vigneshpandi wrote:

Can anyone explain how is this derived?

Attachment:

Probability of the product.JPG

The only way to get x^2 - (by)^2 is if you multiply (x+by) with (x-by). Looking at the expressions given, the only pair that will satisfy this is (x+y) & (x-y)

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

24 Apr 2012, 02:04

How do I know from the phrasing of this problem that, once chosen, an expression is taken off the list (ie not replaced)? When I was solving the problem I assumed that an expression of form (x+y)(x+y) is possible. In that case the number of combinations is 8.

How do I know from the phrasing of this problem that, once chosen, an expression is taken off the list (ie not replaced)? When I was solving the problem I assumed that an expression of form (x+y)(x+y) is possible. In that case the number of combinations is 8.

Question says: "If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random..." so, at least for me, it's quite obvious that there is no replacement whatsoever. Also if it were the case the question would explicitly mention that.
_________________

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

30 Apr 2012, 13:34

How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?

For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?

How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?

For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?

Don't these two problems have the same concept?

Mathematically the probability of picking two expressions simultaneously (at once), or picking them one at a time (without replacement) is the same. _________________

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

01 May 2012, 00:51

Bunuel wrote:

Lstadt wrote:

How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?

For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?

Don't these two problems have the same concept?

Mathematically the probability of picking two expressions simultaneously (at once), or picking them one at a time (without replacement) is the same.

Hello Bunuel,

Maybe telling you how I solved the problem (and got a wrong answer) will explain how I was confused.

Since the wanted product is (x + y ) (x - y), I assumed that I will be picking (x+y) then (x-y) [from the expressions) which means choosing 1/4 then choosing 1/3 and multiplying them to get 1/12 which is wrong. I think I approached it the same was as 144144 (posted above) did.

How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?

For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?

Don't these two problems have the same concept?

Mathematically the probability of picking two expressions simultaneously (at once), or picking them one at a time (without replacement) is the same.

Hello Bunuel,

Maybe telling you how I solved the problem (and got a wrong answer) will explain how I was confused.

Since the wanted product is (x + y ) (x - y), I assumed that I will be picking (x+y) then (x-y) [from the expressions) which means choosing 1/4 then choosing 1/3 and multiplying them to get 1/12 which is wrong. I think I approached it the same was as 144144 (posted above) did.

Thank you very much Bunuel.

It should be 1/4*1/3*2, since you can choose (x+y) first and then (x-y) or vise versa.
_________________

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

23 Jul 2012, 11:29

This is how I analysed this. x-y is a must here since the form asked is x^2-by^2. So probabability of selecting x-y is 1/4 and probability of selecting the rest is 2/3. the probability of both is 1/4*2/3. Is this right???
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

23 Jul 2012, 12:01

144144 wrote:

I understand perfectly the 4C2 way, but i want to understand the others as well... .

I was hoping someone can help me understand this. What is the 4C2 way? C means what?

How I thought of the problem was each equation has a 1/4 chance

1/4 1/4 1/4 1/4 - Two out of the 4 equations = success so the probability of getting one of the equations we need is 2/4 or 1/2 then we have 3 equations left over, and the probability of getting the second equation we need is 1/3 so the probability of getting both equations is 1/2*1/3 = 1/6

However I only got this after you guys said that we needed a x+y, x-y

The b in the question threw me off, I am not sure that if I was given a similar problem some weeks from now I would understand what to do with that b

Last edited by akuma86 on 24 Jul 2012, 09:43, edited 1 time in total.

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

23 Jul 2012, 13:47

Akuma,

4C2 stands for binomial coefficient with parameters 4 and 2. In layman's terms, it says how many unique combinations of two can I derive from a set of 4 items. In mathematical terms, it simply means 4! / (2! x (4-2)!) which simplifies to 4! / (2! x 2!) which is equivalent to 6. Now you just need to figure out how many of the combinations fulfill your criteria - in this case 1 - and place that number over 6 to get the probability. To get the amount of unique pairings of Y numbers in a set of X (where order does not matter) I believe you can always use this shorthand formula

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

09 Nov 2012, 22:18

2

This post received KUDOS

wonderwhy wrote:

If 2 of the 4 expressions x+y, x+5y, x-y, 5x-y are chosen at random, what is the probability that the product will be in the form of x square i (b*y) square if b is an integer?

Answer options: 1/2 1/3 1/4 1/5 1/6

How is this computed?

Total No. of ways of selecting 2 items from 4 items = 4C2 = 6. We can see from the 4 items given that only one pair when multiplied will be of the form (x+ay)(x-ay). So answer is 1/6. E.

Kudos Please... If my post helped.
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

04 Dec 2012, 17:45

2

This post received KUDOS

1

This post was BOOKMARKED

VERY SIMPLE WAY TO SOLVE THIS PROB WITHOUT ANY COMPLEX THEORY N ANY CALCULATION : FROM THE QUESTION STEM , YOU COULD UNDERSTAND VERY EASILY THAT ONLY ONE COMBINATION I.E (X+Y) AND (X-Y) IS THERE TO GET X2-(BY)2 SO WE HAVE TO CHOOSE ONLY (X+Y) AND (X-Y) wE CAN SELECT IT IN TWO WAY I.E . (X+Y) AND THEN (x-Y) OR (X-Y) AND THEN (X+Y) FOR FIRST CASE PROBABILITY =(1/4*1/3) FOR SECOND CASE PROBABILITY =(1/4*1/3) sO TOTAL PROBABILITY =2*((1/4*1/3)=1/6 HOPE IT COULD SIMPLY WORK FOR YOU

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

30 Jun 2013, 14:44

Great explanations in this topic. Again, when explained, easier than it appeared.

As there are 2 steps to this (number of possible outcomes, number of valid pairs); if you're lost at figuring out the latter it is maybe of some use that it is easy to realize that if there are 6 possible outcomes, only 1/2, 1/3 and 1/6 could be valid answers as x/6 can't be simplified to 1/4 or 1/5. Now you only need to figure out if there are 1, 2 or 3 valid pairs, or if time ran out guess 1 out of 3.

Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink]

Show Tags

25 Jul 2013, 13:43

I also got 1/6 but not really understanding why the equation pair is (x+y)(x-y).

My thoughts were that we needed to factor into (x-by)(x+by) where the factored "b" was the squared root of the original "b", or essentially a placeholder for a perfect square.

This led me to think that the only equation pair possible that would work to satisfy the original ask would be (x+5y) and (x-y) since we need two equations with differing signs but also needed an constant infront of the "y" term.