If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 00:36

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 08 Mar 2010
Posts: 20
Schools: Richard Ivey School of Business (University of Western Ontario)
Followers: 0

Kudos [?]: 59 [9] , given: 0

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

### Show Tags

15 Apr 2010, 03:19
9
KUDOS
52
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

59% (01:37) correct 41% (00:39) wrong based on 1067 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?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93389 [18] , given: 10557

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

### Show Tags

15 Apr 2010, 03:41
18
KUDOS
Expert's post
32
This post was
BOOKMARKED
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.

_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 958 [2] , given: 25

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, 21:54
2
KUDOS
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)

No of ways to pick pairs = C(4,2) = 6

No of pairs satisfying condition = 1

Probability = 1/6
_________________
Senior Manager
Joined: 08 Nov 2010
Posts: 417
Followers: 7

Kudos [?]: 106 [1] , given: 161

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

### Show Tags

12 Nov 2010, 10:48
1
KUDOS
Hey guys,

I know its a stupid question, but for some reason i cannot figure it out now... Maybe Im just too tired...

Why cant we do this questions simpler? as in -

1/4*1/3?

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

Thanks.

Sry for the stupidity...
_________________
Manager
Joined: 30 Sep 2010
Posts: 59
Followers: 1

Kudos [?]: 47 [3] , given: 0

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

### Show Tags

12 Nov 2010, 11:12
3
KUDOS
2
This post was
BOOKMARKED
it should be 1/2 * 1/3 = 1/6

First case you can chose either x+y or x-y. So you have 2 options out of 4. second time you have 1 option out of 3
Intern
Joined: 30 Oct 2011
Posts: 11
Followers: 0

Kudos [?]: 6 [0], given: 2

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, 01:04
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.
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93389 [1] , given: 10557

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

### Show Tags

24 Apr 2012, 11:52
1
KUDOS
Expert's post
alexcey wrote:
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.
_________________
Manager
Joined: 08 Jun 2011
Posts: 98
Followers: 1

Kudos [?]: 45 [0], given: 65

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, 12:34
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?

Don't these two problems have the same concept?
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93389 [2] , given: 10557

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, 23:32
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.
_________________
Manager
Joined: 08 Jun 2011
Posts: 98
Followers: 1

Kudos [?]: 45 [0], given: 65

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, 23:51
Bunuel 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.

Thank you very much Bunuel.
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93389 [2] , given: 10557

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, 23:57
2
KUDOS
Expert's post
Bunuel 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.

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.
_________________
Senior Manager
Status: Prevent and prepare. Not repent and repair!!
Joined: 13 Feb 2010
Posts: 274
Location: India
Concentration: Technology, General Management
GPA: 3.75
WE: Sales (Telecommunications)
Followers: 9

Kudos [?]: 89 [0], given: 282

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, 10:29
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+

Intern
Joined: 14 Mar 2012
Posts: 12
GPA: 3.94
Followers: 0

Kudos [?]: 1 [0], given: 0

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, 11:01
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, 08:43, edited 1 time in total.
Intern
Joined: 21 Sep 2010
Posts: 6
Followers: 0

Kudos [?]: 7 [0], given: 9

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, 12:47
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
Moderator
Joined: 02 Jul 2012
Posts: 1231
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 116

Kudos [?]: 1390 [2] , given: 116

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, 21:18
2
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?

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.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Intern
Joined: 01 Dec 2012
Posts: 35
Concentration: Finance, Operations
GPA: 2.9
Followers: 0

Kudos [?]: 48 [2] , given: 8

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, 16:45
2
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
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 25

Kudos [?]: 433 [0], given: 11

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

### Show Tags

07 Dec 2012, 03:34
how many ways to select a pair from 4 expressions?
$$\frac{4!}{2!2!}=6$$

how many pairs to achieve $$x^2-(by)^2$$?
(x+y)(x-y) only or 1 pair

$$\frac{desired}{{all pair possibilities}}=\frac{1}{6}$$
_________________

Impossible is nothing to God.

Manager
Joined: 09 Apr 2013
Posts: 214
Location: United States
Concentration: Finance, Economics
GMAT 1: 710 Q44 V44
GMAT 2: 740 Q48 V44
GPA: 3.1
WE: Sales (Mutual Funds and Brokerage)
Followers: 4

Kudos [?]: 77 [0], given: 40

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

### Show Tags

15 Jun 2013, 12:16
I made a careless error on this problem...

I thought that b(y^2) would be the same as (by)^2.... but nope... i didn't catch the parentheses.
Intern
Joined: 13 Jun 2013
Posts: 8
Followers: 0

Kudos [?]: 1 [0], given: 8

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, 13:44
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.
Manager
Joined: 11 Jan 2011
Posts: 69
GMAT 1: 680 Q44 V39
GMAT 2: 710 Q48 V40
Followers: 0

Kudos [?]: 18 [0], given: 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, 12:43
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 $x^2-(by)^2$ 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.

Thanks,
Rich
Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at   [#permalink] 25 Jul 2013, 12:43

Go to page    1   2    Next  [ 39 posts ]

Similar topics Replies Last post
Similar
Topics:
1 If x = -5 and y = -2, what is the value of 3(x-y)^2 - xy ? 1 11 Mar 2015, 20:23
17 If x,y are numbers >0 and 2^x*5^y=400 8 04 May 2013, 12:42
4 If two of the four expressions x + y , x + 5y , x - y , and 5x - y are 4 21 Mar 2010, 09:53
10 If two of the four expressions x+y, x+5y, x-y, 5x+y are chos 8 04 Oct 2009, 00:57
6 If two of the four expressions x+y, x+5y, x-y, and 5x-y are 12 24 Oct 2007, 21:35
Display posts from previous: Sort by