Author 
Message 
TAGS:

Hide Tags

Manager
Status: Never ever give up on yourself.Period.
Joined: 23 Aug 2012
Posts: 151
Location: India
Concentration: Finance, Human Resources
GMAT 1: 570 Q47 V21 GMAT 2: 690 Q50 V33
GPA: 3.5
WE: Information Technology (Investment Banking)

Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
28 Dec 2012, 06:18
5
This post received KUDOS
12
This post was BOOKMARKED
Question Stats:
70% (02:00) correct 30% (02:37) wrong based on 276 sessions
HideShow timer Statistics
Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3? A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Don't give up on yourself ever. Period. Beat it, no one wants to be defeated (My journey from 570 to 690) : http://gmatclub.com/forum/beatitnoonewantstobedefeatedjourney570to149968.html



VP
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 1358
GRE 1: 1540 Q800 V740

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
28 Dec 2012, 19:34
4
This post received KUDOS
From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4 Therefore probability = 4/12 = 1/3 Option B
_________________
GyanOne  Top MBA Rankings and MBA Admissions Blog
Top MBA Admissions Consulting  Top MiM Admissions Consulting
Premium MBA Essay ReviewBest MBA Interview PreparationExclusive GMAT coaching
Get a FREE Detailed MBA Profile Evaluation  Call us now +91 98998 31738



Intern
Joined: 17 Nov 2012
Posts: 20

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
01 Jan 2013, 13:52
1
This post received KUDOS
GyanOne wrote: From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4
Therefore probability = 4/12 = 1/3
Option B we need to discuss a little bit here. I agree that y is prime. However y is also chosen, so the probability for a prime number chosen is 13/41. And the probability for x divisible by 3 is 4/12 So the result is (13/41)*(4/12). Please correct me if I'm wrong, thanks



VP
Joined: 02 Jul 2012
Posts: 1213
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
01 Jan 2013, 21:37
akhandamandala wrote: GyanOne wrote: From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4
Therefore probability = 4/12 = 1/3
Option B we need to discuss a little bit here. I agree that y is prime. However y is also chosen, so the probability for a prime number chosen is 13/41. And the probability for x divisible by 3 is 4/12 So the result is (13/41)*(4/12). Please correct me if I'm wrong, thanks We do not have to worry about choosing y. We are already given that y is prime. So no matter what number we choose as y, we still have to only find out the probability for choosing x. If otherwise, the question had asked for the number of ways in which xy would be divisible by 3, then the number of ways in which y could be selected would have to have been considered.
_________________
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



VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1372
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
01 Jan 2013, 21:58
Would it have been complex if it were given that B consists of all prime numbers from 1 to 50, inclusive? Will it be \(1/3 + 1/15\) then?
_________________
Prepositional Phrases ClarifiedElimination of BEING Absolute Phrases Clarified Rules For Posting www.UnivScholarships.com



VP
Joined: 02 Jul 2012
Posts: 1213
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 01:56
Marcab wrote: Would it have been complex if it were given that B consists of all prime numbers from 1 to 50, inclusive? Will it be \(1/3 + 1/15\) then? Is that not what the question is already effectively asking??
_________________
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



VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1372
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 02:21
1
This post received KUDOS



VP
Joined: 02 Jul 2012
Posts: 1213
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 02:28
Marcab wrote: The question says that B consists of all integers from 10 to 50, inclusive. True.. But the options to choose Y is essentially a set of prime numbers between 10 and 50...
_________________
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



VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1372
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 02:49
MacFauz wrote: Marcab wrote: The question says that B consists of all integers from 10 to 50, inclusive. True.. But the options to choose Y is essentially a set of prime numbers between 10 and 50... I agree that indirectly it is stating that B consists of prime numbers from 10 to 50, but my query was rather different. Suppose, if it were given that B consists of all prime numbers from 1 to 50 rather than 10 to 50, then what will be the answer. I hope, I am being clear this time.
_________________
Prepositional Phrases ClarifiedElimination of BEING Absolute Phrases Clarified Rules For Posting www.UnivScholarships.com



VP
Joined: 02 Jul 2012
Posts: 1213
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 03:22
Marcab wrote: MacFauz wrote: Marcab wrote: The question says that B consists of all integers from 10 to 50, inclusive. True.. But the options to choose Y is essentially a set of prime numbers between 10 and 50... I agree that indirectly it is stating that B consists of prime numbers from 10 to 50, but my query was rather different. Suppose, if it were given that B consists of all prime numbers from 1 to 50 rather than 10 to 50, then what will be the answer. I hope, I am being clear this time. Oh.. Sorry. I read that as 10 to 50 again... I'm not very sure. But let me give it a try edit: If it is 1 to 50, For required outcome : y is 3 and x is any number or y is not 3 and x is a multiple of 3 \(\frac{1}{15} * 1 + \frac{14}{15} * \frac{4}{12}\) \(= \frac{12}{15*12} + \frac{14*4}{15*12}\) \(= \frac{68}{180} = \frac{17}{45}\) A very ugly answer and lots of calculator work.. There should be an easier more correct way..
_________________
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
Last edited by MacFauz on 02 Jan 2013, 03:38, edited 1 time in total.



Intern
Joined: 17 Nov 2012
Posts: 20

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 03:33
1
This post received KUDOS
MacFauz wrote: akhandamandala wrote: GyanOne wrote: From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4
Therefore probability = 4/12 = 1/3
Option B we need to discuss a little bit here. I agree that y is prime. However y is also chosen, so the probability for a prime number chosen is 13/41. And the probability for x divisible by 3 is 4/12 So the result is (13/41)*(4/12). Please correct me if I'm wrong, thanks We do not have to worry about choosing y. We are already given that y is prime. So no matter what number we choose as y, we still have to only find out the probability for choosing x. If otherwise, the question had asked for the number of ways in which xy would be divisible by 3, then the number of ways in which y could be selected would have to have been considered. Thanks for your reply. I would like to defend my idea: "Set B consists of all the integers between 10 and 50, inclusive" means set B will contain prime and nonprime numbers "If x is a number chosen randomly from Set A" means x is chosen randomly from set A without limitation condition "y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y" means y is also chosen like x but with condition. So y is not a given number, y must be chosen from set B and satisfy the condition "has no factor z such that 1 < z < y". Now take a look from another view: "Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, what is the probability that the product xy is divisible by 3?" The condition for y in this version is removed, so what is your idea in this case ???



VP
Joined: 02 Jul 2012
Posts: 1213
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 03:51
akhandamandala wrote: MacFauz wrote: akhandamandala wrote: From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4
Therefore probability = 4/12 = 1/3
Option B we need to discuss a little bit here. I agree that y is prime. However y is also chosen, so the probability for a prime number chosen is 13/41. And the probability for x divisible by 3 is 4/12 So the result is (13/41)*(4/12). Please correct me if I'm wrong, thanks We do not have to worry about choosing y. We are already given that y is prime. So no matter what number we choose as y, we still have to only find out the probability for choosing x. If otherwise, the question had asked for the number of ways in which xy would be divisible by 3, then the number of ways in which y could be selected would have to have been considered. Thanks for your reply. I would like to defend my idea: "Set B consists of all the integers between 10 and 50, inclusive" means set B will contain prime and nonprime numbers "If x is a number chosen randomly from Set A" means x is chosen randomly from set A without limitation condition "y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y" means y is also chosen like x but with condition. So y is not a given number, y must be chosen from set B and satisfy the condition "has no factor z such that 1 < z < y". Now take a look from another view: "Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, what is the probability that the product xy is divisible by 3?" The condition for y in this version is removed, so what is your idea in this case ??? In such a case, there are 13 ways in which y can be selected to be a multiple of 3 and 28 ways in which y can be selected to not be a multiple of 3. \(\frac{13}{41} * 1 + \frac{28}{41}* \frac{4}{12}\) \(= \frac{67}{123}\) i believe the fact the y is a prime number greater than 10 given in the original question is significant because it clearly says that y CANNOT be a multiple of 3.
_________________
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: 17 Nov 2012
Posts: 20

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 04:18
1
This post received KUDOS
MacFauz wrote: akhandamandala wrote: MacFauz wrote: we need to discuss a little bit here.
I agree that y is prime. However y is also chosen, so the probability for a prime number chosen is 13/41. And the probability for x divisible by 3 is 4/12 So the result is (13/41)*(4/12).
Please correct me if I'm wrong, thanks We do not have to worry about choosing y. We are already given that y is prime. So no matter what number we choose as y, we still have to only find out the probability for choosing x. If otherwise, the question had asked for the number of ways in which xy would be divisible by 3, then the number of ways in which y could be selected would have to have been considered. Thanks for your reply. I would like to defend my idea: "Set B consists of all the integers between 10 and 50, inclusive" means set B will contain prime and nonprime numbers "If x is a number chosen randomly from Set A" means x is chosen randomly from set A without limitation condition "y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y" means y is also chosen like x but with condition. So y is not a given number, y must be chosen from set B and satisfy the condition "has no factor z such that 1 < z < y". Now take a look from another view: "Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, what is the probability that the product xy is divisible by 3?" The condition for y in this version is removed, so what is your idea in this case ??? In such a case, there are 13 ways in which y can be selected to be a multiple of 3 and 28 ways in which y can be selected to not be a multiple of 3. \(\frac{13}{41} * 1 + \frac{28}{41}* \frac{4}{12}\) \(= \frac{67}{123}\) i believe the fact the y is a prime number greater than 10 given in the original question is significant because it clearly says that y CANNOT be a multiple of 3. That's very good. We step by step go to the mutual agreement that "y is a number chosen randomly from Set B" means there are various ways in which y is chosen; so y may be a prime or may be a nonprime number. What is the probability that y will be a prime number?



VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1372
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 04:42
Quote: For required outcome : y is 3 and x is any number or y is not 3 and x is a multiple of 3 MacFauz why have you neglected the 3 in the other one? Can't we take any number just as it was done earlier?
_________________
Prepositional Phrases ClarifiedElimination of BEING Absolute Phrases Clarified Rules For Posting www.UnivScholarships.com



VP
Joined: 02 Jul 2012
Posts: 1213
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Jan 2013, 20:44
@akhandamandala : I believe that in the original question, it is given that "y" is a prime number. So, the probability that y will be prime is 100%. @Marcab : I have already considered a situation where y is 3 and x is a multiple of 3 in the first scenario. Hence, in the second scenario, I am considering only a situation where y is not a multiple of 3 and x is a multiple of 3. If in the second case, "y" is not 3 and "x" is also not a multiple of 3, then the product would not be divisible by 3.
_________________
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: 17 Nov 2012
Posts: 20

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
03 Jan 2013, 02:53
Set B does not contains prime numbers only, that's the set of ALL integers from 10 to 50, and there are 11 (not 13 as aforementioned) prime numbers in the range [10, 50]. In the other hand, y is chosen RANDOMLY from the set B, and y is prime; so P(y is prime) = 1 is not persuasive.
I posted the original quiz on facebook (without the multiple choices), there's a reply with the idea that the probability for y is prime is <1. So far my idea does not stand alone. Why don't you make a survey in which you post this original quiz (omit the multiple choices) on your facebook or blog?



Manager
Status: Never ever give up on yourself.Period.
Joined: 23 Aug 2012
Posts: 151
Location: India
Concentration: Finance, Human Resources
GMAT 1: 570 Q47 V21 GMAT 2: 690 Q50 V33
GPA: 3.5
WE: Information Technology (Investment Banking)

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
03 Jan 2013, 04:28
1
This post received KUDOS
2
This post was BOOKMARKED
If y has no factor z such that 1 < z < y, then y must be prime. Let's look at a few examples to see why this is true: 6 has a factor 2 such that 1 < 2 < 6: 6 is NOT prime 15 has a factor 5 such that 1 < 5 < 15: 15 is NOT prime 3 has NO factor between 1 and 3: 3 IS prime 7 has NO factor between 1 and 7: 7 IS prime Because it is selected from Set B, y is a prime number between 10 and 50, inclusive. The only prime number that is divisible by 3 is 3, so y is definitely not divisible by 3. Thus, xy is only divisible by 3 if x itself is divisible by 3. We can rephrase the question: “What is the probability that a multiple of 3 will be chosen randomly from Set A?” There are 21 – 10 + 1 = 12 terms in Set A. Of these, 4 terms (12, 15, 18, and 21) are divisible by 3 Thus, the probability that x is divisible by 3 is 4/12 = 1/3. The correct answer is B. (Source : Manhattan GMAT Advanced Quant Guide)
_________________
Don't give up on yourself ever. Period. Beat it, no one wants to be defeated (My journey from 570 to 690) : http://gmatclub.com/forum/beatitnoonewantstobedefeatedjourney570to149968.html



Intern
Joined: 17 Nov 2012
Posts: 20

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
03 Jan 2013, 13:04
1
This post received KUDOS
I think this question is very controversial. There would be significant amount of people understand that y is chosen randomly in order to satisfy the condition y is prime, not that y is given as a prime.



NonHuman User
Joined: 09 Sep 2013
Posts: 13850

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
02 Mar 2014, 10:18
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Rice (Jones) Thread Master
Joined: 25 Feb 2014
Posts: 235

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Show Tags
04 Mar 2014, 20:23
y has no factor z such that 1 < z < y this means that y is a prime and it has to lie between 10 and 50 inclusive. Now, since the product of xy is divisible by 3 so that means that we have to look for multiples of 3 in set A since primes from set B are all greater than 3. Set B has 12 elements and out of which 4 are multiples of 3 so probability should be 4/12 = 1/3. Consider giving kudos if my post helped.
_________________
Consider KUDOS if my post helped
I got the eye of the tiger, a fighter, dancing through the fire 'Cause I am a champion and you're gonna hear me roar




Re: Set A consists of all the integers between 10 and 21, inclus
[#permalink]
04 Mar 2014, 20:23



Go to page
1 2
Next
[ 29 posts ]



