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:

Set A consists of all the integers between 10 and 21, inclus [#permalink]

Show Tags

28 Dec 2012, 06:18

2

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

72% (03:02) correct
28% (02:20) wrong based on 220 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?

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/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

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

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]

Show Tags

21 Sep 2015, 13:04

4

This post received KUDOS

itwarriorkarve wrote:

tarunktuteja wrote:

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.

After following all the discussion on this thread, I feel that considering only x is not correct. The question clearly states that y is the probability of selected a prime number from the set B. And then it asks for the probability of product xy being divisible by 3! Clearly we cannot assume that y will always be prime. We need to take it into consideration that we have picked y as a prime number from B and x as a factor of 3 from set A. The events x and y are definitely not mutually exclusive in this case as we need to find the probability for product xy . Also there can be cases where y if not prime could be a multiple of 3 and then xy would still be divisible by 3!

Please highlight if I'm overlooking something, But clearly the options and OA isnt correct.

"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" you are missing the bold face part. It is already given that we are choosing y among the set which is govern by 1<z<y. If this condition was not given, we would have followed your approach.
_________________

Please award kudos if you like my explanation. Thanks

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).

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 non-prime 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 ???

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 non-prime 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.

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 non-prime number. What is the probability that y will be a prime number?

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

1

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/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

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.

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]

Show Tags

16 May 2015, 23:36

1

This post received KUDOS

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.
_________________

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]

Show Tags

21 Sep 2015, 11:53

1

This post received KUDOS

alexeykaplin wrote:

I'm little bit confused here. The way provided is pretty logical and I like it but I have a conundrum with more choosey approach.

In Set A we have 12 integers, these ones divisble by 3 - 12, 15,18, 21 - 4 numbers In set B we have 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 - 11 prime numbers

prob of xy divisible by 3 = (xy's divisble by 3)/(total possible xy's)

xy's divisible by 3= 11x4=44 total possible xy's= 11x12-4=128, I deduct 4 to avoid double counting of 11, 13, 17,19 as these appear in both sets

so, I have 44/128=11/32, slightly more that 1/3.

please, tell me where am I wrong here.

It is already given that y is being chosen from the list of prime numbers, and hence there will be a probability of 1 if we choose any number from y. And for x, we have to find the probability of choosing a multiple of 3. Hence, the final probability is 4/12*1 = 1/3
_________________

Please award kudos if you like my explanation. Thanks

Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]

Show Tags

21 Sep 2015, 12:34

1

This post received KUDOS

tarunktuteja wrote:

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.

After following all the discussion on this thread, I feel that considering only x is not correct. The question clearly states that y is the probability of selected a prime number from the set B. And then it asks for the probability of product xy being divisible by 3! Clearly we cannot assume that y will always be prime. We need to take it into consideration that we have picked y as a prime number from B and x as a factor of 3 from set A. The events x and y are definitely not mutually exclusive in this case as we need to find the probability for product xy . Also there can be cases where y if not prime could be a multiple of 3 and then xy would still be divisible by 3!

Please highlight if I'm overlooking something, But clearly the options and OA isnt correct.

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.

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?
_________________

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.
_________________

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

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 non-prime 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.

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.

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...