GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 30 May 2020, 08:08 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Set A consists of all the integers between 10 and 21, inclus

Author Message
TAGS:

### Hide Tags

Intern  Joined: 14 Jul 2015
Posts: 6
Location: United States
Concentration: Strategy, Operations
GPA: 3.6
WE: Engineering (Energy and Utilities)
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

1
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
Intern  Joined: 09 Jun 2015
Posts: 9
Concentration: Statistics, Technology
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

1
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.
Intern  Joined: 14 Jul 2015
Posts: 6
Location: United States
Concentration: Strategy, Operations
GPA: 3.6
WE: Engineering (Energy and Utilities)
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

3
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.
Senior Manager  S
Joined: 08 Dec 2015
Posts: 277
GMAT 1: 600 Q44 V27
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

I think this is a bad question.. The way its written it seems that you have to chose 4 out of 11 (the multiples of 3 from 10 to 21) and then it will have to match a prime chosen from 10 to 50, so 11 possibilities of out 50. So we get 4/11*11/40.

The question should read that a number WAS CHOSEN from bla bla, so its clear that the action of choosing it from the respective sample has already taken place.

Anyways maybe its just me..
Manager  B
Joined: 16 Mar 2016
Posts: 119
Location: France
GMAT 1: 660 Q47 V33
GPA: 3.25
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

and y has no factor z such that 1 < z < y
Question poorly worded, does the probability includes the drawing of y ?
Director  G
Joined: 26 Oct 2016
Posts: 592
Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

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.
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
_________________
Thanks & Regards,
Anaira Mitch
Manager  B
Joined: 24 May 2014
Posts: 83
Location: India
GMAT 1: 590 Q39 V32
GRE 1: Q159 V151

GRE 2: Q159 V153
GPA: 2.9
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

Though the question is simple, so much prose makes it seem complicated. 'y' has to be a prime number as the integer randomly selected cannot have a factor 'z' of the form 1<z<y.

y=11,13,17,19,23,29,31,37,41,43 (Seeing this sequence it can be determined that none of the values are divisible by 3)

x=10,11,12,13,14,15,16,17,18,19,20,21 (12,15,18,21 are the only integers divisible by 3). Hence probability = 4/12 = 1/3
VP  V
Joined: 27 May 2012
Posts: 1027
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

daviesj wrote:
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

Set y that satisfies the given criteria : { 11,13,17,19,23,29,31,37,41,43,47,}
set x = { 12,15,18,21}

No. of pairs of XY that are divisible by 3 = 44
Total xy = total 12 elements of set x * 11 elements of set y that satisfy the condition , hence total xy = 12 *11

Probability = $$\frac{44}{12*11}$$ = $$\frac{1}{3}$$

Main point of confusion may be when selecting the total number of pairs , remember when selecting one element of Y we have a certain condition that it should be prime , when selecting one element from x we have no such condition. Hence total cases are 12 * 11 .Out of which favorable cases are 4*11 = 44. Hence Probability = $$\frac{44}{12*11}$$ = $$\frac{1}{3}$$

Hope this helps .
_________________
- Stne
Manager  B
Joined: 28 Jan 2018
Posts: 50
Location: Netherlands
Concentration: Finance
GMAT 1: 710 Q50 V36 GPA: 3
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

daviesj wrote:
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

y has no factor z such that 1 < z < y, implied that y is a prime number.

Therefore x must be a multiple of 3 and is between 10 and 21 inclusive, we got the following number: 12,15,18,21

The probability that xy is divisible by 3: $$\frac{4}{12}$$ = $$\frac{1}{3}$$ => Answer (B)
Non-Human User Joined: 09 Sep 2013
Posts: 15021
Re: Set A consists of all the integers between 10 and 21, inclus  [#permalink]

### Show Tags

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.
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 15021
Re: Set A consists of all the integers between 10 and 21, inclusive  [#permalink]

### Show Tags

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, inclusive   [#permalink] 17 Mar 2020, 09:56

Go to page   Previous    1   2   [ 31 posts ]

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