It is currently 21 Oct 2017, 11:17

### 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
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

# Events & Promotions

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

# m15,#10

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
VP
Joined: 18 May 2008
Posts: 1258

Kudos [?]: 527 [0], given: 0

m15,#10 [#permalink]

### Show Tags

28 Nov 2008, 02:10
1
This post was
BOOKMARKED
Set $$S$$ consists of all prime integers less than 10. If two numbers are chosen form set $$S$$ at random, what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen?

(A) $$\frac{1}{3}$$
(B) $$\frac{2}{3}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{7}{10}$$
(E) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

What is wrong with my approach?
The set consists of (2,3,5,7)
Niow acc to soncition the nos satisfying are (5,7), (3,5) and (3,7)
Thus the probabiliy is (1/4*1/4)+(1/4*1/4)+(1/4*1/4)=3/16

But this is not the ans

Kudos [?]: 527 [0], given: 0

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [1], given: 0

Re: m15,#10 [#permalink]

### Show Tags

28 Nov 2008, 03:21
1
This post received
KUDOS
It should be (1/4*1/3)+(1/4*1/3)+(1/4*1/3) = 3/12 = 1/4.

But, the sequence is immaterial here as we are interested in sum. Hence, (5,3) and (3,5) will give the same result. Thus, multiply 1/4 by 2 and 1/2 should be the answer.

Alternatively, I will solve it as follows:

favorable outcome = 3 (as you have listed earlier).

Since, order does not matter, total outcome = 4C2 = 6

Hence, probability = 3/6 = 1/2

Kudos [?]: 279 [1], given: 0

Senior Manager
Affiliations: CFA
Joined: 21 Dec 2008
Posts: 384

Kudos [?]: 121 [1], given: 64

Location: United States (NY)
Schools: Columbia - Class of 2013
GMAT 1: 710 Q45 V43
GMAT 2: 760 Q49 V45
Re: m15,#10 [#permalink]

### Show Tags

21 Dec 2009, 07:59
1
This post received
KUDOS
All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2.

3/4 * 2/3 = 6/12 = 1/2

So C.
_________________

Kudos [?]: 121 [1], given: 64

Math Expert
Joined: 02 Sep 2009
Posts: 41894

Kudos [?]: 129130 [1], given: 12194

Re: m15,#10 [#permalink]

### Show Tags

30 Dec 2010, 02:12
1
This post received
KUDOS
Expert's post
ritula wrote:
Set $$S$$ consists of all prime integers less than 10. If two numbers are chosen form set $$S$$ at random, what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen?

(A) $$\frac{1}{3}$$
(B) $$\frac{2}{3}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{7}{10}$$
(E) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

S={2, 3, 5, 7}

The simplest way would be to realize that we choose half of the numbers (basically we divide the group of 4 into two smaller groups of 2) and since a tie is not possible then the probability that the product of the numbers in either of subgroup is more than that of in another is 1/2 (the probability doesn't favor any of two subgroups).

Answer: C.
_________________

Kudos [?]: 129130 [1], given: 12194

Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 146

Kudos [?]: 222 [1], given: 0

Location: Toronto
Re: m15,#10 [#permalink]

### Show Tags

03 Jan 2011, 13:17
1
This post received
KUDOS
zahuruddin wrote:
i think the ans is 1/3

Pls check whether my understanding is correct

We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)

There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.

So p= 2/4C2 = 2/6 = 1/3 Ans

Hi!

1 isn't a prime number, which is the first reason that you're deriving a different answer.

The second reason is that you've misinterpreted the question - we're not comparing the product of the two numbers we select to the product of the non-primes less than 10; we're comparing the product of the two numbers we select to the product of the other primes less than 10.

So, our set is {2, 3, 5, 7} and we're comparing the products of two pairs of numbers from within the set.

The quickest ways to solve are to either use Bunuel's intuitive approach (we're comparing 2 vs 2 with no ties, so symmetry will lead to half the pairs being greater than the other half) or via brute force. Since there are only 6 possibilities, it doesn't take long to count them out:

2*3 vs 5*7 - not greater
2*5 vs 3*7 - not greater
2*7 vs 3*5 - not greater
3*5 vs 2*7 - greater
3*7 vs 2*5 - greater
5*7 vs 3*5 - greater

Probability = (# of desired outcomes)/(total # of possibilities) = 3/6 = 1/2

Kudos [?]: 222 [1], given: 0

Manager
Joined: 14 Oct 2008
Posts: 160

Kudos [?]: 64 [0], given: 0

Re: m15,#10 [#permalink]

### Show Tags

28 Nov 2008, 03:31
I would go with exactly the same approach as the second one stated by scthakur.

Total outcomes = 2 numbers out of 4 hence 4C2 =6

Possible outcomes = 2 numbers out of 3 hence 3C2 = 3

Probability = 3/6 = 1/2

Whats the QA ?

Kudos [?]: 64 [0], given: 0

Manager
Joined: 18 Nov 2008
Posts: 115

Kudos [?]: 34 [0], given: 0

Re: m15,#10 [#permalink]

### Show Tags

28 Nov 2008, 05:16
scthakur, nice explanation

C

Kudos [?]: 34 [0], given: 0

Intern
Joined: 25 Nov 2008
Posts: 7

Kudos [?]: 10 [0], given: 0

Re: m15,#10 [#permalink]

### Show Tags

07 Dec 2008, 20:14
We now the total possibilities is 4c2

List out the possibilities

Chossen Not choosen
23 57 - no
25 37 -no
27 35 - no

35 27 -yes
37 25 - yes

57 23 - yes

4c2=6

3/6 = 1/2

Kudos [?]: 10 [0], given: 0

Manager
Joined: 29 Nov 2009
Posts: 107

Kudos [?]: 36 [0], given: 5

Location: United States
Re: m15,#10 [#permalink]

### Show Tags

21 Dec 2009, 07:37
The total number of combinations is the following: 4!/(2!*2!) = 6

The only ones that work are these: 3*5, 3*7, 5*7

3/6 = 1/2

Kudos [?]: 36 [0], given: 5

Manager
Joined: 04 Dec 2009
Posts: 69

Kudos [?]: 9 [0], given: 4

Location: INDIA
Re: m15,#10 [#permalink]

### Show Tags

21 Dec 2009, 20:29
Ans : C

there are only two possibal out come and have 50% chance of geting either one
_________________

MBA (Mind , Body and Attitude )

Kudos [?]: 9 [0], given: 4

Manager
Joined: 26 Dec 2009
Posts: 132

Kudos [?]: 12 [0], given: 10

Location: United Kingdom
Concentration: Strategy, Technology
GMAT 1: 500 Q45 V16
WE: Consulting (Computer Software)
Re: m15,#10 [#permalink]

### Show Tags

28 Dec 2010, 10:56
C for me as well.

Kudos [?]: 12 [0], given: 10

Intern
Joined: 16 Dec 2010
Posts: 7

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

Re: m15,#10 [#permalink]

### Show Tags

28 Dec 2010, 21:36
i think the ans is 1/3

Pls check whether my understanding is correct

We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)

There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.

So p= 2/4C2 = 2/6 = 1/3 Ans

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

Senior Manager
Joined: 01 Nov 2010
Posts: 284

Kudos [?]: 85 [0], given: 44

Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Re: m15,#10 [#permalink]

### Show Tags

29 Dec 2010, 00:19
GUD eXPLANATION..
its C.
_________________

kudos me if you like my post.

Attitude determine everything.
all the best and God bless you.

Kudos [?]: 85 [0], given: 44

Manager
Joined: 24 Jul 2009
Posts: 192

Kudos [?]: 47 [0], given: 10

Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Re: m15,#10 [#permalink]

### Show Tags

30 Dec 2010, 00:27
C. The limitations of the set allow you to list the specific product options and compare them. I'm someone who's skilling up in quant and since I know that I make simple mistakes, I opted to actually write out the product options.
_________________

Reward wisdom with kudos.

Kudos [?]: 47 [0], given: 10

Intern
Joined: 28 Dec 2011
Posts: 31

Kudos [?]: 10 [0], given: 21

GMAT 1: 750 Q50 V41
Re: m15,#10 [#permalink]

### Show Tags

30 Dec 2011, 07:06
Counting 1 as a prime number is exactly the mistake I made... Funnily enough, they have an answer choice that corresponds to this particular way of "solving" the question.

Kudos [?]: 10 [0], given: 21

Manager
Joined: 15 Dec 2011
Posts: 182

Kudos [?]: 45 [0], given: 13

Schools: LBS '14 (A)
GMAT 1: 730 Q50 V39
GPA: 3.9
Re: m15,#10 [#permalink]

### Show Tags

30 Dec 2011, 07:14
C

total ways of choosing 2 out of 4 = !4/(!2*!2) i.e. 6
total ways the product can be greater than the other two = (3,5), (5,7), (7,3) = 3
answer=3/6 = 1/2

Kudos [?]: 45 [0], given: 13

Manager
Joined: 08 Sep 2011
Posts: 68

Kudos [?]: 5 [0], given: 5

Concentration: Finance, Strategy
Re: m15,#10 [#permalink]

### Show Tags

30 Dec 2011, 08:30
I solved by calculating how many solutions do not involve 2.

1 - 1/4 * 1 - 1/3 = 6 /12 = 1/2. answer is C

Kudos [?]: 5 [0], given: 5

Manager
Joined: 21 Nov 2010
Posts: 127

Kudos [?]: 5 [0], given: 12

Re: m15,#10 [#permalink]

### Show Tags

30 Dec 2011, 19:05
C.
The Kaplan Instructor explained it pretty well.

Kudos [?]: 5 [0], given: 12

Intern
Joined: 19 Dec 2009
Posts: 36

Kudos [?]: 10 [0], given: 8

Re: m15,#10 [#permalink]

### Show Tags

31 Dec 2011, 05:27
Moss wrote:
All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2.

3/4 * 2/3 = 6/12 = 1/2

So C.

This one is fast and least error prone for this problem.

Thanks.

Kudos [?]: 10 [0], given: 8

Re: m15,#10   [#permalink] 31 Dec 2011, 05:27
Display posts from previous: Sort by

# m15,#10

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.