Author 
Message 
TAGS:

Hide Tags

VP
Joined: 09 Jul 2007
Posts: 1084
Location: London

If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
Updated on: 02 Sep 2014, 20:07
Question Stats:
60% (01:35) correct 40% (02:00) wrong based on 429 sessions
HideShow timer Statistics
If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer? (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by Ravshonbek on 03 Jan 2008, 15:40.
Last edited by Bunuel on 02 Sep 2014, 20:07, edited 3 times in total.
Renamed the topic, edited the question and added the OA.



Director
Joined: 12 Jul 2007
Posts: 854

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
03 Jan 2008, 16:04
I'm going to assume it's B/A and give that a shot. A: 1, 2, 3, 4 B: 4, 6, 8 Total # of outcomes: 4*3 = 12 Choose 4: only 3 makes a noninteger Choose 6: only 4 makes a noninteger Choose 8: only 3 makes a noninteger 3/12 = 1/4 probability of B/A not being an integer. Sorry if this isn't actually the problem



SVP
Joined: 29 Mar 2007
Posts: 2493

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
04 Jan 2008, 00:45
Ravshonbek wrote: If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer? Answers: (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3 A: 1,2,3,4 B: 4,6,8 b/a=i? 4/3, 6/4, 8/3 so: 1/4*1/3 = 1/12 (3) > 1/4 B



Director
Joined: 12 Jul 2007
Posts: 854

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
04 Jan 2008, 07:07
kazakhb wrote: I can't understand why i can't think the way as you guys do, especially "wisconsin", generally he answers every question in this forum, if i was to solve that problem I would need million years( Most people feel like this when they first start studying for the GMAT. It's not that the math is extremely difficult, it's how the math is being tested that's strange and new. When I first started out I spent lots of time on this forum and made sure I really understood the concepts and reasoning behind each question. In my opinion it's not enough to blindly memorize formulas for this test, you need to understand WHY a question is solved the way it's solved. This way, when the GMAT throws you for a loop, you have solid math founded in an understanding that allows you to apply it to a broad spectrum of problems...and not just on problems and problem types you've seen in the past. so keep asking questions and digging deeper on here and before long you'll be a pro! there are also great books out there to help you brush up on math skills. I love the MGMAT series for most quant topics and Veritas Project GMAT for combinatorics. You may want to give them a look as well



Senior Manager
Joined: 19 Nov 2007
Posts: 433

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
04 Jan 2008, 10:51
if the question is modified by taking out the word 'respecitively' from this question, how would one approach it?
_________________
Underline your question. It takes only a few seconds! Search before you post.



SVP
Joined: 07 Nov 2007
Posts: 1738
Location: New York

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
25 Aug 2008, 10:51
Ravshonbek wrote: If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer? Answers: (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3 {1,2,3,4} {4,6,8} b/a not integer 4/3,6/3,8/3 p = 3/12 =1/4
_________________
Your attitude determines your altitude Smiling wins more friends than frowning



Manager
Joined: 27 Oct 2008
Posts: 181

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
27 Sep 2009, 11:06
If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer?
Answers: (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3
Soln: a is from the following set {1,2,3,4} b is from the following set {4,6,8}
Total number of ways of choosing 2 integers, one from each set is = 4* 3 = 12 ways
Now the number of possibilities where b/a is not an integer is for the following outcomes {b,a} => {4,3},{6,4},{8,3} = 3 ways
Hence probability is = 3/12 = 1/4
ans is B



Intern
Joined: 17 Jan 2010
Posts: 26
Schools: SSE, LSE

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
05 Mar 2010, 07:45
x2suresh wrote: Ravshonbek wrote: If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer? Answers: (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3 {1,2,3,4} {4,6,8} b/a not integer 4/3,6/3,8/3 p = 3/12 =1/4 Hey suresh, you probably meant: b/a not integer 4/3,6/ 4,8/3



Intern
Joined: 24 Feb 2012
Posts: 31

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
29 Feb 2012, 11:44
OK, here's an example of how to get an easy question wrong. I assumed (misread) the question to mean that a and b sets are interchangeable. Implying that a and b could be selected from either {1,2,3,4} or {4,6,8}.
I now realize that I had misread the question, but if that was the original question, then: P(b/a is not an integer) = P(b/a not an integer where b in {1234} and a in {468}) + P(b/a not an integer where b in {468} and a in {1234}) = 11/12 + 3/12 = 7/12.
Going back to the question that was asked: P(b/a not an integer where b in {468} and a in {1234}) = 3/12 = 1/4.



Senior Manager
Joined: 20 Aug 2015
Posts: 392
Location: India

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
18 Oct 2015, 23:52
Ravshonbek wrote: If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer? (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3 In such problems, it is better to first write down all the cases that are in cosideration Set A = {1, 2, 3, 4} Set B = {4, 6, 8} Total outcomes: 4*3 = 12 Favourable outcomes: We need B/A, hence considering one element of the set B at a time 4: only 3 satisfies our condition (4 is divisible by 1, 2 and 4) 6: only 4 satisfies our condition (6 is divisible by 1, 2 and 3) 8: only 3 satisfies our condition (8 is divisible by 1, 2 and 4) Favourable outcomes = 3 Probability = 3/12 = 1/4 Option B



Board of Directors
Joined: 17 Jul 2014
Posts: 2730
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
06 Apr 2016, 18:18
Ravshonbek wrote: If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer? (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3 a relatively easy question that can be solved by enumerating all the outcomes: 1st list: 1, 2, 3, 4 2nd list: 4, 6, 8 now, possible options: 4/1 4/2 4/3  noninteger 4/4 6/1 6/2 6/3 6/4  noninteger 8/1 8/2 8/3  noninteger 8/4 total 12 outcomes, out of which 3 are "successful" outcomes. 3/12 = 1/4



Intern
Joined: 28 Apr 2015
Posts: 19
GMAT 1: 600 Q46 V27 GMAT 2: 630 Q47 V30

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
20 Jul 2016, 04:47
Ravshonbek wrote: If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer? (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E) 2/3 Am I the only one who interpreted the question in the following manner? Given that the first set consists of 4 consecutive integers starting with 1, I assumed the set will be something like this {10, 11, 12, 13} or {102,103,104,105} and same for the second set {404, 406, 408}. This made me lose my mind on how to go about



NonHuman User
Joined: 09 Sep 2013
Posts: 7045

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
Show Tags
06 Sep 2017, 23:04
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




Re: If two numbers, a and b, are to be chosen from a set of 4
[#permalink]
06 Sep 2017, 23:04






