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If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
03 Jan 2008, 14:40

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Difficulty:

45% (medium)

Question Stats:

67% (02:23) correct
33% (01:46) wrong based on 128 sessions

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?

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
03 Jan 2008, 23:45

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?

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
04 Jan 2008, 06:07

2

This post received KUDOS

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 VeritasProject GMAT for combinatorics. You may want to give them a look as well

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
25 Aug 2008, 09:51

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

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
27 Sep 2009, 10:06

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

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
05 Mar 2010, 06:45

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

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
29 Feb 2012, 10:44

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.

Re: If two numbers, a and b, are to be chosen from a set of 4 [#permalink]
02 Sep 2014, 14:41

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