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# At least 2/3 of the 40 members of a committee must vote in

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At least 2/3 of the 40 members of a committee must vote in [#permalink]

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05 Dec 2012, 08:05
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At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass?

(A) 19
(8) 17
(C) 16
(D) 14
(E) 13
[Reveal] Spoiler: OA

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Re: At least 2/3 of the 40 members of a committee must vote in [#permalink]

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05 Dec 2012, 08:10
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At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass?

(A) 19
(8) 17
(C) 16
(D) 14
(E) 13

Since, at least 2/3 of the 40 members of a committee must vote in favor, then at least $$\frac{2}{3}*40=26\frac{2}{3}$$ or at least 27 members should vote in favor. Therefore at most 13 members could vote against.

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Re: At least 2/3 of the 40 members of a committee must vote in [#permalink]

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29 Sep 2013, 00:12
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AasaanHai wrote:
Bunuel wrote:
At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass?

(A) 19
(8) 17
(C) 16
(D) 14
(E) 13

Since, at least 2/3 of the 40 members of a committee must vote in favor, then at least $$\frac{2}{3}*40=26\frac{2}{3}$$ or at least 27 members should vote in favor. Therefore at most 13 members could vote against.

What is wrong in the way I am doing this question?

It is given that at least 2/3rd of the 40 members should vote for the resolution. Since the members have to be a positive integer, I have considered 40 as 39 and then divided it by 3. After calculation, 26 members should vote for resolution. The remaining are 14 members who can vote against it; I marked D, which is incorrect.

Well you calculated for 39 members...How about the 1 that you left out....Since the Questions says At least 2/3 of 40 which is 26.666....Now atleast means it can be more and the least possible Integer value is 27....So most 13 members can vote against.
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Re: At least 2/3 of the 40 members of a committee must vote in [#permalink]

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30 Jan 2016, 10:59
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Hi studentsensual,

You seem to understand that you can't have a 'fraction of a person' in these types of questions. Thus, you have to pay attention to the question that is ASKED and think in terms of 'whole persons.'

Here, we're told that AT LEAST 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. Since 2/3 of 40 is 26.666666, but we can't have 26.666666 people, we have to round UP to 27. In contrast, 26 people would NOT be enough, since 26/40 = 13/20 = 65% (and not the 66 2/3% minimum that a resolution needs to be passed).

Knowing that it would take AT LEAST 27 people to pass a resolution, the GREATEST number who could vote AGAINST the resolution in this case would be 40-27 = 13.

[Reveal] Spoiler:
E

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3511 [1], given: 173 Manager Joined: 01 Oct 2010 Posts: 81 Kudos [?]: 25 [0], given: 115 Location: United States (NC) GPA: 2.3 WE: Information Technology (Computer Software) Re: At least 2/3 of the 40 members of a committee must vote in [#permalink] ### Show Tags 28 Sep 2013, 23:32 Bunuel wrote: Walkabout wrote: At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass? (A) 19 (8) 17 (C) 16 (D) 14 (E) 13 Since, at least 2/3 of the 40 members of a committee must vote in favor, then at least $$\frac{2}{3}*40=26\frac{2}{3}$$ or at least 27 members should vote in favor. Therefore at most 13 members could vote against. Answer: E. What is wrong in the way I am doing this question? It is given that at least 2/3rd of the 40 members should vote for the resolution. Since the members have to be a positive integer, I have considered 40 as 39 and then divided it by 3. After calculation, 26 members should vote for resolution. The remaining are 14 members who can vote against it; I marked D, which is incorrect. _________________ --------------------------------------------------------------- Consider to give me kudos if my post helped you. Kudos [?]: 25 [0], given: 115 Manager Joined: 12 Jan 2013 Posts: 218 Kudos [?]: 83 [0], given: 47 Re: At least 2/3 of the 40 members of a committee must vote in [#permalink] ### Show Tags 19 Dec 2013, 02:38 Walkabout wrote: At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass? (A) 19 (8) 17 (C) 16 (D) 14 (E) 13 40/3 = 13.3 (approx).. --> 13.3 x 2 = 26.6, so at least 27 people have to vote in favor for it to pass. ----> 13 at most can vote against for it to pass. Kudos [?]: 83 [0], given: 47 Manager Joined: 12 Jan 2013 Posts: 218 Kudos [?]: 83 [0], given: 47 Re: At least 2/3 of the 40 members of a committee must vote in [#permalink] ### Show Tags 10 Jan 2014, 01:39 Walkabout wrote: At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass? (A) 19 (8) 17 (C) 16 (D) 14 (E) 13 Dividing 40/3 gives us 13.33333, thus 2/3 = 26.6666.. So in order for it to pass, at least 26.6 people need to vote on it, but of course we need an integer so at least 27 people need to vote on it (so, in practice, a bigger portion than 2/3).. That leaves us with 13 people that can vote against and still have it pass. The non-integer throws us off and in some cases makes us round down. I think that's what they primarily want to test us on in this case, if we understand that we should round UP for required votes instead of round down. Kudos [?]: 83 [0], given: 47 Intern Joined: 19 Jan 2014 Posts: 30 Kudos [?]: 19 [0], given: 51 Re: At least 2/3 of the 40 members of a committee must vote in [#permalink] ### Show Tags 21 Feb 2014, 16:21 The way I calculated was. At least 2/3 so it's the minimum than the greatest portion we would have left would be 1/3. Hence (40x1)/3 13,3... = 13 people Kudos [?]: 19 [0], given: 51 Intern Joined: 27 Oct 2015 Posts: 17 Kudos [?]: 6 [0], given: 175 Re: At least 2/3 of the 40 members of a committee must vote in [#permalink] ### Show Tags 26 Jan 2016, 22:59 2/3 of 40 which is 26.666, there is no 0.666 person, so I decided 2/3 of 40 is 26, thus 40-26=14, at least means minimum, not maximum, so minimum possible numbers, then it's logical to consider 26, moreover question asks the greatest number of members, how could I realize wherther to round up or not? Thanks! WoundedTiger wrote: AasaanHai wrote: Bunuel wrote: Since, at least 2/3 of the 40 members of a committee must vote in favor, then at least $$\frac{2}{3}*40=26\frac{2}{3}$$ or at least 27 members should vote in favor. Therefore at most 13 members could vote against. Answer: E. What is wrong in the way I am doing this question? It is given that at least 2/3rd of the 40 members should vote for the resolution. Since the members have to be a positive integer, I have considered 40 as 39 and then divided it by 3. After calculation, 26 members should vote for resolution. The remaining are 14 members who can vote against it; I marked D, which is incorrect. Well you calculated for 39 members...How about the 1 that you left out....Since the Questions says At least 2/3 of 40 which is 26.666....Now atleast means it can be more and the least possible Integer value is 27....So most 13 members can vote against. Kudos [?]: 6 [0], given: 175 Intern Joined: 27 Oct 2015 Posts: 17 Kudos [?]: 6 [0], given: 175 Re: At least 2/3 of the 40 members of a committee must vote in [#permalink] ### Show Tags 31 Jan 2016, 19:31 So, when dealing with individuals, should I every time to round the numbers up? Thanks! EMPOWERgmatRichC wrote: Hi studentsensual, You seem to understand that you can't have a 'fraction of a person' in these types of questions. Thus, you have to pay attention to the question that is ASKED and think in terms of 'whole persons.' Here, we're told that AT LEAST 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. Since 2/3 of 40 is 26.666666, but we can't have 26.666666 people, we have to round UP to 27. In contrast, 26 people would NOT be enough, since 26/40 = 13/20 = 65% (and not the 66 2/3% minimum that a resolution needs to be passed). Knowing that it would take AT LEAST 27 people to pass a resolution, the GREATEST number who could vote AGAINST the resolution in this case would be 40-27 = 13. Final Answer: [Reveal] Spoiler: E GMAT assassins aren't born, they're made, Rich Kudos [?]: 6 [0], given: 175 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10120 Kudos [?]: 3511 [0], given: 173 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: At least 2/3 of the 40 members of a committee must vote in [#permalink] ### Show Tags 31 Jan 2016, 23:50 Hi studentsensual, If a question expects you to 'round up' or 'round down', then you have to make sure to choose the answer that focuses on the specifics of the given prompt and answers the question that is asked: Here, the specific information stated that AT LEAST 2/3 (meaning at least 66 2/3%) of the members must vote in favor of a resolution for it to pass. Since 26/40 is only 65%, then 26 people would NOT be enough. Thus, you have to round that number UP to 27 (because that's what the question requires). There could conceivably be questions that would require you to round down, but that's not what THIS question required. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: At least 2/3 of the 40 members of a committee must vote in [#permalink]

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04 Feb 2016, 06:00

At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass?
(A) 19
(8) 17
(C) 16
(D) 14
(E) 13

A club collected exactly $599 from its members. If each member contributed at least$12, what is the greatest number of members the club could have?
(A) 43
(B) 44
(C) 49
(D) 50
(E) 51

599/12=49.912
Since n represents individual people, it must be a whole number; the greatest possible value of n is thus 49, not 50

EMPOWERgmatRichC wrote:
Hi studentsensual,

If a question expects you to 'round up' or 'round down', then you have to make sure to choose the answer that focuses on the specifics of the given prompt and answers the question that is asked:

Here, the specific information stated that AT LEAST 2/3 (meaning at least 66 2/3%) of the members must vote in favor of a resolution for it to pass. Since 26/40 is only 65%, then 26 people would NOT be enough. Thus, you have to round that number UP to 27 (because that's what the question requires). There could conceivably be questions that would require you to round down, but that's not what THIS question required.

GMAT assassins aren't born, they're made,
Rich

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Re: At least 2/3 of the 40 members of a committee must vote in [#permalink]

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04 Feb 2016, 12:11
Hi studentsensual,

These two questions provide you with different types of information/situations, so they require that you round in different 'directions.' In the second prompt, we're told that each member contributed AT LEAST $12, but that the total collected was$599. In this situation, the total number of members CANNOT be 50, since 50($12) =$600 - and that total is just a little TOO HIGH. As such, we have to round DOWN.

GMAT assassins aren't born, they're made,
Rich
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# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3511 [0], given: 173 Board of Directors Status: Aiming MBA Joined: 18 Jul 2015 Posts: 2756 Kudos [?]: 910 [0], given: 67 Location: India GPA: 3.65 WE: Information Technology (Health Care) Re: At least 2/3rd ofthe 40 members of a committee.. [#permalink] ### Show Tags 04 Sep 2016, 10:18 Manonamission wrote: At least 2/3rd ofthe 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass? (A) 19 (B) 17 (C) 16 (D) 14 (E) 13 Atleast 2/3 votes in favor mean 2/3 of 40 ~27. Hence, Max against would be 40-27 = 13. hence E _________________ How I improved from V21 to V40! ? Kudos [?]: 910 [0], given: 67 Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3097 Kudos [?]: 1114 [0], given: 327 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: At least 2/3rd ofthe 40 members of a committee.. [#permalink] ### Show Tags 04 Sep 2016, 13:09 Manonamission wrote: At least 2/3rd of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass? (A) 19 (B) 17 (C) 16 (D) 14 (E) 13 Vote in favour - $$\frac{2}{3} * 40$$ = $$26 \frac{2}{3}$$ ~ 27 No of voters who can vote against the resolution and still get it passed is (40 - 27) 13 Hence answer will be (E) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Kudos [?]: 1114 [0], given: 327 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10120 Kudos [?]: 3511 [0], given: 173 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: At least 2/3rd ofthe 40 members of a committee.. [#permalink] ### Show Tags 04 Sep 2016, 20:59 Hi Manonamission, One of they 'keys' to this question is that you can't have a 'fraction of a person.' Thus, you have to pay attention to the question that is ASKED and think in terms of 'whole persons.' Here, we're told that AT LEAST 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. Since 2/3 of 40 is 26.666666, but we can't have 26.666666 people, we have to round UP to 27. In contrast, 26 people would NOT be enough, since 26/40 = 13/20 = 65% (and not the 66 2/3% minimum that a resolution needs to be passed). Knowing that it would take AT LEAST 27 people to pass a resolution, the GREATEST number who could vote AGAINST the resolution in this case would be 40-27 = 13. Final Answer: [Reveal] Spoiler: E GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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At least 2/3 of the 40 members of a committee must vote in [#permalink]

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09 Oct 2017, 06:36
At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass?

(A) 19
(8) 17
(C) 16
(D) 14
(E) 13

At least $$\frac{2}{3}$$ of 40 people must vote for a resolution to pass.

$$\frac{2}{3}$$ * 40 = 26$$\frac{2}{3}$$, or 26.67 persons

In this problem, if you are torn between rounding up to 27 or down to 26, one way to check is mentioned briefly by EMPOWERgmatRichC above: reverse the process, though I will include both numbers in question. Divide both numbers by 40.

$$\frac{26}{40}$$ = .65, or 65%

$$\frac{27}{40}$$ = .675, or 67.5%

26 is not enough: 65 percent is not "at least" 66.67 percent. It doesn't matter that 27 is a bit too much.

You cannot have a fraction of a person, and 27 clears the hurdle of at least 66.67%, while 26 does not.

If 27 must vote "yes," only 40 - 27 = 13 can vote "no."

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Re: At least 2/3 of the 40 members of a committee must vote in [#permalink]

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12 Oct 2017, 17:57
At least 2/3 of the 40 members of a committee must vote in favor of a resolution for it to pass. What is the greatest number of members who could vote against the resolution and still have it pass?

(A) 19
(8) 17
(C) 16
(D) 14
(E) 13

We are given that at least 2/3 of the members must vote IN FAVOR of a resolution in order for it to pass; however, we need to determine the greatest number of members who could vote AGAINST the resolution and still cause its passage. Remember, in a vote there are only two options, voting in FAVOR and voting AGAINST. Thus, we know the following:

2/3 of total voters need to vote in FAVOR for it to pass; this means that 1/3 of total voters can vote AGAINST for it to pass.

To finish the problem, we can set up the following equation:

1/3 x 40 = total votes AGAINST to have resolution pass

1/3 x 40 = 40/3 = 13 1/3 voters

Since we need the resolution TO PASS, we must round this number down to 13. Thus, 13 voters can vote against the resolution and still have it pass.

Alternate Solution:

Notice that 2/3 x 40 = 80/3 = 26 2/3 and we have to round this up to 27 because we can’t have a fraction of a person. Therefore, we need at least 27 voters to vote IN FAVOR of the resolution to pass it. This means that we can have at most 40 – 27 = 13 individuals voting AGAINST it, and still it will pass. Therefore, the maximum number of voters who can vote against it and still have it pass is 13.

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Re: At least 2/3 of the 40 members of a committee must vote in   [#permalink] 12 Oct 2017, 17:57
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