Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 19 Oct 2012
Posts: 344
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35 GMAT 2: 710 Q50 V38
GPA: 3.81
WE: Information Technology (Computer Software)

Set Q contains n nonzero numbers. [#permalink]
Show Tags
02 Oct 2013, 10:41
5
This post was BOOKMARKED
Question Stats:
57% (01:07) correct 43% (01:25) wrong based on 192 sessions
HideShow timer Statistics
Set Q contains n nonzero numbers. Does set Q contain an even number of negative numbers? (1) n is even. (2) The product of all the numbers in Q is 100,000
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Citius, Altius, Fortius



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4675

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
02 Oct 2013, 12:40
vabhs192003 wrote: Set Q contains n nonzero numbers. Does set Q contain an even number of negative numbers?
(1) n is even.
(2) The product of all the numbers in Q is 100,000 I'm happy to help with this. We know none of the numbers in Set Q equal zero, but they could be positive or negative, whole or fractions or decimals. The question is: Does set Q contain an even number of negative numbers?Statement #1: n is evenThis simply tells us the overall number of elements in the set is even, but we have absolutely no idea what the breakdown might be between positive and negative. This statement doesn't help us at all. Alone and by itself, it's insufficient. Statement #2: The product of all the numbers in Q is 100,000Here, the value of the product doesn't matter  what matters is that the product is positive. You see If we multiply two negative numbers, (2)(2) = +4, we get a positive product. If we multiply three negative numbers, (2)(2)(2) = 8, we get a negative product. If we multiply four negative numbers, (2)(2)(2)(2) = +16, we get a positive product. If we multiply five negative numbers, (2)(2)(2)(2)(2) = 32, we get a negative product. If we multiply six negative numbers, (2)(2)(2)(2)(2)(2) = +64, we get a positive product. In general, the product of an even number of negative numbers is positive, and the product of an odd number of negative numbers is negative. Either way, multiplying that product by any number of positive factors will not change the sign. Thus, if the product of the number in Q is a positive number, then the number of negative members of the set must be even. It could be an all positive set, with zero negative numbers (zero is an even number!), or it could have all positive except for two negatives, or four negatives, or etc. Thus, this statement, alone and by itself, it's sufficient. OA = (B) Does all this make sense? Mike
_________________
Mike McGarry Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)



Senior Manager
Joined: 19 Oct 2012
Posts: 344
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35 GMAT 2: 710 Q50 V38
GPA: 3.81
WE: Information Technology (Computer Software)

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
02 Oct 2013, 12:54
mikemcgarry wrote: vabhs192003 wrote: Set Q contains n nonzero numbers. Does set Q contain an even number of negative numbers?
(1) n is even.
(2) The product of all the numbers in Q is 100,000 I'm happy to help with this. We know none of the numbers in Set Q equal zero, but they could be positive or negative, whole or fractions or decimals. The question is: Does set Q contain an even number of negative numbers?Statement #1: n is evenThis simply tells us the overall number of elements in the set is even, but we have absolutely no idea what the breakdown might be between positive and negative. This statement doesn't help us at all. Alone and by itself, it's insufficient. Statement #2: The product of all the numbers in Q is 100,000Here, the value of the product doesn't matter  what matters is that the product is positive. You see If we multiply two negative numbers, (2)(2) = +4, we get a positive product. If we multiply three negative numbers, (2)(2)(2) = 8, we get a negative product. If we multiply four negative numbers, (2)(2)(2)(2) = +16, we get a positive product. If we multiply five negative numbers, (2)(2)(2)(2)(2) = 32, we get a negative product. If we multiply six negative numbers, (2)(2)(2)(2)(2)(2) = +64, we get a positive product. In general, the product of an even number of negative numbers is positive, and the product of an odd number of negative numbers is negative. Either way, multiplying that product by any number of positive factors will not change the sign. Thus, if the product of the number in Q is a positive number, then the number of negative members of the set must be even. It could be an all positive set, with zero negative numbers (zero is an even number!), or it could have all positive except for two negatives, or four negatives, or etc. Thus, this statement, alone and by itself, it's sufficient. OA = (B) Does all this make sense? Mike HI Mike. Gee!! Thanks for that detailed solution. I got this question wrong in my 1st go but then caught the exceptional case of there being 0 negative numbers too (which is EVEN). I posted this questions for others to try out, but its good to have a detailed desc from an expert for others to refer. Thanks again.
_________________
Citius, Altius, Fortius



Math Expert
Joined: 02 Sep 2009
Posts: 44654

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
03 Oct 2013, 01:00



Intern
Joined: 15 Jul 2013
Posts: 5

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
12 Nov 2013, 06:05
It is probable for the statement 2 that the elements are all positive, how come statement 2 is sufficient then?



Math Expert
Joined: 02 Sep 2009
Posts: 44654

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
12 Nov 2013, 07:29



Manager
Joined: 17 Jul 2013
Posts: 89

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
13 May 2014, 02:01
1
This post received KUDOS
1
This post was BOOKMARKED
gilda wrote: It is probable for the statement 2 that the elements are all positive, how come statement 2 is sufficient then? Hi, To answer this we know that if the product of all numbers is positive then it means following two cases: 1. there are even number of negative numbers to result in positive product 2. there are all positive numbers in Q none is negative number, hence the count of the negative number is ZERO which is an even number. Hope this helps ....



Board of Directors
Joined: 17 Jul 2014
Posts: 2745
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
04 Oct 2016, 06:57
what a brilliant question! I forgot that 0 is an even number shame on me. B should be the answer.



Manager
Joined: 26 Mar 2016
Posts: 77
Location: Greece
GPA: 2.9

Set Q contains n nonzero numbers. [#permalink]
Show Tags
Updated on: 17 Nov 2016, 12:13
(1) Clearly not sufficient, we cannot find out if there is an even or an odd number of negative #s. (2) 100,000= 2^5 * 5^5 >0. Therefore negative numbers must be 0,2 or 4 in order for the product to be a positive number. (B).
_________________
+1 Kudos if you like the post
Originally posted by syahasa2 on 09 Oct 2016, 04:39.
Last edited by syahasa2 on 17 Nov 2016, 12:13, edited 1 time in total.



Intern
Joined: 24 Sep 2016
Posts: 34

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
05 Nov 2016, 23:51
mikemcgarry wrote: vabhs192003 wrote: Set Q contains n nonzero numbers. Does set Q contain an even number of negative numbers?
(1) n is even.
(2) The product of all the numbers in Q is 100,000 I'm happy to help with this. We know none of the numbers in Set Q equal zero, but they could be positive or negative, whole or fractions or decimals. The question is: Does set Q contain an even number of negative numbers?Statement #1: n is evenThis simply tells us the overall number of elements in the set is even, but we have absolutely no idea what the breakdown might be between positive and negative. This statement doesn't help us at all. Alone and by itself, it's insufficient. Statement #2: The product of all the numbers in Q is 100,000Here, the value of the product doesn't matter  what matters is that the product is positive. You see If we multiply two negative numbers, (2)(2) = +4, we get a positive product. If we multiply three negative numbers, (2)(2)(2) = 8, we get a negative product. If we multiply four negative numbers, (2)(2)(2)(2) = +16, we get a positive product. If we multiply five negative numbers, (2)(2)(2)(2)(2) = 32, we get a negative product. If we multiply six negative numbers, (2)(2)(2)(2)(2)(2) = +64, we get a positive product. In general, the product of an even number of negative numbers is positive, and the product of an odd number of negative numbers is negative. Either way, multiplying that product by any number of positive factors will not change the sign. Thus, if the product of the number in Q is a positive number, then the number of negative members of the set must be even. It could be an all positive set, with zero negative numbers (zero is an even number!), or it could have all positive except for two negatives, or four negatives, or etc. Thus, this statement, alone and by itself, it's sufficient. OA = (B) Does all this make sense? Mike I marked the answer as E because i thought B is not sufficient. We can have two possibilities in B : 1. When all the numbers are positive and none are negative. 2. When even numbers are negative. So how can B be sufficient here??
_________________
“Most of the important things in the world have been accomplished by people who have kept on trying when there seemed to be no hope at all.”



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3399
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
06 Nov 2016, 05:25
TheMechanic wrote: Set Q contains n nonzero numbers. Does set Q contain an even number of negative numbers?
(1) n is even.
(2) The product of all the numbers in Q is 100,000 FROM STATEMENT  I (NOT POSSIBLE)Q is a set of even number of nonzero numbers... This means there are 2 possibilities 1. There can be even no of ve numbers 2. There can be even no of +ve numbers.. Thus we can not comment about the constituent of the set QFROM STATEMENT  I (POSSIBLE)Since product of the elements is positive , there must be even number of non negative numbers in set Q Hence , Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked, answer will be (B)
_________________
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 )



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4675

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
07 Nov 2016, 14:06
emmafoster wrote: I marked the answer as E because i thought B is not sufficient. We can have two possibilities in B : 1. When all the numbers are positive and none are negative. 2. When even numbers are negative.
So how can B be sufficient here??
Dear emmafoster, I'm happy to respond, my friend. As you correctly point out, there are two possibilities: (a) there are some negative numbers in Set Q, and number of negative numbers is even (b) there are no negative numbers in Set Q Here's the missing piece: zero is an even number! Thus, if there are no negative numbers, then the number of negative numbers is zero, and this is still even! Technically, I could make the 100% mathematically correct statement that the number of walruses in the Magoosh office is an even number! Since quite obviously the number is zero, this statement is technically true because zero is even. Thus, whether the number of negative numbers in Set Q is 2, 4, 6, etc. or whether it is simply 0, in all cases the number of negative numbers has to be even, and statement #2 is sufficient. OA = (B) Does all this make sense? Mike
_________________
Mike McGarry Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2298

Re: Set Q contains n nonzero numbers. [#permalink]
Show Tags
07 Dec 2017, 08:47
TheMechanic wrote: Set Q contains n nonzero numbers. Does set Q contain an even number of negative numbers?
(1) n is even.
(2) The product of all the numbers in Q is 100,000 We are given that set Q contains n nonzero numbers  all numbers except for zero. We need to determine whether set Q contains an even number of negative numbers. Statement One Alone: n is even. Knowing that n is even is not enough information to determine whether set Q contains an even number of negative numbers. For instance if n = 2, set Q could contain 2 negative numbers and 0 positive numbers or 1 negative number and 1 positive number. Statement one is not sufficient to answer the question. We can eliminate answer choices A and D. Statement Two Alone: The product of all the numbers in Q is 100,000. In analyzing statement two, we may recall that when multiplying negative numbers, the ONLY WAY to produce a positive product is by multiplying together an even number of negative numbers. Since we know that the product of all the numbers in Q is 100,000, n must be even. Statement two alone is sufficient to answer the question. Answer: B
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: Set Q contains n nonzero numbers.
[#permalink]
07 Dec 2017, 08:47






