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Set Q contains n non-zero numbers.

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Set Q contains n non-zero numbers. [#permalink]

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Set Q contains n non-zero 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
[Reveal] Spoiler: OA

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Re: Set Q contains n non-zero numbers. [#permalink]

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vabhs192003 wrote:
Set Q contains n non-zero 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 even
This 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 100,000
Here, 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 :-)
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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 02 Oct 2013, 11:54
mikemcgarry wrote:
vabhs192003 wrote:
Set Q contains n non-zero 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 even
This 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 100,000
Here, 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. :-D
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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 03 Oct 2013, 00:00
vabhs192003 wrote:
Set Q contains n non-zero 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


Similar questions to practice:
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if-there-are-more-than-two-numbers-in-a-certain-list-is-128114.html
sequence-s-consists-of-24-nonzero-integers-if-each-term-in-98728.html

Hope this helps.
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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 12 Nov 2013, 05:05
It is probable for the statement 2 that the elements are all positive, how come statement 2 is sufficient then?

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Re: Set Q contains n non-zero numbers. [#permalink]

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gilda wrote:
It is probable for the statement 2 that the elements are all positive, how come statement 2 is sufficient then?


The question asks whether Q contains an even number of negative numbers. (2) says that the product of all the numbers in Q is positive. Now, if the number of negative numbers in Q were odd, then the product of all the numbers in Q would have been negative, thus the number of negative numbers in Q is not odd, therefore it must be even.

Hope it's clear.
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Re: Set Q contains n non-zero numbers. [#permalink]

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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 ....

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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 04 Oct 2016, 05:57
what a brilliant question!
I forgot that 0 is an even number :(

shame on me.

B should be the answer.

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Set Q contains n non-zero numbers. [#permalink]

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New post 09 Oct 2016, 03:39
(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).
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Last edited by syahasa2 on 17 Nov 2016, 11:13, edited 1 time in total.

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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 05 Nov 2016, 22:51
mikemcgarry wrote:
vabhs192003 wrote:
Set Q contains n non-zero 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 even
This 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 100,000
Here, 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??
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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 06 Nov 2016, 04:25
TheMechanic wrote:
Set Q contains n non-zero 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 non-zero 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 Q

FROM 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)

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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 07 Nov 2016, 13:06
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 :-)
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Re: Set Q contains n non-zero numbers. [#permalink]

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New post 07 Dec 2017, 07:47
TheMechanic wrote:
Set Q contains n non-zero 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 non-zero 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
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Re: Set Q contains n non-zero numbers.   [#permalink] 07 Dec 2017, 07:47
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