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# Sequence of nonzero numbers (test from mba.com)

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Re: Sequence of nonzero numbers (test from mba.com) [#permalink]  29 Nov 2010, 13:41
I understand! we should have a total of 5 pairs, which contains 3 negative pairs!
The five pairs are: (1,-3); (-3,2); (2,5); (5,-4); and (-4,-6).
The three negative pairs are: (1,-3); (-3,2); (5,-4)

Thanks a lot Bunuel!!!!!
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Numbers [#permalink]  01 May 2011, 13:45
For a finite sequence of non-zero numbers, the number of variation in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. what is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?

A) One
B) Two
C) three
D) Four
E) Five
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Re: Numbers [#permalink]  01 May 2011, 17:06
number of variations = 3 [C]
1. 1* -3
2. -3 * 2
3. 5 * -4
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Re: Numbers [#permalink]  01 May 2011, 22:11
Straight 3
Hence C.
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Re: Numbers [#permalink]  02 May 2011, 05:39
Solution:
1,-3 YES NEGATIVE
-3,2 YES NEGATIVE
2,5
5,-4 YES NEGATIVE

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Re: Sequence of nonzero numbers (test from mba.com) [#permalink]  20 Jul 2011, 04:09
tt11234 wrote:
the question is refering to consecutive terms, don't you need to put the sequence in order -from teh lowest to the highest and then see the consecutive numbers. my answer is that there is 0 variation in sigh becasue the only pairs of consecutive numbers are -4 -3 and 1 2 and none of the pairs has negative sign when you multiply them. i am really confused with the way that gmac asks the question...can someone clarify it a little more? thanks!

I have the exact same question... I read some replies that the question doesnt ask that ..but doesnt sequence and consecutive mean in order ascending or descending ....I fail to understand
Re: Sequence of nonzero numbers (test from mba.com)   [#permalink] 20 Jul 2011, 04:09
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