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Finite sequence....

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by gmatpup » Fri Nov 11, 2011 9:16 am
For a finite sequence of nonzero numbers, the number if variations in sign if 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

Answer is D
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by shankar.ashwin » Fri Nov 11, 2011 9:46 am
I can see only 3 sign changes.

1*-3
-3*2
5*-4
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by gmatpup » Fri Nov 11, 2011 9:55 am
that's what I thought the answer was too!! So I am not sure how to go about this question. It is from gmatprep.
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by Anurag@Gurome » Mon Nov 28, 2011 9:11 pm
edirik wrote:I am unable to understand the question stem, can you please check if there is a typo. It seems like there are more "if"s than needed.
The question is: For a finite sequence of nonzero numbers, the number of variations 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

1 * -3 = -3 (negative)
-3 * 2 = -6 (negative)
5 * -4 = -20 (negative)

So, the number of variations are 3, so the correct answer should be C.
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