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by crackgmat007 » Mon Aug 10, 2009 8:37 pm
For a finite set 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 consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?

a. 1
b. 2
c. 3
d. 4
e. 5

OA - D
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by tohellandback » Mon Aug 10, 2009 8:44 pm
to me it looks like the answer should be C

1*-3- negative
-3*2- negative
5*-4 negative
The powers of two are bloody impolite!!
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by crackgmat007 » Mon Aug 10, 2009 9:48 pm
tohellandback wrote:to me it looks like the answer should be C

1*-3- negative
-3*2- negative
5*-4 negative
dont we have to arrange the number in consecutive order?
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by tohellandback » Mon Aug 10, 2009 9:50 pm
crackgmat007 wrote:
tohellandback wrote:to me it looks like the answer should be C

1*-3- negative
-3*2- negative
5*-4 negative
dont we have to arrange the number in consecutive order?
IMO a sequence doesn't necessarily have to be the natural order.
it says "consecutive terms of the sequence". so IMO the numbers are in some order already.
The powers of two are bloody impolite!!
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