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by sukhman » Tue Oct 15, 2013 9:44 am
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 is in sign for the sequence: 1, -3, 2, 5, -4, -6?
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by Brent@GMATPrepNow » Tue Oct 15, 2013 9:47 am
sukhman wrote: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 is in sign for the sequence: 1, -3, 2, 5, -4, -6?
A. 1
B. 2
C. 3
D. 4
E. 5

We're asked to look at every pair of consecutive terms. If the product of that pair is negative, this counts as one variation.

Let's examine the pairs of consecutive terms:

1 and -3: product is negative
-3 and 2: product is negative
2 and 5: product is positive
5 and -4: product is negative
-4 and -6: product is positive

Since 3 pairs of consecutive terms have negative products, the correct answer is C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by theCodeToGMAT » Tue Oct 15, 2013 9:50 am
1 x -3 = "-3"
-3 x 2 = "-6"
2 x 5 = 10
5 x -4 = "-20"
-4 x -6 = 24

Answer [spoiler]{3}[/spoiler]
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