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GMAT Problem Solving Numbers

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GMAT Problem Solving Numbers

by harshpamnani » Thu Jun 23, 2011 5:22 pm
Q.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?

This question says,consecutive terms.Means they should be in a sequence. If I put the above numbers in a sequence then it would be -6,-4,-3,1,2,5.Now,question says, number of variations r defined as number of pairs of consecutive terms.In terms of pairs of positive and negative,I have only one pair: -3,1.So answer should be 1 pair, but corect answer is 3 pairs, which could be right in a sense if we don't use consecutive sequence.Please guide me,why the answer is 3.
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by VivianKerr » Thu Jun 23, 2011 5:33 pm
According to the question: "number of variations" = # of pairs of consecutive terms IN THE SEQUENCE for which the product is (-).

The sequence is: 1, -3, 2, 5, -4, -6.

The word "consecutive" in this question has to do with two terms that are placed next to each other in a sequence. You cannot rearrange the order of the sequence, because them you are creating a new sequence.

Let's find the product of all the "consecutive" terms:

1 x -3 = -3
-3 x 2 = -6
2 x 5 = 10
5 x -4 = -20
-4 x -6 = 24

Three of the products are negative, so the answer is 3 pairs.
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by MBA.Aspirant » Thu Jun 23, 2011 5:42 pm
the sequence doesn't have to be in ascending order.

It can go like: 1,2,3,4,5 or 5,4,3,2,1 or a Fibonacci one where the term equal the sum of the 2 preceding terms: 3, -1, 2, 1, 3. There're many ways to do it

for the sequence mentioned, there're 3 times the product is negative:

1, -3, 2, 5, -4, -6.

the pairs are (1,-3), (-3,2) and (5,-4)
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