Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

For a finite sequence of non zero numbers, the number of [#permalink]

Show Tags

12 Sep 2005, 16:29

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

75% (01:37) correct
25% (00:39) wrong based on 216 sessions

HideShow timer Statistics

For a finite sequence of non zero numbers, the number of variations in the 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?

if we order the sequence in increasing order -6,-4,-3,1,2,5 i see only one pair for which the product of two consecutive numbers is negative
i would go with 1

if we order the sequence in increasing order -6,-4,-3,1,2,5 i see only one pair for which the product of two consecutive numbers is negative i would go with 1

Well, yes, but the question does not ask you to do that.

I first mistook the words to mean consecutive pairs as
(1, -3)
(-3, 2)
(2, 5)
and variation of sign only in consecutive pairs, so the lengthiest variation of sign in this case is 2. But i was wrong.

For a finite sequence of non zero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms for the sequence for which the product of the two consecutive term is negative . What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?

For a finite sequence of non zero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms for the sequence for which the product of the two consecutive term is negative . What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?

For a finite sequence of non zero 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

This is the way I would solve the problem: What they're basically asking is the number of times the sign changes when each term in the sequence is multiplied as a pair. For example: 1*-3 = -3 Base -3 * 2 = -6 No variation 2 * 5 = 10 Variation 5 * -4 = -20 Variation -4 * -6 = Variation

Altogether, the sign varies three times, therefore the answer is C. I hope this answer helps, and that it was a relatively quick way to solve it. It took me about one minute. If there is an alternate way to solve this problem, I would love to hear everyone's comments.

For a finite sequence of non zero 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

This is the way I would solve the problem: What they're basically asking is the number of times the sign changes when each term in the sequence is multiplied as a pair. For example: 1*-3 = -3 Base -3 * 2 = -6 No variation 2 * 5 = 10 Variation 5 * -4 = -20 Variation -4 * -6 = Variation

Altogether, the sign varies three times, therefore the answer is C. I hope this answer helps, and that it was a relatively quick way to solve it. It took me about one minute. If there is an alternate way to solve this problem, I would love to hear everyone's comments.

I dont think we need to consider variation in the way explained above. Variations simply would mean to consider consecutive terms whose product is -ve. Can someone confirm/explain/provide clarity on this.

1st pair is (1,-3), which has a product of -3 - satisfies the condition 2nd Pair is (-3,2),which has a product of -6 - satisfies the condition 3rd Pair is (2,5), which has a + product of 10, thus does not satisfy the condition 4th Pair is (5,-4), which has a product of -20 - satisfies the condition 5th pair is (-4,-6), product 24 , thus does not satisfy the condition

For a finite sequence of non zero numbers, the number of variations in the 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. 1 B. 2 C. 3 D. 4 E. 5

The questions basically asks: how many pairs of consecutive terms are there in the sequence such that the product of these consecutive terms is negative.

Re: Sequence of nonzero numbers (test from mba.com) [#permalink]

Show Tags

17 Jul 2010, 17:48

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!

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!

Couple of things:

1. Even if put the terms in ascending order {-6, -4, -3, 1, 2, 5} still one pair of consecutive terms will make negative product: -3*1=-1=negative. But if you are right, then ANY sequence of the non-zero integers which have both negative and positive numbers will have variation of 1 and the question does not make sense any more.

2. A sequence by definition is already an ordered list of terms. So if we are given the sequence of 10 numbers: 5, 6, 0, -1, -10, -10, -10, 3, 3, -100 it means that they are exactly in that order and not in another.

Re: Sequence of nonzero numbers (test from mba.com) [#permalink]

Show Tags

29 Nov 2010, 12:41

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)

Re: Sequence of nonzero numbers (test from mba.com) [#permalink]

Show Tags

20 Jul 2011, 03:09

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: For a finite sequence of non zero numbers, the number of [#permalink]

Show Tags

02 Jan 2014, 03:03

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: For a finite sequence of non zero numbers, the number of [#permalink]

Show Tags

14 Dec 2014, 12:00

Hi Brunel,

Quick question. In the sequence below, would consecutive terms not be: 1, 2, -3, 4, 5, -6? Normally consecutive terms would be: 1,2,3,4,5,... etc, but here we are given negatives:(. This threw off the definition slightly.

Therefore the number of consecutive -ve pairs would be:

2* (-3) = -ve -4*5 = -ve 5* (-6) = -ve

I get the same answer as you do, but I am just wondering if my approach is correct. As well, if 1, 2, -3, 4, 5, -6 is GMAT's correct definition of consecutive integers, please let me know.

Thanks, Regards, Yela

Bunuel wrote:

For a finite sequence of non zero numbers, the number of variations in the 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. 1 B. 2 C. 3 D. 4 E. 5

The questions basically asks: how many pairs of consecutive terms are there in the sequence such that the product of these consecutive terms is negative.

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...