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A simple prob but i got it wrong!! [#permalink] New post 12 Sep 2005, 17:29
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For a finite sequence of non zero numbers, the no of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of 2 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
[Reveal] Spoiler: OA
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 [#permalink] New post 12 Sep 2005, 20:16
Agree with duttsit's approach. Going with 3.
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 [#permalink] New post 12 Sep 2005, 20:20
I got 3 as well... same reason above... most challenging part of this question is the wording
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 [#permalink] New post 13 Sep 2005, 07:00
what a waste of words... i feel bad for those who take this test with english as their second language
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 [#permalink] New post 13 Sep 2005, 08:27
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
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 [#permalink] New post 13 Sep 2005, 09:36
BG wrote:
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.
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 [#permalink] New post 13 Sep 2005, 13:27
Yeah, the OA is 3.

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.
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Dangerous Sequence [#permalink] New post 13 Jun 2008, 17:58
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?
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Re: Dangerous Sequence [#permalink] New post 13 Jun 2008, 18:03
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Vavali wrote:
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?



Ans 3

Variation 1 : 1*-3
Variation 2 : -3*2
Variation 3 : 5*-4

I like your outrageous titles for all the questions :-)
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Re: Dangerous Sequence [#permalink] New post 14 Jun 2008, 00:43
rpmodi wrote:
Vavali wrote:
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?



Ans 3

Variation 1 : 1*-3
Variation 2 : -3*2
Variation 3 : 5*-4

I like your outrageous titles for all the questions :-)


8-) just trying to add a little flavour to the mix.

kudos to you my friend +1
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GPrep # 1 [#permalink] New post 28 Nov 2009, 02:57
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


OA:
[Reveal] Spoiler:
C


Can someone explain the step by step procedure to solve and what exactly is being asked?

Thanks!
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Re: GPrep # 1 [#permalink] New post 28 Nov 2009, 07:59
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.
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Re: GPrep # 1 [#permalink] New post 30 Nov 2009, 03:36
veenahe wrote:
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.
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Re: GPrep # 1 [#permalink] New post 30 Nov 2009, 06:58
Here the sequence is 1, -3, 2, 5, -4, -6

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

Thus 3 pair satisfy the equation.

Thus OA is C
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Re: GPrep # 1 [#permalink] New post 30 Nov 2009, 07:17
IMO C

1*-3=-3----->1 -ve
-3*2=-6----->1 -ve
2*5=10------>0 +ve
5*-4=-20---->1 -ve
-4*-6=24---->0+ve

total 3
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Sequence of nonzero numbers (test from mba.com) [#permalink] New post 17 Jul 2010, 10:21
This is from the test downloaded on mba.com.
Can't get the right answer, my (A) is wrong.
Attachment:
sequence of nonzero numbers.JPG
sequence of nonzero numbers.JPG [ 27.72 KiB | Viewed 3710 times ]

P.S. Sorry, i had to post it to GMAT Problem Solving (PS)
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Re: Sequence of nonzero numbers (test from mba.com) [#permalink] New post 17 Jul 2010, 16:16
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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.

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

So there are 3 pairs of consecutive terms.

Answer: C.

Hope it's clear.
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Re: Sequence of nonzero numbers (test from mba.com) [#permalink] New post 17 Jul 2010, 18:48
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!
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Re: Sequence of nonzero numbers (test from mba.com) [#permalink] New post 17 Jul 2010, 19:09
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!


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.

Hope it's clear.
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Hey guy, can you please help me with this PS [#permalink] New post 29 Nov 2010, 12:36
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 term is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6 ?

-One, -Two, -Three, -Four, -Five.
Hey guy, can you please help me with this PS   [#permalink] 29 Nov 2010, 12:36
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