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For a finite sequence of non zero numbers, the number of [#permalink]
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12 Sep 2005, 17:29
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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
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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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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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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]
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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]
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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: GPrep # 1 [#permalink]
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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]
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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]
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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]
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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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Re: Sequence of nonzero numbers (test from mba.com) [#permalink]
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17 Jul 2010, 16:16



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Re: Sequence of nonzero numbers (test from mba.com) [#permalink]
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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]
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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 nonzero 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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Re: Sequence of nonzero numbers (test from mba.com) [#permalink]
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29 Nov 2010, 13:41
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)
Thanks a lot Bunuel!!!!!



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Re: Numbers [#permalink]
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01 May 2011, 17:06
number of variations = 3 [C] 1. 1* 3 2. 3 * 2 3. 5 * 4



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Re: Numbers [#permalink]
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02 May 2011, 05:39
Solution: 1,3 YES NEGATIVE 3,2 YES NEGATIVE 2,5 5,4 YES NEGATIVE
Answer: THREE



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Re: Sequence of nonzero numbers (test from mba.com) [#permalink]
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20 Jul 2011, 04:09
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



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Re: For a finite sequence of non zero numbers, the number of [#permalink]
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02 Jan 2014, 04:03
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Re: For a finite sequence of non zero numbers, the number of [#permalink]
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14 Dec 2014, 13:00
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.
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: For a finite sequence of non zero numbers, the number of [#permalink]
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14 Dec 2014, 13:33
Hi, Can you please explain how 1 and 3 can be consecutive??




Re: For a finite sequence of non zero numbers, the number of
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14 Dec 2014, 13:33



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