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For a finite sequence of non zero numbers, the number of [#permalink]
22 Feb 2012, 05:33

00:00

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Difficulty:

25% (low)

Question Stats:

71% (01:35) correct
28% (00:48) wrong based on 181 sessions

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 ?

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. 1 B. 2 C. 3 D. 4 E. 5

this problem is already posted in the forum. My doubt is every body multiplying the negative number with positive number to find the variations. but the question asked for "number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative." for ex:1,-3 are not consecutive . please explain

You are probably mixing consecutive terms in a sequence and consecutive integers: 1 and -3 are not consecutive integers, but they are consecutive terms in the sequence given. See complete solution below.

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

Given sequence: {1, -3, 2, 5, -4, -6}

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: For a finite sequence of non zero numbers [#permalink]
02 Sep 2012, 07:58

We can take two consecutive numbers of this sequence and the product of those two numbers has to be negative. There are 5 pairs, we can build: (1, -3), (-3, 2), (2, 5), (5, -4), (-4, -6)

Re: For a finite sequence of non zero numbers, the number of [#permalink]
03 Sep 2012, 21:42

Bunuel wrote:

TomB 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. 1 B. 2 C. 3 D. 4 E. 5

this problem is already posted in the forum. My doubt is every body multiplying the negative number with positive number to find the variations. but the question asked for "number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative." for ex:1,-3 are not consecutive . please explain

You are probably mixing consecutive terms in a sequence and consecutive integers: 1 and -3 are not consecutive integers, but they are consecutive terms in the sequence given. See complete solution below.

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

Given sequence: {1, -3, 2, 5, -4, -6}

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.

So there are 3 pairs of consecutive terms of the sequence for which the product is negative.

Answer: C.

Hope it's clear.

I have answered correctly, but my pairs were: (2, -3) (-4,5) (5,-6). My question is, Bunuel, why do we consider (1-3) as pair while (5;-6) not? Thanks.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: For a finite sequence of non zero numbers, the number of [#permalink]
04 Sep 2012, 02:10

Expert's post

ziko wrote:

Bunuel wrote:

TomB 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. 1 B. 2 C. 3 D. 4 E. 5

this problem is already posted in the forum. My doubt is every body multiplying the negative number with positive number to find the variations. but the question asked for "number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative." for ex:1,-3 are not consecutive . please explain

You are probably mixing consecutive terms in a sequence and consecutive integers: 1 and -3 are not consecutive integers, but they are consecutive terms in the sequence given. See complete solution below.

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

Given sequence: {1, -3, 2, 5, -4, -6}

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.

So there are 3 pairs of consecutive terms of the sequence for which the product is negative.

Answer: C.

Hope it's clear.

I have answered correctly, but my pairs were: (2, -3) (-4,5) (5,-6). My question is, Bunuel, why do we consider (1-3) as pair while (5;-6) not? Thanks.

Again, we are told that "the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence ..." 1 and -3 are consecutive terms in the sequence while 5 and -6 are not.
_________________

Re: For a finite sequence of non zero numbers, the number of [#permalink]
04 Sep 2012, 02:58

Thank you Bunuel, i got it, i did not realised that 1, -3, 2, 5, -4, -6 is a given finite sequence, for some reason i understood it as a set. Although now i see that if it were a set then the answer would be 0, since there are no pair with negative signs in a normal consequtive sequence.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: For a finite sequence of non zero numbers, the number of [#permalink]
04 Sep 2012, 03:05

Expert's post

ziko wrote:

Thank you Bunuel, i got it, i did not realised that 1, -3, 2, 5, -4, -6 is a given finite sequence, for some reason i understood it as a set. Although now i see that if it were a set then the answer would be 0, since there are no pair with negative signs in a normal consequtive sequence.

1. Even if we consider 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 in this case, ANY sequence of 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: For a finite sequence of non zero numbers, the number of [#permalink]
20 Jul 2013, 23:58

So the only thing different about this question is that people might re-arrange the sequence and that's what you are not supposed to do?
_________________

Re: For a finite sequence of non zero numbers, the number of [#permalink]
16 Sep 2013, 08:54

Bunuel wrote:

fozzzy wrote:

So the only thing different about this question is that people might re-arrange the sequence and that's what you are not supposed to do?

People might do a lot of things. The point is to read the stem carefully.

Ok it took me like 5 reads to understand what the question is about. I understood Bunuel's explanation (straight forward) but didn't get that GMAT declared a fancy way of saying the product of each pair of integers... I wonder how many of these does it take to drop you off your seat!