BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

The number of variations in sign...PS

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Sun Nov 14, 2010 11:01 pm

The number of variations in sign...PS

by adwaitkasbekar » Tue Dec 21, 2010 11:01 pm
Can anyone please explain how this problem is to be solved?

For a finite set 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 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
Top
Source: — Problem Solving |
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Tue Dec 21, 2010 11:27 pm
adwaitkasbekar wrote:Can anyone please explain how this problem is to be solved?

For a finite set 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 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
if I am not mistaken this problem is mixed symbolism and number properties (+ve/-ve)*(+ve/-ve)

if we consider (1;-3) (-3;2) (2;5) (5;-4) (-4;-6) then we get in total 3 (-ve)

IOM C
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Sun Nov 14, 2010 11:01 pm

by adwaitkasbekar » Tue Dec 21, 2010 11:30 pm
the answer is indeed C.

I got confused by the language.

Thanks for helping me out.
Top
Master | Next Rank: 500 Posts
Posts: 181
Joined: Sat Oct 16, 2010 1:57 pm
Thanked: 4 times

by N:Dure » Tue Dec 21, 2010 11:31 pm
adwaitkasbekar wrote:
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 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
I think the bolded part is what he's looking for. So pairs of consecutive terms in the sequence whose product is negative will be (1, -3) (-3, 2) (5,-4)

Number of variations = 3
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Sat Apr 14, 2012 6:55 am
Location: Bangalore, India
GMAT Score:430

by NathaliaC » Mon Jul 02, 2012 9:13 am
This language is indeed very confusing. In my case, while solving I rewrote the sequence in order because the problem mentions "sequence of consecutive terms", so I rewrote to -6, -4, -3, 1, 2, 5. This led to only 1 product of consecutive terms that is negative (-3 * 1). Wrong answer.
-Nathália
Top
Post new topic Post Reply
• Page 1 of 1