IMO Emgmt_gmat wrote:63) S is finite set of numbers. Does S contain more negative numbers than positive numbers?
a. Product of all numbers in S is -1200
b. There are 6 numbers in S
Please explain..
It is a question of data sufficiency so should be posted in data sufficiency forummgmt_gmat wrote:63) S is finite set of numbers. Does S contain more negative numbers than positive numbers?
a. Product of all numbers in S is -1200
b. There are 6 numbers in S
Please explain..
shashank.ism wrote:It is a question of data sufficiency so should be posted in data sufficiency forummgmt_gmat wrote:63) S is finite set of numbers. Does S contain more negative numbers than positive numbers?
a. Product of all numbers in S is -1200
b. There are 6 numbers in S
Please explain..
Well for the solution of this problem.
a. Product of all numbers in S is -1200. Lets factorize the number
1200 = 5x5x3x2x2x2x2
so we don't know the total numbers etc here so we can't say whether has more negative numbers than positive numbers ---- insufficient
b. 6 numbers in S. so for product to be -ve there may be one -ve no. or 3 -ve number or 5 -ve number.
so it is also insufficient.
combined also same case B. persist we can't say how many nos. ar -ve so insufficient.
Hi,cat er ina wrote: but guys explain this..together statement 1 and 2 why is not sufficient??? because it seems to me that there have to be 3 negative and three positive numbers since the product has to be negative!!! so not considering the order +*-*-*+*-*+ or +*-*+*-*+*- anyway it s 3 pos numbers and 3 neg numbers
Stuart Kovinsky wrote:Hi,cat er ina wrote: but guys explain this..together statement 1 and 2 why is not sufficient??? because it seems to me that there have to be 3 negative and three positive numbers since the product has to be negative!!! so not considering the order +*-*-*+*-*+ or +*-*+*-*+*- anyway it s 3 pos numbers and 3 neg numbers
we could also have 1 negative and 5 positives, or 5 negatives and 1 positive.
As long as there's an odd number of negative terms, we'll get a negative product.
Here's how we can analyze with minimal math:
Q: Does finite set S contain more negative than positive terms?
What do we know? Not much - just that there's a fixed number of terms in S. To the statements!
(1) The product of the terms is negative (we don't care what the product is, just the sign).
To get a negative product, you simply need an odd number of negative terms. (1) tells us nothing about the number of terms: insufficient.
(2) We know the number of terms, but nothing about sign: insufficient.
Together: We know that we have 6 terms, including an odd number of negative terms. So, the breakdowns could be:
1 negative, 5 positives ("no" answer to the question);
3 negatives, 3 positives ("no" answer to the question); or
5 negatives, 1 positive ("yes" answer to the question).
Since we can get both a "no" and a "yes", still insufficient: choose E.
* * *
As an aside, we can't definitively factorize the numbers for 1200, since there's no reason why we can't use fractions or repeat the same numbers multiple times (although we don't need to factorize for this question, so it's not a big deal).
