Finite sets?

Let S contain five numbers
s = (a,b,c,d,e)

now

I) product of all numbers = -ve

this is possible if
any one is - ve
any 3 are -ve

If one is -ve, so no. of negative(1) < no. of +ve(4)

if 3 are -ve then no. of -ve (3)> no.of+ve(2)

However in our assumption,we assumed that S contains 5 numbers, in case S contains 6

so,if no. of -ve = 3, no. of -ve = no of +ve

so we cannot precisely say anything from
(I)


(II) not suff


combine I and II


Even if there are 6 no. in the set and we know their product is -ve, we cannot say precisely that whether
1 no is -ve,2 no's are -ve or 3 no's are -ve

in case 3 no's are -ve, no of -ve = no of +ve

however in other two cases,
no. of -ve<no of +ve

Hence combining both also Insuff!!
So E

Hope this helps

Karan

Stmt 1 says that the product of n numbers is a -ve number.
The product of even number of negative signs is a +ve sign.
Ie. there could be any odd number of -ve numbers in the set.
So insufficient.

Stmt 2 says that there are 6 numbers.
not very useful
insufficient.


combining them.
lets consider 6 numbers whose product is a -ve number.
There could be 5 -ve numbers and 1 +ve number OR
5 +ve numbers and 1 -ve number.

Insufficient.

hence E