From 1 list could be

a) even = 1,2 product greatest smallest + ve , final product + ve


b) odd = -3 -2 -1 product greatest smallest + ve , final product - ve


Hence insufficient knock off AD

From 2 list could be


1 ,2 final product + ve

1, -2 final product - ve


Hence insufficient knock of B left with CE


Combining

List is even and product greatest smallest + ve

List
1,2 product greatest smallest + ve , final product + ve

-4 -3 -2 -1 product greatest smallest + ve , final product + ve


Hence C

Hi sujaysolanki
Sorry, I cannot understand what ve means....Also, can you explained more detailed,
Thanks

Hi again sujaysolanki,
Ok, I kinda' see what you did in there, but I think you approach is limited to those specific numbers that you picked....How about when you have a list of 6 numbers for instance. When taken together (1) and (2), le's say you have a number of 6 integers and the product of the greatest and the smalest is positive. Now, you have 4 other numbers , which 3 of them could be positive and the 4th one negative, in this case you final product can be negative....See here I don't get it. I just need a stronger argument for this solution.
Thanks

I got it now.
Thanks

This is because
1) as per condition 1st the only 2 possibilities is either both numbers are +ve or -ve
2) as per 2nd cdt the whether the two are +ve or -ve the Product will be +ve as the no. Of integers is even

Hence c is the corect option