Answer should be C.

Let's first simplify the question.The question asks whether the product of integers is +ve.

When is the product of integers +ve? When the number of negative integers is even.Whether the number of positive integers is odd does not matter, because (+ve * +ve )= +ve and (+ve *+ve * +ve)=+ve.

lets assume the integers in the list are -1,2,3 [odd -ve integers].Then the product too will be -ve. Now assume the integers in the list are -2,-1,1,2,3[even =ve's].The product is +ve.

So,The question boils down to :whether no of -ve integers is even ?

1) This says that the product of smallest and largest integer is +ve.Lets assume a= smallest integer and b= largest integer Then ab >0 only when both have same sign,i.e,either both are +ve or both are -ve.

But can we conclude that the product of all integers is +ve?NO

let's say a = -10 and b=-2.The product ab =20. If there is another integer between a and b,say c,where c =-5.Then the product will be -ve.But if there are even number of -ve integers between a and b,then the product is +ve.

Hence insufficient.

2) This is clearly insufficient.Total no of integers being even gives us no info about whether -ve integers are odd or even.

eg {-1,2,3,4}=-ve product {-2,-1,3,4}= positive product

Insufficient.


Combining 1 and 2.

We know that both the smallest and the largest integer have the same sign (from 1).We also know that the total number of integers is even.

So,if we choose smallest and largest = -10 and -2 respectively.

Then list of integers in the list could be {-10,-2} or {-10,-6,-5,-2}. Both give +ve product.

You can't have {-10,-5,-2} in which case total no of integers would be odd [a contradiction of 2].

Hence sufficient.

OA?

danjuma wrote:A certain lists consists of several different integers. Is the product of all the integers in the list positive?

1. The product of the greatest and smallest of the integers in the list is positive.

2.There is an even number of integers in the list.

Rishab has given a very good explanation, Answer should be C,