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(00:24)
correct 50%
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It seems a simple question to me, but I think the answer is incorrect. Can anyone please help?
Is the product of all of the elements in Set S are negative?
1) All of the elements in Set S are negative 2) There are 5 negative numbers in set S.
Here is the explanation for my answer which I think is B - i.e Statement 2 alone is sufficient.
Statement 1 --> Alone is not sufficient because we don't the number of elements in Set S. There can be ODD or EVEN number of elements which will be lead to Positive and Negative Product.
Statement 2--> Is Sufficient. Because the product will be negative as there are ODD number (5) of elements in the set.
But this is what MGMAT says: If any of the elements in set S equals Zero, then the product of the set will be zero which is neither +ve nor -ve. How can any of the element in Set S be Zero? Statement 2 says that there are 5 negative numbers and 0 is neither
But MGMAT reckons the correct answer is C. Can anyone please tell me why and how?
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It seems a simple question to me, but I think the answer is incorrect. Can anyone please help?
Is the product of all of the elements in Set S are negative?
1) All of the elements in Set S are negative 2) There are 5 negative numbers in set S.
Here is the explanation for my answer which I think is B - i.e Statement 2 alone is sufficient.
Show more
I think the OA is perfectly fine . St1 : all elements in set S are negative: we dont know how many elements are there .. if there are even no. of elements then the product is +ve . ......... hence insufficient
st2: there are 5 negative numbers: Dont Assume that the set contains only 5 numbers .. the set might contain 20 numbers out of which 5 are negative. B could have been right, if the question specifically mentioned non zero numbers. it doesnt
1&2 together: tells us that all elements are negative( st 1) and there are 5 negative numbers. thus the product of 5 negative numbers will be always be negative.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.