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First of all: the product of four consecutive even integers can be either 0 (when one of the eve integers is zero) or positive (when all integers are positive or when all integers are negative).

(1) The sum of these integers is positive but smaller than 20:

If the greatest term is \(\ge 8\), e.g. {2, 4, 6, 8 }, then the sum will be more than or equal to 20.

If the smallest term \(\le -4\), e.g. {-4, -2, 0, 2 }, then the sum won't be positive.

Hence this statement gives only TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } (\(sum \lt 20\) and \(product =0\)). The product of the terms of either of 2 sets is zero, so the answer to the question is NO. Sufficient.

(2) The product of the middle two of these integers is positive. The product could be zero as well as positive. For example consider {0, 2, 4, 6 } (\(product = 0\)) and {2, 4, 6, 8 } (\(product=positive\)). Not sufficient.

First of all: the product of four consecutive even integers can be either 0 (when one of the eve integers is zero) or positive (when all integers are positive or when all integers are negative).

(1) The sum of these integers is positive but smaller than 20:

If the greatest term is \(\ge 8\), e.g. {2, 4, 6, 8 }, then the sum will be more than or equal to 20.

If the smallest term \(\le -4\), e.g. {-4, -2, 0, 2 }, then the sum won't be positive.

Hence this statement gives only TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } (\(sum \lt 20\) and \(product =0\)). The product of the terms of either of 2 sets is zero, so the answer to the question is NO. Sufficient.

(2) The product of the middle two of these integers is positive. The product could be zero as well as positive. For example consider {0, 2, 4, 6 } (\(product = 0\)) and {2, 4, 6, 8 } (\(product=positive\)). Not sufficient.

Answer: A

For (1), i would just assume, just to simplify, the integers to be x, x+2, x+4, x+6. So, their sum will be (4x + 12). Now (4x + 12) < 20. This now follows the only two possible sets for x = 0 and x = -2, as pointed out by you.

Bunuel wrote:

TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } (\(sum \lt 20\) and \(product =0\))

I just feel better playing with the equations than the numbers directly.
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First of all: the product of four consecutive even integers can be either 0 (when one of the eve integers is zero) or positive (when all integers are positive or when all integers are negative).

(1) The sum of these integers is positive but smaller than 20:

If the greatest term is \(\ge 8\), e.g. {2, 4, 6, 8 }, then the sum will be more than or equal to 20.

If the smallest term \(\le -4\), e.g. {-4, -2, 0, 2 }, then the sum won't be positive.

Hence this statement gives only TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } (\(sum \lt 20\) and \(product =0\)). The product of the terms of either of 2 sets is zero, so the answer to the question is NO. Sufficient.

(2) The product of the middle two of these integers is positive. The product could be zero as well as positive. For example consider {0, 2, 4, 6 } (\(product = 0\)) and {2, 4, 6, 8 } (\(product=positive\)). Not sufficient.

Answer: A

I went the equation route too. Felt comfortable with that.

Let the numbers be 2n , 2n+2 , 2n+4 , 2n+6 ( 4 consecutive even int)

1 says that sum is +ve but less than 20.

Hence 0 < 8n + 12 < 20 It gives -1.5 < n < 1

Hence n = 0,-1. So we have 0 , 2 , 4 , 6 --> Prod = 0 or -2,0,2,4 ---> Prod = 0

Hence st 1 is suff

St 2: We only know middle numbers prod > 0. So prod could be > or = 0. Not suff

I think this is a high-quality question and I don't agree with the explanation. In many of your explanations, Zero is considered "neither positive nor negatice". Infact the answer hinges on this point.

I think this is a high-quality question and I don't agree with the explanation. In many of your explanations, Zero is considered "neither positive nor negatice". Infact the answer hinges on this point.

Here you have considered Zero as postive, why?

Where in the solution is 0 considered positive? That's not so.
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