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Re M1727 [#permalink]
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16 Sep 2014, 00:02
Official Solution: 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
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Re: M1727 [#permalink]
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23 Oct 2015, 01:18
Bunuel wrote: Official Solution:
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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Re: M1727 [#permalink]
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30 Apr 2016, 07:01
vivekgautam1 wrote: Bunuel wrote: Official Solution:
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



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Re M1727 [#permalink]
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06 Sep 2016, 00:03
I think this is a highquality 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?



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Re: M1727 [#permalink]
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06 Sep 2016, 00:31



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Re: M1727 [#permalink]
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27 Aug 2017, 11:43
It says in the second stem that the product is positive. Zero is neither positive nor negative? Correct me if i am wrong



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Re M1727 [#permalink]
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22 Nov 2017, 22:02
I think this is a highquality question and the explanation isn't clear enough, please elaborate. Shouldn't the set be {2, 2, 4, 8} instead of considering the 0 in the set as 0 is NOT an even integer?
By this logic also, statement 1 seems is correct. I marked 1 by this logic.
Please explain. Should we consider 0 or not?



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Re: M1727 [#permalink]
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22 Nov 2017, 23:11
jasanisanket24 wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. Shouldn't the set be {2, 2, 4, 8} instead of considering the 0 in the set as 0 is NOT an even integer?
By this logic also, statement 1 seems is correct. I marked 1 by this logic.
Please explain. Should we consider 0 or not? ZERO:1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself. Check below for more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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