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M27-20

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M27-20  [#permalink]

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New post 16 Sep 2014, 01:27
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

69% (00:59) correct 31% (00:51) wrong based on 29 sessions

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Re M27-20  [#permalink]

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New post 16 Sep 2014, 01:27
Official Solution:


(1) \(x*|y|\) is a prime number. Since only positive numbers can be primes, then: \(x*|y|=positive\) hence \(x=positive\). Sufficient.

(2) \(x*|y|\) is a non-negative integer. Notice that we are told that \(x*|y|\) is non-negative, not that it's positive, so \(x\) can be positive as well as zero. Not sufficient.


Answer: A
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Re: M27-20  [#permalink]

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New post 28 Feb 2017, 14:53
|y| means y can be -y or +y. so x can be -x when y is negative and x is positive when y is positive. How is A the answer? I think it is E
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Re: M27-20  [#permalink]

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New post 28 Feb 2017, 21:19
Shiridip wrote:
|y| means y can be -y or +y. so x can be -x when y is negative and x is positive when y is positive. How is A the answer? I think it is E


Absolute value of a number is always non-negative. So, |y| (absolute value of y), regardless whether y itself is negative or positive, will be non-negative.

Please re-read the solution again:
(1) \(x*|y|\) is a prime number. Since only positive numbers can be primes, then: \(x*|y|=positive\) hence \(x=positive\). Sufficient.
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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

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Re: M27-20 &nbs [#permalink] 28 Feb 2017, 21:19
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