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M16-14

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M16-14  [#permalink]

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New post 15 Sep 2014, 23:58
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

58% (00:46) correct 42% (00:56) wrong based on 144 sessions

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Re M16-14  [#permalink]

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New post 15 Sep 2014, 23:58
1
Official Solution:


Statement (1) by itself is sufficient. Both \(a\) and \(b\) are odd (otherwise \(ab\) would be even). Therefore, \(\frac{a}{b}\) is not an even integer. If \(\frac{a}{b}\) were an even integer, \(a\) would be even.

Statement (2) by itself is insufficient. Consider \(a = 4\), \(b = 2\) (the answer to the question is "yes") and \(a = 5\), \(b = 3\) (the answer to the question is "no").


Answer: A
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M16-14  [#permalink]

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New post Updated on: 18 Dec 2015, 09:07
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. Do we need to be told in the question prompt that a>b?

Nevermind, I get it now too. Either it will yield a non-integer or an odd integer, NEVER an even integer!

Originally posted by jvh42 on 01 Dec 2015, 09:22.
Last edited by jvh42 on 18 Dec 2015, 09:07, edited 1 time in total.
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M16-14  [#permalink]

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New post 18 Dec 2015, 09:00
hi Bunuel, I get that both a and b are odd from statement 1. But what tells us that a/b actually yields an integer? I chose E because I found no proof that a was divisible by b to begin with. I am thinking wrong here?
Thanks
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Re: M16-14  [#permalink]

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New post 18 Dec 2015, 09:03
Actually, never mind my question, I get it. Thanks
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Re: M16-14  [#permalink]

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New post 04 Jan 2016, 12:24
I think, (E) should be the answer.
(A) would be the answer provided the condition a>b is given in question.

Nevertheless, got the Q. Thanks for everything Gmat club!
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Re: M16-14  [#permalink]

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New post 05 Jan 2016, 00:56
MK1480 wrote:
I think, (E) should be the answer.
(A) would be the answer provided the condition a>b is given in question.

Nevertheless, got the Q. Thanks for everything Gmat club!


The first statement, regardless whether a>b or not, gives a NO answer to the question: - a/b is NOT an even integer. Thus it's sufficient.
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Re: M16-14  [#permalink]

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New post 13 Mar 2017, 12:12
I solved it using the general rule for odd and even numbers.

Multiplication:
-------------
(i)odd * odd = odd
(ii)odd * even= even
(iii)even* even = even

Addition/subtraction
-------------------
(iv)Odd +/- Odd = Even
(v)Even +/- Odd= Odd
(vi)Even +/- Even = Even

Applying the above rule to the question,

1 stmt) given ab is odd, that means a and b are odd numbers - applies rule (i) ONLY. so a/b yields either a fraction or an odd integer but not a even integer. So this stmt is sufficient, as it gives an answer no.

2nd stmt) a and b can be either odd or even - applies rule (iv) or (vi). So can't conclude. Not sufficient.

Hope this helps!
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Re: M16-14 &nbs [#permalink] 13 Mar 2017, 12:12
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