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M22-22

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M22-22  [#permalink]

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New post 16 Sep 2014, 00:17
12
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

27% (01:20) correct 73% (01:34) wrong based on 210 sessions

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Re M22-22  [#permalink]

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New post 16 Sep 2014, 00:17
3
1
Official Solution:


Note that we are not told that \(a\), \(b\) and \(c\) are integers.

(1) \(a-c-b\) is even. If the variables are integers then \(a+b+c\) will be even but if they are not, for example if \(a=3.5\), \(b=1\) and \(c=0.5\) (\(a-c-b=2=even\)), then \(a+b+c=5=odd\). Not sufficient.

(2) \(\frac{a-c}{b}\) is odd. The same here: if the variables are integers then \(a+b+c\) will be even but if they are not, for example if \(a=3.5\), \(b=1\) and \(c=0.5\) (\(\frac{a-c}{b}=3=odd\)), then \(a+b+c=5=odd\). Not sufficient

(1)+(2) \(a+b+c\) may or may not be even (again if variables are integers: YES but if \(a=3.5\), \(b=1\), \(c=0.5\) answer is NO). Not sufficient.


Answer: E
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Re: M22-22  [#permalink]

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New post 23 Dec 2014, 15:27
Would the answer be D if it was given that a, b, and c are integers?
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New post 24 Dec 2014, 06:23
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New post 26 Jun 2015, 04:03
Here's how I would look at Statement A,

a - c - b = even
a + c + b - 2c - 2b = even
a + c + b - 2(c+b) = even

For this statement to be true,
a + c + b = even, since only "even - even = even"

Am I missing something?
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New post 26 Jun 2015, 05:06
mukulc wrote:
Here's how I would look at Statement A,

a - c - b = even
a + c + b - 2c - 2b = even
a + c + b - 2(c+b) = even

For this statement to be true,
a + c + b = even, since only "even - even = even"

Am I missing something?


Yes.

I guess you haven't read the solution. Very first sentence there says: "note that we are not told that a, b and c are integers".
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Re: M22-22  [#permalink]

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New post 12 Mar 2016, 22:50
For statement 1, is it correct to say that if A+B+C=Even and A,B,C are integers, then A-B-C is also even under any order of A,B,C (i.e., also even C-A-B or B-A-C)?
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New post 13 Mar 2016, 07:22
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New post 01 Aug 2016, 11:02
I think this is a high-quality question and I agree with explanation. Good one!
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New post 22 Aug 2016, 04:36
I think this is a high-quality question and I agree with explanation. Bombastic Question Lol great trap! You people never fail to surprise me!
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New post 01 Dec 2016, 06:33
Sigh, how many more times am I going to forget to check if I'm dealing with integers. Quickest and proudest answer D, thinking I'm all that smart, just to be blown by the "might not be integers" detail.
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New post 30 Sep 2017, 11:30
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Here is a method I used to solve this question in the most efficient manner that I haven't seen discussed here.

Stem:\(A+B+C= Even\)
(1)\(A-B-C=E\)
This becomes \(A=E+B+C\)
Important property: In a term of the form (Even number) + X, the (Even number) plays no role in the Even-Odd nature of the term
In turn, this becomes A=B+C,
Plugging into the equation we get:\(B+C+B+C=E?\)this becomes \(2B+2C=E\)
If A & B are integers, this is sufficient, but A & B can both be .1

(2)\(\frac{A-C}{B}=O\)
Or \(O*B=A-C\)
Important property: In a term of the form (Odd number)*(X), the (Odd number) plays no role in the Even-Odd nature of the term
\(B=A-C\)
Plugging into the equation we are left with \(2A=E?\)
If A is an integer, this is sufficient, but A can be .1

Therefore E
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Re: M22-22  [#permalink]

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New post 24 Jul 2018, 06:02
wow bamboozled.
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Re M22-22  [#permalink]

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New post 19 Nov 2018, 05:26
I think this is a high-quality question and I agree with explanation. A very good question.

Completely stumped ignoring the fact that they are not necessarily integers. Chose D after solving the question for about 4 mts.
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Re M22-22 &nbs [#permalink] 19 Nov 2018, 05:26
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