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

Q: is \(a+b+c=even\)?

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

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

Re: Is A+B+C even? 1) A-C-B is even 2) (A-C)/B is odd [#permalink]

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16 Dec 2011, 22:49

E it is.

I tried to solve algebraically. However, I have a question.

If it were mentioned that all three numbers were integers, how would the answer change? How to figure this out algebraically? (Without having to pick numbers) _________________

Re: Is A+B+C even? 1) A-C-B is even 2) (A-C)/B is odd [#permalink]

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18 Dec 2011, 23:27

jitbec wrote:

E.

What is the point of mentioning just the answer that you figured out? The OA is already known. If you could provide an effective method to solve the problem, it would be highly appreciated.

Re: Is A+B+C even? 1) A-C-B is even 2) (A-C)/B is odd [#permalink]

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18 Dec 2011, 23:43

Lucky2783 wrote:

i dont an algebric way to do it but here is a simple permutation on A,B,C "if given is that A, B , C are integers" .

then option 1 A-B-C is even only in below cases A-C-B E-E-E E-O-O (E=Even, O=Odd) O-O-E O-E-O in all cases A+ B +c will be even .

stmt 2: A-B/C is odd

A/C - B/C should be odd. E O O E

A/C can be even in both cases E/E or E/O eg: 4/2 or 6/3. B/C can be even in both cases E/E or E/O eg: 4/2 or 6/3.

in the same way we can do for Odd case .

i.e. this stmt is insuff.

Hi I am not quite clear about your explanation for statement 2. I think you have interchanged two of the variables too. Could you please elaborate? Would be grateful.

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