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Re: GmatClub Test: Number Properies - I (DS) [#permalink]
07 May 2010, 10:21
ykaiim wrote:
Is 2(a+b-c) an odd integer?
1. a, b and c are consecutive numbers 2. b=a+c
The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)
Can someone tell me where I am wrong?
You forgot to multiply the -1 by 2
2(a+b-c) 2(0+1-2) 2(-1) = -2
or 2a + 2b - 2c 2(0) + 2(1) - 2(2) 2-4 = -2
Is the question correct? Multiplying anything by 2 will never give an odd so the question states the answer. In which case you wouldn't need to test 1 or 2
Re: GmatClub Test: Number Properies - I (DS) [#permalink]
07 May 2010, 16:07
lagomez wrote:
ykaiim wrote:
Is 2(a+b-c) an odd integer?
1. a, b and c are consecutive numbers 2. b=a+c
The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)
Can someone tell me where I am wrong?
You forgot to multiply the -1 by 2
2(a+b-c) 2(0+1-2) 2(-1) = -2
or 2a + 2b - 2c 2(0) + 2(1) - 2(2) 2-4 = -2
Is the question correct? Multiplying anything by 2 will never give an odd so the question states the answer. In which case you wouldn't need to test 1 or 2
I agree with lagomez. any number x 2 = Even Something must be wrong
Re: GmatClub Test: Number Properies - I (DS) [#permalink]
07 May 2010, 17:11
2
This post received KUDOS
Expert's post
ykaiim wrote:
Is 2(a+b-c) an odd integer?
1. a, b and c are consecutive numbers 2. b=a+c
The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)
Can someone tell me where I am wrong?
\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).
Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).
Statement (2) is clearly not sufficient. _________________
Re: GmatClub Test: Number Properies - I (DS) [#permalink]
07 May 2010, 17:31
Bunuel wrote:
ykaiim wrote:
Is 2(a+b-c) an odd integer?
1. a, b and c are consecutive numbers 2. b=a+c
The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)
Can someone tell me where I am wrong?
\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).
Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).
Statement (2) is clearly not sufficient.
What numbers can be substituted for A, B, and C and when multiplied by 2 gives you an odd integer?
Re: GmatClub Test: Number Properies - I (DS) [#permalink]
07 May 2010, 17:49
1
This post received KUDOS
Expert's post
lagomez wrote:
Bunuel wrote:
ykaiim wrote:
Is 2(a+b-c) an odd integer?
1. a, b and c are consecutive numbers 2. b=a+c
The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)
Can someone tell me where I am wrong?
\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).
Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).
Statement (2) is clearly not sufficient.
What numbers can be substituted for A, B, and C and when multiplied by 2 gives you an odd integer?
Note that stem does not say that a, b and c are integers.
So for example: a=3/2, b=0, c=0 --> \(2(a+b-c)=3\). As I said if \(a+b-c=\frac{odd}{2}\), then \(2(a+b-c)=2*\frac{odd}{2}=odd\). This option is ruled out by statement (1), which says that a, b and c are integers.
Re: GmatClub Test: Number Properies - I (DS) [#permalink]
08 May 2010, 03:56
Bunuel wrote:
ykaiim wrote:
Is 2(a+b-c) an odd integer?
1. a, b and c are consecutive numbers 2. b=a+c
The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)
Can someone tell me where I am wrong?
\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).
Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).
Statement (2) is clearly not sufficient.
totally right I assumed that a, b and c were integers
Re: Is 2(a+b-c) an odd integer? 1. a, b and c are consecutive [#permalink]
08 Nov 2013, 11:38
If we substitute b = a + c in 2(a+b-c), we get 4a. Now, whatever the value of a, the outcome will always be even. This statement should be sufficient too. In my opinion, the answer should be D.
I am not sure that whether what I have done is correct but, this is what came to my mind when I was attempting this question; I marked D. _________________
--------------------------------------------------------------- Consider to give me kudos if my post helped you.
Re: Is 2(a+b-c) an odd integer? 1. a, b and c are consecutive [#permalink]
09 Nov 2013, 01:47
Expert's post
AasaanHai wrote:
If we substitute b = a + c in 2(a+b-c), we get 4a. Now, whatever the value of a, the outcome will always be even. This statement should be sufficient too. In my opinion, the answer should be D.
I am not sure that whether what I have done is correct but, this is what came to my mind when I was attempting this question; I marked D.
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