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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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