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is y = 2?

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Source: — Data Sufficiency |
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by belize » Wed Aug 05, 2009 8:09 am
(1) y ≠ 1
y could be 2, 3, 4, or etc.
eliminate A and D

(2) x+2 is multiple of y
if y = 2, and x is multiple of y, then x is even
x + 2 is still even, and it is multiple of y
so y = 2 is valid

Choice B is correct
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by karan81 » Wed Aug 05, 2009 8:34 am
Shouldn't the corect answere be C.
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by belize » Wed Aug 05, 2009 9:06 am
You are right. I didn't consider the possibility of y = 1 in statement (2), so C should be the answer.

I always forget about 0 or 1.
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by shahdevine » Wed Aug 05, 2009 9:23 am
belize wrote:(1) y ≠ 1
y could be 2, 3, 4, or etc.
eliminate A and D

(2) x+2 is multiple of y
if y = 2, and x is multiple of y, then x is even
x + 2 is still even, and it is multiple of y
so y = 2 is valid

Choice B is correct
no disrespect, but the logic here does not seem strong. for statement 2, what if y is not 2? The question is asking you to verify whether y is 2...you cannot assume its 2 and solve backwards without trying y's that are not 2.

I'm very curious...could someone plz post OA...
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by belize » Wed Aug 05, 2009 9:49 am
I thought that you can validate a statement by plugging in what the question is asking for.
For this question, I test statement (2) first by using y=2 and then by a couple other numbers that are not 2.

When y = 2 and x is a multiple of y, then x+2 is always even and is therefore a multiple of y.

When y=1, x (could be any integer) is a multiple of y, and x+2 is also a multiple of y. So y can also be 1.

When y=3, x is a multiple of y, but x+2 is not.

Since y could be 1 or 2, statement (2) is insufficient. But (1) and (2) combined is, so I corrected my answer to C. Is this correct? Perhaps my thinking is still illogical. Please advise if there is a more efficient way to approach this question. Thanks!
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by ShikenO/kau » Wed Aug 05, 2009 9:53 am
Only even consecutive numbers can have 1 and 2 as common factors....
All odd consecutive numbers can have only 1 as common factor.
Statement I
Not sufficient
Nothing can be inferred from y#1

StatementII
Not sufficient
Here they have given that x and x+2 have common factor as y...
So when we either consider it as even consecutive or odd consecutive number
If it is odd consecutive number the common factor will be 1
If x and x+2 are even consecutive numbers the value of y can be either 1 or 2....
So insufficient...

Combining both the statements we get y=2.....
As x and x+2 can be only even consecutive numbers....
and y#1 gives y=2....
Hence the answer is C
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by apoorva.srivastva » Wed Aug 05, 2009 11:51 am
great explanation

OA is C
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