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variables

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by naaga » Mon Feb 23, 2009 2:38 am
. If x ≠ 1, is y equal to x + 1?

(1) y-2/x-1 = 0

(2) y^2 = (x + 1)^2


folks I got the answer D

but the OA is A

can anyone explain
Last edited by naaga on Mon Feb 23, 2009 4:11 am, edited 1 time in total.
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Source: — Data Sufficiency |

by DanaJ » Mon Feb 23, 2009 2:52 am
The equation in stmt 1 is fulfilled only if y-2 = 0, giving us y = 2. This does not help regarding the relationship between x and y, so 1 is insufficient.
The second equation helps establish that y = x + 1 or that y = -(x + 1). Since we cannot tell which case it really is, then this stmt is insufficient as well.

Put the two together and you get two cases from stmt 2:
a. y = x + 1 = 2 makes x = 1, which does not fit with the initial constraint that x is not 1
b. y = -(x + 1) = 2, making x = -1, which does fit with x being different from 1.

So I'd go with C.... Maybe I'm missing smth over here, I get a completely different answer....
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by bluementor » Mon Feb 23, 2009 3:04 am
DanaJ wrote:The equation in stmt 1 is fulfilled only if y-2 = 0, giving us y = 2. This does not help regarding the relationship between x and y, so 1 is insufficient.
But we know that x is not equal to 1, and if y = 2, y will never be equal to x + 1. Isn't this sufficient to answer the question stem?

-BM-
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by naaga » Mon Feb 23, 2009 4:17 am
the answer A satisfies the equation,

( y-2 )/ (x -1 ) = 0

Y = 2 , obviously x doesn't equal to 1, with statement we can answer.

why B cannot satisfies the equation ?
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Re: variables

by sureshbala » Mon Feb 23, 2009 5:24 am
naaga wrote:. If x ≠ 1, is y equal to x + 1?

(1) y-2/x-1 = 0

(2) y^2 = (x + 1)^2


folks I got the answer D

but the OA is A

can anyone explain
Statement I alone is sufficient....

Look at this....

Given that y-2/x-1 = 0.

So y-2 =0 , i.e y =2.

Now, since y=2, y will be equal x+1 provided x=1. But in the statement it is clearly mentioned that x is not equal to 1. So x+1 is not equal to 2, where as y=2. So we can conclude that y is not equal to x+1.

Hence I is sufficient.

Coming to statement II, since y^2 = (x+!)^2, we have y= (x+1) or -(x+1). So II is not sufficient.

Hence A
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by DanaJ » Mon Feb 23, 2009 6:06 am
bluementor: you are so right... Your answer actually made me laugh because it was so plain and simple, while I'm over here solving zounds of equations... :))
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