DS y>0?

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by neoreaves » Sun Mar 21, 2010 2:11 am
1) X > 7/2Y .... This doesnt give information no y >0. Y can be positive, negative or zero

2) x > |y| ...again we only know that x is positive. However, we done know anything about y


c) combining both doesnt give us anything about y. So IMO E

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by rockeyb » Sun Mar 21, 2010 3:02 am
If x>0, is y>0?

1). x>7/2y
2). x>|y|


lets rephrase question : If x is + ve is y +ve ?

Statement 1: x> 7/2y

xy > 7/2

we know x = +ve and since 1 says xy is +ve y has to be +ve .

sufficient .

Statement 2 : x > |y|

Now we can write this as x > y and x > - y

We know x is +ve but y cab be +ve OR -ve and still satisfy statement 2 .

Not sufficient .


my answer A.
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by neoreaves » Sun Mar 21, 2010 4:02 am
Statement 1: x> 7/2y

xy > 7/2

I would disagree here.

You can only multiply a positive number on both sides. We don't know if y is positive or negative. If it was positive then i would agree xy > 7/2. However if y was negative then xy < 7/2 (Reversal of inequality sign).


Because we don't know if y is positive or negative. So 1 should be insufficient.

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by harshavardhanc » Sun Mar 21, 2010 7:09 am
rockeyb wrote:If x>0, is y>0?

1). x>7/2y
2). x>|y|


lets rephrase question : If x is + ve is y +ve ?

Statement 1: x> 7/2y

xy > 7/2

we know x = +ve and since 1 says xy is +ve y has to be +ve .

sufficient .
rockeyb,

you cannot do this cross-multiplication. Reason : you are not sure whether y is +ve or -ve.

If it were -ve, then the cross-multiplication will look like xy < 7/2.

Get the point which I'm trying to make? You need to find out whether Y is +ve or -ve. But in your calculations for statement 1 , you have already assumed it to be +ve and then proved it to be +ve. That's circular reasoning.


IMO, both statement 1 and 2 are insufficient to tell that Y is +ve or -ve. Hence, E
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by rockeyb » Sun Mar 21, 2010 8:03 am
My apologies mate .

I missed the point , thanks for pointing that out .

sorry again guys laps of concentration , I need time off ;)
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