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Is x > y?

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by sanju09 » Thu Mar 24, 2011 2:25 am
Is x > y?

(1) x/ (3 y) > 1/3.

(2) -x + p < -y + p.


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Source: — Data Sufficiency |
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by HSPA » Thu Mar 24, 2011 5:30 am
B for me here
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by nikhilsrl » Thu Mar 24, 2011 6:46 am
1) x/3y > 1/3 when evaluated this gives x>y - sufficient
2) -x+p< -y+p
adding -p on both sides -> -x < -y
multiplying by -1 on both sides -> x>y - sufficient

So D.
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by kmittal82 » Thu Mar 24, 2011 7:20 am
IMHO, its (B)

(1)
Multiplying both sides by 3
x/y > 1

However, this doesn't mean x > y, since y could be negative

(2)
-x + p < -y + p
subtracting p from both
-x < -y
Multiplying by -1 both sides, signs change
x > y
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by Night reader » Thu Mar 24, 2011 10:59 am
x>y ?

st(1) we cannot multiply by 3y as we don't know if y (<), (>) 0, BUT we can multiply by 3 --> x/y >1.
It's possible to have x,y>0 OR x,y<0 for both variables (x and y) simultaneously, also |x|>|y|.
If x,y>0 then x>y. If x,y<0 x<y Not Sufficient;

st(2) -x+p < -y+p, -x < -y, x>y Sufficient

IOM B

sanju09 wrote:Is x > y?

(1) x/ (3 y) > 1/3.

(2) -x + p < -y + p.


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by MAAJ » Thu Mar 24, 2011 3:32 pm
IMO (B)

x/y > 1

IF +x/+y > 1 THEN +x > +y (e.g. 8/2 > 1)

IF -x/-y > 1 THEN -x < -y (e.g. -8/-2 > 1)

So Statement 1 is insufficient.
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