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Is X>Y DS question

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Is X>Y DS question

by dzelkas » Sat Aug 25, 2007 7:29 pm
I am confused about this question

Is X>Y?
(i) x/(3y)>1/3
(ii) -x+p<-y+p

the answer is (b) where (ii) alone is sufficient
I understand (ii) where we get x>y
but in (i)

don't we also get x>y?

if it is:

x/(3y)>1/3 if we multiplu both sides by 3Y
don't we get the answer x>y?
I think I am just making silly algebra mistake

thanks
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Source: — Data Sufficiency |
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by agps » Sun Aug 26, 2007 1:17 am
all i can think of is that maybe they assume you can't be sure if Y is diferent from 0?

but i would have anwered D.
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still no go

by dzelkas » Sun Aug 26, 2007 9:19 am
The response they say is that in (1) x and y could be either positive or negative.. still don't get it
anyone can help out on this one?
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by beny » Sun Aug 26, 2007 9:58 am
Be careful when multiplying both sides of an inequality by a variable. If you multiply both sides of an inequality by a negative number, you must flip the inequality. Likewise, if you multiply both sides of an inequality by a variable that represents a negative number, you must flip the inequality. Thus, if x and y are both negative, and x/3y>1/3, multplying both sides by 3y would yield x<y.

i.e.
x=2 and y=1, x>y
x=-2, and y=-1, x<y
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by dzelkas » Sun Aug 26, 2007 10:09 am
thanks! its crazy, I knew the rule to do it in the (ii) but didn't even think about it in the first part..
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my response

by ack » Tue Sep 11, 2007 9:02 am
This is how I would think of this problem

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

x/(3y) is equivalent to (1/3)*(x/y) so from statement 1 we get
(1/3)*(x/y) > 1/3 . We can divide both sides by 1/3 (no variables in 1/3) and we get (x/y) >1 . We can't cross multiply because we don't know the signs of x and y which implies that both x and y could be positive or both x and y could be negative (they have to have the same sign since x/y > 1). So the statement is insufficient

2. -x+p < -y + p

Here we can take away p from both sides by subracting it from both sides and we are left with -x < -y . Multiplying both sides by -1 we get x>y

ack
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