Is xy

Question

Is xy <0?

1). x^3y^4/xy <0

2). |x| - |y| < |x-y|



can you please help? my answer is E

venksune · Answer

I am getting E too in the first pass. terribly tired today, couldnt elaborate.

crackets · Answer

i asked other people and some get B but i disagree. here is what i did:


For (1) the equation can hold true if both are negative OR if x is positive and y is negative (but not when x is negative and y is positive). So in the question stem, if both are negative, then xy>0 but if x is positive and y is negative, then xy<0. not sufficient. 

For (2) in order to hold true, x has to be smaller than y. and if x is smaller than y, both works: a negative and positive number. assuming x is greater than y, then the equation doesn't hold true.
since both, negative and positive values work for (2) to become true and assuming x is smaller than y, i still cannot answer the question because i cannot get one single value as a result.

so i stick with E. can anyone prove me wrong?

Anonymous · Answer

my answer is A.

(1) x^3y^4/xy < 0
=> x^2y^3 < 0
=> (xy)(xy) (y) < 0
LHS is a multiple of (-ve)*(-ve)(y).
Therefore y < 0.
Now question stem says xy<0. We have proved y<0 . therefore x must be >0.

Hence proved xy < 0 using (1) alone.

You guys hvae already ruled out (2). I just substituted values to disprove (2).

Hence A is my answer. Whats OA pls?

sdanquah · Answer

Statement 1:( X^12y)/xy<0 only if xy <0. thus statement 1 is sufficient
Stement 2 is also sufficient because for |x| - |y| < |x-y|
it means x and Y must have different signs and that makes xy<0. thus it is also sufficient.

Thus the answer shd be D if I rewrote statement 1 correctly.

PracticeMore · Answer

Is xy <0?

1). x^3y^4/xy <0

2). |x| - |y| < |x-y|

1. This can be simplified as (xy)^3*y /(xy)<0

so xy and (xy)^3 will have same sign so we can divide them.

(xy)^2*y < 0 ===> y <0 , However xy can be greater than 0 or less than 0.

2. here we have three checkpoints: x, y , x-y for the inequality

a. when x > 0, y >0, x>y
eqn resolves to x-y<x-y , which is not true.

b. when x>0, y>0, x<y
eqn resolves to x-y<-(x-y) => x < y which is true.
so xy > 0

c. when x >0, y < 0
eqn resolves to x + y < x- y => y <0 , so xy < 0

Here itself we have a conflicting value for xy, so it can't be B

d. when x < 0, y < 0 , x >y

-x + y < x- y => x > y which is true. so xy > 0

So from A and B, y < 0 , but still xy can not be decided.

So my answer is E.

crackets · Answer

haha i like this forum.

to anyone who chose A, remember that a negative number cubed, stays negative. 



i still stick with E. :!:

Is xy <0? 1). x^3y^4/xy <0 2). |x| - |y| < |x-y|

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