Is x^2

Question

Is x^2 < x - |y|?

(1) y > x
(2) x < 0

Bunuel · Accepted Answer

No need to consider y<=0 and y>0 cases.

Is x^2 < x - |y|?

Is \(x^2 < x - |y|\)? --> Is \(x^2+|y| < x\)?

(1) y > x. Since \(y > x\), then obviously \(|y| > x\). Now, since \(|y| > x\), then \(|y|+x^2=|y|+(nonnegative \ value)\) will be also greater than \(x\). Therefore \(x^2+|y|>x\) and the answer to the question is NO. Sufficient.

(2) x < 0. This implies that \(x^2>0\). Thus \(|y|+x^2=(nonnegative \ value)+positive>0>x\). Thus the answer to the question is NO. Sufficient.

Answer: D.

Hope it's clear.

Zarrolou · Answer

is x^2 < x - |y|? 
Split the equation in two cases:
 I)  y\geq{}0 ,  x^2<x-y 
 II)  y<0 ,  x^2<x+y

(i) y >x 
or  -y<-x  (we need to have the same operator to sum inequalities). Sum this to each case:
 I)  x^2+(-y)<(-x)+x-y  or  x^2<0  Never true
 II)  x^2+(-y)<(-x)+x+y  or  x^2<2y , but since in this case we are considering  y<0  that equation is never true
 x^2<-ve  Never true.
Sufficient

(ii) x < 0 
 |y|<x-x^2 , the equation  x-x^2  for values of  x<0  is negative.
 |y|<-ve , never true
Sufficient
IMO D

nt2010 · Answer

IMO answer is B D

is x^2 < x - |y|?

From Stmt 1) y > x

Lets use numbers y = 1 and x = 0 when applied to x^2 < x - |y| results as 0 < - 1 ~~Valid~~ Not Valid
y = 5 x = 4 => when applied to x^2 < x - |y| results 16 < -1 Not Valid

Hence - Sufficient

From Stmt 2) x < 0

Substituting various values for x and any value for y in x^2 < x - |y| results x^2 is NOT < x - |y|

Hence - Sufficient.

What is the OA?

Zarrolou · Answer

Hi nt2010,

they are both not valid

WholeLottaLove · Answer

Hi Zarrolou!

Could you run me through this:

(i) y >x
or -y<-x (we need to have the same operator to sum inequalities). Sum this to each case:
I)x^2+(-y)<(-x)+x-y or x^2<0 Never true
II)x^2+(-y)<(-x)+x+y or x^2<2y, but since in this case we are considering y<0 that equation is never true
x^2<-ve Never true.
Sufficient

Firstly, why do we plug in -y<-x? I get where that comes from, but why do we use it here?

Secondly, where do we get x^2+(-y)<(-x)+x-y from? In other words, how does x^2 < x - |y| translate to x^2+(-y)<(-x)+x-y?

Thanks!

Zarrolou · Answer

The most important passage is the first one: x^2 < x - |y| from this I divide the function into 2 cases: y>0 => x^2 x^2 or <) 2)I divided the first equation into two cases : y>0 and y<0 Now I take case y>0 : x^2<+x-y , sum (literally sum) the equation of statement 1 -y<-x , and I obtain this: x^2+(-y)<(-x)+x-y , now I semplify its form into x^2<0 . So in this scenario the equation is never true, 0 cannot be greater than a squared number. Now I repeat the same passages for the other case y<0 : x^2<+x+y , sum the equation of statement 1 -y<-x , and I obtain this: x^2+(-y)<(-x)+x+y , and I get x^2<2y . But since in this scenario I am considering y<0 the right term 2y will be negative. So once more, can a negative number be greater than a squared number? x^2<-ve ? No, so also in this scenario the answer to the main question is NO. To summarize what we did here: we have divided the main question into two cases, we have studied each case on its own, and since the answer to the main question is the same (NO in both) we can say that statement 1 is sufficient

dave785 · Answer

I like D.

Point 1 is sufficient because X^2 is always positive, and x - |y| if y > x is always negative. 
Point 2 is also sufficient for pretty much the same reason.

WholeLottaLove · Answer

That took my a while but I think I got it. Kudos to you! Thanks!

karjan07 · Answer

x^2-x >|y|

x(1-x)>0  since RHS is always > 0

1>X>0

Statement 2 seems sufficient but I am unable to use statement 1 to conclude this logic

WholeLottaLove · Answer

One more thing...

I think I solved for #II but I ran into a snag.

I used the method Zarrolou showed my by summing up the inequalities. So, for #2...

x^2<x-|y| So...

I. x^2 < x- y
OR
II. x^2 < x + y

AND

x < 0

Thus:

x^2 < x-y,
x<0

x^2 + x < x-y ==> x^2 < -y which isn't possible because x^2 will always be positive or zero

AND

x^2 < x+y
x<0

(using the same process as above) ==> x^2 < y Which is possible. For example, what if x^2 = 2 and y =4? Then x^2 could in fact be less than y.

Where did I go wrong?

Zarrolou · Answer

That is not possible.
Remember that you are in the case  y<0

x^2 < y  here y will be negative so  x^2<-ve => not possible

Hope it's clear

WholeLottaLove · Answer

Ok, I think I understand. We have two cases for |y|...y and -y. In the case of x^2 < y... it is derived from the inequality x^2 < x -(-y) right?

Zarrolou · Answer

The passages above were correct, so I think you got the point.

Yes, from  x^2<x+y  you sum  x<0  and obtain  x^2<y , that's correct

WoundedTiger · Answer

Bunuel sama,

I had to ask this....
Why are u so awesome!!!!
Respect.

Posted from my mobile device

grant1377 · Answer

How I solved the problem.

Is x^2 < x - |y|?

(1) y > x
(2) x < 0

Is x^2 < x - |y|

Is |y| < x - x^2

|y| >= 0 , and x - x^2 is only >= 0 when 1<= x <= 0

(1) y>x - Does x - x^2 decrease or increase for some value 1< a < 0?

Test:
x = 1/2 , then x - x^2 = 1/4. Since y > 1/2, answer NO, sufficient

(2) x < 0 - Any value x < 0 produces a negative value for x - x^2. Since |y| > 0, answer NO, sufficient.

goodyear2013 · Answer

Is x^2 < x – |y|?
(1) y > x
(2) x < 0

OE

Show Spoiler

(1): y > x. This means that x – |y| is negative. In contrast, x^2 cannot be negative—it can equal zero, but any value of x other than zero yields a positive value of x^2. Since a negative number will always be less than any positive number (or zero), answer to question is definitely “"no.”"
Sufficient
(2): x < 0. x^2 >0. Since an absolute value such as |y| is always either a positive number or zero, then x– |y| equals either a negative minus a positive or a negative minus zero. In either case, result is negative.
Since a positive number is always greater than a negative number, x^2 > x – |y|, and answer to question is definitely “"no.”"
Sufficient

This is good one.

Bunuel · Answer

Merging similar topics. Please refer to the solutions above.

WoundedTiger · Answer

Sol: We need to find whether x^2< x- |y| or x^2+|y|/ 0 and hence B statement itself becomes sufficient St 1. y >x (Considering only values of x>0 ) y=3, x=2 LHS>RHS y=1/2 x= 1/10 again LHS >RHS y=3,x= 0 agains LHS>RHS Hence the inequality will not be true under given set of conditions. Ans D I GOT IT WRONG TWICE BEFORE GETTING IT RIGHT

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Is x^2 < x - |y|?

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