Is 1/(x - y) < y - x? (1) y is positive (2) x is negative

(1) y greater than 0 :
example 1 : x = 0.5 and y = 0.2 => 1/0.7>-0.3
example 2 : x = 0.2 and y = 0.5 => -1/0.3<0.7
INSUFFICIENT

(2) x less than 0 is the same problem as before !
INSUFFICIENT

(Both) x negative, y positive but whose greater than the other (in absolute value of course)?
example 1: x = -2 and y = 3 => 1/(-5) < 1
example 2: x = -3 and y = 2 => 1/(-5) = -0.2 > -1

ANS: E.

Hope it is clear

Example 2 is incorrect. RHS is y-x which will be +5 not -1.

The answer should be c not e

Is \(\frac{1}{x-y}<y - x\)?

(1) y is positive, clearly insufficient, as no info about \(x\);
(2) x is negative, also insufficient, as no info about \(y\);

(1)+(2) Since y is positive and x is negative, then \(y>x\). We can re-write this as \(x-y<0\), as well as \(y-x>0\). Evaluate LHS and RHS from the question: \((LHS=\frac{1}{x-y})<0\), and \((RHS=y-x)>0\), therefore \((LHS=negative)<(RHS=positive)\). Sufficient.

Answer: C.

OR: Is \(\frac{1}{x-y}<y - x\) --> is \(\frac{1}{x-y}+x-y<0\) --> is \(\frac{1+(x-y)^2}{x-y}<0\)?

(1) \(y\) is positive, clearly insufficient, as no info about \(x\);
(2) \(x\) is negative, also insufficient, as no info about \(y\);

(1)+(2) Is \(\frac{1+(x-y)^2}{x-y}<0\)? Now, nominator in this fraction is always positive (1 plus some non-negative number), but denominator is always negative as \(x-y=negative-positive=negative\) (for example: -3-2=-5).So we would have is \(\frac{positive}{negative}<0\)? Which is true. Sufficient.

Answer: C.

Hope it helps.

1 / ( x - y) < ( y - x ) -- > 1< (y-x)*(x-y) -- > 1<-x^2+2xy-y^2 -- > the question is -1> x^2-2xy+y^2 ?

With regard to the stmt (1) it follows from -1> x^2-2xy+y^2 that x^2+y^2 is always greater than -2xy for any term of x and y. Hence, x^2-2xy+y^2>-1. Even if x were equal to zero, the result will not change. SUFF.

The same solution, i.e. x^2-2xy+y^2>-1, is true also for the case when x is negative according to stmt (2). Here, again it does not matter whether x and y are positive or negative because x^2+y^2 is always greater than -2xy. Here we do not need to know the term of y, anyway the result will not change. Hence stmt (2) - SUFF.
Hence, the ans. is D.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it.

So you can not multiply 1/(x-y)<(y-x ) by x-y and write 1<(y-x)*(x-y) because you don't know whether x-y is positive or negative: if it's positive then you should write 1<(y-x)*(x-y) but if its negative then you should flip the sign and write 1>(y-x)*(x-y)

OA for this question is C not D, refer to the posts above for a solution.

Slightly different approach, I hope everyone can see the attached images with solutions.

Dabral

In order to separate your x's and y's from each other, you need to know which term is larger. This is because you need to know whether x-y or y-x is negative (one of them will be, unless they are equal). Taking both statements together, you know that the two variables are not equal, and you can manipulate the inequality, keeping in mind that you need to flip the sign when multiplying or dividing by a negative number.

I explained it in kind of a backwards way. The first thing you need to know in that problem is whether x and y are equal. No single statement tells you that.

Sorry for the double post, but it seems I can't edit my previous post while viewing the forum with Tapatalk.

Sent from my HTC Glacier using Tapatalk 2

Why can't we do the following rephrase?

1/(x-y) < (y-x)

-1/(y-x) < (y-x)

-1 < (y-x)^2

since the RHS is a square, irrespective of x and y values the equation is satisfied.

The point is that we don't know whether y-x is positive or negative. If it's positive, then yes we'd have -1 < (y-x)^2 but if it's negative, then we'd have -1 > (y-x)^2 (flip the sign when multiplying by negative value).

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

The question can be re written as

\(\frac{1}{(x-y)} +( x-y )<0\)

Or
\(\frac{1+ (x-y)^2}{(x-y)} <0\)
For the expression to be less than zero,numerator and denominator should be of opp sign
Now\((x-y)^2\geq{0}\)

So numerator is positive ,therefore if denominator is negative then answer to the question is yes or else no

st 1 x is positive let x=1 and y=-10 then x-y=9answer is no
But if x=1 and y=3 then answer to the question is yes

St 1 not sufficient .. Option A and D ruled out

St 2 y is negative
If y= -3 and x =-8 then ans to the question is yes
If y=-3 and x =8 then answer to the q. Is no

St 2 alone not sufficient

Combining we see that x-y=+ve

So ans is C

WoundedTiger, it's better to do one step more. In data sufficiency if you have very simple statements you need to understand what you are looking for.

So, a little bit different solution:

The question can be re written as

\(\frac{1}{(x-y)} +( x-y )<0\)

Or
\(\frac{1+ (x-y)^2}{(x-y)} <0\)

Since \(1+(x-y)^2>0\), denominator must be negative: \(x-y<0\) or \(x<y\). Hence, we need to find: is \(x\) less than \(y\)?

(1) Don't know anything about \(y\). Insufficient
(2) Don't know anything about \(x\). Insufficient

(1)+(2) Since \(x>0\) and \(y<0\), we have \(x>y\). Sufficient

The correct answer is C.

I tried this way.

we have 1/(x-y) < y-x

1) says x is positive, we have no value of Y, not sufficient

2) says y is negative, no value of x defined, not sufficient.

1+2 , x is +ve and y is -ve => 1/+ve < -ve => +ve < -ve, not possible.

Answer is C

I remember giving you the below links couple of times before:
Inequality tips

Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach

Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

https://gmatclub.com/forum/inequalities-trick-91482.html
https://gmatclub.com/forum/data-suff-ine ... 09078.html
https://gmatclub.com/forum/range-for-var ... 09468.html
https://gmatclub.com/forum/everything-is ... 08884.html
https://gmatclub.com/forum/graphic-appro ... 68037.html

Hope this helps.

good memory, Bunuel you did. And I always forget where the stuff is thanks!

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