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Can someone explain the quickest way to solve this

Let \(A=(x-y)\), then we have :

\(\frac{1}{A} < -A\)

if A>0, then A^2<-1. So no solution if A<0, then A^2>-1. So all A<0 is a solution

Therefore, if we can impose conditions on x,y such that (x-y) is >0, then we know the answer is always "No" conversely if (x-y) < 0 then we know the answer is always "Yes"

(1) not sufficient as it only talks of y (2) not sufficient as it only talks of x

(1)+(2) y positive & x negative means (x-y)<0. So this inequality is always true. SUFFICIENT

(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

(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.

Can someone explain the quickest way to solve this

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.

Can someone explain the quickest way to solve this

The quickest way to solve this is to note that one side of this inequality is always positive, and the other side is always negative. If x > y, then the left side is positive, and the right side is negative, so this will always be false. On the other hand, if y > x, then the left side is negative, while the right side is positive, so the inequality will always be true.

(1) Tells you nothing about x. Insufficient. (2) Tells you nothing about y. Insufficient. Together: y is positive and x is negative, so y > x, and the inequality is true. Sufficient - (C).

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.

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. _________________

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