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Re: Inequality...involving reciprocals [#permalink]
15 Jan 2012, 05:35

2

This post received KUDOS

Expert's post

Apoorva81 wrote:

Is 1/x-y < y - x ?

(1) y is positive. (2) x is negative.

can you please provide detailed explanation..??

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) We have y=positive and x=negative, thus y>x (this can be rewritten as y-x>0 or 0>x-y). Now: LHS=\frac{1}{x-y}=\frac{1}{negative}=negative, and RHS=y-x=positive thus \frac{1}{x-y}=negative<y-x=positive. Sufficient.

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 _________________

Re: Inequality...involving reciprocals [#permalink]
09 May 2014, 07:38

Bunuel wrote:

Apoorva81 wrote:

Is 1/x-y < y - x ?

(1) y is positive. (2) x is negative.

can you please provide detailed explanation..??

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) We have y=positive and x=negative, thus y>x (this can be rewritten as y-x>0 or 0>x-y). Now: LHS=\frac{1}{x-y}=\frac{1}{negative}=negative, and RHS=y-x=positive thus \frac{1}{x-y}=negative<y-x=positive. Sufficient.

Answer: C.

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.

Re: Inequality...involving reciprocals [#permalink]
09 May 2014, 08:25

1

This post received KUDOS

Expert's post

rishiroadster wrote:

Bunuel wrote:

Apoorva81 wrote:

Is 1/x-y < y - x ?

(1) y is positive. (2) x is negative.

can you please provide detailed explanation..??

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) We have y=positive and x=negative, thus y>x (this can be rewritten as y-x>0 or 0>x-y). Now: LHS=\frac{1}{x-y}=\frac{1}{negative}=negative, and RHS=y-x=positive thus \frac{1}{x-y}=negative<y-x=positive. Sufficient.

Answer: C.

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. _________________