Is (x^-1-y^-1)/(x^-2-y^-2)1?

Question

Is  \frac{x^{-1}-y^{-1}}{x^{-2}-y^{-2}}>1

(1) x + y is greater than 0
(2) x and y are positive integers and each is greater than 2

TheMechanic · Answer

B the original equation would solve to: \frac{xy}{y+x} > 0 Stmt1: X+Y > 0 ; but then XY could be negative (x=4, y=-2) or positive (x=2, y=2) - > Implying Not sufficient. Stmt2: Sufficient. Seems like too easy. Am I missing something here?! :roll:

abhi84 · Answer

B

The inequality can be simplified to "Is xy/x+y > 1?"

From A, x+y > 0, but this could be a result of both x and y +ve or one of the negative and lower value than the other one. The result of xy/x+y can be > 1 or < 1 depending on both the conditions

From B, x >2 and y>2. Therefore xy/x+y always greater than 1.

brownbear24 · Answer

Because the equation does not state that X and Y must be different numbers, why could you not just put in X=3, Y=3 and allow the equation to cancel down to zero?

Bunuel · Answer

x and y are some specific numbers and the questions asks whether xy/(x+y)>1. You cannot arbitrarily say that x=y=3 and proceed. Also, it's not clear what you mean by the highlighted part above.

rhine29388 · Answer

first simplify the question stem,then plug some values as per the statements.we can see that correct answer is option B
Simplified q (xy)/(x+y) > 1

mdacosta · Answer

Hello - can someone give a more detailed step-by-step on the xy/x+y > 1 part?

I'm getting: (X^2 - Y^2)/(X-Y) => (X+Y)(X-Y)/(X-Y) => X+Y

Thanks!

chetan2u · Answer

Hi,
 x^{-2} - y^{-2} = \frac{1}{x^2}-\frac{1}{y^2} = \frac{y^2-x^2}{(xy)^2} ...
and  x^{-1}-y^{-1} = \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} 
SO  \frac{x^{-1}-y^{-1}}{x^{-2}-y^{-2}}>1  will be  \frac{y-x}{xy}/\frac{y^2-x^2}{(xy)^2}=\frac{y-x}{xy} * \frac{(xy)^2}{y^2-x^2} .............
 \frac{y-x}{xy} * \frac{(xy)^2}{y^2-x^2} = \frac{xy}{x+y} .............

etienneg · Answer

I think what  tbmiers  meant was, why could X and Y not be the same number, cancel each other out and the equation become 0/0=undefined?
Statement 1 proves that this is not the case - but the question got me thinking.

Is my logic incorrect that by simplifying the quadratic equation we are removing a possible answer?
And following that logic the answer is actually (C) - we need Stmt 1 to know that X≠Y

chetan2u · Answer

Hi,
you have made an observation which may be valid in some scenario, but is not the case here..
x=y does not give you infinity or 0/0... you have to cancel out the common terms in both numerator and denominator and then only should look at the denominator..

another example
 if the equation is  \frac{x^2-y^2}{x-y}  , we do not require to mention x should not be equal to y, because you have a term in numerator that will cancel out x-y

etienneg · Answer

Thanks  chetan2u  for clearing that up.

Can you perhaps give an example of where my logic would indeed be valid?
(sorry if this goes off topic)

chetan2u · Answer

Hi,
its perfectly fine to clear doubts..

If we have a equation where we are left in the end with a term in the denominator, it will be important to mention the same..

example..
 \frac{x+y}{x} , here x
eq{0} ...
OR \frac{x^2+y^2}{x-y}  , here x
eq{y} ...
OR  \frac{x^2+y^2-2xy}{x^2-y^2} , here x
eq{-y} ... 
WHY? 
  \frac{x^2+y^2-2xy}{x^2-y^2} = \frac{(x-y)^2}{(x-y)(x+y)} =\frac{x-y}{x+y}

MathRevolution · Answer

If we modify the original condition and the question, we can write the question as, Is (x^-1-y^-1)/(x^-2-y^-2)>1? --> (1/x-1/y)/(1/x^2-1/y^2)>1?. If we multiply the left side with (xy)^2 to the numerator and to the denominator, we get xy(y-x)/(y^2-x^2)>1?. If we modify again, the question becomes xy(y-x)/(y+x)(y-x)>1?. If x and y are the same number, the denominator becomes 0, which is not possible. Hence, we need to divide each side and get xy/(y+x)>1?. If x+y>0, we get xy>x+y?, xy-x-y>0?, xy-x-y+1>1?, (x-1)(y-1)>1?. And, we have x>2 and y>2. The answer is always yes and the condition is sufficient. Hence, the correct answer is B. 
 
- Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

bumpbot · Answer

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Is (x^-1-y^-1)/(x^-2-y^-2)>1?

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