If x and y are nonzero integers, is

[VJesus12]

If x and y are nonzero integers, is $$\left(x^{-1}+y^{-1}\right)^{-1}\ >\left(x^{-1}\cdot y^{-1}\right)^{-1}\ ?$$

(1) x = 2y

(2) x + y > 0

[spoiler]OA=A[/spoiler]

Source: Manhattan GMAT

[Jay@ManhattanReview]

If x and y are nonzero integers, is $$\left(x^{-1}+y^{-1}\right)^{-1}\ >\left(x^{-1}\cdot y^{-1}\right)^{-1}\ ?$$

(1) x = 2y

(2) x + y > 0

[spoiler]OA=A[/spoiler]

Source: Manhattan GMAT

We have to determine whether $$\left(x^{-1}+y^{-1}\right)^{-1}\ >\left(x^{-1}\cdot y^{-1}\right)^{-1}\ ?$$ .

From $$\left(x^{-1}+y^{-1}\right)^{-1}\ >\left(x^{-1}\cdot y^{-1}\right)^{-1}\ ?$$ , we have

1/(1/x + 1/y) > 1/(1/xy)
xy/(x + y) > xy

We cannot cancel xy from both the sides since we do not know the sign of xy. If xy is positive, we can do so; we would then have 1/(x + y) > 1

However, if xy is negative, we can still cancel xy, but we will have to reverse the sign of inequality; thus, the result would be 1/(x + y) < 1.

Since we do not know the sign of xy, we leave 1/(1/x + 1/y) > 1/(1/xy) as it is.

Let's take each statement one by one.

(1) x = 2y

Given x = 2y, we have xy/(x + y) > xy => 2y^2 /(3y) > 2y^2 => 1/3y > 1. Note that I canceled 2y^2 without considering the sign of y. We can do so since it is immaterial whether y is positive or negative. y^2 would always be positive.

Thus, we have to determine whether 1/3y > 1.

Case 1: y is negative

Thus, 1/3y > 1 => 1/(-3|y|) > 1 => A negative number > A positive number. The answer is yes.

Case 2: y is positive

Thus, 1/3y > 1 => 1/(3|y|) > 1 => A less than 1 number > 1. The answer is yes.

Sufficient.

(2) x + y > 0

From xy/(x + y) > xy, we have

Case 1: Say x = y = 1; thus x + y = 2 > 0

Thus, xy/(x + y) > xy => (1*1)/(1 + 1) > 1*1 => 1/2 <1. The answer is No.

Case 2: Say x = 3 and y = -1; thus x + y = 2 > 0

Thus, xy/(x + y) > xy => (3*-1)/(3 - 1) > 3*-1 => -3/2 > -3 => -1.5 > -3. The answer is Yes.

No unique answer. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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[fskilnik@GMATH]

If x and y are nonzero integers, is $$\left(x^{-1}+y^{-1}\right)^{-1}\ >\left(x^{-1}\cdot y^{-1}\right)^{-1}\ ?$$

(1) x = 2y

(2) x + y > 0
Source: Manhattan GMAT

$$x,y\,\, \ne 0\,\,\,{\rm{ints}}\,\,\,\left( * \right)$$
$$\frac{1}{{\frac{1}{x} + \frac{1}{y}}}\,\,\mathop > \limits^? \,\,\frac{1}{{\frac{1}{{xy}}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\boxed{\,\,\,\,\frac{{xy}}{{y + x}}\,\,\,\mathop > \limits^? \,\,\,xy\,\,\,}$$

$$\left( 1 \right)\,\,x = 2y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{{2{y^2}} \over {3y}}\,\,\,\mathop > \limits^? \,\,\,2{y^2}\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)\,\,\,{y^2} > \,\,0} \,\,\,\,\,{2 \over {3y}}\,\,\,\mathop > \limits^? \,\,\,2$$
$$\left\{ \matrix{
\,y \le - 1\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\, \hfill \cr
\,y \ge 1\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

$$\left( 2 \right)\,\,x + y > 0\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {1;1} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {3; - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.