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by gmat_for_life » Sat Mar 12, 2016 12:31 pm
If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?

(1) x = 2y

(2) x + y > 0

The official answer is A. However, if the above algebraic expression is simplified to xy(1-x-y)>0? option A seems insufficient. If we assume x and y as -2 and -1, the simplified equation is greater than 0. However if we assume x and y as 2,1, the same equation is less than 0. Could you guys please let me know where I'm going wrong?

Regards,
Amit
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Source: — Data Sufficiency |
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by John fran kennedi » Sat Mar 12, 2016 9:47 pm
Could you please write down how you did simplify it ?
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by GMATGuruNY » Sun Mar 13, 2016 4:29 am
gmat_for_life wrote:If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?

(1) x = 2y

(2) x + y > 0
(xˉ¹ + yˉ¹)ˉ¹ > (xˉ¹yˉ¹)ˉ¹ ?

(1/x + 1/y)ˉ¹ > (1/x * 1/y)ˉ¹ ?

[ (x+y)/xy ]ˉ¹ > xy ?

xy/(x+y) > xy ?

Statement 1:
Substituting x=2y into xy/(x+y) > xy, we get:
(2y)(y) / (2y+y) > (2y)(y) ?

2y² / 3y > 2y² ?

y² / 3y > y² ?

Since y is NONZERO, y² > 0, allowing us to divide both sides by y²:
1/3y > 1 ?

1/y > 3 ?

Since y is an INTEGER, it is not possible that 1/y > 3.
Thus, the answer to the question stem is NO.
SUFFICIENT.

Statement 2:
Substituting x=1 and y=1 into xy/(x+y) > xy, we get:
(1*1)/(1+1) > (1*1) ?
1/2 > 1 ?
In this case, the answer to the question stem is NO.

Substituting x=-1 and y=3 into xy/(x+y) > xy, we get:
(-1*3)/(-1+3) > (-1*3) ?
-3/2 > -3 ?
In this case, the answer to the question stem is YES.

Since the answer is NO in the first case but YES in the second case, INSUFFICIENT.

The correct answer is A.
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by Matt@VeritasPrep » Thu Mar 17, 2016 11:04 pm
We could also use

Is (xˉ¹ + yˉ¹)ˉ¹ > (xˉ¹yˉ¹)ˉ¹ ?

Is (xˉ¹ + yˉ¹)ˉ¹ - (xˉ¹yˉ¹)ˉ¹ > 0 ?

Is 1/(1/x + 1/y) - 1/(1/xy) > 0?

Is xy/(x+y) - xy > 0 ?

Is xy * (1/(x+y) - 1) > 0?

Given S1, we have

Is 2y² * (1/3y - 1) > 0 ?

Is 2y² * (1 - 3y)/3y > 0 ?

Since 2y² > 0, we can divide both sides to reach

Is (1 - 3y)/3y > 0 ?

This will be positive if (1 - 3y) and 3y share a sign. But if y > 1/3, then the first is negative and the second is positive, and if 0 > y the first is positive and the second is negative. So for any nonzero integer, we'll have conflicting signs, and (1 - 3y)/3y CANNOT be positive; SUFFICENT.

Given S2, we have

Is xy * (1/(x+y) - 1) > 0? (from our simplified stem)

Is xy * (1/positive - 1) > 0?

If (x + y) = 1, then 1/(x+y) - 1 is 0, so we can't have a result > 0. But if (x + y) = 2, then 1/(x+y) - 1 is negative, so the question becomes "Is xy also negative?" We don't have enough information to say, so this statement is INSUFFICIENT.
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