If x + y 0, is 1/(x + y)

Question

If x + y ≠ 0, is 1/(x + y) < 2?

(1) x^3 = y^3
(2) 1/x < 2

GMATinsight · Answer

Question: is 1/(x + y) < 2? Statement 1 : x^3 = y^3 i.e. x = y (But they can be integers or fractions both) Case 1: x = y = 1, 1/(x + y) < 2 Case 2: x = y = 0.1, 1/(x + y) > 2 NOT SUFFICIENT Statement 2 : 1/x < 2 no information about y and x may be an integer or a non-integer both (also they could be +ve and -ve both) hence NOT SUFFICIENT Combining the statements i.e. x = y (But they can be integers or fractions both) but from statement-2, we get that x must be greater than 0.5 if x >0 Also, Case 1: if x and y are negative then 1/(x + y) < 2 Case 2: if x = y > 0.5, (x + y) > 1 and 1/(x + y) < 2 SUFFICIENT Answer: Option C

gmatophobia · Answer

IMO C

Bunuel · Answer

Official Solution:

If  x + y ≠ 0 , is  \frac{1}{x + y} < 2 ? 
 
(1)  x^3 = y^3 
 
Taking the cube root of the above gives  x = y . However, this is not sufficient to determine whether  \frac{1}{x + y} < 2 . For instance, consider  x = y = 1  and  x = y = \frac{1}{2} .
 
(2)  \frac{1}{x} < 2 
 
This statement is clearly insufficient, since no information about  y  is provided.
 
(1)+(2) From (1), since  x = y , the question becomes: is  \frac{1}{2x} < 2 ? This can be further rewritten as: is  \frac{1}{x} < 4 ? Since (2) states that  \frac{1}{x} < 2 , it follows that  \frac{1}{x}  must also be less than 4.Sufficient.

Answer: C

Bunuel · Answer

Practice harder version of this question  HERE .

bumpbot · Answer

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If x + y 0, is 1/(x + y) < 2? (1) x^3 = y^3 (2) 1/x < 2

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