Is x

Question

Is x < y?

(1) 1/x < 1/y
(2) x/y < 0

chetan2u · Answer

lovely Q (1) \frac{1}{x} < \frac{1}{y} .. \frac{1}{x} - \frac{1}{y}<0........................\frac{y-x}{xy}<0 so xy and y-x have opposite signs.. case I: xy >0, y-x<0....y0...y>x insuff (2) \frac{x}{y}< 0 Just tells us that one is positive and one is negative any of the two x and y can be negative so xx suff C

BrentGMATPrepNow · Answer

Target question : Is x < y? Statement 1 : 1/x < 1/y Let's TEST some values. There are several values of x and y that satisfy statement 1. Here are two: Case a : x = -1 and y = 1. In this case, the answer to the target question is YES, x is less than y Case b : x = 2 and y = 1. In this case, the answer to the target question is NO, x is not less than y Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2 : x/y < 0 This tells us that x and y have OPPOSITE signs, but we don't have any information about which variable is positive and which variable is negative. So, there's no way to tell whether x < y Since we can't answer the target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Statement 2 tells us that x and y have OPPOSITE signs Statement 1 tells us that 1/x < 1/y. If x and y have OPPOSITE signs, then we can be certain that x is NEGATIVE and y is POSITIVE. Take: 1/x < 1/y Multiply both sides by x to get: 1 > x/y Multiply both sides by y to get: y > x Perfect! The answer to the target question is YES, x is less than y Since we can answer the target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent RELATED VIDEO https://www.youtube.com/watch?v=Od98BOicIq4

asbaj · Answer

solving for (2) first x/y is negative -> x and y are of opposite signs case 2a. -> x is +, y is - is x < y => NO case 2b. -> x is i, y is + is x < y => YES Therefore not sufficient solving for (1) 1/x < 1/y lets take couple of cases 1a) both of same sign we can rewrite this as x>y and that is sufficient to say "No" 1b) both of diff sign lets break it down, case 1b_1. -> x is +, y is - notice this is not possible as 1/x is a positive term so 1/x > 1/y and that violates the constraint case 1b_2. -> x is i, y is + 1/x (-ve term) < 1/y (positive term) don't flip sign when taking reciprocal x < y => Sufficient to say "Yes" since we have both Y and N as answer but no info on the signs of x & y, this is insufficient to answer combining, Awesome, this was the missing piece! opposite signs tells us 1a) is out. We are left with case 1b_2 and that's sufficient to answer.

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Is x < y? (1) 1/x < 1/y (2) x/y < 0

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