nazar Posted June 18, 2008 Share Posted June 18, 2008 If x and y are real numbers, is (x^2 -y^2)^5>0? (1) (x+y)^2 =25 (2) x-y>0 OA: Later Quote Link to comment Share on other sites More sharing options...
Makumajon Posted June 18, 2008 Share Posted June 18, 2008 LHS = (x+y)^5*(x-y)^5 (1) x+y=+5 or -5. And do not know which one is greater: x or y? Insufficient. (2) x>y. Do not know whether they are (+, +) or (-,-). Insufficient. Combining: Insufficient. Suppose x=3, y=2. LHS>0. Suppose x=-2, y = -3. LHS E Quote Link to comment Share on other sites More sharing options...
chanojmarian Posted June 18, 2008 Share Posted June 18, 2008 (x^2 -y^2)^5 = ((x-y)(x+y))^5=(x-y)^5*(x+y)^5 Statement 1 Alone not enough Statement 2 Alone not enough Together:(x-y)^5>0 Also (x+y)^4>0 But we do not know if the remaining term (x+y)>0 since x+y can +0r -5 derving from (1). So the answer is E. Quote Link to comment Share on other sites More sharing options...
rupali41 Posted March 20, 2014 Share Posted March 20, 2014 E is the answer Quote Link to comment Share on other sites More sharing options...
Brent Hanneson Posted October 14, 2014 Share Posted October 14, 2014 If x and y are real numbers, is (x^2 -y^2)^5>0? (1) (x+y)^2 = 25 (2) x-y>0 r Target question:Is (x^2 - y^2)^5 > 0? IMPORTANT: (x^2 - y^2)^5 will be positive (i.e., > 0) if and only if (x^2 - y^2) > 0. So, we can REPHRASE the target question.... REPHRASED target question:Is x^2 - y^2 > 0? Statement 1: (x+y)^2 = 25 This tells us that EITHER x+y = 5 or x+y = -5 There are several possible values for x and y that satisfy this condition. Here are two: case a: x = 3 and y = 2, in which case x^2 - y^2 > 0 case b: x = 0 and y = -5, in which case x^2 - y^2 Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: x - y > 0 In other words, x is GREATER than y There are several possible values for x and y that satisfy this condition. Here are two: case a: x = 3 and y = 2, in which case x^2 - y^2 > 0 case b: x = 0 and y = -5, in which case x^2 - y^2 Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined There are still several possible values for x and y that satisfy BOTH conditions. Here are two: case a: x = 3 and y = 2, in which case x^2 - y^2 > 0 case b: x = 0 and y = -5, in which case x^2 - y^2 Since we cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT Answer = E Cheers, Brent Quote Link to comment Share on other sites More sharing options...
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