Is x 0? (1) (x^2 - 1)(x^3) 0 (2) x^2

Question

Is x > 0?

(1) (x^2 - 1)(x^3) > 0

(2) x^2 < 1

chetan2u · Answer

1) (x^2-1)(x^3)>0 If x>0.....x^3>0 and x^2-1>0 or x^2>1...X>1 If x<0...x^3<0 and x^2-1<0 or x^2<1.... So x>-1 Both cases possible Insufficient 2) x^2<1 -10, x>1 but statement II says otherwise it says X is <1, so X can not be greater than 0 So x<0 Sufficient C

jennysussna · Answer

For statement one I combined to have X^5-X^3>0 as I thought that was easier to test than solve out the inequality. Can you clarify on the last part? What do you mean statement II means otherwise? How do you go from having 00? I tested -(1/2) and it was true for both statements, so it seems that x is negative.

chetan2u · Answer

Hi.. I wrote the opposite signs by mistake.. So (x^2-1)(x^3)>0... So two case 1) both are positive x^3>0, x^2>1..X>1 2) both negative x^3<0, x^2<1...-1

T1101 · Answer

chetan2u Just to be sure i fully understand it: S(1) + S(2): results in the following range for x --> -10, am I correct???

chetan2u · Answer

Yes combined it tells us -1<X<0

Princ · Answer

OA:C (1) \quad (x^2 - 1)(x^3) > 0 It leads to two cases Case 1 (x^2-1)>0 AND x^3>0 This leads to x>1 OR Case 2 (x^2-1)<0 AND x^3<0 This leads to -11 OR -1 0? No

CAMANISHPARMAR · Answer

This is a yes/no Q.

Start with statement 2 as it is easier to evaluate. According to statement 2 x could be a positive fraction such that x lies between 0 & 1 or a negative fraction which lies between -1 and 0. Clearly we have two answers (a Yes & a No) to the same questions hence statement 2 alone is not sufficient to answer whether x>0? Strike out option B & D.

Statement 1 alone would give the same result as in statement 2; if x is a negative fraction which lies between -1 and 0 then we have a no but if x >1 then we have a yes. Hence insufficient.

Combining 1 and 2 we get only one answer that is x is a negative fraction which lies between -1 and 0. And we have a No to the Q. Ans C is correct.

BrentGMATPrepNow · Answer

Target question : Is x > 0? Statement 1 : (x² - 1)(x³) > 0 Let's TEST some values. There are several values of x that satisfy statement 1. Here are two: Case a : x = 2. In this case, the answer to the target question is YES, x is greater than 0 Case b : x= -2. In this case, the answer to the target question is NO, x is not greater than 0 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2 : x² < 1 Let's TEST some values. There are several values of x that satisfy statement 2. Here are two: Case a : x = 0.5. In this case, the answer to the target question is YES, x is greater than 0 Case b : x= -0.5. In this case, the answer to the target question is NO, x is not greater than 0 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Statement 2 tells us that x² < 1 Subtract 1 from both sides to get: x² - 1 < 0 In other words, x² - 1 is negative Statement 1 tells us that (x² - 1)(x³) > 0 Divide both sides of the inequality by (x² - 1) to get: x³ < 0 If x³ < 0, then we can be certain that x is negative The answer to the target question is NO, x is not greater than 0 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

minustark · Answer

1) Either (x^2 - 1) and x^3 are both positive or both negative. For x>0, its always true. Now, when x < 0, x^3 will be negative, and to make x^2 -1 <0, we have to take x as -1 < x < 0 (x cannot be greater than 0 as in that case x^3 will be positive and the inequality will be less than 0). x can be either positive or negative.
2) -1 < x < 1. Not sufficient.

Together, x lies in the range -1< x< 0. Sufficient.
C is the answer

Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1

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