*.Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
*Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
*BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
*EACH statement ALONE is sufficient to answer the question asked.
*Statements (1) and (2) TOGETHER are NOT sufficient to answer the question.
Is x<0
1. x^3(1-x^2)<0
2.(x^2-1)>0

• *BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
  here is sol details Url---> http://postimg.org/image/jftrtznqx/
• 8 years agoHelpfull: Yes(17) No(3)
• Statements (1) and (2) TOGETHER are NOT sufficient to answer the question
  
• 8 years agoHelpfull: Yes(5) No(5)
• Guys,pls ans this Q with proper explanation..
• 8 years agoHelpfull: Yes(3) No(0)
• x^3(1-x^2)0 which is false
  let x=0 -1>0 which is false
  let x=-1 0>0 which is false
  
  Therefore Statement (1) is sufficient & not statement (2) alone.
• 8 years agoHelpfull: Yes(3) No(4)
• statement 1 alone sufficient..
• 8 years agoHelpfull: Yes(2) No(6)
• statement 1
  either x30
  
  each statement alone is sufficient to answer the question
• 8 years agoHelpfull: Yes(2) No(1)
• Neither of the statement is sufficient as x is not less than 0.
• 8 years agoHelpfull: Yes(2) No(0)
• If we analyse Statement 1: LHS can be less than 0 for two cases:
  - Case 1: x < 0 and |x| < 1 (so that x^3 will be -ve and (1-x^2) will be +ve)
  - Case 2: x > 0 and |x| > 1 (so that x^3 will be +ve and (1-x^2) will be -ve)
  
  Now on analyzing Statement 2: x^2 > 1 => |x| > 1 => case 2 above gets satisfied. Hence we need both the statements.
• 8 years agoHelpfull: Yes(1) No(0)
• @Naveen, why you are taking & condition, It may possible to take or condition also.In given ques it was not said that take and for both statemens.
  And Finaly you answered x belongs to (1, infinity) which is also wrong. Just put x=2 in statement 1 and see the outcomes.
• 8 years agoHelpfull: Yes(0) No(1)
• Dear bro, anurag u hv to work more on Data sufficiency...
  If u put x=2
  1)-24 0
  Both conditions followed...
• 8 years agoHelpfull: Yes(0) No(0)
• @Anurag if u put x=2
  Then 1) -24 2)3
  Both conditions followed i.e 1).0
• 8 years agoHelpfull: Yes(0) No(0)
• Is x
• 8 years agoHelpfull: Yes(0) No(0)
• but naveen kumar when we put x=-1 then 2 condition will fail it will become zero.
• 8 years agoHelpfull: Yes(0) No(0)
• @ Ankush the value of x lies between 1 and infinity...
  
• 8 years agoHelpfull: Yes(0) No(0)
• take x as -1 & solve
  
  both r not sufficient
• 8 years agoHelpfull: Yes(0) No(1)
• X>0
  x^3-x^5 >0 if x is negative
  but its provided that x^3(1-x^2)0 does not define the sign of x
  => only statement 1 is sufficient and reliable to provide solution
  Option A
• 8 years agoHelpfull: Yes(0) No(0)
• Statement 1 alone is sufficient .
• 7 years agoHelpfull: Yes(0) No(0)
• just put x=2 and check.
  1.-24
  2.3>0
  both are true but still x is positive.
  hence both are not sufficient to answer the question.
• 7 years agoHelpfull: Yes(0) No(0)
• For me (C)
  
  Stat 1
  
  x^3(1-x^2) 0
  [ x > 0 and (x^2 -1)>0 ] or [ x < 0 and (x^2 -1) 1 or -1 < x < 0
  
  INSUFF
  
  Stat 2
  x^2-1 1 or -1 < x < 0 ] and -1 < x < 1
  -1 < x < 0
  
  SUFF
• 6 years agoHelpfull: Yes(0) No(0)
• BODHAYAN NANDI If we take x
• 6 years agoHelpfull: Yes(0) No(0)