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Is x negative? [#permalink]
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21 Dec 2009, 13:01
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Is x negative? (1) x^3(1x^2) < 0 (2) x^21 < 0
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Re: Need Help with OG 11 ed DS 154 [#permalink]
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21 Dec 2009, 15:12
gi wrote: Hi All, I'm new to this forum. I'm working through some OG DS questions and I find that some of the explanations to ans are not clear or take too long to solve in 2 mins. Q:DS154 Is x negative? 1) x^3(1x^2) < 0 2) x^21 < 0 Thanks (1) \(x^3(1x^2)<0\), two cases (one of the multiples negative, another positive): A. \(x^3>0\) and \(1x^2<0\); \(x^3>0\) > \(x>0\) 0 \(1x^2<0\) > \(x<1\) or \(x>1\) (1)0(1) Hence the range for A is \(x>1\). B. \(x^3<0\) and \(1x^2>0\) \(x^3<0\) > \(x<0\) 0 \(1x^2>0\) > \(1<x<1\) (1) 0(1) Hence the range for B is \(1<x<0\). Two ranges: \(x>1\) OR \(1<x<0\). Not sufficient. (2) \(x^21 < 0\) > \(1<x<1\). Not sufficient. (1)+(2) Inersection of the ranges from (1) and (2) is \(1<x<0\), hence \(x<0\) is true. Sufficient. Answer: C.
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Re: Need Help with OG 11 ed DS 154 [#permalink]
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21 Dec 2009, 15:21
IMO C:
A) X^2(1X^2) <0 so X^3<0 or X^2>1 x could be positive or negative
B) X^21<0 (X1)(X+1)<0 so X<1 or X<1 x could be positive or negative
Combined X is negative for both statements to hold true.
OA?



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Re: Need Help with OG 11 ed DS 154 [#permalink]
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21 Dec 2009, 15:33
C; like Bunuel's details. Inequalities without coffee can not be so much fun!!!
Bunuel,
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Re: Need Help with OG 11 ed DS 154 [#permalink]
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21 Dec 2009, 15:52



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Re: Inequalities [#permalink]
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10 Apr 2010, 11:07
IMO C. 1, is equivalent to x(x1)(x+1) >0 ( x^2 is always>0 and i have inverted the sign) this gives x>1 and 1<x<0 , thus not sufficient 2, gives 1< x<1 not sufficient. if you combine the result we get 1<x<0 , thus x is negative and thus C. Another way, since we know both are not sufficient individually. 2nd equation gives x^2 1 <0 => 1 x^2 >0 now in first equation, if its ve then x^3 must be ve....thus both taken together sufficient.
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Re: Inequalities [#permalink]
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10 Apr 2010, 11:12
First I go to Stmt 2: Stmt 2 shows that 1 Stmt1 not sufficient too!!
With the help of stmt 2, if we put the negative value of x, the stmt 1 becomes true!!
Hence "C" is the answer in my opinion!!!
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Re: Inequalities [#permalink]
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10 Apr 2010, 11:37
Thanks guys for the quick solution..Heres a question..
For st1, if I take a +ve number that is \(>=2\) the inequality holds true, then why do we need to consider st2?
Look forward to the replies..



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Re: Inequalities [#permalink]
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10 Apr 2010, 11:41
abhi758 wrote: Thanks guys for the quick solution..Heres a question..
For st1, if I take a +ve number that is \(>=2\) the inequality holds true, then why do we need to consider st2?
Look forward to the replies.. Thats wrong approach towards DS question. x>=2 satisfy the equation 1 , what about x = 1/2, that is also satisfying the equation.Then how we can be sure whether x is + or ... sufficiency comes when we are sure this statement as a whole ans the question and there is no other possibility which will be against it,. Before starting another DS question do get this concept, else no use of DS questions. letme know if you got it else i will try to explain in other way.
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Re: Inequalities [#permalink]
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11 Apr 2010, 23:41
Thanks for bringing this to my notice. If you could explain the first st1 working in the solution provided by you: Quote: 1, is equivalent to x(x1)(x+1)>0 ( x^2 is always>0 and i have inverted the sign)
this gives x>1 and 1<x<0 , thus not sufficient Unable to understand the working for \(x^3(1x^2)<0\).



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Re: Inequalities [#permalink]
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12 Apr 2010, 00:51
\(x^3(1x^2) = x* x^2 * (x^2 1)* (1 ) < 0\) since x^2 >0 , it will not affect the sign \(x*(x^21) * (1) < 0\) => \(x*(x^21) * > 0\) when we multiple both sides by 1, > becomes < eg. 1 > 2 , but 1 < 2 => \(x*(x1)*(x+1) * > 0\)
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Re: Inequalities [#permalink]
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12 Apr 2010, 07:38
Thanks gurpreet for your time and patience!
Just a ques on the last part..
After: \(x(x1)(x+1)>0\) you get => \(x>1\) AND you get => \(x>1\) OR \(1<x\) AND next would you get => \(x>0\) OR \(x<0\)??
This might sound a little basic but really confused with the whole thing..



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Re: Inequalities [#permalink]
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12 Apr 2010, 10:03
No, its not right. Read this inequalitiestrick91482.html
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Re: Inequalities [#permalink]
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12 Apr 2010, 10:55
IMO E
ST 1 both a pos and a neg # satisfiy this equation. Not suf
ST 2....samething.
Cobined we can't conclude the sign of X
also, keep in mind that the ques doesn't states if X is 0 or 1
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Re: Inequalities [#permalink]
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12 Apr 2010, 11:36
Alternative approach by taking numbers. Statement1: \(x^3(1x^2) < 0\) From this we can say \(x <> 0\) and \(x <>+1 or 1\). Consider x=2 Then the equation results in \(8*3 <0\). Satisfy the equation. Consider x= 1/2 then the equation results in \(1/8*(3/4) >0\). Does not satisfy the equation. Consider x=2 Then the equation results in \(8*3 >0\). Does not satisfy the equation. Consider x= 1/2 then the equation results in \(1/8*(3/4) <0\).Satisfy the equation. This statement is not sufficient to conclude. Statement2: \(x^21 <0\) From the statement we can say that \(x <>=1 or 1\) Consider x=2 then the equation results in \(3 >0\). Does not satisfy the equation. Consider x=1/2 then the equation results in \(3/4 <0\). Satisfy the equation. Consider x=2 then the equation results in \(3 >0\). Does not satisfy the equation. Consider x=1/2 then the equation results in \(3/4 <0\). Satisfy the equation. This statement alone is not sufficient. Combining both the statements when x is ve that is 1/2 , it satisfies both the equations. Ans is C.
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Re: Inequalities [#permalink]
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14 Apr 2010, 02:18
1. x^3(1x^2) <0 x^3 x^5<0 x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).
2. x^2<1
1 < x < 1
which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).
the answer is A



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Re: Inequalities [#permalink]
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14 Apr 2010, 03:09
abhi758 wrote: Is X negative?
1. \(x^3(1x^2) < 0\) 2. \(x^21 < 0\)
Please explain your answer.. Is \(x<0\)? (1) \(x^3(1x^2) < 0\) Two cases: A. \(x>0\) and \(1x^2<0\), which means \(x<1\) or \(x>1\) > \(x>1\); B. \(x<0\) and \(1x^2>0\), which means \(1<x<1\) > \(1<x<0\). So \(x^3(1x^2) < 0\) holds true for two ranges of \(x\): \(x>1\) and \(1<x<0\). If \(x\) is in the first range answer to the question is NO, but if \(x\) is in the second range answer to the question is YES. Two different answers. Not sufficient. (2) \(x^21<0\) > \(1<x<1\). x can be positive as well as negative. Not sufficient. (1)+(2) \(x>1\) or \(1<x<0\) AND \(1<x<1\) > intersection of the ranges from (1) and (2) is \(1<x<0\). \(x\) is negative. Sufficient. Answer: C. aiyedh wrote: 1. x^3(1x^2) <0 x^3 x^5<0 x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).
2. x^2<1
1 < x < 1
which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).
the answer is A Red part is not correct. \(x^3 < x^5\) does not hold true for all positive \(xes\). For example \(x=0.5\) > \(x^3>x^5\). Also \(x\) can be negative in the range \(1<x<0\) and \(x^3 < x^5\) will hold true.
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Re: Inequalities [#permalink]
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17 Apr 2010, 12:25
abhi758 wrote: Is X negative?
1. \(x^3(1x^2) < 0\) 2. \(x^21 < 0\)
Please explain your answer.. here, is small note: as soon as we see one number can be either positive or negative we know the statement is not sufficient. so statement 1) \(x^3(1x^2) < 0\) , here \(x^3\) can +ve or ve => not suff. similarly; \(x^21 < 0\) => (x1)(x+1)<0 ; here also x1 or x+1 can +ve or ve not suff. combining both : since \(x^21 < 0\) or 1 \(x^2 > 0\) hence from statement 1 \(x^3 < 0\) => x< 0 => suff.



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is x negative [#permalink]
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30 Apr 2013, 20:57
is x negative 1) X^3X^5<0 2) X^21 <0 while solving stmt 1 X^3(1X^2)<0 X^3(X+1)(1X)<0 Then i took the roots as 1,0,+1 on the number line to find the range as per vertias prep graph approach but am getting ranges as 0>X<1 and x<1 which is wrong .please do correct me
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Re: is x negative [#permalink]
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30 Apr 2013, 21:18
skamal7 wrote: is x negative 1) X^3X^5<0 2) X^21 <0
while solving stmt 1 X^3(1X^2)<0 X^3(X+1)(1X)<0 Then i took the roots as 1,0,+1 on the number line to find the range as per vertias prep graph approach but am getting ranges as 0>X<1 and x<1 which is wrong .please do correct me You will have to rearrrange the equation to read x(x+1)(x1)>0. You can drop the x^2 as it is always positive and x is not equal to zero. Now if you plot the roots, it will give you 1,0 and 1. Thus, the valid ranges will be x>1 OR 1<x<0. Insufficient. From F.S 2, we know that x<1 > 1<x<1. Insufficient. On taking both fact statements together, for 1<x<0, both the conditions are fulfilled. Sufficient. C.
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