1) 5m-10>0
=>5m>10
=>m>2
=>m>0
sufficient
2) (m^3)-m>0
=>m^3>m
possible only when m>0
because if m<0, m^3 becomes<m. (eg. -8<-2) also m cant be 0
=>m>0
sufficient
hence D
Is m positive?
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scoobydooby
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Ian Stewart
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Just to clarify here, if m^3 > m, that doesn't necessarily mean that m > 0. It means that either m > 1, or that -1 < m < 0.scoobydooby wrote: 2) (m^3)-m>0
=>m^3>m
possible only when m>0
because if m<0, m^3 becomes<m. (eg. -8<-2) also m cant be 0
Of course, we're told in the stem that m is an integer, so it isn't possible that -1 < m < 0, and that's the reason Statement 2 is sufficient, and the answer is D. If we didn't know that m was an integer, the answer would be A here.
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Vemuri
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You make a good point Ian.Ian Stewart wrote:Just to clarify here, if m^3 > m, that doesn't necessarily mean that m > 0. It means that either m > 1, or that -1 < m < 0.scoobydooby wrote: 2) (m^3)-m>0
=>m^3>m
possible only when m>0
because if m<0, m^3 becomes<m. (eg. -8<-2) also m cant be 0
Of course, we're told in the stem that m is an integer, so it isn't possible that -1 < m < 0, and that's the reason Statement 2 is sufficient, and the answer is D. If we didn't know that m was an integer, the answer would be A here.
But, I have a question here related to inequality. In the second stem, it is given that m^3-m>0 ==> m(m^2-1)>0. This means that either m>0 or m^2>1 i.e. m<-1 or m>1. Isn't this correct? If yes, then I guess the answer changes. Obviously, I seem to be doing something wrong here. Would you please care to clarify?
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cramya
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Vemuri,
Factorizing is a good way to arrive at possible solutions just so we dont eliminate any solutions by cancelling.
m (m^2-1) > 0
Case 1:
m>0 and m^2-1>0 i.e m^2>1
I am sure u got this one
Case 2:
m<0 and m^2-1 < 0 i.e m^2 < 1 i.e |m| < 1
Any m between -1 and 0 will satisfy this condition.
Hope this helps!
Regards,
CR
Factorizing is a good way to arrive at possible solutions just so we dont eliminate any solutions by cancelling.
m (m^2-1) > 0
Case 1:
m>0 and m^2-1>0 i.e m^2>1
I am sure u got this one
Case 2:
m<0 and m^2-1 < 0 i.e m^2 < 1 i.e |m| < 1
Any m between -1 and 0 will satisfy this condition.
Hope this helps!
Regards,
CR
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cubicle_bound_misfit
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Hi Ian,
I don't understand one thing. Can this be solved like this ??
STMT 2 says
m^3 - m>0
i.e. m(m^2-1)>0
i.e. m(m+1)(m-1)>0
now three consecutive negative integres can not be multiplied to give a product which is positive.
Hence m has to be >0 .
Is that right approach?
I don't understand one thing. Can this be solved like this ??
STMT 2 says
m^3 - m>0
i.e. m(m^2-1)>0
i.e. m(m+1)(m-1)>0
now three consecutive negative integres can not be multiplied to give a product which is positive.
Hence m has to be >0 .
Is that right approach?
Cubicle Bound Misfit
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Vemuri
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Thank you all for posting your thoughts. Let me summarize the various ways the second statement can be queried to answer the question.
Option I:
m^3-m>0 ==> m^3>m. This is possible only when m>0. If m<0, then m^3 will be < m (which negates the statement). --> as scoobydooby mentioned.
Option II:
When I change the statement to factors, m(m^2-1)>0 ==> m(m-1)(m+1)>0. These are consecutive numbers & the statement will be satisfied only when all the factors are > 0.
Option III:
When I change the statement to inequalities, i.e. m(m^2)>0 ==> m>0 & m^2-1>0. This can be further subdivided into 2 for the statement to be > 0:
Check 1: If m>0, then m^2-1 also has to be >0, i.e. m^2>1 (or) m>1 or m<-1. But, m<-1 cannot be possible because the statment should be positive.
Check 2: If m<0, then m^2-1 also has to be <0, i.e. m^2<1 (or) -1<m<1. Since the question says that m is an integer, the only value of m in this case is =0. But, the statement says that m(m^2-1)>0. So, this check is not valid.
Hence, statment II is sufficient to answer the question.
Option I:
m^3-m>0 ==> m^3>m. This is possible only when m>0. If m<0, then m^3 will be < m (which negates the statement). --> as scoobydooby mentioned.
Option II:
When I change the statement to factors, m(m^2-1)>0 ==> m(m-1)(m+1)>0. These are consecutive numbers & the statement will be satisfied only when all the factors are > 0.
Option III:
When I change the statement to inequalities, i.e. m(m^2)>0 ==> m>0 & m^2-1>0. This can be further subdivided into 2 for the statement to be > 0:
Check 1: If m>0, then m^2-1 also has to be >0, i.e. m^2>1 (or) m>1 or m<-1. But, m<-1 cannot be possible because the statment should be positive.
Check 2: If m<0, then m^2-1 also has to be <0, i.e. m^2<1 (or) -1<m<1. Since the question says that m is an integer, the only value of m in this case is =0. But, the statement says that m(m^2-1)>0. So, this check is not valid.
Hence, statment II is sufficient to answer the question.
