December 17, 2018 December 17, 2018 06:00 PM PST 07:00 PM PST Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong. December 17, 2018 December 17, 2018 10:00 PM PST 11:00 PM PST From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.
Author 
Message 
Manager
Joined: 13 May 2017
Posts: 101
Location: Finland
Concentration: Accounting, Entrepreneurship
GPA: 3.14
WE: Account Management (Entertainment and Sports)

Is m < m^2 < m^3 , for all real values of m?
[#permalink]
Show Tags
25 Nov 2018, 10:10
Question Stats:
46% (01:26) correct 54% (01:12) wrong based on 82 sessions
HideShow timer Statistics
Is m < \(m^2 < m^3\) , for all real values of m? (1) \(m < m^3\) (2) \(m^3 > m^2\)
Official Answer and Stats are available only to registered users. Register/ Login.



GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 549

Re: Is m < m^2 < m^3 , for all real values of m?
[#permalink]
Show Tags
25 Nov 2018, 11:45
rencsee wrote: Is m < \(m^2 < m^3\) , for all real values of m?
(1) \(m < m^3\) (2) \(m^3 > m^2\)
Very nice problem, congrats! (kudos!) \(?\,\,\,:\,\,\,m < {m^2} < {m^3}\) \(\left( 1 \right)\,\,\,m < {m^3}\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,m = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,m =  {1 \over 2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\left( {{m^2} < {m^3}\,\,{\rm{is}}\,\,{\rm{false}}} \right)\,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,\,{m^3} > {m^2}\,\,\,\,\, \Rightarrow \,\,\,\,{m^2}\left( {m  1} \right) > 0\,\,\,\,\, \Rightarrow \,\,\,\,m > 1\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,m\,\,\left( { > \,0} \right)\,\,} \,\,\,\,{m^2} > m\,\,\,\,\left( {{\rm{and}}\,\,\left( 2 \right)} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\) The correct answer is therefore (B). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
_________________
Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12questions Mock included) to gain a 50% discount!



Manager
Joined: 26 Apr 2011
Posts: 59
Location: India
GPA: 3.5
WE: Information Technology (Computer Software)

Re: Is m < m^2 < m^3 , for all real values of m?
[#permalink]
Show Tags
26 Nov 2018, 01:28
rencsee wrote: Is m < \(m^2 < m^3\) , for all real values of m?
(1) \(m < m^3\) (2) \(m^3 > m^2\) Please elaborate more on the solution. Confused on the approach to solve this problem
_________________
Do What Your Heart Says....



GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 549

Is m < m^2 < m^3 , for all real values of m?
[#permalink]
Show Tags
26 Nov 2018, 03:36
MrCleantek wrote: Please elaborate more on the solution. Confused on the approach to solve this problem Hi MrCleantek ! My pleasure. In (1) I have presented what we call a BIFURCATION: Two different VIABLE answers to the question asked (= each scenario satisfies both the question stem prestatements and the statement itself). This is the proper mathematical way to guarantee insufficiency. It avoids the common "it´s clear..." that permits falling into intuition traps! In (2) I have shown that \(m^2 > m\) and that was enough, because the other part of the inequalityfocused is given in the statement itself. My rationale was the following: I put \(m^2\) as a common nonzero factor (otherwise statement (2) would be contradicted: 0 > 0 is false), hence \(m^2 > 0\) and, therefore \(m > 1\). Afterwards I have multiplied both sides by \(m > 0\) (we know m>1), the inequality is mantained and we get the one we wished for! An alternate approach (quite similar) would be the following: \({m^3} > {m^2}\,\,\,\,\mathop \Rightarrow \limits_{\left( {m\, \ne \,0} \right)}^{:\,\,{m^2}\, > \,\,0} \,\,\,m > 1\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,m\,\,\left( { > \,0} \right)} \,\,\,{m^2} > m\) I hope everything is clear now. Regards and success in your studies, Fabio.
_________________
Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12questions Mock included) to gain a 50% discount!



Manager
Joined: 26 Apr 2011
Posts: 59
Location: India
GPA: 3.5
WE: Information Technology (Computer Software)

Re: Is m < m^2 < m^3 , for all real values of m?
[#permalink]
Show Tags
27 Nov 2018, 02:05
fskilnik wrote: MrCleantek wrote: Please elaborate more on the solution. Confused on the approach to solve this problem Hi MrCleantek ! My pleasure. In (1) I have presented what we call a BIFURCATION: Two different VIABLE answers to the question asked (= each scenario satisfies both the question stem prestatements and the statement itself). This is the proper mathematical way to guarantee insufficiency. It avoids the common "it´s clear..." that permits falling into intuition traps! In (2) I have shown that \(m^2 > m\) and that was enough, because the other part of the inequalityfocused is given in the statement itself. My rationale was the following: I put \(m^2\) as a common nonzero factor (otherwise statement (2) would be contradicted: 0 > 0 is false), hence \(m^2 > 0\) and, therefore \(m > 1\). Afterwards I have multiplied both sides by \(m > 0\) (we know m>1), the inequality is mantained and we get the one we wished for! An alternate approach (quite similar) would be the following: \({m^3} > {m^2}\,\,\,\,\mathop \Rightarrow \limits_{\left( {m\, \ne \,0} \right)}^{:\,\,{m^2}\, > \,\,0} \,\,\,m > 1\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,m\,\,\left( { > \,0} \right)} \,\,\,{m^2} > m\) I hope everything is clear now. Regards and success in your studies, Fabio. I got it now. Thanks...
_________________
Do What Your Heart Says....



GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 549

Re: Is m < m^2 < m^3 , for all real values of m?
[#permalink]
Show Tags
27 Nov 2018, 18:56
MrCleantek wrote: I got it now. Thanks...
Great, MrCleantek! I am glad to know. See you in other posts! (Thanks for the kudos.) Regards, Fabio.
_________________
Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12questions Mock included) to gain a 50% discount!




Re: Is m < m^2 < m^3 , for all real values of m? &nbs
[#permalink]
27 Nov 2018, 18:56






