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If m^3n^2=300, then the lowest possible value of m is
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01 Apr 2013, 00:22
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If m^3n^2=300, then the lowest possible value of m is between (A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5
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Re: If m^3n^2=300, then the lowest possible value of m is
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01 Apr 2013, 00:51




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Re: If m^3n^2=300, then the lowest possible value of m is
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01 Apr 2013, 02:02
Bunuel wrote: guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5 \(m^3n^2=300\) > \(m=\sqrt[3]{n^2300}\). To minimize m we should minimize n^2. The lowest value of n^2 is 0, thus the lowest value of m is \(m_{min}=\sqrt[3]{300}\). m is less than 5 (since (5)^3=125) and more than 10 (since (10)^3=1000). Answer: C. Such an Easy approach , Bunuel . I succumbed to the time pressure . I wish I could think like you



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Re: If m^3n^2=300, then the lowest possible value of m is
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01 Apr 2013, 06:40
guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5 \(m^3n^2=300\) So, \(m^3 = n^2  300\) For \(m^3\) to be minimum, \((n^2  300)\) must be minimum For \((n^2  300)\) to be minimum, \(n^2\) must be minimum, so \(n^2\) = 0 So \(m^3\) = 300 So m = 6. ..... So m lies between 10 and 5



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Re: If m^3n^2=300, then the lowest possible value of m is
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25 Apr 2013, 05:57
Bunuel, pls help. how are we considering m as minimum with cube root of 300, is that not an unreal number (negative root)? I was considering the cube root for the lowest positive value of m
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Re: If m^3n^2=300, then the lowest possible value of m is
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Re: If m^3n^2=300, then the lowest possible value of m is
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25 Apr 2013, 07:53
sdas wrote: Bunuel, pls help. how are we considering m as minimum with cube root of 300, is that not an unreal number (negative root)? I was considering the cube root for the lowest positive value of m To add to what Bunnuel said. Try to think in reverse. You can always multiply a negative number 3 times to get an odd number, but you cannot multiply a negative number 2 times to get a negative number
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Re: If m^3n^2=300, then the lowest possible value of m is
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10 Jul 2014, 13:07
The way I intepreted this problem, is m can be any negative number. M does not have to be an integer. I immediately chose the greatest negative range as the answer because I figured I could offset it with some (n^2) to equal 300. For example if I chose m to be 20 than (m^(3)) would be 8000. and I would find a number that for (n2) that is equal to 7970.
Bunuel whats wrong with this logic?



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Re: If m^3n^2=300, then the lowest possible value of m is
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11 Jul 2014, 01:07
\(m^3  n^2 = 300\) Adjusting the ve signs \(n^2 = 300 + m^3\) \(5^3 = 125; & 10^3 > 300\) So least value of m should be between 5 & 10 Answer = C
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Re: If m^3n^2=300, then the lowest possible value of m is
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11 Jul 2014, 04:09
bankerboy30 wrote: The way I intepreted this problem, is m can be any negative number. M does not have to be an integer. I immediately chose the greatest negative range as the answer because I figured I could offset it with some (n^2) to equal 300. For example if I chose m to be 20 than (m^(3)) would be 8000. and I would find a number that for (n2) that is equal to 7970.
Bunuel whats wrong with this logic? Hi Bankerboy30, In your case, you would need to find n such that square of n would equal 7700 (3008000). Now, we know that square of a real number cannot be negative and we don't deal with imaginary numbers in GMAT. So, you need to go by a limitation that square of n can be minimum ZERO, not less than that. If you use that, you will get the answer as Bunuel got. Does it help? AEL



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Re: If m^3n^2=300, then the lowest possible value of m is
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05 Aug 2015, 19:56
Please tag exponents Thank you guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5
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Re: If m^3n^2=300, then the lowest possible value of m is
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08 Apr 2016, 18:23
guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5 m will be minimum when n=0, otherwise by deducting a positive number, the negative will get even bigger. m^3 = 300 ok... 5x5x5=125..so clearly can be lower than 5. D and E are out. 10x10x10=1000 clearly not lower than 10. A and B out. C remains.



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Re: If m^3n^2=300, then the lowest possible value of m is
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Re: If m^3n^2=300, then the lowest possible value of m is
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18 Jul 2018, 09:39
Bunuel wrote: guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5 \(m^3n^2=300\) > \(m=\sqrt[3]{n^2300}\). To minimize m we should minimize n^2. The lowest value of n^2 is 0, thus the lowest value of m is \(m_{min}=\sqrt[3]{300}\). m is less than 5 (since (5)^3=125) and more than 10 (since (10)^3=1000). Answer: C. Bunuel hello there how are you ? can you please explain how after this \(m^3n^2=300\) you get this \(m=\sqrt[3]{n^2300}\). Exponent 3 is outside of darical sign and exponent 2 is inside radical sign



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Re: If m^3n^2=300, then the lowest possible value of m is
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18 Jul 2018, 09:42
dave13 wrote: Bunuel wrote: guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5 \(m^3n^2=300\) > \(m=\sqrt[3]{n^2300}\). To minimize m we should minimize n^2. The lowest value of n^2 is 0, thus the lowest value of m is \(m_{min}=\sqrt[3]{300}\). m is less than 5 (since (5)^3=125) and more than 10 (since (10)^3=1000). Answer: C. Bunuel hello there how are you ? can you please explain how after this \(m^3n^2=300\) you get this \(m=\sqrt[3]{n^2300}\). Exponent 3 is outside of darical sign and exponent 2 is inside radical sign \(m^3n^2=300\); \(m^3=n^2300\); \(m=\sqrt[3]{n^2300}\).
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If m^3n^2=300, then the lowest possible value of m is
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18 Jul 2018, 10:03
Bunuel from here \(m^3=n^2300\) how do you get ths \(m=\sqrt[3]{n^2300}\) what are you doing such that exponent 3 goes to the right please help



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Re: If m^3n^2=300, then the lowest possible value of m is
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Re: If m^3n^2=300, then the lowest possible value of m is
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18 Jul 2018, 11:12
guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5 \(m^3n^2=300\) \(m^3 + 300 = n^2\)To minimize the value of \(m^3\), we must minimize the right side of the equation in blue. Since the square of a value cannot be negative, the least possible option for the right side is 0: \(m^3 + 300 = 0\) \(m^3 = 300\) Since \(5^3 = 125\) and \(10^3 = 1000\), m must be between 10 and 5.
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Re: If m^3n^2=300, then the lowest possible value of m is
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18 Jul 2018, 23:46
guerrero25 wrote: If m^3n^2=300, then the lowest possible value of m is between
(A) 20 and 15 (B) 15 and 10 (C) 10 and 5 (D) 5 and 0 (E) 0 and 5 \(n^2\) =\(m^3\)+300 since L.H.S (+), R.H.S should be positive so option (a), (b) is wrong from (c) (d) and (e) lowest possible value of m we can get from option c



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If m^3n^2=300, then the lowest possible value of m is
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24 Jul 2018, 01:33
When we look at the answer choices, we can see that A and B is too high to get 300 even if we take n=0. Thus eliminate A and B. Since we need the lowest value of m assume that n=0 (it will help minimize m as the higher absolute value of a negative number the lower that number, thus the case, in which m^3= 300 gives the possible lowest value for m ). We get m = 6, smth, which is in the range of 10 and 5. Hence C.
Hope it helps!




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