If m^3-n^2=-300, then the lowest possible value of m is

Question

If m^3-n^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

Bunuel · Accepted Answer

m^3-n^2=-300  -->  m=\sqrt {n^2-300} . 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 {-300} .

m is less than -5 (since (-5)^3=-125) and more than -10 (since (-10)^3=-1000).

Answer: C.

guerrero25 · Answer

Such an Easy approach , Bunuel . I succumbed to the time pressure . I wish I could think like you :wink:

agrchandan · Answer

m^3-n^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

sdas · Answer

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

Bunuel · Answer

Even roots from  negative number  is undefined on the GMAT:  \sqrt {negative}=undefined , for example  \sqrt{-25}=undefined .

Odd roots have the same sign as the base of the root. For example,  \sqrt {125} =5  and  \sqrt {-64} =-4 .

AbhiJ · Answer

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

bankerboy30 · Answer

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?

Bunuel whats wrong with this logic?

PareshGmat · Answer

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

aneternallearner · Answer

Hi Bankerboy30,

In your case, you would need to find n such that square of n would equal -7700 (300-8000). 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

BrushMyQuant · Answer

Please tag exponents
Thank you

mvictor · Answer

m will be minimum when n=0, otherwise by deducting a positive number, the negative will get even bigger.
m^3 = 300
ok...
-5x-5x-5=-125..so clearly can be lower than -5. D and E are out.
-10x-10x-10=-1000 clearly not lower than -10. A and B out.
C remains.

stonecold · Answer

Here the value of M must lie between -6 and -7 Go ahead push that C

dave13 · Answer

Bunuel hello there

how are you ?

can you please explain how after this m^3-n^2=-300 you get this m=\sqrt {n^2-300} . Exponent 3 is outside of darical sign and exponent 2 is inside radical sign

Bunuel · Answer

m^3-n^2=-300 ;

m^3=n^2-300 ;

m=\sqrt {n^2-300} .

dave13 · Answer

Bunuel from here m^3=n^2-300 how do you get ths m=\sqrt {n^2-300}

what are you doing such that exponent 3 goes to the right

please help

Bunuel · Answer

Take the cube root. The same way we get  x=\sqrt {y}  from  x^3 = y .

GMATGuruNY · Answer

m^3-n^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.

C

shridhar786 · Answer

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

Gayluk · Answer

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!

Hope it helps!

If m^3-n^2=-300, then the lowest possible value of m is

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If m^3-n^2=-300, then the lowest possible value of m is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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