If n = (-72)^(1/3) then the value of n is

Question

If  n = (-72)^{(\frac{1}{3})}  then the value of n is

A. -9 < n < -8
B. -8 < n < -7
C. -7 < n < -6
D. -6 < n < -5
E. -5 < n < -4

BrentGMATPrepNow · Accepted Answer

Find some "nice" third powers (aka cube roots)

(-1)^{(\frac{1}{3})} = -1 
 (-8)^{(\frac{1}{3})} = -2 
 (-27)^{(\frac{1}{3})} = -3 
 (-64)^{(\frac{1}{3})} = -4 
 (-125)^{(\frac{1}{3})} = -5

Since -72 is BETWEEN -64 and -125,  (-72)^{(\frac{1}{3})}  will be BETWEEN -4 and -5

Answer: E

Bunuel · Answer

Similar questions: https://gmatclub.com/forum/new-tough-an ... l#p1029224 https://gmatclub.com/forum/the-value-of ... 86290.html

stonecold · Answer

Let m= (-72)^1/3
Hence m^3 = -72

Lets check

-1^3 => -1
-2^3 => -8
-3^3 => -27
-4^3 => -64
-5^3 => -125

Hence m must be between (-5,-4) 
Hence E.

sashiim20 · Answer

n = (-72)^{(\frac{1}{3})}

Cubing both sides, we get;

n^3 = (-72)

-4^3 = -64

-5^3 = -125

Therefore value of n has to be between, -5 and -4.

-5 < n < -4 . Answer (E) ...

pushpitkc · Answer

Since  n = (-72)^{(\frac{1}{3})}  ,n is definitely negative.

72, when prime-factorized will give  2^3 * 3^2 
Cube root of  2^3 (8)is 2
Cube root of  3^2 (9) is a little more than 2

Hence  (-72)^{(\frac{1}{3})}  is a little more than -4
This comes in the range  -5 < n < -4(Option E)

gmathopeful19 · Answer

IMO, E.

Rewritten the question becomes what is the cube root of 72?

4^3 = 64, so N is greater than 4

5^3 = 125, so N is less than 5

E) N is between -4 and -5

EMPOWERgmatRichC · Answer

Hi All,

While you will almost certainly have to deal with a square-root a couple of times on Test Day (in both Arithmetic and Geometry questions), cube-roots are relatively rare (you probably won't see them at all on Test Day). That having been said, cube-roots are based on the same general math concepts as square-roots.

For example... 
The square-root of 4 is +2 ... since (2)(2) = 4 
The cube-root of 27 is +3 ... since (3)(3)(3) = 27 
The cube-root of -27 is -3 ... since (-3)(-3)(-3) = -27

In this question, we're asked for the cube-root of -72, but the solution will NOT be an integer (notice how the answer choices are all RANGES - that's a hint that N is not an integer AND that we don't actually have to find the exact value of N; we just have to figure out what numbers it is between). Thus, we have to look for cubes of negative integers:

(-3)^3 = -27 
(-4)^3 = -64 
(-5)^3 = -125

Since -72 is between -64 and -125, the cube-root of -72 has to be between -4 and -5

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

We need to find two integers, one of which,, when raised to the 3rd power, is less than -72, and the other of which, when raised to the 3rd power, is greater than -72.

Since -4^3 = -64 and -5^3 = -125, and since -72 is between those two numbers, n must be between -4 and -5.

Answer: E

bumpbot · Answer

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If n = (-72)^(1/3) then the value of n is

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If n = (-72)^(1/3) then the value of n is

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