Vincen wrote:Is the nth root of n greater than the cube root of 3?

1) The nth root of n is equal to the 4th root of 4
2) The nth root of n is equal to the square root of 2

IMPORTANT: This question illustrates a situation in which we need not perform any calculations. Instead, we need only recognize that we COULD perform calculations, which would allow us to determine whether or not a statement is sufficient.

Target question: Is (nth root of n) greater than (cube root of 3)?

Statement 1: The nth root of n is EQUAL TO the 4th root of 4
"EQUAL TO" is key here.
Since we COULD determine the exact value of the 4th root of 4 (which is equal to nth root of n), and since we COULD determine the exact value of the cube root of 3, we could definitely determine whether (nth root of n) is greater than (cube root of 3)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The nth root of n is EQUAL TO the square root of 2
Once again, we have "EQUAL TO"
So, we COULD determine the exact value of the √2 (which is equal to nth root of n), and we COULD determine the exact value of the cube root of 3.
So, we could definitely determine whether (nth root of n) is greater than (cube root of 3)
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

Vincen wrote:Is the nth root of n greater than the cube root of 3?

1) The nth root of n is equal to the 4th root of 4
2) The nth root of n is equal to the square root of 2

We need to determine whether ^n√n > ^3√3.

Statement One Alone:

The nth root of n is equal to the 4th root of 4

Since ^n√n = ^4√4, the question becomes: is ^4√4 > ^3√3?

Let's raise both sides to the 12th power:

(^4√4)^12 > (^3√3)^12 ?

4^3 > 3^4 ?

64 > 81 ?

We see that the answer is no. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The nth root of n is equal to the square root of 2

Thus, we know that:

Since ^n√n = √2, the question becomes: is √2 > ^3√3?

Let's raise both sides to the 6th power:

(√2)^6 > (^3√3)^6 ?

2^3 > 3^2 ?

8 > 9 ?

We see that the answer is no again. Statement two alone is also sufficient to answer the question.

Answer: D