\(\frac{x^(n^2-n+2)}{x^(n-2)(n+1)}\)
=\(\frac{x^(n^2-n+2)}{x^(n^2-n-2)}\)
=\(x^(n^2-n+2-n^2+n+2)\)
=\(x^4\)

So we only need the value of x to determine the answer.

Statement 1:
As we found out above, the value is independent of the value of n. We do not know that value of x. If x=1 then answer is 1 and if x=2 answer is 16.
Not sufficient

Statement 2:
This directly gives us the value of x and the answer will be \(2^4 =16\).
Sufficient.

Answer B

Hit kudos if you like the explanation.

Formatted correctly otherwise this is very confusing:

\(\frac{x^{n^2−n+2}}{x^{(n−2)(n+1)}} = ?\)

(1) \(n=5\)
(2) \(x=2\)

To solve, take the equation and simplify by first multiplying out the denominator:

\(= \frac{x^{n^2-n+2}}{x^{n^2-n-2}}\)

Separate out the X using rules of exponents:

\(= \frac{(x^{n^2})( x^{-n}) (x^{2})}{(x^{n^2}) ( x^{-n}) (x^{-2})}\)

Flip the negative exponents:

\(= \frac{(x^{n^2})( x^{n}) (x^{2})(x^{2})}{(x^{n^2}) ( x^{n})}\)

Cancel out and simplify:

\(= \frac{(x^{2})(x^{2})}{1}\)

\(= x^{4}\)

Thus if we know \(x\), we will have solved the problem.

\(\frac{x^{n^2−n+2}}{x^{(n−2)(n+1)}} = ?\)


\(\frac{x^{n^2−n+2}}{x^{(n^2+n-2n-2)}}\)


\(\frac{x^{n^2−n+2}}{x^{(n^2-n-2)}}\)

\(x^(n^2−n+2 - n^2+n+2)\)

\(x^(2+2)\)

\(x^(4)\)

Which means if we have the value of x we should be able to find the answer.

(1) \(n=5\)

Clearly not sufficient as we are not aware of the value of x

Hence, (1) =====> is NOT SUFFICIENT

(2) \(x=2\)

\(x=2\)

Substitute this value in above equation.

\(x^(4)\)

\(2^(4)\)

\(= 16\)


Hence, (2) =====> is SUFFICIENT

Hence, Answer is B
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Lets simplify the fraction : \(\frac{x^{(n^2-n+2)}}{x^{(n-2)(n+1)}}\)

\(\frac{x^{(n^2-n+2)}}{x^{(n^2-n-2)}}\)

\(x^{(n^2-n+2)-(n^2-n-2)}\)

\(x^{(n^2-n+2- n^2+n+2)}\)

\(x^4\)

So we just need the value of \(x\).

(1) \(n=5\)

Does not provide value of \(x\). Hence I is Not Sufficient.

(2) \(x=2\)

Gives the value of x.

\(x^4 = 2^4 = 16\)

Hence II is Sufficient.

Answer (B)...

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