If n is an integer such that 1/9

Question

If  n  is an integer such that  \frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2} , what is the median of all possible values of  n ?

A. -3
B. -2
C. 0
D. 2
E. 3

Bunuel · Accepted Answer

Official Solution:

If  n  is an integer such that  \frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2} , what is the median of all possible values of  n ?

A. -3
B. -2
C. 0
D. 2
E. 3

Given that  n  is an integer, for the inequality  \frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}  to hold true,  \frac{1}{n^2-1}  must be  \frac{1}{8} ,  \frac{1}{7} ,  \frac{1}{6} ,  \frac{1}{5} ,  \frac{1}{4} , or  \frac{1}{3} . Hence,  n^2-1  must be 8, 7, 6, 5, 4, or 3. Consequently,  n^2  must be 9, 8, 7, 6, 5, or 4. However,  n^2  cannot be 8, 7, 6, or 5, because in these cases  n  won't be an integer. Therefore,  n^2  is either 9 or 4, making  n  equal to -3, 3, -2, or 2. Therefore, the median of all possible values of  n  is the average of the two middle terms, thus  \frac{-2+2}{2} = 0 .

Answer: C

gmatophobia · Answer

\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}

Taking reciprocal, we get

2 < n^2 - 1 < 9

Adding 1 on both sides we get

3 < n^2 < 10

Therefore, n can take the following values: {-3, -2, 2 , 3}

Median =  \frac{-2 + 2 }{ 2 }= 0

Option C

Dp5Shivam · Answer

Therefore, n can take the following values: {-3, -2, 2 , 3}

where did we get these values from ? pls help.

Bunuel · Answer

n^2 = 9 or 4, so n is -3, 3, -2, or 2.

If n is an integer such that 1/9 < 1/(n^2 - 1) < 1/2, what is the

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