This one is quite easily solved if you take a minute and figure out what the problem means. First, you have 6 consecutive odd integers - the median of this set will always be EVEN. Thus, you can eliminate 3 answers right off the bat, and you're leftover with just 10 and 30 to test.

If the median is 10, then the set is {5, 7, 9, 11, 13, 15}. This would make \(n^2 = 5\) and \(n = \sqrt{5}\). Since \(7\sqrt{5}\) is not 15, we know the median is not 10 and must be 30. But just for completeness, let's check this one too.

If the median is 30, then the set is {25, 27, 29, 31, 33, 35}. \(n^2 = 25\), \(n = 5\), and \(7n = 35\). Therefore the median is 30.

Answer: D

Lowest is \(n^2\)

So, \(N = { n^2 , n^2 + 2 , n^2 + 4 , n^2 + 6 , n^2 + 8 , n^2 + 10 }\)

Or, \(n^2 + 10 = 7n\)

Or, \(n^2 - 7n + 10 = 0\)

Or, \(n^2 - 2n - 5n + 10 = 0\)

Or, n ( n - 2) -5 ( n - 2) = 0

So, n = 2 , 5

Now, n can never be 2 , else we will be getting even numbers...

\(N = { n^2 , n^2 + 2 , n^2 + 4 , n^2 + 6 , n^2 + 8 , n^2 + 10 }\)

Or, N = { 25 , 27 , 29 , 31 , 33 , 35 }

Hence median = \(\frac{(29 + 31)}{2}\) =30

Hence answer will be (D) 30
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Solution:

Since n^2 is the smallest value and 7n is the largest value, n^2 < 7n. Also, n can’t be negative; otherwise, the right hand side of n^2 < 7n would have been negative, whereas the left hand side is always nonnegative. So n must be positive (n can’t be 0 either since all integers in set N are odd). Therefore, when we divide both sides by n, we have n < 7. That is, n is 1, 3, or 5.

If n = 1, then n^2 = 1 would be the smallest value and we would have set N = {1, 3, 5, 7, 9, 11}. However, we see that, in this case, the largest value would be 11, which is not equal to 7n.

If n = 3, then n^2 = 9 would be the smallest value and we would have set N = {9, 11, 13, 15, 17, 19}. However, we see that, in this case, the largest value would be 19, which is not equal to 7n.

If n = 5, then n^2 = 25 would be the smallest value and we would have set N = {25, 27, 29, 31, 33, 35}. We see that, in this case, the largest value, 35, is indeed equal to 7n. Lastly, we see that the median of set N is (29 + 31)/2 = 60/2 = 30.

Alternate Solution:

Since there are six consecutive odd integers in set N and since n^2 is the smallest element of N, the largest element in N must be n^2 + 10. If we don’t see this, we can simply write N as N = {n^2, n^2 + 2, n^2 + 4, n^2 + 6, n^2 + 8, n^2 + 10}. We are also told that the largest element in N is equal to 7n, so we can write:

n^2 + 10 = 7n

n^2 - 7n + 10 = 0

(n - 2)(n - 5) = 0

n = 2 or n = 5

Since we are told that n^2 is odd, n cannot equal 2; thus, it must be true that n = 5. Hence, the set N can be expressed as N = {25, 27, 29, 31, 33, 35). The median of this set is (29 + 31)/2 = 30.

Answer: D
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Odd consecutive integers will have even median hence the right ans choice will be either 10 or 30

If the median is 10 then it lies between 9 and 11 6th integer will be 15 which is not a multiple of 7 so the right and choice is 30 hence D is correct ans

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