IMO B
when N=1 A=1
if it's 2N then the first number will be 2. the range will be from 2-40, having the median 20....
don't know if it's right though....

OA???

2N=5A-3 only when A is an odd integer. Thus the first 20 values of N will be from 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56....96.

therefore the median will be the avg of two middle terms (since there are an even amount of numbers). 46+51/2 = 48.5

dtweah wrote:In the equation 2N=5A-3, where N>0, A>0 and N and A take on finitely many values, what is the median of the first consecutive 20 values of N?

A). 15
B) 20
C) 38.5
D) 48.5
E) 49

For this question to make sense, we need more information. Did it originally read "what is the median of the first 20 consecutive integer values of N"?

Without a modifier for "consecutive", there's no unique answer to the question. If we know that we're looking for consecutive integer values of N, then dwilliams' solution is perfect.

Stuart Kovinsky wrote:

dtweah wrote:In the equation 2N=5A-3, where N>0, A>0 and N and A take on finitely many values, what is the median of the first consecutive 20 values of N?

A). 15
B) 20
C) 38.5
D) 48.5
E) 49

For this question to make sense, we need more information. Did it originally read "what is the median of the first 20 consecutive integer values of N"?

Without a modifier for "consecutive", there's no unique answer to the question. If we know that we're looking for consecutive integer values of N, then dwilliams' solution is perfect.

2N =5A-3
2N +8 =5A-3 +8
2(N+4)=5(A+1)
Regardless of A, N+4 must be evenly divisble by 5 if the equation is to hold. Then N must be of the sequence 1, 6, 11, ..... The nth term of this sequence is 5N-4. The median of the first 20 terms is the average of the 10th and 11th which can be found to be 48.5.
Choose D.