After finding the average of $35$ scores, a student carelessly included the average with the $35$ scores and found the 
average of these $36$ numbers. The ratio of the second average to the true average was
$\textbf{(A) }1:1\qquad \textbf{(B) }35:36\qquad \textbf{(C) }36:35\qquad \textbf{(D) }2:1\qquad  \textbf{(E) }\text{None of these}$
Assume the $35$ scores are the first $35$ natural numbers: $1, 2, 3, \dots 35.$
The average of the scores is $\frac{\frac{(35)(36)}{2}}{35} = 18.$
If we add $18$ to the first $35$ numbers, the new average is $\frac{648}{36} = 18$ still.
The two averages are the same, therefore the answer is $\textbf{(A) }1:1.$
-edited by coolmath34