1971 AHSME Problems/Problem 16


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

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