$1 \cdot 117$ A student, after finding the average of 35 scores, carelessly mixed this average with the 35 scores and found the average of these 36 numbers. The ratio of the second average to the correct average is
(A) $1: 1$.
(B) $35: 36$.
(C) $36: 35$.
(D) $2: 1$.
(E) None of these.
(22nd American High School Mathematics Examination, 1971)
[Solution] Let the 35 scores be denoted as $x_{1}, x_{2}, x_{3}, \cdots, x_{35}$, and their average is $\bar{x}$. Therefore, $35 \bar{x}=x_{1}+x_{2}+x_{3} \cdots+x_{35}$.
The average of the 36 numbers is
$$
\frac{\left(x_{1}+x_{2}+x_{3}+\cdots+x_{35}+\bar{x}\right)}{36}=\frac{(35 \bar{x}+\bar{x})}{36}=\frac{36 \bar{x}}{36}=\bar{x} \text {, }
$$

Thus, the required ratio is $1: 1$.
Therefore, the answer is $(A)$.