2007 AMC 10B Problem 16

Only percents are given, so the class size does not matter. Pick an easy one: $10$ students, which means $1$ junior and $9$ seniors.

1. The class average is $84$, so all $10$ scores add to
$$10 \times 84 = 840.$$
2. The $9$ seniors average $83$, so their scores add to
$$9 \times 83 = 747.$$
3. Whatever is left belongs to the junior:
$$840 - 747 = 93.$$
4. All juniors got the same score, so each junior scored $93$. (With $20$ students, $2$ juniors, you get $1680 - 1494 = 186$, which is $93$ each again.)

The answer is $\boxed{\textbf{(C)}\ 93}$.

An average times the number of people gives the total, and totals can be added and subtracted while averages cannot. Choosing a convenient class size turns the percents into real people. Another way to see it: each of the $9$ seniors is $1$ point below $84$, so together they are $9$ points short, and the one junior must make up all $9$ points by scoring $84 + 9 = 93$.

Balance the points above and below $84$. Out of every $10$ students, $9$ seniors are each $1$ point below average, a total shortfall of $9$ points. The $1$ junior must be $9$ points above average to balance it, so the junior scored $84 + 9 = 93$.

The class size does not matter, so suppose there are $10$ students: $1$ junior and $9$ seniors.

The class total is $10 \cdot 84 = 840$ points. The seniors contribute $9 \cdot 83 = 747$ points. The junior's score is the rest:
$$840 - 747 = 93.$$

Since every junior scored the same, each junior received $93$.

The answer is $\boxed{\textbf{(C)}\ 93}$.