2016 AMC 10A Problem 7

The six known values sum to $60 + 100 + 40 + 50 + 200 + 90 = 540$. The mean of all seven values is $x$, so
$$\frac{540 + x}{7} = x \implies 540 + x = 7x \implies 6x = 540 \implies x = 90.$$

Verify the other two conditions with $x = 90$. Sorted, the data are $40, 50, 60, 90, 90, 100, 200$: the fourth value (median) is $90$, and $90$ is the only value appearing twice, so it is the mode. Everything is consistent.

The answer is $\boxed{\textbf{(D)}\ 90}$.

Of the three statistics, the mean is the only one that yields an equation directly in $x$; median and mode depend on where $x$ lands. Solve with the mean, then confirm with the others. The mode condition is automatically satisfied by any $x$ equal to one of the existing values, and the median needs $x$ to sit in the middle of the sorted list, which $90$ does.

The mode must be one of the listed values since $x$ appears only once otherwise, so $x \in \{40, 50, 60, 90, 100, 200\}$; the median of the seven values must be the middle one, which forces $x$ to be $60$ or $90$. Only $90$ makes the total $630 = 7 \cdot 90$.

• Solving (540 + x)/7 = x incorrectly as 540 = 7x, or averaging only the six known values and getting 90 by luck without verifying.
• Assuming the median is the average of the middle two values (that is for an even count); with seven values it is the fourth.