2016 AMC 10A Problem 7

Of mean, median and mode, only the mean gives an equation directly, because the median and mode depend on where $x$ lands. So start with the mean.

1. Add the six known values: $60 + 100 + 40 + 50 + 200 + 90 = 540$.
2. With $x$ included there are $7$ values with total $540 + x$. The mean is $x$, so the total must be $7$ times $x$:
$$540 + x = 7x.$$
3. Take $x$ from both sides: $540 = 6x$, so $x = 90$.
4. Check the median. Put the seven values in order: $40, 50, 60, 90, 90, 100, 200$. The middle (4th) value is $90$. Good.
5. Check the mode. The value $90$ appears twice and every other value once, so the mode is $90$. Good.

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

When several conditions must all hold, use the one that gives a clean equation first, then test the answer against the rest. Here the mean fixes $x$, and the median and mode are quick checks. Remember that $x$ is part of the data, so it belongs in the total.

Start from the mode instead. Every listed value appears once, so for $x$ to be the mode it must equal one of them: $40, 50, 60, 90, 100$ or $200$. The median must also be $x$, so $x$ has to land in the middle of the ordered list, which only $60$ and $90$ can do. Now test the mean: with $x = 60$ the total is $600$, not $7 \cdot 60 = 420$; with $x = 90$ the total is $630 = 7 \cdot 90$. So $x = 90$.

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 $\boxed{\textbf{(D)}\ 90}$.