2010 AMC 10B Problem 12

Percents are easier to picture with a class of $100$ students, so $x\%$ is just the number of students who changed.

• Start: $50$ say Yes, $50$ say No.
• End: $70$ say Yes, $30$ say No.

1. Let $v$ be the number who switch from Yes to No. The Yes group loses those $v$ students and still has to grow from $50$ to $70$, so $v + 20$ students must switch from No to Yes.
2. The total number who changed is
$$x = v + (v + 20) = 2v + 20.$$
So $x$ is smallest when $v$ is smallest and largest when $v$ is largest.

Smallest $x$.

1. Take $v = 0$: nobody leaves Yes, and $20$ students move from No to Yes. Then $x = 20$. This clearly works, and $x$ cannot be less, because $v$ cannot be negative.

Largest $x$.

1. Everyone who switches to No ends up in the final No group, which has only $30$ students. So $v \le 30$.
2. Everyone who switches to Yes started in the No group, which has $50$ students. So $v + 20 \le 50$, which again says $v \le 30$.
3. Take $v = 30$: all $50$ No students switch to Yes, and $30$ Yes students switch to No. Check: Yes ends with $50 - 30 + 50 = 70$, and No ends with $30$. Then $x = 2 \cdot 30 + 20 = 80$.

The difference between the largest and smallest values is $80 - 20 = 60$.

The answer is $\boxed{\textbf{(D)}\ 60}$.

The same overall change (Yes up by $20$) can come from very different amounts of switching, because students can switch in both directions and cancel each other out. Pick one of the two flows as your unknown, write everything in terms of it, and then find how big and how small it can be. For the largest value, check every group the switchers leave and every group they join.

Think about the two extreme stories directly. Least switching: only the $20$ needed students move from No to Yes, so $x = 20$. Most switching: all $50$ No students become Yes, and then $30$ Yes students must become No to bring Yes back down to $70$. That is $50 + 30 = 80$. The difference is $60$.