2010 AMC 10B Problem 12

Let $u\%$ of students switch from No to Yes and $v\%$ switch from Yes to No. The final Yes percentage is $50 + u - v = 70$, so $u = v + 20$. The total switching is
$$x = u + v = 2v + 20.$$

Minimum: $v = 0$ gives $x = 20$. This is achievable: $20\%$ of the original No group changes to Yes and everyone else stays put.

Maximum: $v$ is limited by the end-of-year No group, which is only $30\%$, so $v \le 30$. Also $u = v + 20 \le 50$ (the original No group), again $v \le 30$. Taking $v = 30$ gives $u = 50$: every original No becomes Yes, and $30\%$ of the original Yes group becomes No. Then $x = 80$.

The difference is $80 - 20 = 60$.

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

A net change of $20\%$ can be produced by many gross flows. Parametrizing by one flow ($v$) and finding its range from the group sizes gives the whole range of $x$. Always check both constraints (source group and destination group) and confirm that the extreme is actually attainable.

Think of the two extreme scenarios directly: the least disruption is $20\%$ switching No to Yes; the most is a full swap of the $50\%$ No group to Yes while the largest possible $30\%$ of the Yes group flips to No, for $80\%$.

• Assuming the answer is just the 20% net change, or letting more students switch Yes-to-No than the 30% who answer No at the end.
• Claiming a maximum of $100\%$ by having everyone switch, which would produce $50\%$ Yes at the end, not $70\%$.