2000 AMC 10 Problem 3

1. Each day Jenny eats $20\%$ of what is in the jar that morning, so $80\%$ stays. And $80\% = \tfrac45$.
2. The easiest way is to go backward from the end. At the end of day 2 there are $32$ beans. These are $\tfrac45$ of what was there at the start of day 2.
3. If $\tfrac45$ of a number is $32$, then $\tfrac15$ of it is $8$, and the whole number is $5 \cdot 8 = 40$. So day 2 started with $40$ beans.
4. Do the same step again. $40$ is $\tfrac45$ of the original amount. Then $\tfrac15$ of the original is $10$, and the original is $50$.
5. Check going forward: $20\%$ of $50$ is $10$, leaving $40$. Then $20\%$ of $40$ is $8$, leaving $32$. It works.

The answer is $\boxed{\textbf{(B)}\ 50}$.

"Eat $20\%$ of what is there" means multiply by $\tfrac45$ each day. The second day's $20\%$ is taken from a smaller pile, so it is fewer beans than the first day's. To undo "multiply by $\tfrac45$" you divide by $\tfrac45$, which is the same as multiplying by $\tfrac54$. Adding $20\%$ back does not undo it.

Use algebra. Call the original amount $x$. After two days there are $\tfrac45 \cdot \tfrac45 \cdot x = \tfrac{16}{25}x$ beans. Setting $\tfrac{16}{25}x = 32$ gives $x = 32 \cdot \tfrac{25}{16} = 50$.

Each day Jenny keeps $100\% - 20\% = 80\% = \tfrac45$ of whatever is in the jar that morning. Two days in a row multiply the original count $x$ by $\tfrac45$ twice:
$$\left(\tfrac45\right)^2 x = \tfrac{16}{25}\,x = 32.$$
So $x = 32 \cdot \tfrac{25}{16} = 2 \cdot 25 = 50$.

Check: $50 \to 40 \to 32$.

The answer is $\boxed{\textbf{(B)}\ 50}$.