# Difference between revisions of "1999 AMC 8 Problems/Problem 18"

## Problem

At Central Middle School the $108$ students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of $15$ cookies, lists this items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.

They learn that a big concert is scheduled for the same night and attendance will be down $25\%$. How many recipes of cookies should they make for their smaller party?

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$

## Solution

### Solution 1

If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. But with the smaller attendance, you will only need $100\% - 25\% = 75\%$ of these cookies, or $75\% \cdot 216 = 0.75\cdot 216 = 162$ cookies.

$162$ cookies requires $\frac{162}{15} = 10.8$ batches. However, since half-batches are forbidden, we must round up to get $\left\lceil \frac{162}{15} \right\rceil = 11$ batches, and the correct answer is $\boxed{E}$.

### Solution 2

If there were $108$ students before, with the $25\%$ of students missing, there will be $75\%$ of $108$ students left. This is $75\% \cdot 108 = 0.75 \cdot 108 = 81$ students. These students eat $81 \cdot 2 = 162$ cookies. Follow the logic of the second paragraph above to find that there needs to be $11$ batches, and the correct answer is $\boxed{E}$.

 1999 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions