2009 AMC 8 Problems/Problem 23

Problem

On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

Solution

If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans.

$\begin{align*} x^2+(x+2)^2 &= 394\\ x^2+x^2+4x+4 &= 394\\ 2x^2 + 4x - 390 &= 0\\ x^2 + 2x - 195 &= 0\\ (x+15)(x-13) &=0 \end{align*}$

Because $x=-15,13$ , and there cannot be a negative amount of students, there are $13$ girls, $13+2=15$ boys, and $13+15=\boxed{\textbf{(B)}\ 28}$ students.

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2009 AMC 8 Problems/Problem 23

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