Difference between revisions of "2009 AMC 8 Problems/Problem 23"

Revision as of 12:44, 7 September 2024

Problem

On the last day of school, Mrs. Wonderful 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 1

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.

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

From here, we can see that $x = 13$ as $13^2 + 26 = 195$ , so there are $13$ girls, $13+2=15$ boys, and $13+15=\boxed{\textbf{(B)}\ 28}$ students.

Solution 2 (Don't need quadratic equation)

Consider the solutions, there are two more boys than girls, so if there are 26 students, we have 14 boys and 12 girls. $4^2+2^2 according to bglah blah hkjlhkjhkjlhkjh

@@ Line 20: / Line 20: @@
 Consider the solutions, there are two more boys than girls, so if there are 26 students, we have 14 boys and 12 girls. $4^2+2^2 according to bglah blah hkjlhkjhkjlhkjh
-==See Also==
-{{AMC8 box|year=2009|num-b=22|num-a=24}}
-{{MAA Notice}}

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Difference between revisions of "2009 AMC 8 Problems/Problem 23"

Revision as of 12:44, 7 September 2024

Problem

Solution 1

Solution 2 (Don't need quadratic equation)