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Difference between revisions of "2009 AMC 8 Problems/Problem 23"
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==Problem==
 
==Problem==
  
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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 <math>400</math> 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?  
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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 <math>400</math> 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?
  
 
<math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 </math>
 
<math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 </math>
  
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==Solution==
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==Solution 1==
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IF there are <math>x</math> girls, then there are <math>x+2</math> boys. She gave each girl <math>x</math> jellybeans and each boy <math>x+2</math> jellybeans, for a total of <math>x^2 + (x+2)^2</math> jellybeans. She gave away <math>400-6=394</math> jellybeans.
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If there are <math>x</math> girls, then there are <math>x+2</math> boys. She gave each girl <math>x</math> jellybeans and each boy <math>x+2</math> jellybeans, for a total of <math>x^2 + (x+2)^2</math> jellybeans. She gave away <math>400-6=394</math> jellybeans.
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
x^2+(x+2)^2 &= 394\\
 
x^2+(x+2)^2 &= 394\\
 
x^2+x^2+4x+4 &= 394\\
 
x^2+x^2+4x+4 &= 394\\
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2x^2 + 4x - 390 &= 0\\
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2x^2 + 4x &= 390\\
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x^2 + 2x - 195 &= 0\\
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x^2 + 2x &= 195\\
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(x+15)(x-13) &=0
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
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Because <math>x=-15,13</math>, there are <math>13</math> girls, <math>15</math> boys, and <math>13+15=\boxed{\textbf{(B)}\ 28}</math> students.
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From here, we can see that <math>x = 13</math> as  <math>13^2 + 26 = 195</math>, so there are <math>13</math> girls, <math>13+2=15</math> boys, and <math>13+15=\boxed{\textbf{(B)}\ 28}</math> students.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=22|num-a=24}}
 
{{AMC8 box|year=2009|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:04, 20 October 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.

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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