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Difference between revisions of "2009 AMC 8 Problems/Problem 23"
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<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>
 +
 +
==Solution==
 +
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*}
 +
x^2+(x+2)^2 &= 394\\
 +
x^2+x^2+4x+4 &= 394\\
 +
2x^2 + 4x - 390 &= 0\\
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x^2 + 2x - 195 &= 0\\
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(x+15)(x-13) &=0
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\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.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=22|num-a=24}}
 
{{AMC8 box|year=2009|num-b=22|num-a=24}}

Revision as of 16:42, 25 December 2012

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

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 $x=-15,13$, there are $13$ girls, $15$ 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
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