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

(Solution)
m (Solution (Does not need quadratic equation))
 
Line 11: Line 11:
 
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\\
2x^2 + 4x - 390 &= 0\\
+
2x^2 + 4x &= 390\\
x^2 + 2x - 195 &= 0\\
+
x^2 + 2x &= 195\\
(x+15)(x-13) &=0
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
Because <math>x=-15,13</math>, and there cannot be a negative amount of students, there are <math>13</math> girls, <math>13+2=15</math> boys, and <math>13+15=\boxed{\textbf{(B)}\ 28}</math> students.
+
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 03:18, 31 October 2020

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\\ 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 (ProblemsAnswer KeyResources)
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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