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

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==Solution 2 (Don't need quadratic equation)==
 
==Solution 2 (Don't need quadratic equation)==
  
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Consider the solutions, there are two more boys than girls, so if there are 26 students, we have 14 boys and 12 girls. <math>4^2+2^2</math> is end by <math>0</math>, so it is not the answer we need.
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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
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Try all of them and we will get the answer is <math>\boxed{\textbf{(B)}\ 28}</math>.
 
  
 
==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}}

Revision as of 12:43, 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

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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