Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Problem)
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
==Problem==
 
==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 <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?  
+
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. She had 6 jolly ranchers and howe rejaaaaaaaa vfhakdjh fs;dfHow many studenfffffffdats were in ffffffffffffffffdsfher class?ffffffffffffffffffff
  
 
<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>
Line 19: Line 19:
 
==Solution 2 (Don't need quadratic equation)==
 
==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. <math>4^2+2^2</math> is end by <math>0</math>, so it is not the answer we need.
+
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
−
 
 
−
Try all of them and we will get the answer is <math>\boxed{\textbf{(B)}\ 28}</math>.
 
−
 
 
−
==See Also==
 
−
{{AMC8 box|year=2009|num-b=22|num-a=24}}
 
−
{{MAA Notice}}
 

Revision as of 13:44, 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. She had 6 jolly ranchers and howe rejaaaaaaaa vfhakdjh fs;dfHow many studenfffffffdats were in ffffffffffffffffdsfher class?ffffffffffffffffffff

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_23&oldid=226707"