Difference between revisions of "2009 AMC 8 Problems/Problem 23"
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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 | + | 2x^2 + 4x &= 390\\ |
| − | x^2 + 2x | + | x^2 + 2x &= 195\\ |
| − | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | + | 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}} | ||
Revision as of 02:18, 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 
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?
Solution
If there are 
girls, then there are 
boys. She gave each girl jellybeans and each boy jellybeans, for a total of 
jellybeans. She gave away 
jellybeans.
From here, we can see that 
as 
, so there are 
girls, 
boys, and 
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. 
