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}} |
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 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 |
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