2009 AMC 8 Problem 23

Below is the professionally curated solution for Problem 23 of the 2009 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 8 solutions, or check the answer key.

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Concepts:quadraticfactoring

Difficulty rating: 1600

23.

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 400400 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?

2626

2828

3030

3232

3434

Solution:

Let bb be the number of boys in the class and gg be the number of girls. From the problem, we get that b=g+2.b = g + 2.

If each boy gets bb jelly beans, then Mrs. Wonderful will give out a total of b2b^2 jelly beans to all the boys. Similarly, she will give out g2g^2 jelly beans to all the girls.

Therefore, b2+g2=4006(g+2)2+g2=3942g2+4g+4=394g2+2g195=0(g+15)(g13)=0. \begin{gather*} b^2 + g^2 = 400 - 6 \\ (g + 2)^2 + g^2 = 394 \\ 2g^2 + 4g + 4 = 394 \\ g^2 + 2g - 195 = 0 \\ (g + 15)(g - 13) = 0. \end{gather*} Since gg cannot be negative, we get that g=13.g = 13. This means that b=15,b = 15, so b+g=28.b + g = 28.

Thus, B is the correct answer.

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