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Algebra AMC 8 Math Olympiad USA Math Olympiad

Quadratic equation Problem | AMC 8, 2009 | Problem 23

Try this beautiful problem from Algebra based on Quadratic equation fro AMC-8(2009) problem no 23.You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra based on Quadratic equation.

Algebra based on Quadratic equation – AMC-8, 2009 – Problem 23


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  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?

  • $34$
  • $28$
  • $25$

Key Concepts


Algebra

quadratic equation

Factorization

Check the Answer


Answer:$28$

AMC-8 (2009) Problem 23

Pre College Mathematics

Try with Hints


Let the number of girls be x

so the number of boys be x+2

Can you now finish the problem ……….

She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

can you finish the problem……..

Let the number of girls be x

so the number of boys be x+2

she gave each girl x jellybeans and each boy x+2 jellybeans,

Therefore she gave total number of  jelly beans to girls be \(x^2\)

Therefore she gave total number of  jelly beans to boys be \((x+2)^2\)

 She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

\(\Rightarrow x^2 + x^2 +4x+4=394\)

\(\Rightarrow 2x^2 +4x-390=0\)

\(\Rightarrow x^2 +2x -195=0\)

\((x+15)(x-13)=0\)

i.e x=-15 , 13 (we neglect negetive as number of students can not be negetive )

Therefore x=13 i.e number os girls be 13 and number of boys be 13+2=15

total number of students ne 13+15=28

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