How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Quadratic equation Problem | AMC 8, 2009 | Problem 23

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

Factorization

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

# Knowledge Partner

### Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.