# Dividing Fare for Jeepney

One of the topics in Mathematics that has a huge application is quadratic equations. From projectile motion, prize optimization, profit estimates, solving speed, age problems, and more.

Today, let’s talk about one specific problem that has a direct application of quadratic function. This problem is usually encountered in entrance exams for Science high schools, UPCAT, Math challenge, civil service exam, and just when you need it in real life scenario.

Problem:

A group of people has a Halloween trip to the scariest Museum in Spookyland,  the bus driver asked them to pay $300.00 for the entire trip. 5 backed out before the trip and the remaining passengers would have to pay$10 more. How many people were in the group originally?

Solution:

Let x: be the number of people originally in the group.

Let y: be the original amount of money the passengers would pay.

The original amount of money the people should pay together can be expressed as,

$y=\displaystyle\frac{300}{x}$ ♥

Without the 5 that backed out, the number of people which is x will become x-5. The amount of money the passengers left should pay will increase from y to y+10. Constructing that to an equation we have,

$y+10=\displaystyle\frac{300}{x-5}$ ♥♥

The question asked is number of people originally or x. By substituting  ♥ to ♥♥, we have

$y+10=\displaystyle\frac{300}{x-5}$

$\displaystyle\frac{300}{x}+10=\displaystyle\frac{300}{x-5}$

Multiplying  $x(x-5)$ to both sides of equation we have,

$x(x-5)(\displaystyle\frac{300}{x}+10)=(\displaystyle\frac{300}{x-5})x(x-5)$

$300(x-5)+10x(x-5)=300x$

$300x-1500+10x^2-50x=300x$

$10x^2-50x-1500=0$

$x^2-5x-150=0$

$(x-15)(x+10)=0$

$x_1=15,x_2=-10$

x=15 is the only solution, -10 is just an extraneous root.

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.