• May 29th 2008, 05:49 PM
jasmine7
I really need help....
Could anyone explain how to do the problem because my math teacher didn't even explain how to do this kind of problem at school.

The Computer Club is going on a field trip to a computer museum. The trip costs $240, to be paid for equally by each club member. The day before the trip, however, four students decide not to go. This increases the cost of the trip by$2 for each remaining student. How many students are going to the computer museum?
• May 29th 2008, 07:40 PM
TheEmptySet
Quote:

Originally Posted by jasmine7
I really need help....
Could anyone explain how to do the problem because my math teacher didn't even explain how to do this kind of problem at school.

The Computer Club is going on a field trip to a computer museum. The trip costs $240, to be paid for equally by each club member. The day before the trip, however, four students decide not to go. This increases the cost of the trip by$2 for each remaining student. How many students are going to the computer museum?

Let S be the number of students going on the trip, and let C be the cost per student.

So then we can get one equation

$S\cdot C=240$

Now we can use the rest of the info to write another equation

we know that four less students are going so thats S-4 and that it will cost two dollars more per student C+2 and this still needs to equal the cost of the trip. So we get

$(S-4)(C+2)=240 \iff SC +2S-4C-8=240$

Now we can isolate a variable from the first equation to get

$S=\frac{240}{C}$ and sub this into the 2nd equationt to get

$\frac{240}{C}\cdot C +2\cdot \frac{240}{C}-4C-8=240$

Reducing gives

$240 + \frac{480}{C}-4C=248$

Now multiply the equation by C to clear the fractions and get

$240C+480-4C^2=248C \iff 0=4C^2+8C-480$

$0=4[C^2+2C-120] \iff 0=4(C-10)(C+12)$

So C is 10 or -12 Since the cost shouldn't be negative. I wish someone would pay me to go to a museum :D C is 10. Plugging back into the first equation we get S=24.

I hope this helps. Good luck.

(Rock)