# Thread: any help would be apprieciated

1. ## any help would be apprieciated

It has been 12 years since I have taken an algrebra class, it is taking a while to get back into it.

The question I have is: At the north campus of performing arts school, 10% of the students are Music majors. At the south campus, 90% of the student are Music majors. The campuses are merged into one east campus. If 42% of the 1000 students at the east campus are Music majors, How many students did the nort and saouth campuses have before the merger?

Thank you in advance if you are able to help!

2. 42% of 1000 are music majors. That's 420 people.

10% are from the north campus.

90% are from the south campus.

10% of 420 and 90% of 420.

3. ## doesn't check

yeah I have tried that method but the answer doesn't check out correctly.

4. Let x=students of north campus and y=students of south campus

42% of 1000 students = 420 students

We will have the equations :
x+y=1000

$\displaystyle \frac{10}{100}x+\frac{90}{100}y=420$ .

x=1000-y and replace it to the second equation, find y and then x.

We will have x=North campous students=600 and y=South campous students=400

5. Originally Posted by tiphanie
...
The question I have is: At the north campus of performing arts school, 10% of the students are Music majors. At the south campus, 90% of the student are Music majors. The campuses are merged into one east campus. If 42% of the 1000 students at the east campus are Music majors, How many students did the nort and saouth campuses have before the merger?

Thank you in advance if you are able to help!
Let x denote the number of students at the NC and y the number of students at the SC.

Then you know:

$\displaystyle \left| \begin{array}{rcl}x+y&=&1000\\ 0.1x+0.9y &=& 420\end{array}\right.$

Solve for x and y. I've got x = 600, y = 400

EDIT: ...too late - again!