At the north campus of a performing arts​ school, ​20% of the students are music majors. At the south​ campus, 70​% of the students are music majors. The campuses are merged into one east campus. If ​30% of the 1000 students at the east campus are music​ majors, how many students did the north and south campuses have before the​ merger?

Respuesta :

Answer:

The north campus had [tex]800[/tex] while the south campus had [tex]200[/tex].

Step-by-step explanation:

The question stated that there are [tex]1000[/tex] students in the merged campus. That's the sum of the number of students in the north and the south campus before the merger.

Let [tex]x[/tex] denote the number of students in the north campus before the merger. The other [tex](1000 -x)[/tex] students would all belong to the south campus before the merger.

Before the merger, [tex]20\%[/tex] of the students of the north campus are of music majors. In terms of [tex]x[/tex], that's [tex](20\%)\cdot x = 0.20\,x[/tex] students.

On the other hand, [tex]70\%[/tex] of the students of the south campus are of music majors. That corresponds to [tex](70\%)\cdot (1000 - x) = 0.70\cdot (1000 - x) = 700 - 0.70\, x[/tex] students.

With a similar logic, the number of music students in the merged east campus will be [tex]30\% \times 1000 = 300[/tex].

The question implies that the sum of students in the two campuses should be equal to the number of music students in the east campus. That is:

[tex]0.20\, x+ (700 - 0.70\, x) = 300[/tex].

Solve for [tex]x[/tex]:

[tex]x = 800[/tex].

In other words, there are [tex]800[/tex] students in the north campus and [tex](1000 - 800) = 200[/tex] students in the south campus before the merger.

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