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ycow [4]
3 years ago
15

At the north campus of a performing arts​ school, ​20% of the students are music majors. At the south​ campus, 70​% of the stude

nts 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?
Mathematics
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

The north campus had 800 while the south campus had 200.

Step-by-step explanation:

The question stated that there are 1000 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 x denote the number of students in the north campus before the merger. The other (1000 -x) students would all belong to the south campus before the merger.

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

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

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

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:

0.20\, x+ (700 - 0.70\, x) = 300.

Solve for x:

x = 800.

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

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