Problem 67 - Algebra and Trigonometry

Question

At the north campus of a performing arts school, 10% of the students are music majors. At the south campus, 90% of the students 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 north and south campuses have before the merger?

Step-by-Step Solution

Answer

The north campus had 680 students and the south campus had 320 students prior to the merger.

Step 1: Define the variables

Let's represent the number of students from the north campus by the variable \(n\) and the number of students from the south campus by \(s\). The total number of students is 1000.

Step 2: Setup the weighted average equation

Using the conditions given: 10% of north campus students and 90% of south campus students are music majors. Also, 42% of the total merged students are music majors. Therefore we can form the equation: \(0.10n + 0.90s = 0.42 * 1000\).

Step 3: Use the total number of student equation

We also know that the total number of students from both the campuses equals 1000, hence \(n + s = 1000\).

Step 4: Solve the equations

We can solve these equations together to find the values of \(n\) and \(s\). Multiply the equation \(n + s = 1000\) with 0.10 so that we can subtract from the first equation. This gives us \(0.10n + 0.10s = 100\).

Step 5: Subtract equations to get the value of s

Now, subtract \(0.10n + 0.10s = 100\) from \(0.10n + 0.90s = 420\). This will give you \(s = 320\).

Step 6: Substitute the value of s in the second equation

Put \(s = 320\) in \(n + s = 1000\) yields \(n = 680\).

Key Concepts

Systems of Equations

When confronted with problems involving multiple quantities and conditions, systems of equations become a valuable tool. In this scenario, two unknowns exist: the number of students from the north campus and the number from the south campus. A system of equations allows us to solve for these unknowns using predefined relationships.

• First Equation: Relates to the percentage of music majors in the merged campus.
• Second Equation: Represents the total number of students.

Both equations are interconnected, providing distinct information about the same set of populations. By solving them together, we leverage these connections and derive specific values for each unknown variable.

Percentage Calculation

Percentage calculations are ubiquitous in everyday scenarios, such as determining discounts or calculating grades. In this exercise, percentages inform us about the proportion of music majors on each campus.

• North Campus: 10% are music majors.
• South Campus: 90% are music majors.
• Merged East Campus: 42% of 1000 students are music majors.

These percentages instruct us on how to set up our equations to reflect real-world conditions accurately. Understanding these percentages is crucial to setting up the problem correctly and interpreting the results meaningfully.

Problem Solving

Problem-solving requires a systematic approach, especially in mathematical puzzles. This problem begins with understanding what is asked: to find the number of students on each campus before the merger.
1. Identify Known & Unknown Variables: We started by representing the number of students from each campus with variables. 2. Translate Problem Data into Equations: Based on given percentages, our system of equations is crafted. 3. Solve Step-by-Step: With a clear strategy, like using substitution or elimination, we zero in on the solution. Approaching problems methodically minimizes errors and offers a clearer pathway to your answers.

Algebraic Equations

Algebraic equations are the backbone for representing and solving real-world problems. They condense complex relationships into manageable mathematical expressions.

• In our exercise, the first equation, \(0.10n + 0.90s = 420\), reflects the distribution of music majors.
• The second equation, \(n + s = 1000\), captures the total number of students from combined campuses.

Algebraic operations like addition, subtraction, and substitution allow us to solve these equations efficiently. This process not only yields precise answers but also strengthens comprehension of the scenario at hand.