Problem 50 - Algebra and Trigonometry

Question

Classroom Use A small school has 100 students who occupy three classrooms: A, B, and C. After the first period of the school day, half the students in room A move to room B, one-fifth of the students in room B move to room C, and one- third of the students in room C move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room?

Step-by-Step Solution

Answer

Rooms A, B, and C have 20, 50, and 30 students, respectively.

Step 1: Identify Variables

Let the number of students in rooms A, B, and C initially be \( a \), \( b \), and \( c \) respectively. We know that \( a + b + c = 100 \).

Step 2: Apply Changes after First Period

After the first period, half the students from room A move to room B, leaving \( \frac{a}{2} \) in room A and increasing those in B to \( b + \frac{a}{2} \). One-fifth of students from room B (\( \frac{b}{5} \)) move to room C, so room B now has \( b + \frac{a}{2} - \frac{b}{5} \) students and room C has \( c + \frac{b}{5} \). Lastly, one-third of students from room C move to room A, leaving \( c - \frac{c}{3} \) in C and moving \( \frac{c}{3} \) to room A.

Step 3: Set Equations for Unchanged Room Number

Since the number of students in each room remains the same after these movements, set up the equations:1. \( \frac{a}{2} + \frac{c}{3} = a \)2. \( b + \frac{a}{2} - \frac{b}{5} = b \)3. \( c + \frac{b}{5} = c \)

Step 4: Solve for a, b, and c

Solve each equation:1. Simplifying \( \frac{a}{2} + \frac{c}{3} = a \) gives: \( \frac{c}{3} = \frac{a}{2} \) leading to \( c = \frac{3a}{2} \).2. The equation \( \frac{a}{2} - \frac{b}{5} = 0 \) simplifies to \( b = \frac{5a}{2} \).3. The equation \( \frac{b}{5} = 0 \) simplifies directly to give \( c = b \).Using these relationships in conjunction with \( a + b + c = 100 \): Replace \( c \) with \( b \) and solve: \( a + \frac{5a}{2} + \frac{3a}{2} = 100 \). Simplifying yields \( 5a = 100 \), which gives \( a = 20 \).From \( b = \frac{5a}{2} \), we find \( b = 50 \), hence \( c = 30 \).

Step 5: Conclusion

The final count of students in each room is \( a = 20 \) in room A, \( b = 50 \) in room B, and \( c = 30 \) in room C.

Key Concepts

Linear Equations

Linear equations are fundamental in understanding various mathematical problems, including the Classroom Arrangement problem. A linear equation is an algebraic equation where each term is either a constant or the product of a constant and a single variable. When graphed, these equations form a straight line. Linear equations have the form of\[ ax + by + cz = d \]In the Classroom Arrangement problem, we use linear equations to represent the total number of students before and after the movements. Each variable (\(a\), \(b\), \(c\)) represents the number of students in rooms A, B, and C, respectively. Our initial equation \(a + b + c = 100\) serves as a constraint showing that the total student count remains constant.Linear equations help us set relationships between variables. For instance, if half of the students in room A move to room B, this change needs to be accounted for in the equations, keeping the total constant. The balance ensures that the number of students doesn’t change after transitions, forming equations like \(\frac{a}{2} + \frac{c}{3} = a\). This setup helps solve for individual variable values by substitution or elimination methods.

Student Distribution

In managing a classroom, understanding student distribution is essential for optimal learning environments. Student distribution refers to how students are spread out across different areas. In the given classroom problem, students are distributed among three rooms: A, B, and C.Initially, we assume that the total distribution is known to be 100 students overall. This means the allocation across three rooms is represented by \(a, b, c\) such that \(a + b + c = 100\). After specific movements —such as half the students from room A moving to room B— the distribution changes temporarily. However, the condition that the total counts in each room remain unchanged after movements compels the need for careful calculation. This balance allows instructors to determine if equal or equitable resources and attention are being provided to each group or class within a school.Effective student distribution helps:- Manage class sizes efficiently- Provide balanced educational resources- Create equitable learning opportunitiesThis problem illustrates how mathematical reasoning with variable allocation can help make practical decisions in learning environments.

Problem Solving Steps

Breaking down a problem into smaller, manageable steps is a crucial strategy in problem solving. In the Classroom Arrangement problem, we utilize a step-by-step approach to achieve accurate results.

• Step 1 - Identify Variables: Define the variables (\(a\), \(b\), \(c\)) for the count of students in rooms A, B, and C.
• Step 2 - Apply Changes: Analyze changes after movements between rooms. Update values based on conditions given: half from A move to B, one-fifth from B to C, and one-third from C to A.
• Step 3 - Set Equations: Form equations to reflect that despite movements, room populations remain consistent.
• Step 4 - Solve Equations: Use algebraic methods to simplify and solve these equations, determining correct variables satisfying all conditions.
• Step 5 - Conclusion: Validate calculations logically. Establish that the calculated numbers uphold the initial problem constraints of consistent room populations.

These steps guide students through logical reasoning and ensure problems are dissected thoroughly to arrive at a systematic and reliable solution.