Problem 49 - Elementary Algebra

Question

For Problems \(33-50\), set up an equation and solve the problem. (Objective 2 ) In a survivor competition, the Pachena tribe can shuck 300 oysters in 10 minutes less time than it takes the Tchaika tribe. If the Pachena tribe shucks oysters at a rate of 5 oysters per minute faster than the Tchaika tribe, find the rate of each tribe.

Step-by-Step Solution

Answer

The Tchaika tribe shucks oysters at 10 oysters/min, and the Pachena tribe at 15 oysters/min.

Step 1: Understand the given information

First, determine what we know from the problem. The Pachena tribe can shuck 300 oysters in 10 minutes less than the Tchaika tribe. They shuck oysters 5 oysters per minute faster than the Tchaika tribe. Let's denote the rate of shucking for the Tchaika tribe as \( r \) oysters per minute. Thus, the rate for the Pachena tribe will be \( r + 5 \) oysters per minute.

Step 2: Set up equations based on time

The time it takes for the Tchaika tribe to shuck 300 oysters is \( \frac{300}{r} \) minutes. The time it takes for the Pachena tribe to shuck 300 oysters is \( \frac{300}{r+5} \) minutes. According to the problem, the Pachena tribe takes 10 minutes less, so we have the equation:\[ \frac{300}{r} = \frac{300}{r+5} + 10 \]

Step 3: Solve the equation for the rate

Clear the fractions by multiplying every term by \( r(r+5) \):\[ 300(r+5) = 300r + 10r(r+5) \]Expand and simplify the equation:\[ 300r + 1500 = 300r + 10r^2 + 50r \]Combine like terms:\[ 10r^2 + 50r + 1500 = 0 \]

Step 4: Simplify and factor the quadratic equation

Divide the entire equation by 10 to simplify:\[ r^2 + 5r - 150 = 0 \]Factor the quadratic equation:\( (r - 10)(r + 15) = 0 \).

Step 5: Find the possible rates and interpret the results

The solutions for \( r \) are \( r = 10 \) and \( r = -15 \). A negative rate doesn't make sense in this context, so we use \( r = 10 \). Therefore, the Tchaika tribe's rate is 10 oysters per minute, and the Pachena tribe's rate is \( 10 + 5 = 15 \) oysters per minute.

Key Concepts

Rate and Speed Problems

Algebra word problems often involve rate and speed, which are crucial in solving real-world scenarios. In our problem, the rates of two tribes shucking oysters are the variables at play.
The Tchaika tribe shucks oysters at a rate we denote as \(r\) oysters per minute. The Pachena tribe shucks at a rate of \(r + 5\) oysters per minute. These rates give us insights into the speed and efficiency of each tribe.
When dealing with rate and speed problems, it's essential to remember these common mathematical relationships:

• Rate: The number of items processed per unit of time.
• Time: The total duration taken to process a set number of items is calculated as \( \text{Time} = \frac{\text{Total Items}}{\text{Rate}} \).
• Rate Differentials: When comparing two parties, like the tribes, calculating the difference in their rates is critical for solving the problem.

A vital step in rate problems is establishing the relationship between time, speed, and quantities. It's also crucial to frame equations based on these relationships, allowing for solving unknowns effectively.

Quadratic Equations

Quadratic equations are a staple in algebra, revealing how different variables relate to each other. They typically take the form \( ax^2 + bx + c = 0 \). In our exercise, the equation we derived was: \[ 10r^2 + 50r + 1500 = 0 \]
Quadratics are based on the notion of polynomial equations with a degree of two. Solving them helps predict various real-world scenarios, like rates in this case.
Understanding quadratics involves knowing:

• Terms and Coefficient: These include the square term, linear term, and constant.
• Solutions or roots: Where the function values equal zero.
• Graphical representation: They form parabolas on graphs, and solutions intersect the x-axis.

Sometimes, these equations yield two solutions. Interpretation is essential because not all solutions apply logically to the problem context.

Problem Solving Steps

Effective problem solving requires a structured approach, especially for algebraic word problems. Here's a breakdown based on the exercise:
1. **Understand the Problem:** Grasp all the details, such as what the rates represent, and define your variables. In our example, we constructed the rates \(r\) and \(r + 5\).
2. **Translate Words into Equations:** Transform worded problems into mathematical expressions. The challenge was to equate time differences based on rates.3. **Simplify and Solve the Equation:** This involved clearing fractions and reducing the quadratic equation.
4. **Interpret the Results:** Check which solutions are logical based on the problem's context. For our problem, only a positive rate made sense.
These steps ensure clarity and accuracy while solving math problems and help you manage complex data or conditions effectively.

Factoring Quadratics

Factoring is an efficient method to solve quadratic equations. It involves breaking down the equation into two simpler expressions (binomials) that multiply to give the original quadratic equation.
In our exercise, we simplified the quadratic to \(r^2 + 5r - 150 = 0\) and found its factors as \((r - 10)(r + 15) = 0\).
This step shows how assuming solutions from the factored form reveals the possible values for \(r\). Factoring quadratics effectively can dramatically save time and simplify calculations.
Here's how to factor a quadratic:

• Look for two numbers that multiply to the constant term and add to the linear coefficient.
• Express the quadratic as a product of these numbers and their respective variable parts.
• Solve each factor separately to find the solutions.

This method turns a potentially complex quadratic equation into manageable parts, allowing you to identify viable answers quickly.