Problem 36 - Beginning and Intermediate Algebra

Question

Solve each problem. Round answers to the nearest tenth as needed. Joel can work through a stack of invoices in 1 hr less time than Noel can. Working together they take \(1 \frac{1}{2} \mathrm{hr} .\) How long would it take each person working alone?

Step-by-Step Solution

Answer

Noel: 3.6 hours, Joel: 2.6 hours

Step 1: Define the Variables

Let the time it takes Noel to work through the stack of invoices alone be represented as \(t\) hours. Since Joel can finish the work 1 hour faster than Noel, Joel's time will be \(t - 1\) hours.

Step 2: Set Up the Equation

Since they can finish the work together in 1.5 hours (which is \(\frac{3}{2}\) hours), we can set up an equation using the work rates. Noel’s work rate is \(\frac{1}{t}\) and Joel’s work rate is \(\frac{1}{t - 1}\). Their combined work rate is \(\frac{1}{1.5} = \frac{2}{3}\). Therefore, the equation is:\[ \frac{1}{t} + \frac{1}{t-1} = \frac{2}{3} \]

Step 3: Solve the Equation

To solve the equation, first find a common denominator for the fractions on the left side, which is \(t(t-1)\).Multiply each term by \(t(t-1)\):\[ t(t-1) \times \frac{1}{t} + t(t-1) \times \frac{1}{t-1} = t(t-1) \times \frac{2}{3} \]This simplifies to:\[ (t-1) + t = \frac{2t(t-1)}{3} \]Combine like terms on the left side:\[ 2t - 1 = \frac{2t(t-1)}{3} \]

Step 4: Clear the Fraction

Multiply both sides by 3 to clear the fraction:\[ 3(2t - 1) = 2t(t-1) \]This expands to:\[ 6t - 3 = 2t^2 - 2t \]

Step 5: Form a Quadratic Equation

Rearrange the equation to form a quadratic equation:\[ 2t^2 - 8t + 3 = 0 \]

Step 6: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula \( t = \frac{-b \text{ ± } \text{√}(b^2 - 4ac)}{2a} \), where \(a = 2\), \(b = -8\), and \(c = 3\):\[ t = \frac{8 \text{ ± } \text{√}((-8)^2 - 4 \times 2 \times 3)}{2 \times 2} \]This simplifies to:\[ t = \frac{8 \text{ ± } \text{√}(64 - 24)}{4} \]\[ t = \frac{8 \text{ ± } \text{√}40}{4} \]\[ t = \frac{8 \text{ ± } 2\text{√}10}{4} \]\[ t = 2 \text{ ± } 0.5\text{√}10 \]

Step 7: Calculate the Time

Evaluate both possibilities for \(t\):\[ t_1 = 2 \text{ + } 0.5 \text{√} 10 \approx 3.6 \]\[ t_2 = 2 \text{ - } 0.5 \text{√} 10 \approx 0.4 \]Since Noel working alone in 0.4 hours is not realistic, the valid solution is \( t \text{≈ } 3.6\). Therefore, it takes Noel approximately 3.6 hours to complete the work alone. For Joel, use \( t - 1 \approx 3.6 - 1 \approx 2.6\). Joel can complete the work alone in 2.6 hours.

Key Concepts

Solving Quadratic Equations

Quadratic equations are a fundamental part of algebra. They are mathematical expressions that can be written in the form \[ ax^2 + bx + c = 0 \]. These equations can be solved using several methods, including factoring, completing the square, and the quadratic formula.
In our problem, the quadratic equation formed is \[ 2t^2 - 8t + 3 = 0 \].
We solve it using the quadratic formula:
\[ t = \frac{-b \text{ ± } \text{√}(b^2 - 4ac)}{2a} \].
Here, \(a = 2\), \(b = -8\), and \(c = 3\). Substitute these values into the formula:
\[ t = \frac{8 \text{ ± } \text{√}(64 - 24)}{4} \], \[ t = \frac{8 \text{ ± } 2\text{√}10}{4} \].
This simplifies to \[ t = 2 \text{ ± } 0.5\text{√}10 \].
Evaluate both possibilities for \(t\):
\[ t_1 = 2 \text{ + } 0.5 \text{√}10 \approx 3.6 \],
\[ t_2 = 2 \text{ - } 0.5 \text{√}10 \approx 0.4 \].
Since 0.4 hours doesn't make sense for the problem context, we discard it. Thus, \( t \approx 3.6\).
Understanding how to use the quadratic formula effectively is crucial in solving many algebraic problems.

Time and Work

Time and work problems are common in algebra because they deal with how individuals or systems complete tasks. The main idea is to figure out the rate at which work is done.
Here's the breakdown of our variables:

• Let \( t \) be the time Noel needs to complete a task alone.
• Joel can do the same work in \( t - 1 \) hours.
• Together, they can complete the work in 1.5 hours.

Rate problems often boil down to understanding the rate per unit of time. Noel's rate is \( \frac{1}{t} \), while Joel's rate is \( \frac{1}{t-1} \).
When working together:
\[ \frac{1}{t} + \frac{1}{t-1} = \frac{2}{3} \].
Solving such problems involves setting up the correct equation and solving it accurately, usually leading to a quadratic equation.

Algebraic Equations

Algebraic equations are mathematical statements that use variables, constants, and arithmetic operations. The key feature of these equations is that they show equality, allowing us to find unknown values.
Consider our problem:
The combined work equation is \[ \frac{1}{t} + \frac{1}{t-1} = \frac{2}{3} \].
To eliminate fractions, find a common denominator, which in this case, is \( t(t-1) \). This transforms the equation:
\[ t(t-1) \times \frac{1}{t} + t(t-1) \times \frac{1}{t-1} = t(t-1) \times \frac{2}{3} \],
simplifying to:
\[ t-1 + t = \frac{2t(t-1)}{3} \],
\[ 2t - 1 = \frac{2t(t-1)}{3} \].
Clearing fractions by multiplying by 3 gives:
\[ 6t - 3 = 2t^2 - 2t \].
Rearranging terms, we get the quadratic equation:
\[ 2t^2 - 8t + 3 = 0 \].
Understanding how to manipulate and solve algebraic equations is vital in mathematics.

Educational Mathematics

Studies in educational mathematics focus on enhancing learning and teaching. The goal is to help students understand concepts deeply and apply them effectively. This exercise integrates various mathematical concepts ensuring well-rounded learning.
Some reasons why solving such problems is valuable:

• It combines real-world scenarios with mathematical theory.
• It helps students enhance problem-solving skills.
• It teaches how to break complex problems into manageable steps.

Educational mathematics aims not only at helping students to solve problems but also to understand the underlying principles. This knowledge aids in tackling more advanced concepts as they progress in their studies.