Problem 45 - Elementary Algebra

Question

Jerry can lay a tile floor in 3 hours less time than Jake. If they work together, the floor takes 2 hours. How long would it take Jerry to lay the floor by himself?

Step-by-Step Solution

Answer

Jerry can lay the floor by himself in 3 hours.

Step 1: Define Variables

Let \( t \) be the time it takes Jake to lay the floor by himself. Hence, Jerry can complete the job in \( t - 3 \) hours because he works 3 hours faster than Jake.

Step 2: Express as Rates

Jake's rate of work is \( \frac{1}{t} \) floors per hour, and Jerry's rate of work is \( \frac{1}{t-3} \) floors per hour.

Step 3: Combine Rates

Working together, their combined rate is \( \frac{1}{t} + \frac{1}{t-3} = \frac{1}{2} \) since they complete the floor in 2 hours.

Step 4: Set Up Equation

The equation becomes \( \frac{1}{t} + \frac{1}{t-3} = \frac{1}{2} \).

Step 5: Find a Common Denominator

The common denominator for the fractions is \( t(t-3) \). Multiply each term by the common denominator to clear the fractions: \( t(t-3)\left(\frac{1}{t} + \frac{1}{t-3}\right) = t(t-3)\times\frac{1}{2} \).

Step 6: Simplify and Solve

Simplify the equation: \( (t-3) + t = \frac{t(t-3)}{2} \) becomes \( 2t - 3 = \frac{t^2 - 3t}{2} \). Multiply through by 2 to eliminate the fraction: \( 4t - 6 = t^2 - 3t \). Rearrange to form a quadratic equation: \( t^2 - 7t + 6 = 0 \).

Step 7: Factor the Quadratic

Factor the quadratic equation: \( (t-6)(t-1) = 0 \).

Step 8: Solve for t

Set each factor equal to zero: \( t-6 = 0 \) or \( t-1 = 0 \). Thus, \( t = 6 \) or \( t = 1 \).

Step 9: Interpret the Solution

Since Jake cannot take 1 hour (as it would make Jerry take -2 hours), the only feasible solution is \( t = 6 \). Hence, Jerry takes \( 6 - 3 = 3 \) hours.

Key Concepts

Algebraic Equations

When dealing with word problems, establishing algebraic equations is crucial. Think of them as mathematical representations of the scenario described in words. In this problem, we first define variables to represent the time taken by Jerry and Jake. Let's break it down:

• Jake's time to complete the work by himself is represented as \( t \).
• Since Jerry works faster—specifically, 3 hours faster—his time is expressed as \( t - 3 \).

This setup helps us use algebraic techniques to express and solve the problem. By forming an equation that incorporates the rates at which Jerry and Jake work, we can solve for \( t \), the unknown variable representing Jake's time. From there, Jerry's time to do the work can be easily derived. Breaking down word problems into such equations makes it more manageable and solvable by applying algebraic rules.

Rates of Work

In this problem, understanding rates of work is key because Jerry and Jake's ability to finish the task depends on their respective rates. Here, the rate is the fraction of work done per unit time (in this case, hours).

• Jake's rate is \( \frac{1}{t} \) floors per hour.
• Jerry's rate is \( \frac{1}{t-3} \) floors per hour because he is faster.

When they work together, their combined work rate becomes the sum of their individual rates. Thus, their joint rate is given by adding their rates: \( \frac{1}{t} + \frac{1}{t-3} = \frac{1}{2} \). This equation represents their collective ability to complete the floor in 2 hours. Rates of work can effectively model and quantify how tasks are completed when people work both individually and together.

Quadratic Equations

In solving this work problem, we come across a quadratic equation. These types of equations are characterized by their highest variable exponent being 2. When we reach step 6 of the solution, we work with the equation \( t^2 - 7t + 6 = 0 \).

Quadratic equations can often be solved by:

• Factoring
• Using the quadratic formula
• Completing the square

Here, factoring is the method used. We find two numbers that multiply to 6 and add to -7, which are -6 and -1. This results in the factors \((t-6)(t-1) = 0\).

By setting each factor equal to zero:\[ t - 6 = 0 \quad \text{or} \quad t - 1 = 0 \] we find \( t = 6 \) or \( t = 1 \). Since \( t = 1 \) leads to an illogical negative time for Jerry, \( t = 6 \) is the valid solution. Therefore, Jerry takes 3 hours to complete the task alone. Quadratic equations are often used in scenarios where relationships are curved (non-linear), making them valuable tools in various algebraic applications.