Problem 21 - College Algebra with Corequisite Support

Question

For exercises 17 and 18, use this scenario: A retired woman has $$\$ 50,000$$ to invest but needs to make $$\$ 6,000$$ a year from the interest to meet certain living expenses. One bond investment pays \(15 \%\) annual interest. The rest of it she wants to put in a CD that pays \(7 \%\). Fiora starts riding her bike at 20 \(\mathrm{mi} / \mathrm{h}\). After a while, she slows down to \(12 \mathrm{mi} / \mathrm{h}\), and maintains that speed for the rest of the trip. The whole trip of \(70 \mathrm{mi}\) takes her \(4.5 \mathrm{~h}\). For what distance did she travel at \(20 \mathrm{mi} /\) \(\mathrm{h} ?\)

Step-by-Step Solution

Answer

Fiora traveled 40 miles at 20 mi/h.

Step 1: Define Variables

Let \( x \) represent the distance traveled at 20 mi/h, and \( y \) represent the distance traveled at 12 mi/h. We know the total distance is 70 miles and the total time is 4.5 hours.

Step 2: Create an Equation for Distance

Since the total distance traveled is 70 miles, we have the equation:\[ x + y = 70 \]

Step 3: Create an Equation for Time

For time, we consider the relationship \( \text{time} = \frac{\text{distance}}{\text{speed}} \). So the time for each segment of the trip can be expressed as:\[ \frac{x}{20} + \frac{y}{12} = 4.5 \]

Step 4: Solve the System of Equations

First, we solve the distance equation for one of the variables, say \( y \):\[ y = 70 - x \]Now, substitute \( y = 70 - x \) into the time equation:\[ \frac{x}{20} + \frac{70 - x}{12} = 4.5 \]

Step 5: Clear Fractions in Time Equation

To remove the fractions, multiply the entire equation by the least common multiple of 20 and 12, which is 60:\[ 60 \left( \frac{x}{20} + \frac{70 - x}{12} \right) = 60 \times 4.5 \]Simplifying gives:\[ 3x + 5(70 - x) = 270 \]

Step 6: Solve for x

Distribute and combine like terms:\[ 3x + 350 - 5x = 270 \]Combine terms:\[ -2x = -80 \]Solve for \( x \):\[ x = 40 \]

Step 7: Verify Solution

Substitute \( x = 40 \) back into the first equation to verify:\[ x + y = 70 \Rightarrow 40 + y = 70 \Rightarrow y = 30 \]Check time:\[ \frac{40}{20} + \frac{30}{12} = 2 + 2.5 = 4.5 \] hours, confirming the solution is correct.

Key Concepts

Distance Problems

Distance problems often involve calculating the total distance traveled using different speeds over various segments of a journey. In essence, these problems ask how far someone traveled at each speed. Let’s visualize this with the scenario involving Fiora's bike ride.

We know the total distance Fiora traveled was 70 miles, split between two different speeds: 20 mi/h and 12 mi/h. Each distance segment has a corresponding time of travel, forming the basis of our distance calculation equation. The equation to find the total distance is: \( x + y = 70 \), where \( x \) is the distance traveled at 20 mi/h and \( y \) at 12 mi/h.

Distance problems require careful identification of variables and understanding how they form part of the whole journey. Here, the sum of the distances at both speeds equals the total distance traveled. These problems teach us to break down journeys into segments we can quantify or measure.

Rate Problems

Rate problems involve solving how long it takes to travel a certain distance at a particular speed. They frequently appear alongside distance problems because they offer a dynamic view of a journey. In our example, rate problems provided us with the insight to calculate the time taken for each speed Fiora traveled.

A crucial formula here is time expressed as \( \text{time} = \frac{\text{distance}}{\text{speed}} \). By independently inserting each speed for the corresponding distance segment, we create the equation \( \frac{x}{20} + \frac{y}{12} = 4.5 \), representing the total travel time of 4.5 hours.

Rate problems deftly connect distance with time, showcasing how they interplay to complete a total journey. They help students apply logical reasoning through equation manipulation, such as clearing denominators and substituting known values, which is critical for accurately solving the problem.

Investment Scenarios

Investment scenarios are common in mathematics and economics curricula, designed to teach students how to manage savings and necessary returns. In the context of the retiree’s investment problem, understanding how to balance different investment returns to meet specific income needs is key.

Here, the woman has $50,000 to invest. She places some of this in a bond yielding 15% interest, while the remainder goes into a certificate of deposit (CD) earning 7%. Her goal is to generate $6,000 annually from these investments. To achieve this, students must learn to set up equations representing these financial scenarios: the sum of the investments adds up to the principal $50,000, while the resulting interests should meet the income target.

Such problems typically involve forming systems of equations reflecting both principal allocation and interest return calculations. Investment scenarios serve as vital practical exercises, merging mathematical reasoning with real-world economic insights, helping students comprehend the dual need for both growth and security in investments.