Problem 74
Question
\(\quad\) A retired teacher has a choice of two investment plans: An insured fund that pays \(11 \%\) interest A risky investment that pays a \(13 \%\) return If the same amount invested at the higher rate would general an extra \(\$ 150\) per year than it would at the lower rate, how much does the teacher have to invest?
Step-by-Step Solution
Verified Answer
The teacher needs to invest $7500.
1Step 1: Understanding the Problem
The teacher has two investment options: one with an interest rate of 11% and another with a 13% interest rate. If the teacher invests the same amount in both options, the investment at the higher rate yields an extra $150 annually compared to the lower rate. Our goal is to find the amount the teacher invests.
2Step 2: Setting Up the Equations
Let's denote the amount to be invested by the teacher as \( x \). The return from the 11% option is \( 0.11x \) and from the 13% option is \( 0.13x \). The difference in income from these two investments is given to be $150. Thus, we can set up the equation \( 0.13x - 0.11x = 150 \).
3Step 3: Simplifying the Equation
Simplify the equation \( 0.13x - 0.11x = 150 \) to find the unknown. You get \( 0.02x = 150 \).
4Step 4: Solving for the Investment Amount
To solve for \( x \), divide both sides of the equation by 0.02: \( x = \frac{150}{0.02} \).
5Step 5: Calculating the Result
Calculate \( x = \frac{150}{0.02} = 7500 \). Thus, the teacher needs to invest \( \$7500 \) in either investment to meet the condition given in the problem.
Key Concepts
Investment CalculationsAlgebraic EquationsFinancial Mathematics
Investment Calculations
Investment calculations are the foundation of understanding how much you should invest and what returns to expect. In our problem, we dealt with two investment options:
- An insured fund with an 11% interest rate
- A risky investment with a 13% interest rate
To figure out how much a retired teacher should invest, the problem sets a scenario where the extra income from the higher interest rate needs to be $150 more than the lower one. This involves selecting an appropriate amount to reach this goal of additional earnings.
The core of investment calculations is accurately determining potential returns. This involves:
- Identifying the potential interest rate or return
- Computing the actual dollar gain from that percentage
- Comparing the differing outcomes to find extra income, how long it might take to achieve goals, or in our case, how much must be invested.
This process helps you not just solve word problems, but also make informed decisions about personal or business investments.
Algebraic Equations
Algebra provides a powerful tool for solving interest rate problems, like the one faced by our teacher. In this example, algebraic equations are used to set up and solve the problem.We began by letting the amount the teacher invests be represented by a variable, such as \( x \). Using the interest rates:- The return from the 11% interest is represented as \( 0.11x \).- Similarly, the return from the 13% rate is \( 0.13x \).The difference equation \( 0.13x - 0.11x = 150 \) demonstrates the equation setup. Simplifying this step-by-step, we derived \( 0.02x = 150 \).Algebra doesn't just give us a way to express the problem; it allows us to systematically find solutions:- We isolate the variable \( x \) by performing operations on both sides.- Finally, dividing by 0.02 gives us the amount to invest, which resolves the given scenario.This shows how algebra seamlessly helps turn a word problem into a solvable equation.
Financial Mathematics
Financial mathematics connects algebraic skills to real-life financial decisions. Here, we used it to solve a practical problem, ensuring our finished calculations were correct and meaningful.
By setting differences in expected returns, financial mathematics lets you:
- Calculate necessary investments to achieve desired returns
- Plan for future financial goals with precision
- Compare different investment options to find the most suitable one
Other exercises in this chapter
Problem 73
Evaluate each expression. See Example \(9 .\) $$ (-3-\sqrt{25})^{2} $$
View solution Problem 73
Insert either \(a\) symbol to make a true statement. $$ 0 . \overline{016} \quad \frac{2}{125} $$
View solution Problem 74
Solve each equation. If the equation is an identity or a contradiction, so indicate. See Example 10. $$ 5(y+2)+7-3 y=2(y+9) $$
View solution Problem 74
Simplify by combining like terms. See Example 5 . $$\frac{3}{16} x-\frac{5}{4} x$$
View solution