Problem 69 - Precalculus Mathematics for Calculus

Question

A woman invests a total of \(\$ 20,000\) in two accounts, one paying \(5 \%\) and the other paying \(8 \%\) simple interest per year. Her annual interest is \(\$ 1180 .\) How much did she invest at each rate?

Step-by-Step Solution

Answer

$14,000 at 5%, $6,000 at 8%.

Step 1: Define Variables

Let \( x \) be the amount invested at the \( 5\% \) interest rate. Then, the amount invested at the \( 8\% \) rate is \( 20000 - x \).

Step 2: Setup Interest Equations

The equation for interest from the \( 5\% \) account is \( 0.05x \), and from the \( 8\% \) account, it is \( 0.08(20000 - x) \).

Step 3: Write the Total Interest Equation

According to the problem, the total interest received is \( 1180 \). Therefore, the equation is: \[ 0.05x + 0.08(20000 - x) = 1180 \].

Step 4: Solve for "x"

Expand and simplify the equation: \( 0.05x + 1600 - 0.08x = 1180 \). Simplifying further gives: \(-0.03x + 1600 = 1180 \). Subtract \( 1600 \) from both sides to obtain: \(-0.03x = -420 \). Divide both sides by \(-0.03\) to isolate \( x \): \( x = 14000 \).

Step 5: Calculate the Other Amount

Since \( x = 14000 \), the amount invested at \( 8\% \) is \( 20000 - 14000 = 6000 \).

Step 6: Verify the Solution

Substitute \( x = 14000 \) and \( 20000 - x = 6000 \) back into the interest equations: - Interest from \( 14000 \) at \( 5\% \) is \( 0.05 \times 14000 = 700 \).- Interest from \( 6000 \) at \( 8\% \) is \( 0.08 \times 6000 = 480 \).- Total interest \( = 700 + 480 = 1180 \), which matches the given total interest.

Key Concepts

Simple Interest

Simple interest is a way to calculate the interest charged or earned on an investment over a specific period of time. It is very important for solving investment word problems like this one. The formula for simple interest is: \[ I = P imes r imes t \]where:

• \( I \) is the interest,
• \( P \) is the principal amount (the initial sum of money),
• \( r \) is the rate of interest per period,
• \( t \) is the time the money is invested or borrowed for, in years.

In our problem, we are given different rates for two different investments and need to find out how much is invested in each to reach a total interest of $1180. By setting up individual interest equations for each account, you apply this formula to calculate the exact amounts. Understanding simple interest helps to solve the puzzle of how your investments grow over time or what you'll pay in case of a loan.

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the power of one. They form a straight line when graphed, but when it comes to solving investment problems, they represent the relationships between different quantities involved.
In this exercise, we are tasked to find out how much was invested at each rate to accrue a specific amount of total interest. The equation \[ 0.05x + 0.08(20000 - x) = 1180 \]
is an example of a linear equation. It perfectly represents the total interest from the two accounts since both interest amounts depend linearly on the amounts invested.
By rearranging the equation, we break it down into simpler parts to find the value of \( x \), the amount invested at \( 5\% \). Using a linear equation means we can efficiently find the unknowns without much hassle, allowing for a quick solution to real-world problems.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations that represent a particular value or set of values. This concept is essential for formulating and solving equations in investment word problems.
When we define variables and set up interest equations in the problem, we are creating algebraic expressions. For instance, \( 0.05x \) is an expression representing the interest from one part of the investment, while \( 0.08(20000 - x) \) represents interest from the other part.
To find out how much was invested at each interest rate, you manipulate these expressions: simplify them, combine like terms, and solve linear equations. Grasping algebraic expressions enable you to build a framework that simplifies complex relationships into easier-to-handle components.
With practice, such skills become essential tools for solving investment-related queries and beyond.