Problem 72 - Precalculus: Mathematics for Calculus

Question

Investments A man invests his savings in two accounts, one paying \(6 \%\) and the other paying \(10 \%\) simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is \(\$ 3520 .\) How much did he invest at each rate?

Step-by-Step Solution

Answer

He invested $16000 at 10% and $32000 at 6%.

Step 1: Understand the Problem

We need to determine how much money was invested in two accounts: one earning \(6\%\) interest and the other earning \(10\%\) interest. The amount invested at \(6\%\) is twice the amount invested at \(10\%\). The total interest earned annually from both accounts is \$3520.

Step 2: Set up the Variables

Let's denote \( x \) as the amount invested at \(10\%\). Then, the amount invested at \(6\%\) is \(2x\) because it is twice the amount invested at \(10\%\).

Step 3: Formulate the Equation

Based on simple interest, the interest from the \(6\%\) account is \(0.06 \times 2x = 0.12x\) and from the \(10\%\) account is \(0.10 \times x = 0.10x\). The total interest is \$3520, so we write the equation: \[ 0.12x + 0.10x = 3520 \]

Step 4: Solve the Equation

Combine like terms in the equation: \[ 0.22x = 3520 \]To find \(x\), divide both sides by \(0.22\): \[ x = \frac{3520}{0.22} \approx 16000 \]

Step 5: Determine the Investments

Since \( x \) is the amount invested at \(10\%\), it is \\(16000. The amount invested at \(6\%\) is \(2x\), which equals \(2 \times 16000 = \\)32000\).

Key Concepts

Investment Problems

Investment problems often involve figuring out how money is divided among different investment options, each with its specific interest rate and risk profile. In the given exercise, our task was to discover how much was placed in two separate accounts, each yielding different rates and having contrasting risk levels.

When dealing with such problems, it is essential to grasp the total amount of interest generated and understand how the money was allocated between the accounts. The problem highlighted that twice as much was deposited in a lower-risk account with a lower return rate of 6%, while less was invested in a higher-risk account offering 10% interest.

In real-world scenarios, investment decisions like these are common. Investors balance risk against potential returns, often prioritizing safety for a portion of their funds while seeking higher returns with another portion. To solve these problems, it's crucial to set up a clear structure and use logical reasoning to determine the allocation of resources.

Interest Rates

Interest rates are the percentage at which money grows over time when invested or borrowed. In the context of simple interest, the rate refers to the annual compensation for the use of the invested capital. The exercise provided two separate interest rates: 6% and 10%, for two different accounts.

Simple interest calculates the growth of money linearly, based on the principal amount. The formula to compute simple interest is \( I = P imes r imes t \), where \( I \) is the interest generated, \( P \) is the principal amount, \( r \) is the interest rate, and \( t \) is the time in years.

• The 6% account grows by 6% annually. Thus, the interest generated is a straightforward 6% of the principal over a year.
• Similarly, the 10% account yields interest at a rate of 10% on the invested capital.

Understanding interest rates helps potential investors make informed decisions by evaluating how much they will earn over time and deciding on the risk level they are comfortable with.

Algebraic Equations

Algebraic equations are mathematical statements used to represent relationships between variables and constants. In investment problems, equations can be employed to find unknowns, such as how much money is invested at different rates of return.

In this exercise, we expressed the interest earnings from the two accounts as an algebraic equation. The total interest was given, which allowed us to create an equation based on the known and unknown amounts:

• The amount at 6% was noted as \(2x\).
• The amount at 10% was signified as \(x\).
• The equation \(0.12x + 0.10x = 3520\) was derived from adding together the interests from both accounts.

By solving the equation, variables can be isolated, enabling us to determine specific amounts, as demonstrated when finding \(x\) to be 16000. Solving algebraic equations is a crucial step in many investment problems, providing clarity and a structured approach to finding solutions.