Problem 52 - Algebra for College Students

Question

Solve each investment problem. Sheryl won \(\$ 60,000\) on a slot machine in Las Vegas. She invested part of the money at \(2 \%\) simple interest and the rest at \(3 \%\). In one year, she earned a total of \(\$ 1600\) in interest. How much was invested at each rate?

Step-by-Step Solution

Answer

She invested \( 20000 \) at \( 2\% \) and \( 40000 \) at \( 3\% \).

Step 1: Define Variables

Let \( x \) be the amount invested at \( 2\% \) and \( 60000 - x \) be the amount invested at \( 3\% \). We are asked to find these amounts.

Step 2: Write the Interest Equations

The interest earned from the amount invested at \( 2\% \) is \( 0.02x \). The interest from the amount invested at \( 3\% \) is \( 0.03(60000 - x) \). The total interest earned is \( 1600 \). Thus, the equation is: \[ 0.02x + 0.03(60000 - x) = 1600 \]

Step 3: Distribute and Simplify

Distribute the \( 0.03 \) in the equation: \[ 0.02x + 0.03 \times 60000 - 0.03x = 1600 \] Simplify it to: \[ 0.02x + 1800 - 0.03x = 1600 \]

Step 4: Combine Like Terms

Combine the terms involving \( x \): \[ 0.02x - 0.03x = 1600 - 1800 \] Simplify it to: \[ -0.01x = -200 \]

Step 5: Solve for x

Divide both sides by \( -0.01 \) to solve for \( x \): \[ x = \frac{-200}{-0.01} = 20000 \]

Step 6: Calculate the Amount Invested at 3%

Subtract \( x \) from \( 60000 \) to find the amount invested at \( 3\% \): \[ 60000 - 20000 = 40000 \]

Key Concepts

headline of the respective core concept

Simple interest is a fundamental concept in finance and investment. It is calculated using the formula: \[ \text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time} \] The 'Principal' is the initial amount invested, 'Rate' is the percentage of interest per year, and 'Time' is the duration the money is invested for. In our exercise:

• Sheryl invested different portions of her money at two rates: 2% and 3%.

The total interest earned from both investments after one year was $1600. By breaking down the problem, we identified how much was invested at each rate.

headline of the respective core concept

Linear equations are used to solve problems involving relationships between quantities. In our exercise, we set up an equation to find out how much Sheryl invested at 2% and 3% interest respectively. First, we defined variables. Let \( x \) be the amount invested at 2%, so the rest of the amount, \( 60000 - x \), was invested at 3%. The equation set up was: \[ 0.02x + 0.03(60000 - x) = 1600 \] Linear equations helped us establish the relationship between the amounts invested at different interest rates.

headline of the respective core concept

Variable definition is a crucial step in solving algebraic problems. It involves assigning symbols to unknown quantities. In this exercise:

• We let \( x \) represent the amount invested at 2% interest.
• The remaining amount \( 60000 - x \) represents the investment at 3% interest.

This step helped in transforming the word problem into a workable algebraic equation, making it easier to solve.

headline of the respective core concept

Distribution of terms in algebra involves multiplying each term inside a parenthesis by the factor outside. In our equation: \[ 0.02x + 0.03(60000 - x) = 1600 \] We distributed 0.03 to both terms inside the parenthesis: \[ 0.02x + 0.03 \times 60000 - 0.03x = 1600 \] This simplification is crucial in combining like terms and solving the equation efficiently. Correct distribution led us to: \[ 0.02x + 1800 - 0.03x = 1600 \] By distributing terms properly, we made progress toward isolating the variable.