Problem 35 - Elementary Algebra

Question

Sally invested a certain sum of money at \(9 \%\), twice that sum at \(10 \%\), and three times that sum at \(11 \%\). Her total yearly interest from all three investments was \(\$ 310\). How much did she invest at each rate?

Step-by-Step Solution

Answer

Sally invested $500 at 9%, $1000 at 10%, and $1500 at 11%.

Step 1: Define the Variables

Let \( x \) be the amount of money invested at 9%. Then, twice that amount, \( 2x \), is invested at 10%, and three times that amount, \( 3x \), is invested at 11%.

Step 2: Calculate Individual Interests

Calculate the interest from each investment:- Interest from \( x \) at 9%: \( 0.09x \)- Interest from \( 2x \) at 10%: \( 0.10 \times 2x = 0.20x \)- Interest from \( 3x \) at 11%: \( 0.11 \times 3x = 0.33x \)

Step 3: Setup the Equation

Sum up all the interests and equate to the total interest:\[ 0.09x + 0.20x + 0.33x = 310 \]

Step 4: Simplify the Equation

Combine like terms in the equation:\[ 0.62x = 310 \]

Step 5: Solve for \( x \)

Divide both sides of the equation by 0.62:\[ x = \frac{310}{0.62} \]Calculating this gives \( x = 500 \).

Step 6: Find Each Investment

Now that \( x = 500 \), calculate each investment:- At 9%: \( x = 500 \)- At 10%: \( 2x = 1000 \)- At 11%: \( 3x = 1500 \)

Key Concepts

Investment interest calculations

When we talk about investment interest calculations, it's all about understanding how much money you make from your investments. Here, Sally has invested different amounts at varied interest rates.
Specifically:

• Sally invested a certain amount of money at 9% interest.
• She also invested twice that amount at 10% interest.
• Finally, three times that initial amount was invested at 11% interest.

Calculating interest means figuring out how much extra money those percentages earn Sally every year. For each investment, you multiply the principal amount by the interest rate to get the yearly interest.
Adding those interests together gives you the total interest Sally earns from all her investments:

• From the 9% investment: she earns 0.09 times the amount invested at 9%.
• From the 10% investment: she earns 0.20 times the amount at 10%.
• From the 11% investment: she earns 0.33 times the amount at 11%.

Add these interests for the total yearly interest, which in this case is $310.

Variable definition and manipulation

In algebra, defining and manipulating variables is crucial. Here's how it applied to Sally's investment scenario:

• First, we needed a variable to represent the unknown amount. We chose "x" to be the amount invested at 9%.
• Since Sally invested twice that amount at 10%, that part can be represented as "2x."
• Finally, three times the original amount was invested at 11%, making it "3x."

Knowing how to define variables allows us to create expressions and equations that represent real-life situations.
This setup helps simplify the problem by letting us deal with numbers and percentage rates systematically, avoiding confusion. Manipulating these variables correctly builds a foundation for forming and solving equations.

Equation simplification and solving

Solving an algebraic equation starts with setting up the equation correctly based on the problem at hand. In Sally's case, after defining the variables, we create an equation representing the total interest:

• The equation assembled is: \[ 0.09x + 0.20x + 0.33x = 310 \]

Simplifying the Equation

Combine like terms on the left side. This means adding up all expressions with the variable "x":

• Combine to get: \[ 0.62x = 310 \]

Solving the Equation

Now, isolate "x" by dividing both sides by 0.62:

• This gives us: \[ x = \frac{310}{0.62} \]

The solution, \( x = 500 \), represents the initial amount invested at 9%. With this value, you can easily find the amounts invested at the other rates:

• At 10%: \( 2x = 1000 \).
• At 11%: \( 3x = 1500 \).

This methodical process of simplifying and solving helps make complex problems much more manageable.