Lesson Objectives
• Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
• Learn the six-step method used for solving applications of linear equations
• Learn how to solve word problems that involve the simple interest formula "I = prt"

## How to Solve Simple Interest Word Problems

In our last lesson, we reviewed the six-step method used to solve a word problem that involves a linear equation in one variable.

### Six-step method for Solving Word Problems with Linear Equations in One Variable

1. Read the problem and determine what you are asked to find
2. Assign a variable to represent the unknown
• If more than one unknown exists, we express the other unknowns in terms of this variable
3. Write out an equation that describes the given situation
4. Solve the equation
5. State the answer using a nice clear sentence
6. Check the result
• We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.

### Solving Simple Interest Word Problems » I = prt

When we work with investment problems in Algebra, we will encounter two types of interest: simple interest and compound interest. The main difference between the two comes from the fact that compound interest will earn interest on interest. This means as the account balance grows, interest is paid on the current account balance, not the original amount invested (known as the principal). When we work with simple interest word problems, interest is only earned on the principal or amount initially invested. In other words, when we have simple interest, we do not earn interest on interest. Although this makes our calculation much simpler, it is not often used in real life.
With simple interest word problems, we will use the simple interest formula:
I = prt
I » simple interest earned
p » principal or amount invested
r » rate (interest rate as a decimal)
t » time (normally in years, but could be months, days, or whatever given time frame)
Let's look at some examples.
Example 1: Solve each word problem.
Malorie finds a house for a sale price of $155,000. She meets with her bank and finds a 30-year simple interest mortgage. If Malorie accepts the mortgage, she would pay$232,500 in simple interest over the life of the loan. How much is the interest rate of the mortgage?
Step 1) After reading the problem, it is clear that we need to find the interest rate of the mortgage.
Step 2) We have one unknown, the interest rate for the mortgage. In this case, we can just use r (rate) from the simple interest formula to represent our unknown interest rate.
Step 3) To write an equation, let's first solve our simple interest formula for r (rate): $$\text{r}=\frac{\text{I}}{\text{pt}}$$ Now, we can plug in what we know:
I = 232,500
p = 155,000
t = 30 $$\text{r}=\frac{232{,}500}{155{,}000 \cdot 30}$$ Step 4) Solve the equation: $$\text{r}=\frac{232{,}500}{155{,}000 \cdot 30}$$ $$\text{r}=0.05$$ Step 5) Since r represented the rate of simple interest on the mortgage as a decimal, we can state our answer as:
Malorie's proposed mortgage has an interest rate of 5%.
Step 6) We can read back through the problem to check our answer. Here, we can just plug into the simple interest formula: $$\text{I}=\text{prt}$$ $$232{,}500=155{,}000 \cdot 0.05 \cdot 30$$ $$232{,}500 = 232{,}500 \hspace{.2em}✓$$ Example 2: Solve each word problem.
After speaking with a retirement professional, James decides to invest in two different accounts. A bond fund, that pays 5% annual simple interest, and a savings account that pays 2% annual simple interest. James first decides to invest $29,000 into the bond fund. If his goal is to make 3% annual simple interest, how much additional money should he invest in the savings account? Step 1) After reading the problem, it is clear that we need to find the amount that James needs to invest in the savings account in order to have a combined interest rate of 3%. Step 2) We have one unknown, the amount that is to be invested in the savings account. let x = amount of money in dollars that will be invested in the savings account Step 3) To write an equation, let's organize our information in a table. Since time is 1, we will omit this column. I p r B. F.$1450$29,000.05 S. A..02xx.02 James wants an overall interest rate of 3% for the two investments. Currently, we know James will invest$29,000 in the bond fund and earn 5% annual simple interest. If we plug this into the simple interest formula:
I = 29,000(.05)(1)
I = 1450
This amount will be added to the 2% interest obtained from investing an unknown amount (x) in a savings account:
Using our simple interest formula for the savings account yields:
I = x (.02)(1)
I = .02x
If we sum the two, we think about the simple interest earned from the two investments:
I = 1450 + .02x
Since we want I or the simple interest earned to be 3% of the principal invested, we can set up the following equation:
.03(x + 29,000) = .02x + 1450
In other words, 3% multiplied by the principal invested (x + 29,000) will be equal to the interest earned from the savings account (.02x) and the interest earned from the bond fund (1450).
Step 4) Solve the equation:
.03(x + 29,000) = .02x + 1450
.03x + 870 = .02x + 1450
We can clear the decimals by multiplying both sides of the equation by 100:
3x + 87,000 = 2x + 145,000
3x - 2x = 145,000 - 87,000
x = 58,000
Step 5) Since x represented the amount that needed to be invested in the savings account, we can state our answer as:
James needs to invest $58,000 in the savings account in order to achieve a simple interest rate of 3% from the two investments. Step 6) We can read back through the problem to check our answer. Think about the simple interest earned from his two investments: I = .02(58,000) + .05(29,000) I = 1160 + 1450 I = 2610 The simple interest earned from the two investments is 2610. If we solve the simple interest formula for r (rate): $$\text{I} = \text{prt}$$ $$\text{r}=\frac{\text{I}}{\text{pt}}$$ Since time is 1, we can rewrite our equation as: $$\text{r}=\frac{\text{I}}{\text{p}}$$ Since his goal is 3% interest, let's plug everything in and check: $$.03=\frac{2610}{29,\hspace{-.1em}000 + 58,\hspace{-.1em}000}$$ $$.03=\frac{2610}{87,\hspace{-.1em}000}$$ $$.03=.03 \hspace{.2em}✓$$ #### Skills Check: Example #1 Solve each word problem. Lauren earns$550 per year in annual simple interest from a $10,000 investment. She invested part of the$10,000 in a savings account at 5% annual simple interest and the remainder in bonds paying 6% annual simple interest. How much did Lauren invest at each rate?

A
5%: $5000, 6%:$5000
B
5%: $6000, 6%:$4000
C
5%: $7000, 6%:$3000
D
5%: $8000, 6%:$2000
E
5%: $1000, 6%:$9000

Example #2

Solve each word problem.

Ben is saving money for his welding certification. He deposited some money in a savings account paying 5% annual simple interest. Additionally, he deposited $1200 less than that amount in a CD paying 4% annual simple interest. Ben earns a total of$141 per year in interest from the two investments. How much did he invest at each rate?

A
5%: $8600, 4%:$7400
B
5%: $2400, 4%:$1200
C
5%: $7600, 4%:$6400
D
5%: $5000, 4%:$3800
E
5%: $2100, 4%:$900

Example #3

Solve each word problem.

April invested in a real estate fund that pays 6% annual simple interest. Additionally, she invested $6000 more than three times as much in corporate bonds that pay 5% annual simple interest. If April’s total interest per year is$825, how much did she invest at each rate?

A
5%: $13,500, 6%:$2500
B
5%: $2000, 6%:$7000
C
5%: $8000, 6%:$13,500
D
5%: $19,000, 6%:$4000
E
5%: $2800, 6%:$11,200