Simple Interest Math Worksheet with Problems and Word Problems

Problem Overview:

The worksheet involves calculating simple interest for various loans and solving word problems related to simple interest. The formula for simple interest is:

\[
\text{Simple Interest (SI)} = \text{Principal (P)} \times \text{Rate (R)} \times \text{Time (T)}
\]

Where:
- \( P \) is the principal amount.
- \( R \) is the annual interest rate (in decimal form).
- \( T \) is the time period in years.

Solution to Each Problem:

#### Part 1: Determine the Simple Interest for These Loans

1. $450 at 7% for 2 years
\[
\text{SI} = 450 \times 0.07 \times 2 = 63
\]
Answer: $63

2. $5,200 at 4% for 3 years
\[
\text{SI} = 5200 \times 0.04 \times 3 = 624
\]
Answer: $624

3. $1,300 at 5% for 6 years
\[
\text{SI} = 1300 \times 0.05 \times 6 = 390
\]
Answer: $390

4. $5,400 at 3.5% for 6 months
\[
\text{SI} = 5400 \times 0.035 \times \frac{6}{12} = 5400 \times 0.035 \times 0.5 = 94.5
\]
Answer: $94.5

5. $600 at 4% for 9 months
\[
\text{SI} = 600 \times 0.04 \times \frac{9}{12} = 600 \times 0.04 \times 0.75 = 18
\]
Answer: $18

6. $24,000 at 5.5% for 5 years
\[
\text{SI} = 24000 \times 0.055 \times 5 = 6600
\]
Answer: $6600

7. $15,600 at 3% for 2 years
\[
\text{SI} = 15600 \times 0.03 \times 2 = 936
\]
Answer: $936

8. $1,200 at 5.5% for 4 years
\[
\text{SI} = 1200 \times 0.055 \times 4 = 264
\]
Answer: $264

9. $1,600 at 4.5% for 9 months
\[
\text{SI} = 1600 \times 0.045 \times \frac{9}{12} = 1600 \times 0.045 \times 0.75 = 54
\]
Answer: $54

10. $12,000 at 2.2% for 5 years
\[
\text{SI} = 12000 \times 0.022 \times 5 = 1320
\]
Answer: $1320

#### Part 2: Solve Each Simple Interest Word Problem

11. A new car valued at $28,000 depreciates at 9% per year. What is the value of the car one year after purchase?
- Depreciation is calculated as a reduction in value, so we use:
\[
\text{Value after 1 year} = \text{Original Value} \times (1 - \text{Depreciation Rate})
\]
\[
\text{Value after 1 year} = 28000 \times (1 - 0.09) = 28000 \times 0.91 = 25480
\]
Answer: $25,480

12. Sara puts $4,000 into an investment yielding 5% annual simple interest; she left the money in for five years. How much interest does Sara get at the end of those five years?
\[
\text{SI} = 4000 \times 0.05 \times 5 = 1000
\]
Answer: $1,000

13. A bank is offering 3.5% simple interest on a savings account. If you deposit $7,500, how much interest will you earn in two years?
\[
\text{SI} = 7500 \times 0.035 \times 2 = 525
\]
Answer: $525

14. $400 interest is earned on a principal of $2,000 at a simple interest rate of 5% interest per year. For how many years was the principal invested?
Using the simple interest formula:
\[
\text{SI} = P \times R \times T
\]
Rearrange to solve for \( T \):
\[
T = \frac{\text{SI}}{P \times R} = \frac{400}{2000 \times 0.05} = \frac{400}{100} = 4
\]
Answer: 4 years

15. In how many years will $1,200 yield an interest of $180 at 3% simple interest?
Using the simple interest formula:
\[
\text{SI} = P \times R \times T
\]
Rearrange to solve for \( T \):
\[
T = \frac{\text{SI}}{P \times R} = \frac{180}{1200 \times 0.03} = \frac{180}{36} = 5
\]
Answer: 5 years

16. Jim invested $4,000 in a bond at a yearly rate of 4.5%. He earned $540 in interest. How long was the money invested?
Using the simple interest formula:
\[
\text{SI} = P \times R \times T
\]
Rearrange to solve for \( T \):
\[
T = \frac{\text{SI}}{P \times R} = \frac{540}{4000 \times 0.045} = \frac{540}{180} = 3
\]
Answer: 3 years

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & 63 \\
2. & 624 \\
3. & 390 \\
4. & 94.5 \\
5. & 18 \\
6. & 6600 \\
7. & 936 \\
8. & 264 \\
9. & 54 \\
10. & 1320 \\
11. & 25480 \\
12. & 1000 \\
13. & 525 \\
14. & 4 \\
15. & 5 \\
16. & 3 \\
\end{array}
}
\]