Simple Interest Practice Worksheet with 18 Problems for Calculating Ending Balance
Worksheet titled "Simple Interest" with 18 problems calculating ending balance using simple interest formula, including principal amounts, interest rates, and time periods in years and months.
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Show Answer Key & Explanations
Step-by-step solution for: Simple Interest worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Simple Interest worksheets
To solve the problems involving simple interest, we use the formula for simple interest:
\[
\text{Simple Interest (SI)} = P \times r \times t
\]
Where:
- \( P \) is the principal amount (initial amount of money),
- \( r \) is the annual interest rate (in decimal form),
- \( t \) is the time the money is invested or borrowed for (in years).
The ending balance is calculated as:
\[
\text{Ending Balance} = P + \text{SI}
\]
Let's solve each problem step by step.
---
- \( P = 35,100 \)
- \( r = 4\% = 0.04 \)
- \( t = 3 \)
\[
\text{SI} = 35,100 \times 0.04 \times 3 = 4,212
\]
\[
\text{Ending Balance} = 35,100 + 4,212 = 39,312
\]
Answer: $39,312
---
- \( P = 150 \)
- \( r = 5\% = 0.05 \)
- \( t = 2 \)
\[
\text{SI} = 150 \times 0.05 \times 2 = 15
\]
\[
\text{Ending Balance} = 150 + 15 = 165
\]
Answer: $165
---
- \( P = 1,400 \)
- \( r = 3\% = 0.03 \)
- \( t = 4 \)
\[
\text{SI} = 1,400 \times 0.03 \times 4 = 168
\]
\[
\text{Ending Balance} = 1,400 + 168 = 1,568
\]
Answer: $1,568
---
- \( P = 450 \)
- \( r = 9\% = 0.09 \)
- \( t = 8 \)
\[
\text{SI} = 450 \times 0.09 \times 8 = 324
\]
\[
\text{Ending Balance} = 450 + 324 = 774
\]
Answer: $774
---
- \( P = 2,400 \)
- \( r = 5\% = 0.05 \)
- \( t = 2 \)
\[
\text{SI} = 2,400 \times 0.05 \times 2 = 240
\]
\[
\text{Ending Balance} = 2,400 + 240 = 2,640
\]
Answer: $2,640
---
- \( P = 600 \)
- \( r = 4.5\% = 0.045 \)
- \( t = 5 \) months = \( \frac{5}{12} \) years
\[
\text{SI} = 600 \times 0.045 \times \frac{5}{12} = 11.25
\]
\[
\text{Ending Balance} = 600 + 11.25 = 611.25
\]
Answer: $611.25
---
- \( P = 15,600 \)
- \( r = 5.5\% = 0.055 \)
- \( t = 6 \)
\[
\text{SI} = 15,600 \times 0.055 \times 6 = 5,148
\]
\[
\text{Ending Balance} = 15,600 + 5,148 = 20,748
\]
Answer: $20,748
---
- \( P = 1,300 \)
- \( r = 4\% = 0.04 \)
- \( t = 3 \)
\[
\text{SI} = 1,300 \times 0.04 \times 3 = 156
\]
\[
\text{Ending Balance} = 1,300 + 156 = 1,456
\]
Answer: $1,456
---
- \( P = 750 \)
- \( r = 5.5\% = 0.055 \)
- \( t = 5 \)
\[
\text{SI} = 750 \times 0.055 \times 5 = 206.25
\]
\[
\text{Ending Balance} = 750 + 206.25 = 956.25
\]
Answer: $956.25
---
- \( P = 850 \)
- \( r = 2.2\% = 0.022 \)
- \( t = 11 \) months = \( \frac{11}{12} \) years
\[
\text{SI} = 850 \times 0.022 \times \frac{11}{12} = 17.0833
\]
\[
\text{Ending Balance} = 850 + 17.0833 = 867.0833 \approx 867.08
\]
Answer: $867.08
---
- \( P = 20,600 \)
- \( r = 5.5\% = 0.055 \)
- \( t = 4 \)
\[
\text{SI} = 20,600 \times 0.055 \times 4 = 4,532
\]
\[
\text{Ending Balance} = 20,600 + 4,532 = 25,132
\]
Answer: $25,132
---
- \( P = 100 \)
- \( r = 2\% = 0.02 \)
- \( t = 9 \) months = \( \frac{9}{12} = 0.75 \) years
\[
\text{SI} = 100 \times 0.02 \times 0.75 = 1.5
\]
\[
\text{Ending Balance} = 100 + 1.5 = 101.5
\]
Answer: $101.50
---
- \( P = 4,000 \)
- \( r = 4\% = 0.04 \)
- \( t = 3 \)
\[
\text{SI} = 4,000 \times 0.04 \times 3 = 480
\]
\[
\text{Ending Balance} = 4,000 + 480 = 4,480
\]
Answer: $4,480
---
- \( P = 7,400 \)
- \( r = 2.2\% = 0.022 \)
- \( t = 9 \) months = \( \frac{9}{12} = 0.75 \) years
\[
\text{SI} = 7,400 \times 0.022 \times 0.75 = 125.7
\]
\[
\text{Ending Balance} = 7,400 + 125.7 = 7,525.7
\]
Answer: $7,525.70
---
- \( P = 1,900 \)
- \( r = 6\% = 0.06 \)
- \( t = 3 \)
\[
\text{SI} = 1,900 \times 0.06 \times 3 = 342
\]
\[
\text{Ending Balance} = 1,900 + 342 = 2,242
\]
Answer: $2,242
---
- \( P = 7,300 \)
- \( r = 5.2\% = 0.052 \)
- \( t = 3 \) months = \( \frac{3}{12} = 0.25 \) years
\[
\text{SI} = 7,300 \times 0.052 \times 0.25 = 94.3
\]
\[
\text{Ending Balance} = 7,300 + 94.3 = 7,394.3
\]
Answer: $7,394.30
---
- \( P = 210 \)
- \( r = 5.9\% = 0.059 \)
- \( t = 3 \)
\[
\text{SI} = 210 \times 0.059 \times 3 = 38.37
\]
\[
\text{Ending Balance} = 210 + 38.37 = 248.37
\]
Answer: $248.37
---
- \( P = 35,100 \)
- \( r = 3.6\% = 0.036 \)
- \( t = 3 \) months = \( \frac{3}{12} = 0.25 \) years
\[
\text{SI} = 35,100 \times 0.036 \times 0.25 = 315.9
\]
\[
\text{Ending Balance} = 35,100 + 315.9 = 35,415.9
\]
Answer: $35,415.90
---
\[
\boxed{
\begin{array}{ll}
1. & \$39,312 \\
2. & \$165 \\
3. & \$1,568 \\
4. & \$774 \\
5. & \$2,640 \\
6. & \$611.25 \\
7. & \$20,748 \\
8. & \$1,456 \\
9. & \$956.25 \\
10. & \$867.08 \\
11. & \$25,132 \\
12. & \$101.50 \\
13. & \$4,480 \\
14. & \$7,525.70 \\
15. & \$2,242 \\
16. & \$7,394.30 \\
17. & \$248.37 \\
18. & \$35,415.90 \\
\end{array}
}
\]
\[
\text{Simple Interest (SI)} = P \times r \times t
\]
Where:
- \( P \) is the principal amount (initial amount of money),
- \( r \) is the annual interest rate (in decimal form),
- \( t \) is the time the money is invested or borrowed for (in years).
The ending balance is calculated as:
\[
\text{Ending Balance} = P + \text{SI}
\]
Let's solve each problem step by step.
---
Problem 1: $35,100 at 4% for 3 years
- \( P = 35,100 \)
- \( r = 4\% = 0.04 \)
- \( t = 3 \)
\[
\text{SI} = 35,100 \times 0.04 \times 3 = 4,212
\]
\[
\text{Ending Balance} = 35,100 + 4,212 = 39,312
\]
Answer: $39,312
---
Problem 2: $150 at 5% for 2 years
- \( P = 150 \)
- \( r = 5\% = 0.05 \)
- \( t = 2 \)
\[
\text{SI} = 150 \times 0.05 \times 2 = 15
\]
\[
\text{Ending Balance} = 150 + 15 = 165
\]
Answer: $165
---
Problem 3: $1,400 at 3% for 4 years
- \( P = 1,400 \)
- \( r = 3\% = 0.03 \)
- \( t = 4 \)
\[
\text{SI} = 1,400 \times 0.03 \times 4 = 168
\]
\[
\text{Ending Balance} = 1,400 + 168 = 1,568
\]
Answer: $1,568
---
Problem 4: $450 at 9% for 8 years
- \( P = 450 \)
- \( r = 9\% = 0.09 \)
- \( t = 8 \)
\[
\text{SI} = 450 \times 0.09 \times 8 = 324
\]
\[
\text{Ending Balance} = 450 + 324 = 774
\]
Answer: $774
---
Problem 5: $2,400 at 5% for 2 years
- \( P = 2,400 \)
- \( r = 5\% = 0.05 \)
- \( t = 2 \)
\[
\text{SI} = 2,400 \times 0.05 \times 2 = 240
\]
\[
\text{Ending Balance} = 2,400 + 240 = 2,640
\]
Answer: $2,640
---
Problem 6: $600 at 4.5% for 5 months
- \( P = 600 \)
- \( r = 4.5\% = 0.045 \)
- \( t = 5 \) months = \( \frac{5}{12} \) years
\[
\text{SI} = 600 \times 0.045 \times \frac{5}{12} = 11.25
\]
\[
\text{Ending Balance} = 600 + 11.25 = 611.25
\]
Answer: $611.25
---
Problem 7: $15,600 at 5.5% for 6 years
- \( P = 15,600 \)
- \( r = 5.5\% = 0.055 \)
- \( t = 6 \)
\[
\text{SI} = 15,600 \times 0.055 \times 6 = 5,148
\]
\[
\text{Ending Balance} = 15,600 + 5,148 = 20,748
\]
Answer: $20,748
---
Problem 8: $1,300 at 4% for 3 years
- \( P = 1,300 \)
- \( r = 4\% = 0.04 \)
- \( t = 3 \)
\[
\text{SI} = 1,300 \times 0.04 \times 3 = 156
\]
\[
\text{Ending Balance} = 1,300 + 156 = 1,456
\]
Answer: $1,456
---
Problem 9: $750 at 5.5% for 5 years
- \( P = 750 \)
- \( r = 5.5\% = 0.055 \)
- \( t = 5 \)
\[
\text{SI} = 750 \times 0.055 \times 5 = 206.25
\]
\[
\text{Ending Balance} = 750 + 206.25 = 956.25
\]
Answer: $956.25
---
Problem 10: $850 at 2.2% for 11 months
- \( P = 850 \)
- \( r = 2.2\% = 0.022 \)
- \( t = 11 \) months = \( \frac{11}{12} \) years
\[
\text{SI} = 850 \times 0.022 \times \frac{11}{12} = 17.0833
\]
\[
\text{Ending Balance} = 850 + 17.0833 = 867.0833 \approx 867.08
\]
Answer: $867.08
---
Problem 11: $20,600 at 5.5% for 4 years
- \( P = 20,600 \)
- \( r = 5.5\% = 0.055 \)
- \( t = 4 \)
\[
\text{SI} = 20,600 \times 0.055 \times 4 = 4,532
\]
\[
\text{Ending Balance} = 20,600 + 4,532 = 25,132
\]
Answer: $25,132
---
Problem 12: $100 at 2% for 9 months
- \( P = 100 \)
- \( r = 2\% = 0.02 \)
- \( t = 9 \) months = \( \frac{9}{12} = 0.75 \) years
\[
\text{SI} = 100 \times 0.02 \times 0.75 = 1.5
\]
\[
\text{Ending Balance} = 100 + 1.5 = 101.5
\]
Answer: $101.50
---
Problem 13: $4,000 at 4% for 3 years
- \( P = 4,000 \)
- \( r = 4\% = 0.04 \)
- \( t = 3 \)
\[
\text{SI} = 4,000 \times 0.04 \times 3 = 480
\]
\[
\text{Ending Balance} = 4,000 + 480 = 4,480
\]
Answer: $4,480
---
Problem 14: $7,400 at 2.2% for 9 months
- \( P = 7,400 \)
- \( r = 2.2\% = 0.022 \)
- \( t = 9 \) months = \( \frac{9}{12} = 0.75 \) years
\[
\text{SI} = 7,400 \times 0.022 \times 0.75 = 125.7
\]
\[
\text{Ending Balance} = 7,400 + 125.7 = 7,525.7
\]
Answer: $7,525.70
---
Problem 15: $1,900 at 6% for 3 years
- \( P = 1,900 \)
- \( r = 6\% = 0.06 \)
- \( t = 3 \)
\[
\text{SI} = 1,900 \times 0.06 \times 3 = 342
\]
\[
\text{Ending Balance} = 1,900 + 342 = 2,242
\]
Answer: $2,242
---
Problem 16: $7,300 at 5.2% for 3 months
- \( P = 7,300 \)
- \( r = 5.2\% = 0.052 \)
- \( t = 3 \) months = \( \frac{3}{12} = 0.25 \) years
\[
\text{SI} = 7,300 \times 0.052 \times 0.25 = 94.3
\]
\[
\text{Ending Balance} = 7,300 + 94.3 = 7,394.3
\]
Answer: $7,394.30
---
Problem 17: $210 at 5.9% for 3 years
- \( P = 210 \)
- \( r = 5.9\% = 0.059 \)
- \( t = 3 \)
\[
\text{SI} = 210 \times 0.059 \times 3 = 38.37
\]
\[
\text{Ending Balance} = 210 + 38.37 = 248.37
\]
Answer: $248.37
---
Problem 18: $35,100 at 3.6% for 3 months
- \( P = 35,100 \)
- \( r = 3.6\% = 0.036 \)
- \( t = 3 \) months = \( \frac{3}{12} = 0.25 \) years
\[
\text{SI} = 35,100 \times 0.036 \times 0.25 = 315.9
\]
\[
\text{Ending Balance} = 35,100 + 315.9 = 35,415.9
\]
Answer: $35,415.90
---
Final Answers:
\[
\boxed{
\begin{array}{ll}
1. & \$39,312 \\
2. & \$165 \\
3. & \$1,568 \\
4. & \$774 \\
5. & \$2,640 \\
6. & \$611.25 \\
7. & \$20,748 \\
8. & \$1,456 \\
9. & \$956.25 \\
10. & \$867.08 \\
11. & \$25,132 \\
12. & \$101.50 \\
13. & \$4,480 \\
14. & \$7,525.70 \\
15. & \$2,242 \\
16. & \$7,394.30 \\
17. & \$248.37 \\
18. & \$35,415.90 \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of calculating simple interest worksheet.
