Let's solve each of these
simple interest word problems step by step. We'll use the
simple interest formula:
$$
I = P \times r \times t
$$
Where:
- $ I $ = Interest earned
- $ P $ = Principal amount (initial investment or loan)
- $ r $ = Annual interest rate (in decimal form)
- $ t $ = Time in years
---
1) A new car, valued at $25,000, depreciates at 9% per year. What is the value of the car one year after purchase?
This is
depreciation, not interest. The car loses 9% of its value each year.
- Depreciation amount:
$$
9\% \text{ of } \$25,000 = 0.09 \times 25,000 = \$2,250
$$
- Value after 1 year:
$$
25,000 - 2,250 = \boxed{\$22,750}
$$
✔ Answer: \$22,750
---
2) Find the simple interest on \$6,900 at $\frac{50}{3}\%$ per year for 2 years.
Convert rate to decimal:
$$
\frac{50}{3}\% = \frac{50}{3} \div 100 = \frac{50}{300} = \frac{1}{6} \approx 0.1667
$$
Now use formula:
$$
I = P \times r \times t = 6900 \times \frac{1}{6} \times 2
$$
$$
= 6900 \times \frac{2}{6} = 6900 \times \frac{1}{3} = 2300
$$
✔ Answer: \$2,300
---
3) How much interest is earned in 5 years on \$3,000 invested at an interest rate of 9% per year?
$$
I = 3000 \times 0.09 \times 5 = 3000 \times 0.45 = \boxed{\$1,350}
$$
✔ Answer: \$1,350
---
4) In how many years will \$1,200 yield an interest of \$180 at 3% simple interest?
Use formula:
$$
I = P \times r \times t \Rightarrow 180 = 1200 \times 0.03 \times t
$$
$$
180 = 36t \Rightarrow t = \frac{180}{36} = \boxed{5} \text{ years}
$$
✔ Answer: 5 years
---
5) Sara invested \$5,000 in a bond at a yearly rate of 4%. She earned \$600 in interest. How long was the money invested?
$$
I = P \times r \times t \Rightarrow 600 = 5000 \times 0.04 \times t
$$
$$
600 = 200t \Rightarrow t = \frac{600}{200} = \boxed{3} \text{ years}
$$
✔ Answer: 3 years
---
6) Find the simple interest on \$300 at 9% per year for 4 years.
$$
I = 300 \times 0.09 \times 4 = 300 \times 0.36 = \boxed{\$108}
$$
✔ Answer: \$108
---
7) At what rate per cent per annum will \$250 amount to \$330 in 4 years?
Total amount = Principal + Interest
So:
$$
\text{Interest} = 330 - 250 = \$80
$$
Now use:
$$
I = P \times r \times t \Rightarrow 80 = 250 \times r \times 4
$$
$$
80 = 1000r \Rightarrow r = \frac{80}{1000} = 0.08 = \boxed{8\%}
$$
✔ Answer: 8% per annum
---
8) In how many years will \$600 double itself at 2% simple interest?
Double means final amount = \$1,200 → Interest = 1200 - 600 = \$600
Use:
$$
I = P \times r \times t \Rightarrow 600 = 600 \times 0.02 \times t
$$
$$
600 = 12t \Rightarrow t = \frac{600}{12} = \boxed{50} \text{ years}
$$
✔ Answer: 50 years
---
9) Find the simple interest on \$600 at 10% per year for 2 years.
$$
I = 600 \times 0.10 \times 2 = 600 \times 0.20 = \boxed{\$120}
$$
✔ Answer: \$120
---
✔ Final Answers Summary:
| Problem | Answer |
|--------|--------|
| 1 | \$22,750 |
| 2 | \$2,300 |
| 3 | \$1,350 |
| 4 | 5 years |
| 5 | 3 years |
| 6 | \$108 |
| 7 | 8% |
| 8 | 50 years |
| 9 | \$120 |
Let me know if you'd like this formatted as a printable worksheet!
Parent Tip: Review the logic above to help your child master the concept of interest word problems worksheet.
