Identifying Word Problems in Simple and Compound Interest
When it comes to solving simple and compound interest problems, even the best of us can make small mistakes that lead to big errors in our answers. These concepts are not only important for exams but also play a vital role in understanding real-life financial scenarios.
| Scenario | Description | Example | Solution |
|---|---|---|---|
| 1. Finding the Interest | Calculate the total interest earned or paid over a specific period. | Common: John deposited $2,000 at 5% simple interest for 3 years. How much interest will he earn? | Solution: SI = $\frac{P \times R \times T}{100}$ SI= $\frac{2000 \times 5 \times 3}{100}$ = 300. John will earn $300 in interest. |
| Unique: Maya invested $3,000 in a bank offering compound interest at 6% annually for 2 years. | Solution: CI = P⋅ (1 + $\frac{R}{100})^{T}$ −P = 3000 $\times$ (1 + $\frac{6}{100})^{2}$ −3000 = 370.80 Maya will earn $370.80 in interest. | ||
| 2. Finding Total Amount | Calculate the total amount (Principal + Interest) after a specific period. | Common: A loan of $1,500 is taken at 10% simple interest for 4 years. What is the total amount to be paid? | Solution: SI = $\frac{P \times R \times T}{100}$ = $\frac{1500 \times 10 \times 4}{100}$ = 600. Total Amount= P+SI =1500+600 =2100 |
| Unique: Alex saved $5,000 in a savings account that offers 8% annual compound interest for 3 years. | Solution: CI = P⋅ (1 + $\frac{R}{100})^{T}$ −P =5000 $\times$ (1 + $\frac{8 }{100})^{3}$ −5000 =1259.71 Total Amount = P+CI = 5000+1259.71 =6259.71 | ||
| 3. Determining the Principal | Determine the initial amount (Principal) when the interest and total amount are known. | Common: The total amount after 2 years at 6% simple interest is $1,120. What was the principal? | Solution: SI = A−P, so 1120 = P + $\frac{P \times 6 \times 2}{100}$ Solving, P = 1000 The principal was $1,000. |
| Unique: Lila earned $150 interest in 1 year at 5% compound interest. What was the principal amount? | Solution: CI = P⋅ (1 + $\frac{R}{100})^{T}$ −P, so 150= P $\times$ (1 + $\frac{5}{100})^{1}$ −P = 3000 The principal was $3000 | ||
| 4. Finding the Rate (%) | Calculate the rate of interest when other variables are known. | Common: A \$1,000 deposit earned \$200 simple interest in 4 years. What was the rate of interest? | Solution: SI = $\frac{P \times R \times T}{100}$ so 200= $\frac{1000 \times R \times 4}{100}$ Solving R=5% |
| Unique: A loan of \$4,000 compounded annually became \$4,840 in 2 years. What was the annual rate of interest? | Solution: CI = P⋅ (1 + $\frac{R}{100})^{T}$ −P, so 840= 4000 $\times$ (1+$\frac{R}{100})^{2}$ −4000. Solving R=10% | ||
| 5. Determining the Time | Calculate the time required for the interest or total amount to reach a specific value. | Common: A principal of \$1,000 earned \$300 in simple interest at 5%. How long was it invested? | Solution: SI = $\frac{P \times R \times T}{100}$ so 300= $\frac{1000 \times 5 \times T}{100}$ Solving T= 6 years. |
| Unique: Chris invested \$2,500 at 10% compound interest, and it grew to \$3,025. How many years was it compounded? | Solution: CI = P⋅ (1 + $\frac{R}{100})^{T}$ −P, so 525= 2500 $\times$ (1+$\frac{10}{100})^{T}$ −2500. Solving T=2 years. | ||
| 6. Comparing Interest Types | Understand the difference in returns between simple and compound interest for the same principal, rate, and time. | Common: Compare the interest earned on $1,000 at 5% for 3 years under simple and compound interest. | Solution: SI = $\frac{1000 \times 5 \times 3}{100}$ =150 CI = 1000 (1 + $\frac{5}{100})^{3}$ −1000, =157.63 CI earns slightly more. |
| Unique: A bank offers two options: 12% simple interest for 5 years or 10% compound interest for 5 years. Which is better for $2,000? | Solution: SI = $\frac{2000 \times 12 \times 5}{100}$ =1200 CI = 2000 (1 + $\frac{10}{100})^{5}$ −2000 =1221.02 Compound interest is better. | ||
| 7. Splitting Time Periods | Handle cases where the rate changes or compounding frequency varies over time. | Common: A principal of \$1,000 earns 5% for the first 2 years and 6% for the next 3 years. What is the total simple interest? | Solution: SI$_{1}$ = $\frac{1000 \times 5 \times 2}{100}$ =100 SI$_{2}$ = $\frac{1000 \times 6 \times 3}{100}$ =180 Total SI= 100+180 = 280 |
| Unique: A loan of $5,000 is compounded at 5% annually for the first year and 8% annually for the second year. What is the final amount? | Solution: A= 5000 $\times$ (1 + $\frac{5}{100})$ (1 + $\frac{8}{100})$ =5834. The final answer is $5,834. |
Keep revisiting these tips, double-check your calculations, and focus on understanding the formulas step by step. Remember, every mistake is a stepping stone to learning. With a little practice and a lot of determination, you’ll ace these problems and even impress yourself with your progress! Happy learning!
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