Grade 11 Simple Interest Comprehensive Study Guide

Fundamental Concepts of Interest

  • Definition of Interest: Interest is an amount that a person gets or pays on top of the original investment or loan. From a borrower's perspective, it is the money paid for the use of money. It can also function as a mechanism for imposing a penalty on a borrower for failing to pay a matured financial obligation at a specific time.

  • Parties Involved in Interest Transactions:

    • Lender or Creditor: Refers to the party lending money or extending credit. This party expects the money to earn income from the transaction. For example, in a bank deposit, the depositor is considered the creditor.
    • Borrower or Debtor: Refers to the party using the money or credit. This party expects future expenses at the cost of using the capital. For example, in a bank deposit, the bank is considered the debtor because it is obliged to pay interest to the depositor.

Core Elements of Interest Calculation

Interest is computed using three primary elements:

  • Principal (PP): The amount of money extended for credit or the amount of money deposited in a bank for safekeeping.
  • Interest Rate (rr): The charged amount for using the money over a certain period. It is commonly expressed as a percentage per year (% per year\% \text{ per year}) but must be converted to decimal form for calculations. Unless specified otherwise, the rate is assumed to be annual.
  • Time (tt): The period covered from the moment the principal is borrowed until its due date. The reference point for time is typically 1 year1 \text{ year} or 12 months12 \text{ months}.
  • Maturity Date: The specific due date for the payment of the principal amount.

Simple Interest Theory and Formulas

  • Simple Interest: Refers to interest that is computed only on the original principal during the entire period or duration of borrowing.

  • Simple Interest Formula:     I=PrtI = Prt     Where:

    • II = Interest amount
    • PP = Principal amount
    • rr = Simple interest rate (in decimal)
    • tt = Time (written in years)
  • Maturity Value (Future Value): This is the sum of the principal and the interest accumulated. It represents the total amount to be paid or received at the end of the term.

  • Maturity Value Formulas:

    • M=P+IM = P + I
    • M=P(1+rt)M = P(1 + rt)
  • Derivation of Maturity Value:

    1. Start with the definition: M=P+IM = P + I
    2. Substitute the interest formula (I=PrtI = Prt): M=P+(Prt)M = P + (Prt)
    3. Apply the distributive property (factor out PP): M=P(1+rt)M = P(1 + rt)

Step-by-Step Calculation Examples

Example 1: Basic Simple Interest and Maturity Value

Scenario: On April 1, 2017, Angela borrowed P200000P200\,000 from Prime Lending at 2%2\% interest payable in 1 year1 \text{ year}.

Calculation for Interest:

  • P=200000P = 200\,000
  • r=0.02r = 0.02
  • t=1t = 1
  • I=200000×0.02×1I = 200\,000 \times 0.02 \times 1
  • I=4000I = 4\,000
  • Observation: Angela must pay an additional P4000P4\,000 as interest.

Calculation for Maturity Value:

  • M=P+IM = P + I
  • M=200000+4000M = 200\,000 + 4\,000
  • M=204000M = 204\,000
  • Observation: Angela must pay a total of P204000P204\,000 after one year.
Example 2: Non-Integer Time Periods

Scenario: Russel borrowed P15000P15\,000 payable after 2 years and 9 months2 \text{ years and } 9 \text{ months} with a simple interest rate of 8%8\%.

Step 1: Convert Time to Years Interest is stated as a yearly rate, so time must be converted:

  • 2 years and 9 months=2912 years2 \text{ years and } 9 \text{ months} = 2 \frac{9}{12} \text{ years}
  • 234 years=2.75 years2 \frac{3}{4} \text{ years} = 2.75 \text{ years}

Step 2: Find Maturity Value

  • P=15000P = 15\,000
  • r=0.08r = 0.08
  • t=2.75t = 2.75
  • M=15000[1+(0.08)(2.75)]M = 15\,000[1 + (0.08)(2.75)]
  • M=15000(1+0.22)M = 15\,000(1 + 0.22)
  • M=15000(1.22)M = 15\,000(1.22)
  • M=18300M = 18\,300
  • Conclusion: Russel needs to repay P18300P18\,300.
Example 3: Comparison of Investment Portfolios

Scenario: JM has P45000P45\,000 and must choose between two annual-rate options:

  • Option A: 5%5\% rate for a 2-year2\text{-year} portfolio.
  • Option B: 4%4\% rate for a 21-month21\text{-month} portfolio.
Example 4: Business Financing

Scenario: Michael borrowed P450000P450\,000 at 3%3\% simple interest. How much will he pay after 5 years5 \text{ years}?

  • P=450000P = 450\,000
  • r=0.03r = 0.03
  • t=5t = 5
  • M=450000[1+(0.03)(5)]M = 450\,000[1 + (0.03)(5)]
  • M=450000(1.15)M = 450\,000(1.15)
  • M=517500M = 517\,500

Practice Problems

  1. Maya: Deposited P18000P18\,000 at 4%4\% interest for 3 years3 \text{ years}.

    • I=18000×0.04×3=P2160I = 18\,000 \times 0.04 \times 3 = P2\,160
  2. Anthony: Paid deep surcharge of P10500P10\,500 on a P70000P70\,000 loan after 5 years5 \text{ years}. Seek the interest rate.

    • 10500=70000×r×510\,500 = 70\,000 \times r \times 5
    • 10500=350000r10\,500 = 350\,000r
    • r=0.03 or 3%r = 0.03 \text{ or } 3\%
  3. Sam: Invested P200000P200\,000 at 8%8\% annual rate. Total money after 6 months6 \text{ months} (0.5 years0.5 \text{ years})?

    • M=200000[1+(0.08)(0.5)]=200000(1.04)=P208000M = 200\,000[1 + (0.08)(0.5)] = 200\,000(1.04) = P208\,000
  4. Annie: Wants to earn P8000P8\,000 at 6.4%6.4\% yearly over 2 years2 \text{ years}. Required deposit?

    • 8000=P×0.064×28\,000 = P \times 0.064 \times 2
    • 8000=0.128P8\,000 = 0.128P
    • P=P62500P = P62\,500
  5. Nina: Earned P4500P4\,500 from P75000P75\,000 over 18 months18 \text{ months} (1.5 years1.5 \text{ years}). Interest rate?

    • 4500=75000×r×1.54\,500 = 75\,000 \times r \times 1.5
    • 4500=112500r4\,500 = 112\,500r
    • r=0.04 or 4%r = 0.04 \text{ or } 4\%
  6. Pauline: Investment of P24000P24\,000 grows to P28500P28\,500 at 15%15\%. Time needed?

    • I=2850024000=4500I = 28\,500 - 24\,000 = 4\,500
    • 4500=24000×0.15×t4\,500 = 24\,000 \times 0.15 \times t
    • 4500=3600t4\,500 = 3\,600t
    • t=1.25 yearst = 1.25 \text{ years} (or 1 year and 3 months1 \text{ year and } 3 \text{ months})
  7. Jasmine: Loan fee of P312.50P312.50 after 3 months3 \text{ months} (0.25 years0.25 \text{ years}) at 10%10\%. Borrowed amount?

    • 312.50=P×0.10×0.25312.50 = P \times 0.10 \times 0.25
    • 312.50=0.025P312.50 = 0.025P
    • P=P12500P = P12\,500