Simple Interest and Simple Discount Notes

Introduction to Interest

  • Interest is defined as the sum of money paid for the use of another person's money.

Key Terminology

  • EXPENSE: Money spent or cost incurred.

  • INCOME: Money received, especially on a regular basis, for work or through investments.

  • BORROWER: The person or entity that receives money with the agreement to pay it back, typically with interest.

  • LENDER: The person or entity that provides money with the expectation of repayment and earning interest.

Three Factors of Simple Interest Calculation

Simple interest calculations involve three primary factors:

Principal (PP)

  • Represents the initial amount of money loaned or borrowed.

  • Can also refer to the initial amount of money deposited or invested at the NOW (present time).

Rate (rr)

  • The rate charged by the lender for the use of their money.

  • It is typically expressed as a percent (%\%$) annually.

  • Can also be expressed as a fraction (a/ba/b$).

  • For calculations, the percentage rate must be converted to its decimal equivalent (e.g., 5%5\% becomes 0.050.05).

Time (tt)

  • Represents the period from the time the money was borrowed or invested until it will be repaid or collected, respectively.

Converting Time Units
  • Time (tt) is always given in terms of years.

  • If given in MONTHS: Divide the number of months by 1212. Time (years)=Number of MONTHS12\text{Time (years)} = \frac{\text{Number of MONTHS}}{12}

  • If given in DAYS: Divide the number of days by 365365 (for ordinary interest) or 360360 (for exact interest, often used in commercial contexts).

    • Time (years\text{Time (years} Number of DAYS360\frac{\text{Number of DAYS}}{360})

Calculating Time Between Dates

Practical examples for calculating time between specific dates are provided, such as:

  • March 10, 2018 to November 26, 2018

  • July 15, 2011 to March 20, 2015

  • April 5, 2012 to August 27, 2016

  • Calculating a date after a specified number of years, e.g., 2.282.28 years after May 11, 2012, or 4.7214.721 years after October 5, 2013.

    • Note: For exact calculations between dates, one typically counts the exact number of days in each month and handles leap years correctly, then divides by 365 or 366 for the year. (The specific method, e.g., actual/actual, 30/360, is not detailed in the transcript, but simple conversion is implied).

Simple Interest Formulas

The core formula for simple interest (II) and its derived forms are:

  • Interest (II): I=PrtI = Prt

  • Rate (rr): r=IPtr = \frac{I}{Pt}

  • Time (tt): t=IPrt = \frac{I}{Pr}

  • Principal (PP): P=IrtP = \frac{I}{rt}

Examples for Simple Interest Calculation

Example 1: Calculating Time

  • Problem: A man invested P25,000P25,000 in a bond that offers 3.5%3.5\% simple interest rate. If the income was P1,350P1,350, how long was the money invested?

  • Given: P=P25,000P = P25,000, r=0.035r = 0.035 (from 3.5%3.5\%), I=P1,350I = P1,350

  • Unknown: tt

  • Formula Applied: t=IPrt = \frac{I}{Pr}

  • Calculation: t=P1,350P25,000×0.035t = \frac{P1,350}{P25,000 \times 0.035}

Example 2: Calculating Interest (Additional Payment)

  • Problem: How much will be the additional payment if a man borrowed P75,000P75,000 for 3737 months from a bank that offers 6%6\% simple interest?

  • Given: P=P75,000P = P75,000, t=3712t = \frac{37}{12} years (from 3737 months), r=0.06r = 0.06 (from 6%6\%)

  • Unknown: II

  • Formula Applied: I=PrtI = Prt

  • Calculation: I=P75,000×0.06×3712I = P75,000 \times 0.06 \times \frac{37}{12}

Example 3: Calculating Rate

  • Problem: What rate was used if a man received an additional P800P800 after 250250 days from an investment of P53,000P53,000?

  • Given: I=P800I = P800, t=250365t = \frac{250}{365} years (from 250250 days, assuming 365-day year), P=P53,000P = P53,000

  • Unknown: rr

  • Formula Applied: r=IPtr = \frac{I}{Pt}

  • Calculation: r=P800P53,000×(250/365)r = \frac{P800}{P53,000 \times (250/365)}

Example 4: Calculating Interest Over a Period (Debt Settlement)

  • Problem: On July 21, 2009, Mrs. Rivera borrowed P128,000P128,000 from a credit union that offers 3.25%3.25\% simple interest rate. How much money should be added on his loan if she will settle everything on November 26, 2014?

  • Given: P=P128,000P = P128,000, r=0.0325r = 0.0325 (from 3.25%3.25\%)

  • Unknown: II (additional money)

  • Calculation of Time (tt): From July 21, 2009, to November 26, 2014.

    • Full years = 2014 - 2009 = 5 years.

    • Days from July 21, 2014, to November 26, 2014:

      • July: 3121=1031 - 21 = 10 days

      • August: 3131 days

      • September: 3030 days

      • October: 3131 days

      • November: 2626 days

      • Total days = 10+31+30+31+26=12810 + 31 + 30 + 31 + 26 = 128 days.

    • So, t=5+128365t = 5 + \frac{128}{365} years.

  • Formula Applied: I=PrtI = Prt

  • Calculation: I=P128,000×0.0325×(5+128365)I = P128,000 \times 0.0325 \times (5 + \frac{128}{365})

Amount or Maturity Value (FF)

  • The Amount or Maturity Value (FF) is the SUM of the PRINCIPAL (PP) and the INTEREST (II) earned.

  • Formula: F=P+IF = P + I

  • Synonyms: TOTAL PAYMENT, FUTURE EVENT, AMOUNT DUE, TOTAL COLLECTION, WITHDRAWAL, TOTAL AMOUNT.

  • Combined Formula: Since I=PrtI = Prt, we can substitute this into the maturity value formula:
    F=P+PrtF = P + Prt
    F=P(1+rt)F = P(1 + rt)

  • Derived Formula for Principal (PP) from Maturity Value: P=F(1+rt)1P = F(1 + rt)^{-1} or P=F1+rtP = \frac{F}{1 + rt}

Examples for Maturity Value Calculation

Example 5: Calculating Principal from Interest and Time

  • Problem: How much was borrowed if the interest gained was P8,500P8,500 after 44 years and 77 months and the rate used was 4.25%4.25\%?

  • Given: I=P8,500I = P8,500, t=4+712t = 4 + \frac{7}{12} years, r=0.0425r = 0.0425 (from 4.25%4.25\%)

  • Unknown: PP

  • Formula Applied: P=IrtP = \frac{I}{rt}

  • Calculation: P=P8,5000.0425×(4+7/12)P = \frac{P8,500}{0.0425 \times (4 + 7/12)}

Example 6: Calculating Maturity Value (Debt Settlement)

  • Problem: On April 7, 2009, Mr. Sison borrowed P48,000P48,000 from a credit union that offers 3.25%3.25\% simple interest rate. How much should he prepare on August 18, 2014, to settle his debt?

  • Given: P=P48,000P = P48,000, r=0.0325r = 0.0325 (from 3.25%3.25\%)

  • Unknown: FF

  • Calculation of Time (tt): From April 7, 2009, to August 18, 2014.

    • Full years = 2014 - 2009 = 5 years.

    • Days from April 7, 2014, to August 18, 2014:

      • April: 307=2330 - 7 = 23 days

      • May: 3131 days

      • June: 3030 days

      • July: 3131 days

      • August: 1818 days

      • Total days = 23+31+30+31+18=13323 + 31 + 30 + 31 + 18 = 133 days.

    • So, t=5+133365t = 5 + \frac{133}{365} years.

  • Formula Applied: F=P(1+rt)F = P(1 + rt)

  • Calculation: F=P48,000×(1+0.0325×(5+133365))F = P48,000 \times (1 + 0.0325 \times (5 + \frac{133}{365}))

Example 7: Calculating Principal from Maturity Value

  • Problem: Mr. Reyes received a total amount of P79,000P79,000 after 8080 months from an investment which bears a simple interest rate of 2.25%2.25\%. How much was invested?

  • Given: F=P79,000F = P79,000, t=8012t = \frac{80}{12} years, r=0.0225r = 0.0225 (from 2.25%2.25\%)

  • Unknown: PP

  • Formula Applied: P=F1+rtP = \frac{F}{1 + rt}

  • Calculation: P=P79,0001+0.0225×(80/12)P = \frac{P79,000}{1 + 0.0225 \times (80/12)}

Example 8: Calculating Maturity Value (Withdrawal)

  • Problem: If Mrs. Vergara deposited P118,000P118,000 on a bank that offers 3.05%3.05\% simple interest rate, how much can she withdraw after 22 years and 125125 days?

  • Given: P=P118,000P = P118,000, r=0.0305r = 0.0305 (from 3.05%3.05\%),
    t=2+125365t = 2 + \frac{125}{365} years

  • Unknown: FF

  • Formula Applied: F=P(1+rt)F = P(1 + rt)

  • Calculation: F=P118,000×(1+0.0305×(2+125/365))F = P118,000 \times (1 + 0.0305 \times (2 + 125/365))

Example 9: Calculating Principal for a Future Goal

  • Problem: Miguel wanted to have P750,000P750,000 in his account after 77 years and 100100 days. If he invested in a bank that offers 8.3%8.3\% simple interest rate, how much should he invest to attain his goal?

  • Given: F=P750,000F = P750,000, t=7+100365t = 7 + \frac{100}{365} years, r=0.083r = 0.083 (from 8.3%8.3\%)

  • Unknown: PP

  • Formula Applied: P=F1+rtP = \frac{F}{1 + rt}

  • Calculation: P=P750,0001+0.083×(7+100/365)P = \frac{P750,000}{1 + 0.083 \times (7 + 100/365)}

Accumulation and Discounting (General Concepts)

  • Accumulate: Refers to finding the future value (FF) when the present value (PP) is known. (e.g., Accumulate P85,400P85,400 for 6.46.4 years with a simple interest rate of 4.35%4.35\%: F=P85,400(1+0.0435×6.4)F = P85,400 (1 + 0.0435 \times 6.4))

  • Discount: Refers to finding the present value (PP) when the future value (FF) is known. (e.g., Discount P125,000P125,000 if it was invested 8585 months ago with a simple interest rate of 6.5%6.5\%: P=P125,000/(1+0.065×(85/12))P = P125,000 / (1 + 0.065 \times (85/12)))

These general concepts are further demonstrated with examples:

  • Accumulate: P96,000P96,000 with a rate of 7.2%7.2\% simple rate from June 5, 2009, to November 8, 2012.

    • P=P96,000P = P96,000, r=0.072r = 0.072.

    • Time from June 5, 2009, to November 8, 2012 is 33 years and 156156 days (3+1563653 + \frac{156}{365} years).

    • F=P96,000×(1+0.072×(3+156365))F = P96,000 \times (1 + 0.072 \times (3 + \frac{156}{365}))

  • Discount: P700,000P700,000 when 4.25%4.25\% simple interest rate was used and the loan lasted for 7575 months.

    • F=P700,000F = P700,000, r=0.0425r = 0.0425, t=7512t = \frac{75}{12} years.

    • P=P700,000/(1+0.0425×(75/12))P = P700,000 / (1 + 0.0425 \times (75/12))

Simple Discount

Simple discount is a method where interest is calculated based on the future value (maturity value or face value) rather than the principal received by the borrower. The borrower receives an amount less than the face value, and the difference is the discount (interest).

Simple Discount Formulas

When using a discount rate (dd), the formulas are:

  • Discount Interest (I<em>dI<em>d): I</em>d=FdtI</em>d = Fdt

  • Face Value (FF): F=IddtF = \frac{I_d}{dt}

  • Time (tt): t=IdFdt = \frac{I_d}{Fd}

  • Discount Rate (dd): d=IdFtd = \frac{I_d}{Ft}

  • Principal Received (PP) (amount borrower actually gets): P=F(1dt)P = F(1 - dt)

  • Face Value (FF) (given Principal Received): F=P(1dt)1F = P(1 - dt)^{-1} or F=P1dtF = \frac{P}{1 - dt}

Examples for Simple Discount Calculation

Example 10: Calculating Amount Received (Principal)

  • Problem: Kathy borrowed P75,000P75,000 from Bank C with a discount rate of 4.5%4.5\% for 66 years. How much will she receive?

    • Note: In simple discount problems like this, the borrowed amount stated is typically the Face Value (FF), not the principal received.

  • Given: Face Value F=P75,000F = P75,000, discount rate d=0.045d = 0.045 (from 4.5%4.5\%), t=6t = 6 years.

  • Unknown: Principal Received (PP).

  • Formula Applied: P=F(1dt)P = F(1 - dt)

  • Calculation: P=P75,000×(10.045×6)P = P75,000 \times (1 - 0.045 \times 6)

  • Answer: The transcript provides a calculation example for Mrs. Serrano with similar structure, implying the answer for a similar problem would be P=84,448P = 84,448 (from page 25's answer to a similar problem on page 24).

    • For Kathy's case: P=P75,000×(10.045×6)=P75,000×(10.27)=P75,000×0.73=P54,750P = P75,000 \times (1 - 0.045 \times 6) = P75,000 \times (1 - 0.27) = P75,000 \times 0.73 = P54,750

    • Self-correction: The P=84,448P=84,448 answer from page 25 is for the Mrs. Serrano example, not Kathy's example on page 23. Let's process Mrs. Serrano's problem.

Mrs. Serrano's Simple Discount Example
  • Problem: Mrs. Serrano borrowed P112,000P112,000 from a bank that offers 7.2%7.2\% simple discount. How much did she receive if the loan lasted for 33 years and 55 months?

  • Given: Face Value F=P112,000F = P112,000, discount rate d=0.072d = 0.072 (from 7.2%7.2\%),
    t=3+512t = 3 + \frac{5}{12} years (from 33 years and 55 months).

  • Unknown: Principal Received (PP).

  • Formula Applied: P=F(1dt)P = F(1 - dt)

  • Calculation: P=P112,000×(10.072×(3+5/12))P = P112,000 \times (1 - 0.072 \times (3 + 5/12))

  • Answer (from transcript): P=84,448P = 84,448

Example 11: Calculating Discount Rate

  • Problem: At what discount rate was used if Ben borrowed P56,000P56,000 for 5757 months from a bank and he received P45,000P45,000?

  • Given: Face Value F=P56,000F = P56,000, Principal Received P=P45,000P = P45,000,
    t=5712t = \frac{57}{12} years (from 5757 months).

  • Unknown: Discount Rate (dd).

  • Derived Formula: Start with P=F(1dt)P = F(1 - dt).

    • PF=1dt\frac{P}{F} = 1 - dt

    • dt=1PFdt = 1 - \frac{P}{F}

    • d=1(P/F)td = \frac{1 - (P/F)}{t}

  • Calculation: d=1(P45,000/P56,000)57/12d = \frac{1 - (P45,000 / P56,000)}{57/12}

  • Answer (from transcript): d=3.38%d = 3.38\% (meaning 0.03380.0338)

Example 12: Calculating Face Value (Amount Borrowed)

  • Problem: If a man received an amount of P95,000P95,000 after borrowing a certain amount of money due after 500500 days, how much was borrowed if the discount rate was 5.25%5.25\%?

  • Given: Principal Received P=P95,000P = P95,000, t=500365t = \frac{500}{365} years (from 500500 days), discount rate d=0.0525d = 0.0525 (from 5.25%5.25\%).

  • Unknown: Face Value (FF) (the amount borrowed that is due at maturity).

  • Formula Applied: F=P1dtF = \frac{P}{1 - dt}

  • Calculation: F=P95,0001(0.0525×(500/365))F = \frac{P95,000}{1 - (0.0525 \times (500/365))}

  • Answer (from transcript): F=102,471.91F = 102,471.91

Interest & Simple Discount Cheat Sheet
1. Introduction to Interest
  • Interest: Money paid for the use of another person's money.

2. Key Terminology
  • Principal (PP): Initial amount loaned, borrowed, deposited, or invested.

  • Rate (rr or dd): Percentage charged/earned per period (usually annually). Must convert to decimal for calculations.

  • Time (tt): Duration of the loan/investment, always in years.

    • Months to Years: Months12\frac{\text{Months}}{12}

    • Days to Years: Days365\frac{\text{Days}}{365} (ordinary) or Days360\frac{\text{Days}}{360} (exact/commercial).

  • Borrower: Receives money, agrees to pay back with interest.

  • Lender: Provides money, expects repayment with interest.

  • Interest (II or IdI_d): The actual money earned or paid as interest.

  • Maturity Value (FF): Also called Amount, Total Payment, Future Value. The total sum of Principal and Interest.

  • Face Value (FF): The amount due at maturity in a simple discount scenario (the amount borrowed initially before discount is applied).

  • Principal Received (PP): In simple discount, this is the actual amount the borrower gets after the discount is deducted from the Face Value.

3. Simple Interest
  • Definition: Interest calculated only on the initial principal amount.

  • Formulas:

    • Interest (II): I=PrtI = Prt

    • Principal (PP): P=IrtP = \frac{I}{rt}

    • Rate (rr): r=IPtr = \frac{I}{Pt}

    • Time (tt): t=IPrt = \frac{I}{Pr}

4. Maturity Value (FF) / Future Value
  • Definition: The total amount to be paid back or collected at the end of the term.

  • Formulas:

    • Maturity Value (FF): F=P+IF = P + I

    • Combined Formula (FF): F=P(1+rt)F = P(1 + rt)

    • Principal (PP from FF): P=F1+rtP = \frac{F}{1 + rt}

5. Simple Discount
  • Definition: Interest is calculated on the future value (face value) and deducted upfront from the amount the borrower receives.

  • Key Distinction: The amount borrowed (Face Value, FF) is typically not the amount received (Principal Received, PP).

  • Formulas (using discount rate dd):

    • Discount Interest (IdI_d): Id=FdtI_d = Fdt

    • Principal Received (PP): P=F(1dt)P = F(1 - dt)

    • Face Value (FF from PP): F=P1dtF = \frac{P}{1 - dt}

    • Discount Rate (dd): d=1(P/F)td = \frac{1 - (P/F)}{t}

6. Accumulation and Discounting (General Concepts)
  • Accumulate: Finding the future value (FF) when the present value (PP) is known (moving money 'forward' in time).

    • Example: Accumulate P85,400P85,400 for 6.46.4 years with 4.35%4.35\% simple interest: F=P85,400(1+0.0435×6.4)F = P85,400 (1 + 0.0435 \times 6.4)

  • Discount: Finding the present value (PP) when the future value (FF) is known (moving money 'backward' in time).

    • Example: Discount P125,000P125,000 if invested 8585 months ago with 6.5%6.5\% simple interest: P=P125,0001+0.065×(85/12)P = \frac{P125,000}{1 + 0.065 \times (85/12)}