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WealthWiseGuys Practical Mathematics

Chapter 10

Simple Interest and Borrowing

Translate time, rates, and fees into the true cost of a short-term loan.

Borrowing can move a purchase into today, but repayment uses tomorrow's money. Learn to calculate simple interest, solve for missing loan terms, and compare offers on the same basis. All people, prices, rates, and loan terms in this chapter are hypothetical; round final money amounts to cents and retain unrounded intermediate values unless stated otherwise.

10.1

Principal, Rate, and Time

Learning objectives

After this section you will be able to:

  1. Calculate simple interest and maturity value.
  2. Convert months and days to fractions of a year.
  3. Distinguish a stated interest rate from a dollar interest charge.

Real-life situation

A bike for a Philadelphia commute

A student considers a short-term loan for a used bicycle. The disclosure separates the amount borrowed from the cost of using it.

A bike for a Philadelphia commute

  • Principal: $600.00
  • Simple annual rate: 8%
  • Term: 9 months
  • Interest: $36.00; repayment: $636.00
Figure 10.1 A bike for a Philadelphia commute (hypothetical)

The concept

Simple interest is a charge for using an original amount of money, called principal. Unlike compound interest, it does not add earlier interest to the base for the next interest calculation. A lender quoting 8% annually is quoting a rate, not an $8 charge on every loan.

The formula I = P × r × t requires matching units. Write an annual percent as a decimal and express time in years. Nine months is 9 ÷ 12 year, so an annual rate applies for only three-fourths of a year.

For day-based loans, the agreement must specify a day-count convention. An actual/365 calculation divides actual elapsed days by 365; an actual/360 calculation divides those same days by 360. Neither denominator may be assumed just because a year usually has 365 days.

Maturity value is principal plus interest when the entire loan is repaid at the end. This model assumes no interim payments or fees. Read the payment terms before applying it to a real borrowing arrangement.

Principal and repayment

Principal$ 600
Interest$ 36
Repayment$ 636

Bars share a zero baseline.

Figure 10.1.V Principal and repayment. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • principal

    The original amount of money borrowed or invested, before interest.

  • simple interest

    Interest calculated only on the original principal, not on interest already earned.

  • interest rate

    The percent used to calculate how much interest is earned or charged.

Key terms

Principal
The original sum borrowed or invested.
Simple interest
Interest calculated on original principal only.
Annual rate
Interest per year expressed as a fraction of principal.
Term
The length of the loan.
Maturity value
Principal plus interest payable at the end of the term.

Formulas & rules

  • Simple interest

    I = P × r × t

  • Maturity value

    M = P + I

  • Month conversion

    t = months ÷ 12

  • Actual/365 time

    t = actual days ÷ 365

Worked examples

Example 1The bicycle loan

Find interest and repayment on $600 at 8% simple annual interest for 9 months.

Solution

  1. 1Convert 8% to r = 0.08.
  2. 2Convert time: t = 9 ÷ 12 = 0.75 year.
  3. 3I = 600 × 0.08 × 0.75 = $36.00.
  4. 4M = 600 + 36 = $636.00.

Answer: Interest is $36.00 and maturity value is $636.00.

Example 2A day-count convention

A $1,200 loan charges 6% simple annual interest for 45 actual days using actual/365. Find the amount due.

Solution

  1. 1Use r = 0.06 and t = 45 ÷ 365 year.
  2. 2I = 1,200 × 0.06 × 45 ÷ 365 = $8.876712… .
  3. 3Round the interest to $8.88.
  4. 4M = 1,200 + 8.88 = $1,208.88.

Answer: The borrower owes $1,208.88.

Practice exercises 10.1

  1. Find interest on $800 at 5% for 6 months.
  2. Find maturity value on $1,500 at 4% for 2 years.
  3. A $900 loan runs 60 actual days at 9%, actual/360. Find interest.
  4. Compare actual/360 and actual/365 interest on $730 at 10% for 36 actual days.
  5. A friend calculates 500 × 0.06 × 4 for a four-month loan. Correct the interest.

Recap quiz 10.1

1. Interest on $200 at 6% for one year is:
2. Three months expressed in years is:
3. A loan has principal $400 and interest $18. What is its maturity value?

Section 10.1 recap

  • Match annual rates with time measured in years.
  • Simple interest uses original principal; maturity value adds principal and interest.
Answer key with explanations: Section 10.1

Practice exercises

  1. t = 6 ÷ 12 = 0.5; I = 800 × 0.05 × 0.5 = $20.00.
  2. I = 1,500 × 0.04 × 2 = $120.00; M = 1,500 + 120 = $1,620.00.
  3. t = 60 ÷ 360 = 1/6; I = 900 × 0.09 × 1/6 = $13.50.
  4. Actual/360: 730 × 0.10 × 36 ÷ 360 = $7.30. Actual/365: 730 × 0.10 × 36 ÷ 365 = $7.20. The first costs $0.10 more.
  5. Four months is 4 ÷ 12 year, not 4 years. I = 500 × 0.06 × 4 ÷ 12 = $10.00, not $120.00.

Recap quiz

  1. $12.00: 200 × 0.06 × 1 = $12.00; repayment would be $212.00.
  2. 0.25: 3 ÷ 12 = 0.25 year.
  3. $418: M = P + I = 400 + 18 = $418.
10.2

Solving for Missing Loan Terms

Learning objectives

After this section you will be able to:

  1. Rearrange the simple-interest formula.
  2. Find principal, annual rate, or time from a stated interest charge.
  3. Check answers by substituting into the original equation.

Real-life situation

Working backward from a charge

A student sees a $45 interest charge on a $1,000 loan lasting nine months. Working backward reveals the annual rate.

Working backward from a charge

  • P = $1,000.00; I = $45.00
  • t = 9 ÷ 12 = 0.75 year
  • r = 45 ÷ (1,000 × 0.75) = 0.06 = 6%
Figure 10.2 Working backward from a charge (hypothetical)

The concept

The simple-interest equation can answer more than the interest question. If three quantities are known, isolate the fourth using inverse operations. Dividing both sides by the same nonzero product preserves equality.

To find principal, divide interest by rate times time. To find the annual rate, divide interest by principal times time. The resulting rate is a decimal; multiply by 100 to express it as a percent.

To find time, divide interest by principal times annual rate. That answer is in years. Multiply years by 12 for months, or by the agreement's day-count denominator for days. A fractional month is a mathematical result, not permission to change a contractual due date.

Check a rearranged formula by substitution. Also check scale: charging $45 on $1,000 over less than a year should imply an annual rate above 4.5%, not below it.

Recover the annual interest rate

  1. 1

    Known quantities

    I = $45; P = $1,000; t = 0.75 year

  2. 2

    Isolate rate

    r = I ÷ (P × t)

  3. 3

    Result

    45 ÷ 750 = 0.06 = 6%

Figure 10.2.V Recover the annual interest rate. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • loan term

    How long a loan lasts before it must be fully repaid.

  • total repayment cost

    The full amount paid over a loan's life — principal plus all interest and fees.

  • APR

    Annual percentage rate — the yearly cost of borrowing, including interest and some fees, shown as a percent.

Key terms

Inverse operation
An operation that undoes another, such as division undoing multiplication.
Unknown
A quantity to be determined.
Decimal rate
A percent divided by 100.
Substitution
Replacing symbols with values to check an equation.

Formulas & rules

  • Find principal

    P = I ÷ (r × t)

  • Find rate

    r = I ÷ (P × t)

  • Find time

    t = I ÷ (P × r)

Worked examples

Example 1Recover the annual rate

A $1,000 loan earns $45 simple interest in 9 months. Find the annual rate.

Solution

  1. 1t = 9 ÷ 12 = 0.75 year.
  2. 2r = 45 ÷ (1,000 × 0.75) = 0.06.
  3. 3Convert 0.06 to 6%.
  4. 4Check: 1,000 × 0.06 × 0.75 = $45.00.

Answer: The annual simple-interest rate is 6%.

Example 2How long was the money borrowed?

A $750 principal incurs $30 interest at 8% simple annual interest. Find the term in months.

Solution

  1. 1P × r = 750 × 0.08 = 60 dollars per year.
  2. 2t = 30 ÷ 60 = 0.5 year.
  3. 3Months = 0.5 × 12 = 6.
  4. 4Check: 750 × 0.08 × 6 ÷ 12 = $30.00.

Answer: The term is 6 months.

Practice exercises 10.2

  1. Find principal if interest is $24 at 6% for 8 months.
  2. A $500 loan costs $15 interest over 6 months. Find the annual rate.
  3. How many months produce $48 interest on $1,200 at 8%?
  4. A $2,000 loan has maturity value $2,090 after 9 months. Find its annual simple rate.
  5. At 9% actual/360, how many days produce $18 interest on $800?

Recap quiz 10.2

1. Which formula finds principal?
2. $20 interest on $400 over half a year implies which annual rate?
3. A calculated term of 1.5 years equals:

Section 10.2 recap

  • Rearrange before substituting values.
  • Rate answers are decimals until converted to percentages; time answers follow the rate unit.
Answer key with explanations: Section 10.2

Practice exercises

  1. t = 8 ÷ 12; P = 24 ÷ (0.06 × 8 ÷ 12) = $600.00.
  2. r = 15 ÷ (500 × 0.5) = 0.06 = 6% annually.
  3. t = 48 ÷ (1,200 × 0.08) = 0.5 year; 0.5 × 12 = 6 months.
  4. I = 2,090 − 2,000 = $90; r = 90 ÷ (2,000 × 0.75) = 0.06 = 6%.
  5. t = 18 ÷ (800 × 0.09) = 0.25 year; days = 0.25 × 360 = 90.

Recap quiz

  1. I ÷ (r × t): Dividing I = P × r × t by r × t isolates P.
  2. 10%: r = 20 ÷ (400 × 0.5) = 0.10 = 10%.
  3. 18 months: 1.5 × 12 = 18 months.
10.3

Fees, Proceeds, and Fair Loan Comparisons

Learning objectives

After this section you will be able to:

  1. Calculate usable proceeds and total borrowing cost.
  2. Compare offers for equal cash needs and equal terms.
  3. Distinguish a simple annualized cost indicator from a disclosed APR.

Real-life situation

Two ways to fund a repair

A neighborhood repair worker needs $1,000 for one year. Both offers provide the same usable cash, but one adds an origination fee.

Two ways to fund a repair

  • Offer A: $1,000 cash; 8% simple interest; no fee
  • Offer B: $1,000 cash; 6% simple interest; $30 fee paid at maturity
  • A repayment: $1,080.00; B repayment: $1,090.00
Figure 10.3 Two ways to fund a repair (hypothetical)

The concept

A low stated rate does not necessarily mean a low total cost. Origination fees, processing fees, and other required charges add to interest. Identify which charges are required, when they are paid, and whether they are financed.

Usable proceeds are the money actually available for the intended purchase. A $1,000 face amount with a $40 fee withheld delivers only $960. If the borrower needs $1,000 cash, that loan is not directly comparable with a loan that delivers the full amount.

For a single repayment, a classroom annualized cost indicator divides total borrowing cost by usable proceeds and by years. It is useful for transparent comparisons under the stated assumptions, but it is not a substitute for a legally defined annual percentage rate. APR calculations and disclosures depend on the cash-flow pattern and applicable rules.

Compare equal usable cash, equal terms, and all required payments. Affordability is a separate question: a mathematically cheaper loan can still be difficult to repay. All fee arrangements here are hypothetical examples, not statements of current lending law.

Compare total repayments, including fees

Offer A$ 1,080
Offer B$ 1,090

Bars share a zero baseline.

Figure 10.3.V Compare total repayments, including fees. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • loan

    Money borrowed that must be repaid, usually with interest.

  • monthly payment

    The amount paid each month toward a loan or installment plan.

  • credit

    The ability to buy now and pay later, or money borrowed that must be repaid.

Key terms

Origination fee
A charge for arranging a loan.
Usable proceeds
Cash the borrower can actually use after withheld charges.
Borrowing cost
Total required outflow minus usable cash received.
APR
A standardized annual cost measure whose calculation follows applicable disclosure rules.
Single-payment loan
A loan repaid in one payment at maturity.

Formulas & rules

  • Net proceeds

    Proceeds = face amount − withheld fees

  • Single-payment cost

    Cost = maturity payment + separate required fees − proceeds

  • Classroom annualization

    Annualized indicator = cost ÷ proceeds ÷ t

Worked examples

Example 1Compare equal proceeds

Compare the two one-year $1,000-cash offers in the figure.

Solution

  1. 1A interest = 1,000 × 0.08 × 1 = $80.00.
  2. 2A maturity payment = 1,000 + 80 = $1,080.00.
  3. 3B interest = 1,000 × 0.06 × 1 = $60.00.
  4. 4B maturity payment = 1,000 + 60 + 30 = $1,090.00.
  5. 5Cost difference = 90 − 80 = $10.00.

Answer: Offer A costs $10.00 less despite its higher stated rate.

Example 2A withheld fee

A one-year loan has face amount $1,000, a $40 withheld fee, and $80 interest due at maturity. Find proceeds, cost, and the classroom annualized indicator.

Solution

  1. 1Proceeds = 1,000 − 40 = $960.00.
  2. 2Maturity payment = 1,000 + 80 = $1,080.00.
  3. 3Cost = 1,080 − 960 = $120.00.
  4. 4Indicator = 120 ÷ 960 ÷ 1 = 0.125 = 12.5%.

Answer: Usable cash is $960.00; cost is $120.00; the indicator is 12.5%, not a claimed legal APR.

Practice exercises 10.3

  1. A $700 face loan withholds $25 and requires $735 at maturity. Find proceeds and cost.
  2. A $500-cash six-month loan requires $525 at maturity with no other fees. Find its classroom annualized indicator.
  3. Compare one-year $800-cash loans: A charges 7% with no fees; B charges 5% plus $20 due at maturity.
  4. A lender deducts $30 from a $600 loan. Does it fully fund a $600 expense?
  5. A one-year $900-cash loan charges 4% interest plus a $15 fee at maturity. Find repayment and total cost.

Recap quiz 10.3

1. A $1,000 face loan withholds $50. Usable proceeds are:
2. Which is essential for a fair comparison?
3. $600 received and $648 repaid one year later means what cost?

Section 10.3 recap

  • Compare total costs, not rates alone.
  • A withheld fee reduces usable cash; a classroom annualized indicator is not automatically the disclosed APR.
Answer key with explanations: Section 10.3

Practice exercises

  1. Proceeds = 700 − 25 = $675.00; cost = 735 − 675 = $60.00.
  2. Cost = 525 − 500 = $25; t = 0.5; indicator = 25 ÷ 500 ÷ 0.5 = 0.10 = 10%.
  3. A cost = 800 × 0.07 = $56.00. B cost = 800 × 0.05 + 20 = $60.00. A costs $4.00 less.
  4. Proceeds = 600 − 30 = $570.00. The shortfall is 600 − 570 = $30.00, so no.
  5. Interest = 900 × 0.04 = $36.00; repayment = 900 + 36 + 15 = $951.00; cost = 951 − 900 = $51.00.

Recap quiz

  1. $950: 1,000 − 50 = $950.
  2. Equal usable cash and term: Equal cash and term keep the comparison focused on total cost.
  3. $48: 648 − 600 = $48 of borrowing cost.

Chapter 10 Organizer and Review

Chapter Organizer

Every section at a glance: key formulas, key terms, and the main idea.

  • 10.1 Principal, Rate, and Time

    Key formulas

    • Simple interest: I = P × r × t
    • Maturity value: M = P + I
    • Month conversion: t = months ÷ 12
    • Actual/365 time: t = actual days ÷ 365

    Key terms

    Principal, Simple interest, Annual rate, Term, Maturity value

    Remember

    Match annual rates with time measured in years.

  • 10.2 Solving for Missing Loan Terms

    Key formulas

    • Find principal: P = I ÷ (r × t)
    • Find rate: r = I ÷ (P × t)
    • Find time: t = I ÷ (P × r)

    Key terms

    Inverse operation, Unknown, Decimal rate, Substitution

    Remember

    Rearrange before substituting values.

  • 10.3 Fees, Proceeds, and Fair Loan Comparisons

    Key formulas

    • Net proceeds: Proceeds = face amount − withheld fees
    • Single-payment cost: Cost = maturity payment + separate required fees − proceeds
    • Classroom annualization: Annualized indicator = cost ÷ proceeds ÷ t

    Key terms

    Origination fee, Usable proceeds, Borrowing cost, APR, Single-payment loan

    Remember

    Compare total costs, not rates alone.

Cumulative Review

Mixed practice from this chapter and the two chapters before it.

  1. Section 10.1

    A friend calculates 500 × 0.06 × 4 for a four-month loan. Correct the interest.

    Show answer

    Four months is 4 ÷ 12 year, not 4 years. I = 500 × 0.06 × 4 ÷ 12 = $10.00, not $120.00.

  2. Section 10.2

    At 9% actual/360, how many days produce $18 interest on $800?

    Show answer

    t = 18 ÷ (800 × 0.09) = 0.25 year; days = 0.25 × 360 = 90.

  3. Section 10.3

    A one-year $900-cash loan charges 4% interest plus a $15 fee at maturity. Find repayment and total cost.

    Show answer

    Interest = 900 × 0.04 = $36.00; repayment = 900 + 36 + 15 = $951.00; cost = 951 − 900 = $51.00.

  4. Review of Section 8.3

    Find contribution when price is $15 and variable cost is $9.

    Show answer

    15 − 9 = $6.00 per unit.

  5. Review of Section 9.3

    Gross is $450 and total employee deductions are $65. Find net pay.

    Show answer

    450 − 65 = $385.00.

All prices, stores, people, and accounts in this chapter are hypothetical and for teaching only. Product references are generic categories; no brand endorses or is endorsed by this textbook.

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