5.3 Simple Interest, Compound Growth, Depreciation, and Repayment Comparisons

Simple Interest and Time-Unit Matching

Simple interest is calculated solely on the original principal amount $P$ invested or borrowed. The dollar amount of interest earned or charged per period remains constant throughout the entire term:

$$I = P \times r \times t$$

The total accumulated amount (maturity value or total repayment) $A$ at the end of the term is the principal plus the interest:

$$A = P + I = P(1 + rt)$$

where:

• $P$ is the initial principal (in dollars),
• $r$ is the interest rate per period expressed as a decimal (e.g., $6\%\text{ p.a.} = 0.06$), and
• $t$ is the length of time expressed in the exact same time unit as $r$.

Matching Rate and Time Units and Rearranging $I = Prt$

Most LANTITE questions quote interest rates per annum ($\text{p.a.}$, meaning per year) while giving the investment duration in months or days. Before multiplying, convert the time $t$ into years:

• For $m\text{ months}$, write $t = \frac{m}{12}\text{ years}$ (e.g., $3\text{ months} = 0.25\text{ years}$, $6\text{ months} = 0.5\text{ years}$, $8\text{ months} = \frac{2}{3}\text{ years}$, $9\text{ months} = 0.75\text{ years}$, and $18\text{ months} = 1.5\text{ years}$).
• You can also rearrange $I = Prt$ to solve for any missing variable: $P = \frac{I}{rt}$, $r = \frac{I}{Pt}$, or $t = \frac{I}{Pr}$. For example, if a $5,000 investment earns $450 simple interest over $2\text{ years}$, the annual interest rate is $r = \frac{450}{5{,}000 \times 2} = \frac{450}{10{,}000} = 0.045 = 4.5\%\text{ p.a.}$

Worked Example 1: Simple Interest on a Short-Term Deposit

A school P&C association invests $16,000 of building-fund reserves in a term deposit paying $4.5\%\text{ p.a.}$ simple interest for $8\text{ months}$. How much interest is earned, and what is the total maturity value?

1. Convert rate and time to matching annual units: $r = 4.5\% = 0.045\text{ per year}$, and $t = \frac{8}{12} = \frac{2}{3}\text{ years}$.
2. Calculate simple interest:

$$I = 16{,}000 \times 0.045 \times \frac{2}{3} = 720 \times \frac{2}{3} = 480$$

The term deposit earns $480 in interest. 3. Find total maturity value: $A = 16{,}000 + 480 = 16{,}480$, or $16,480.

---

Compound Interest and Multi-Period Growth

Under compound interest, the interest earned at the end of each period is added to the principal, so that in every subsequent period you earn interest on your accumulated interest. After $n$ compounding periods at a rate $r$ per period, the total balance $A$ and total compound interest $I$ are:

$$A = P(1 + r)^n \quad \text{and} \quad I = A - P$$

On LANTITE, compound growth questions typically span $2$ or $3$ periods and can be evaluated either by multiplying by $(1 + r)$ step by step or by raising the growth factor $(1 + r)$ to the power $n$.

Worked Example 2: Simple vs. Compound Interest Over 3 Years

Compare $10,000 invested for $3\text{ years}$ at $5\%\text{ p.a.}$ under simple interest versus annual compound interest.

• Simple interest: Each year earns $0.05 \times 10{,}000 = 500$ ($500). Over $3\text{ years}$, $I = 3 \times 500 = 1{,}500$ ($1,500), giving a final balance of $11,500.00.
• Compound interest (growth factor $1 + 0.05 = 1.05$):
  • End of Year 1: $10{,}000 \times 1.05 = 10{,}500.00$, or $10,500.00 ($500.00 interest)
  • End of Year 2: $10{,}500 \times 1.05 = 11{,}025.00$, or $11,025.00 ($525.00 interest)
  • End of Year 3: $11{,}025 \times 1.05 = 11{,}576.25$, or $11,576.25 ($551.25 interest)
• Comparison: Compound interest yields $11{,}576.25 - 11{,}500.00 = 76.25$ ($76.25) more than simple interest.

---

Straight-Line vs. Declining-Balance Depreciation

Schools and businesses track the falling book value of equipment (such as laptops, buses, and printing presses) over time—a process called depreciation. Two models are tested:

1. Straight-line depreciation: The asset loses a constant dollar amount $D$ (or a fixed percentage of its original purchase price $P$) every year:

$$V_n = P - nD$$

1. Declining-balance (compound) depreciation: The asset loses a fixed percentage $r$ of its current book value at the start of each year, multiplying the value by the decay factor $(1 - r)$ each period:

$$V_n = P(1 - r)^n$$

Worked Example 3: Depreciating School Laptops Over 3 Years

A school purchases a fleet of laptops for $25,000. Compare the book value after $3\text{ years}$ at $20\%\text{ p.a.}$ under both methods:

• Method A (Straight-line at $20\%\text{ p.a.}$ of original cost): Annual depreciation is $0.20 \times 25{,}000 = 5{,}000$, or $5,000 per year. After $3\text{ years}$, $V_3 = 25{,}000 - 3(5{,}000) = 10{,}000$ ($10,000).
• Method B (Declining-balance at $20\%\text{ p.a.}$): Retaining $80\%$ ($1 - 0.20 = 0.80$) of its value each year gives:
  • End of Year 1: $25{,}000 \times 0.80 = 20{,}000$ ($20,000)
  • End of Year 2: $20{,}000 \times 0.80 = 16{,}000$ ($16,000)
  • End of Year 3: $16{,}000 \times 0.80 = 12{,}800$, or $12,800 (directly $25{,}000 \times 0.80^3 = 25{,}000 \times 0.512 = 12{,}800$).

---

Hire-Purchase and Payment Plan Comparisons

When buying expensive equipment, buyers often choose between paying upfront in cash or using a hire-purchase (instalment) plan consisting of an initial deposit plus $n$ regular repayments:

$$\text{Total Instalment Price} = \text{Deposit} + (n \times \text{Periodic Repayment})$$ $$\text{Extra Cost (Effective Interest Paid)} = \text{Total Instalment Price} - \text{Upfront Cash Price}$$

Always check whether the instalments are billed monthly ($12\text{ per year}$) or fortnightly ($26\text{ per year}$) over a multi-year contract.

Worked Example 4: Comparing Cash Discount vs. Monthly Instalments

A commercial 3D printer for a school STEM lab has a marked price of $3,200. The supplier offers a $5\%$ discount for immediate upfront payment, OR a payment plan requiring a $15\%$ deposit on the marked price plus $24\text{ monthly repayments}$ of $130. How much more does the payment plan cost than the discounted cash price?

1. Discounted upfront cash price: $3{,}200 \times (1 - 0.05) = 3{,}200 \times 0.95 = 3{,}040$ ($3,040).
2. Payment plan total:
   • Deposit: $0.15 \times 3{,}200 = 480$ ($480)
   • Instalments: $24 \times 130 = 3{,}120$ ($3,120)
   • Total paid on plan: $480 + 3{,}120 = 3{,}600$ ($3,600)
3. Difference: The payment plan costs $3{,}600 - 3{,}040 = 560$ ($560) more than the discounted upfront cash option (and $400 more than the $3,200 marked price).