4  Simple and Compound Interest

4.1 Introduction

Interest is the extra amount paid for the use of money, or the return earned on investments.
This chapter builds a complete toolkit:
- Simple Interest (SI)
- Compound Interest (CI)
- Relations between SI and CI
- Varying rates and compounding frequency
- Growth and depreciation models
- Installments and annuities
- Present value and compound discount
- Effective interest rate
- Standard exam problems and shortcuts

4.2 Core Definitions

  • Principal (P): Original sum of money.
  • Rate (R): Annual rate of interest (% per annum).
  • Time (T): Duration (years, unless stated otherwise).
  • Interest (I): Extra money paid or earned.
  • Amount (A): Total = Principal + Interest.

4.3 Simple Interest (SI)

Formula:
\[ SI = \frac{P \times R \times T}{100} \]

Amount:
\[ A = P\left(1+\frac{RT}{100}\right) \]

Example: Find SI on 6000 at 10% for 4 years.
\(SI=6000\times10\times4/100=2400\). Amount=8400.

4.4 Compound Interest (CI)

Formula:
\[ A=P\left(1+\frac{R}{100}\right)^T, \quad CI=A-P \]

Example: P=6000, R=10%, T=2 yrs.
\(A=6000(1.1)^2=7260\), CI=1260.

4.5 SI vs CI

  • For 1 year: SI = CI.
  • For multiple years: CI > SI.
  • Difference after 2 years:
    \(CI-SI=P(R/100)^2\).
  • Difference after 3 years:
    \(CI-SI=P\big[(1+R/100)^3-(1+3R/100)\big]\).

4.6 Non-Annual Compounding

  • Half-yearly:
    \(A=P(1+R/200)^{2T}\)
  • Quarterly:
    \(A=P(1+R/400)^{4T}\)
  • Monthly:
    \(A=P(1+R/1200)^{12T}\)

Example: 8000 at 12% pa, 1 year, quarterly compounding.
\(A=8000(1.03)^4=9012.55\).

4.7 Growth and Depreciation

  • Growth at \(a\%\):
    \(V=V_0(1+a/100)^T\)
  • Depreciation at \(d\%\):
    \(V=V_0(1-d/100)^T\)

Example: Machine 60000 depreciates 10% pa for 2 yrs → \(60000(0.9)^2=48600\).

4.8 Varying Rates

If rate changes each year, multiply factors:
\[ A=P(1+r_1/100)(1+r_2/100)(1+r_3/100)\cdots \]

Example: 5000 at 10%, 12%, 15% for 3 yrs.
\(A=5000(1.1)(1.12)(1.15)=7161\).

4.9 Installments and Annuities

Loan \(P\) repaid in \(n\) equal annual installments at \(R\%\):

Installment =
\[ \frac{P\cdot(R/100)}{1-(1+R/100)^{-n}} \]

Example: Loan 20000, 2 annual installments, R=10%.
Installment ≈ 11573.

4.10 Present Value and Compound Discount

  • Present Value (PV): Value today of sum \(F\) due in \(T\) years:
    \[ PV=\frac{F}{(1+R/100)^T} \]

  • Compound Discount (CD): \(CD=F-PV\).

Example: True value of 1000 due in 2 yrs at 10%:
PV=826.45. Discount=173.55.

4.11 Effective Interest Rate

Nominal rate compounded \(n\) times/year gives:
\[ R_{eff}=(1+\tfrac{R}{100n})^n-1 \]

Example: 12% pa compounded quarterly →
\((1+0.03)^4-1=12.55\%\).

4.12 Solved Examples

  1. SI on 5000 at 8% for 3 yrs = 1200.
  2. CI on 5000 at 8% for 3 yrs = 1298.56.
  3. CI–SI difference on 10000 at 12% for 2 yrs = 144.
  4. Population 80000 grows 5% pa, 3 yrs → 92610.
  5. Machine 60000 depreciates 10% pa, 2 yrs → 48600.
  6. Find R: 5000→6050 in 2 yrs. \((1+R/100)^2=1.21\), so R=10%.
  7. True value of 2000 due in 3 yrs at 10% = 1502.63.

4.13 Common Traps

  • Mistaking SI for CI.
  • Ignoring compounding frequency.
  • Using difference formula beyond 2–3 yrs.
  • “Per annum” vs “per half-year” confusion.
  • Population/depreciation → always CI.
  • Compound discount ≠ trade discount.

4.14 Practice Set – Level 1

  1. SI on 3000 at 6% for 5 yrs.
  2. CI on 3000 at 6% for 5 yrs.
  3. CI–SI difference on 2000 at 5% for 2 yrs.
  4. Amount on 4000 at 10% for 1 yr, half-yearly compounding.
  5. In how many yrs will 8000 double at 10% CI?

4.15 Practice Set – Level 2

  1. Population 150000 increases 4% pa, 2 yrs later?
  2. Car value 300000 depreciates 15% pa, 2 yrs later?
  3. A sum becomes 12100 in 2 yrs at 10% CI. Find P.
  4. SI on a sum at 8% for 4 yrs = 1600. Find P.
  5. CI–SI difference on 5000 at R% for 2 yrs = 25. Find R.

4.16 Practice Set – Level 3

  1. A sum quadruples in 12 yrs at CI. In how many yrs will it become 16 times?
  2. A sum doubles in 10 yrs at CI. Find R.
  3. CI for 2 yrs at 10% is 210. Find P.
  4. True value of 5000 due in 5 yrs at 8%.
  5. Effective rate for 18% compounded half-yearly.

4.17 Answer Key (Outline)

Level 1: 1) 900; 2) 1010.36; 3) 10; 4) 4410; 5) 8 yrs.
Level 2: 6) 162240; 7) 216750; 8) 10000; 9) 5000; 10) 10%.
Level 3: 11) 24 yrs; 12) ≈7.18%; 13) 1000; 14) 3403; 15) 18.81%.

4.18 Summary

  • SI grows linearly: \(SI=PRT/100\).
  • CI grows exponentially: \(A=P(1+R/100)^T\).
  • SI=CI for 1 yr; CI > SI afterwards.
  • Use correct compounding frequency.
  • Population, depreciation, growth → CI model.
  • PV = Future Value discounted.
  • Effective rate > nominal rate if compounded.
  • Installments and annuities extend CI applications.