Episode 8 — Aptitude and Reasoning / 8.4 — Compound Interest

8.4.a Compound Interest -- Concepts and Formulas

1. What Is Compound Interest?

When you deposit money in a bank, the bank pays you interest. Under Simple Interest, the interest is always calculated on the original principal. Under Compound Interest, the interest earned in the first period is added to the principal, and the interest for the next period is calculated on this new (larger) amount. This is called "interest on interest."

Simple Interest:   Interest base stays the same every period.
Compound Interest: Interest base grows every period.

Intuitive Example:

You invest Rs. 1000 at 10% per annum for 2 years.

Simple Interest:
  Year 1 interest = 10% of 1000 = 100
  Year 2 interest = 10% of 1000 = 100
  Total SI = 200
  Amount = 1200

Compound Interest:
  Year 1 interest = 10% of 1000 = 100   --> New principal = 1100
  Year 2 interest = 10% of 1100 = 110
  Total CI = 210
  Amount = 1210

The difference (CI - SI) = Rs. 10, which is the "interest on interest" (10% of the first year's interest of Rs. 100).

2. Core Formula -- Annual Compounding

When interest is compounded once per year:

A = P (1 + R/100)^T

Where:

  • A = Amount (Principal + Interest) at the end of T years
  • P = Principal (initial sum)
  • R = Rate of interest per annum (in %)
  • T = Time in years

The Compound Interest itself is:

CI = A - P = P [(1 + R/100)^T  -  1]

Worked Example 1: Basic CI Calculation

Problem: Find the compound interest on Rs. 5000 at 8% per annum for 3 years.

Given: P = 5000, R = 8%, T = 3 years

Step 1: Calculate Amount
  A = 5000 (1 + 8/100)^3
  A = 5000 (1.08)^3
  A = 5000 x 1.259712
  A = 6298.56

Step 2: Calculate CI
  CI = A - P = 6298.56 - 5000 = Rs. 1298.56

3. Compounding More Than Once a Year

Interest may be compounded more frequently than annually. The general formula is:

A = P (1 + R/(n x 100))^(n x T)

Where n = number of times interest is compounded per year.

3.1 Semi-Annual (Half-Yearly) Compounding

Interest is compounded twice a year (n = 2). The rate per period is halved; the number of periods is doubled.

A = P (1 + R/200)^(2T)

Equivalently: use Rate = R/2 and Time = 2T in the basic formula.

Worked Example 2: Half-Yearly Compounding

Problem: Find CI on Rs. 10,000 at 12% p.a. compounded half-yearly for 1.5 years.

Given: P = 10000, R = 12%, T = 1.5 years
  Half-yearly rate = 12/2 = 6%
  Number of periods = 2 x 1.5 = 3

  A = 10000 (1 + 6/100)^3
  A = 10000 (1.06)^3
  A = 10000 x 1.191016
  A = 11910.16

  CI = 11910.16 - 10000 = Rs. 1910.16

3.2 Quarterly Compounding

Interest is compounded four times a year (n = 4). Rate per period is R/4; number of periods is 4T.

A = P (1 + R/400)^(4T)

Worked Example 3: Quarterly Compounding

Problem: Find the amount on Rs. 16,000 at 20% p.a. compounded quarterly for 9 months.

Given: P = 16000, R = 20%, T = 9 months = 9/12 = 3/4 year
  Quarterly rate = 20/4 = 5%
  Number of periods = 4 x (3/4) = 3

  A = 16000 (1 + 5/100)^3
  A = 16000 (1.05)^3
  A = 16000 x 1.157625
  A = 18522

  CI = 18522 - 16000 = Rs. 2522

3.3 Monthly Compounding

A = P (1 + R/1200)^(12T)

Summary Table: Compounding Frequencies

CompoundingnRate per PeriodNumber of PeriodsFormula
Annual1RTP(1 + R/100)^T
Semi-Annual2R/22TP(1 + R/200)^(2T)
Quarterly4R/44TP(1 + R/400)^(4T)
Monthly12R/1212TP(1 + R/1200)^(12T)

4. Difference Between SI and CI

This is one of the most important exam concepts. The CI-SI difference arises because CI includes "interest on interest."

4.1 Comparison Table

FeatureSimple Interest (SI)Compound Interest (CI)
Interest calculated onOriginal principal onlyPrincipal + accumulated interest
FormulaSI = P x R x T / 100CI = P[(1+R/100)^T - 1]
Growth patternLinear (constant additions)Exponential (accelerating additions)
Interest each yearSame every yearIncreases every year
For T = 1 yearSI = CI (identical)SI = CI (identical)
For T > 1 yearSI < CICI > SI
Amount formulaA = P(1 + RT/100)A = P(1 + R/100)^T

4.2 CI - SI Difference Formula (2 Years)

CI - SI = P (R/100)^2

Derivation:

SI for 2 years = P x R x 2 / 100 = 2PR/100

CI for 2 years = P(1 + R/100)^2 - P
               = P[1 + 2R/100 + R^2/10000] - P
               = 2PR/100 + PR^2/10000

Difference = CI - SI = PR^2/10000 = P(R/100)^2

4.3 CI - SI Difference Formula (3 Years)

CI - SI = P (R/100)^2 (3 + R/100)

Or equivalently:

CI - SI = P R^2 (300 + R) / 100^3

Worked Example 4: CI - SI Difference

Problem: The difference between CI and SI on a sum of money at 5% per annum for 2 years is Rs. 15. Find the sum.

Using the formula: CI - SI = P (R/100)^2

  15 = P (5/100)^2
  15 = P (1/20)^2
  15 = P / 400
  P  = 15 x 400
  P  = Rs. 6000

Worked Example 5: CI - SI Difference (3 Years)

Problem: Find the difference between CI and SI on Rs. 8000 at 10% per annum for 3 years.

CI - SI = P (R/100)^2 (3 + R/100)
        = 8000 (10/100)^2 (3 + 10/100)
        = 8000 x (0.01) x (3.1)
        = 8000 x 0.031
        = Rs. 248

Verification:
  SI = 8000 x 10 x 3 / 100 = 2400

  CI: A = 8000 (1.1)^3 = 8000 x 1.331 = 10648
      CI = 10648 - 8000 = 2648

  CI - SI = 2648 - 2400 = 248  ✓

5. Effective Rate of Interest

When interest is compounded more than once a year, the effective annual rate is higher than the stated (nominal) rate.

Effective Rate (E) = (1 + R/(n x 100))^n - 1

Multiply by 100 to express as a percentage:
E% = [(1 + R/(n x 100))^n - 1] x 100

Where n is the number of compounding periods per year.

Worked Example 6: Effective Rate

Problem: A bank offers 12% per annum compounded quarterly. What is the effective annual rate?

  R = 12%, n = 4
  Effective Rate = (1 + 12/400)^4 - 1
                 = (1 + 0.03)^4 - 1
                 = (1.03)^4 - 1
                 = 1.12550881 - 1
                 = 0.12550881

  Effective Rate = 12.55% (approx.)

So 12% compounded quarterly is equivalent to 12.55% compounded annually.

Common Effective Rate Table

Nominal RateCompoundingEffective Rate
10%Semi-annual10.25%
10%Quarterly10.38%
12%Semi-annual12.36%
12%Quarterly12.55%
20%Semi-annual21.00%
20%Quarterly21.55%

6. Population Growth and Depreciation

The compound interest model applies directly to population growth and asset depreciation.

6.1 Population Growth

If a population grows at R% per annum:

Population after T years = P (1 + R/100)^T

If the growth rate is different in different years (R1, R2, R3 ...):

Population after 3 years = P (1 + R1/100)(1 + R2/100)(1 + R3/100)

Worked Example 7: Population Growth

Problem: The population of a town is 50,000. It increases by 10% in the first year and 20% in the second year. What is the population after 2 years?

  Population = 50000 (1 + 10/100)(1 + 20/100)
             = 50000 x 1.1 x 1.2
             = 50000 x 1.32
             = 66,000

6.2 Depreciation

If the value of an asset decreases at R% per annum:

Value after T years = V (1 - R/100)^T

Where V is the current value.

Worked Example 8: Depreciation

Problem: A machine worth Rs. 2,00,000 depreciates at 15% per annum. What is its value after 3 years?

  Value = 200000 (1 - 15/100)^3
        = 200000 (0.85)^3
        = 200000 x 0.614125
        = Rs. 1,22,825

6.3 Mixed Growth and Decline

If a population increases by R1% in the first year and decreases by R2% in the second year:

Population after 2 years = P (1 + R1/100)(1 - R2/100)

Worked Example 9: Mixed Growth and Decline

Problem: The population of a village is 10,000. It increases by 5% in the first year and decreases by 10% in the second year. What is the population after 2 years?

  Population = 10000 (1 + 5/100)(1 - 10/100)
             = 10000 x 1.05 x 0.90
             = 10000 x 0.945
             = 9450

7. Finding Rate of Interest

When the amount and principal are known, you can find the rate.

Worked Example 10: Finding Rate

Problem: Rs. 4000 becomes Rs. 4840 in 2 years at compound interest. Find the rate.

  A = P(1 + R/100)^T
  4840 = 4000 (1 + R/100)^2
  (1 + R/100)^2 = 4840/4000 = 1.21
  1 + R/100 = sqrt(1.21) = 1.1
  R/100 = 0.1
  R = 10%

8. Finding Time Period

When the amount, principal, and rate are known, you can find the time.

Worked Example 11: Finding Time

Problem: At what time will Rs. 1000 become Rs. 1331 at 10% per annum compound interest?

  A = P(1 + R/100)^T
  1331 = 1000 (1 + 10/100)^T
  1331/1000 = (1.1)^T
  1.331 = (1.1)^T

  Since (1.1)^3 = 1.331
  T = 3 years

9. CI When Rate Varies Each Year

When the rate is different for each year:

A = P (1 + R1/100)(1 + R2/100)(1 + R3/100) ...

Worked Example 12: Variable Rates

Problem: Find CI on Rs. 25,000 for 3 years if the rate is 4% for the first year, 5% for the second year, and 6% for the third year.

  A = 25000 (1 + 4/100)(1 + 5/100)(1 + 6/100)
  A = 25000 x 1.04 x 1.05 x 1.06
  A = 25000 x 1.04 x 1.05 x 1.06

  Step-by-step:
    25000 x 1.04 = 26000
    26000 x 1.05 = 27300
    27300 x 1.06 = 28938

  CI = 28938 - 25000 = Rs. 3938

10. CI for Fractional Time Periods

When T is not a whole number, say T = 2 years and 4 months = 2 + 1/3 years:

A = P (1 + R/100)^2 x (1 + (1/3 x R)/100)

General rule: Compound for the whole-number part, then use simple interest for the fractional part.

For T = n + p/q years:
A = P (1 + R/100)^n x (1 + (p/q) x R/100)

Worked Example 13: Fractional Time

Problem: Find CI on Rs. 10,000 at 12% per annum for 2 years and 4 months.

  T = 2 years + 4/12 years = 2 + 1/3 years

  A = 10000 (1 + 12/100)^2 x (1 + (1/3 x 12)/100)
  A = 10000 (1.12)^2 x (1 + 4/100)
  A = 10000 x 1.2544 x 1.04
  A = 10000 x 1.304576
  A = 13045.76

  CI = 13045.76 - 10000 = Rs. 3045.76

11. Installments (Equal Annual Payments)

If a borrower repays a loan in equal annual installments at compound interest:

P = X/(1+R/100) + X/(1+R/100)^2 + ... + X/(1+R/100)^n

Where X is the annual installment amount and n is the number of installments.

For 2 equal installments:

P = X/(1+R/100) + X/(1+R/100)^2

Worked Example 14: Installments

Problem: A sum of Rs. 2100 is borrowed at 10% compound interest and is to be paid back in 2 equal annual installments. Find the installment amount.

  Let each installment = X

  2100 = X/1.1 + X/(1.1)^2
  2100 = X/1.1 + X/1.21
  2100 = (1.21X + 1.1X) / (1.1 x 1.21)
  2100 = 2.31X / 1.331
  2100 x 1.331 = 2.31X
  2795.1 = 2.31X
  X = 2795.1 / 2.31
  X = Rs. 1210

Summary of All Formulas

+-----------------------------------------------------+----------------------------------------------+
| Scenario                                            | Formula                                      |
+-----------------------------------------------------+----------------------------------------------+
| Amount (annual compounding)                         | A = P(1 + R/100)^T                          |
| Compound Interest                                   | CI = A - P = P[(1+R/100)^T - 1]             |
| Semi-annual compounding                             | A = P(1 + R/200)^(2T)                       |
| Quarterly compounding                               | A = P(1 + R/400)^(4T)                       |
| Monthly compounding                                 | A = P(1 + R/1200)^(12T)                     |
| CI - SI difference (2 years)                        | P(R/100)^2                                   |
| CI - SI difference (3 years)                        | P(R/100)^2 x (3 + R/100)                    |
| Population growth                                   | P_future = P_now(1 + R/100)^T               |
| Depreciation                                        | V_future = V_now(1 - R/100)^T               |
| Variable rates (R1, R2, R3)                         | A = P(1+R1/100)(1+R2/100)(1+R3/100)         |
| Effective annual rate                               | E = (1 + R/n*100)^n - 1                     |
| Fractional time (n + p/q years)                     | A = P(1+R/100)^n x (1 + (p/q)R/100)        |
+-----------------------------------------------------+----------------------------------------------+