Episode 8 — Aptitude and Reasoning / 8.4 — Compound Interest

8.4.c Compound Interest -- Solved Examples

Attempt each problem on your own first, then check the solution.

Basic Level (Problems 1-8)

Problem 1: Basic CI Calculation

Find the compound interest on Rs. 12,000 at 10% per annum for 2 years, compounded annually.

Solution:

Given: P = 12000, R = 10%, T = 2

A = P(1 + R/100)^T
A = 12000 (1 + 10/100)^2
A = 12000 (1.1)^2
A = 12000 x 1.21
A = 14520

CI = A - P = 14520 - 12000 = Rs. 2520

Answer: Rs. 2520

Problem 2: Finding Amount

What will Rs. 25,000 amount to in 3 years at 8% per annum compound interest?

Solution:

Given: P = 25000, R = 8%, T = 3

A = 25000 (1.08)^3
A = 25000 x 1.259712
A = Rs. 31,492.80

Answer: Rs. 31,492.80

Problem 3: CI for 1 Year (SI = CI)

Find CI on Rs. 6000 at 15% for 1 year.

Solution:

For 1 year, CI = SI = P x R x T / 100
CI = 6000 x 15 x 1 / 100
CI = Rs. 900

Answer: Rs. 900

Note: For 1 year, compound and simple interest are identical.

Problem 4: Finding Principal from Amount

A sum of money amounts to Rs. 8820 in 2 years at 5% per annum compound interest. Find the sum.

Solution:

A = P(1 + R/100)^T
8820 = P (1.05)^2
8820 = P x 1.1025
P = 8820 / 1.1025
P = Rs. 8000

Answer: Rs. 8000

Problem 5: Simple CI - SI Difference (2 Years)

The difference between CI and SI on a certain sum at 10% per annum for 2 years is Rs. 50. Find the sum.

Solution:

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

50 = P (10/100)^2
50 = P x 1/100
P = 50 x 100
P = Rs. 5000

Answer: Rs. 5000

Problem 6: Population Growth (Basic)

The population of a town is 20,000. It grows at 5% per annum. Find the population after 2 years.

Solution:

Population = 20000 (1 + 5/100)^2
           = 20000 (1.05)^2
           = 20000 x 1.1025
           = 22,050

Answer: 22,050

Problem 7: Depreciation (Basic)

A car worth Rs. 5,00,000 depreciates at 10% per annum. Find its value after 2 years.

Solution:

Value = 500000 (1 - 10/100)^2
      = 500000 (0.9)^2
      = 500000 x 0.81
      = Rs. 4,05,000

Answer: Rs. 4,05,000

Problem 8: CI When Rate Is Given as Fraction

Find CI on Rs. 6400 at 12.5% per annum for 2 years.

Solution:

12.5% = 1/8, so multiplier = 1 + 1/8 = 9/8

After Year 1: 6400 x 9/8 = 7200
After Year 2: 7200 x 9/8 = 8100

CI = 8100 - 6400 = Rs. 1700

Answer: Rs. 1700

Medium Level (Problems 9-16)

Problem 9: Half-Yearly Compounding

Find CI on Rs. 20,000 at 10% per annum compounded half-yearly for 1.5 years.

Solution:

Half-yearly rate = 10/2 = 5%
Number of periods = 2 x 1.5 = 3

A = 20000 (1 + 5/100)^3
A = 20000 (1.05)^3
A = 20000 x 1.157625
A = 23152.50

CI = 23152.50 - 20000 = Rs. 3152.50

Answer: Rs. 3152.50

Problem 10: Quarterly Compounding

Find the amount on Rs. 32,000 at 8% per annum compounded quarterly for 1 year.

Solution:

Quarterly rate = 8/4 = 2%
Number of periods = 4 x 1 = 4

A = 32000 (1 + 2/100)^4
A = 32000 (1.02)^4

(1.02)^2 = 1.0404
(1.02)^4 = (1.0404)^2 = 1.08243216

A = 32000 x 1.08243216
A = Rs. 34,637.83 (approx.)

Answer: Rs. 34,637.83

Problem 11: Finding Rate from CI of Successive Years

The compound interest on a sum for the 1st year is Rs. 1500 and for the 2nd year is Rs. 1575. Find the rate of interest.

Solution:

The difference between 2nd year CI and 1st year CI
= Interest earned on the 1st year's CI

Extra interest = 1575 - 1500 = 75

Rate = (Extra interest / 1st year CI) x 100
     = (75 / 1500) x 100
     = 5%

Answer: 5%

Problem 12: Finding Time

In how many years will Rs. 2000 amount to Rs. 2662 at 10% per annum compound interest?

Solution:

A = P(1 + R/100)^T
2662 = 2000 (1.1)^T
(1.1)^T = 2662/2000 = 1.331

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

Answer: 3 years

Problem 13: CI - SI Difference (3 Years)

Find the difference between CI and SI on Rs. 20,000 at 8% per annum for 3 years.

Solution:

CI - SI = P(R/100)^2 (3 + R/100)
        = 20000 x (8/100)^2 x (3 + 8/100)
        = 20000 x 0.0064 x 3.08
        = 20000 x 0.019712
        = Rs. 394.24

Verification:
  SI = 20000 x 8 x 3 / 100 = 4800
  CI: A = 20000(1.08)^3 = 20000 x 1.259712 = 25194.24
      CI = 25194.24 - 20000 = 5194.24
  CI - SI = 5194.24 - 4800 = 394.24  ✓

Answer: Rs. 394.24

Problem 14: Variable Rates

Find CI on Rs. 50,000 for 3 years if the rate is 10% for the 1st year, 12% for the 2nd year, and 15% for the 3rd year.

Solution:

A = 50000 (1 + 10/100)(1 + 12/100)(1 + 15/100)
A = 50000 x 1.10 x 1.12 x 1.15

Step by step:
  50000 x 1.10 = 55000
  55000 x 1.12 = 61600
  61600 x 1.15 = 70840

CI = 70840 - 50000 = Rs. 20,840

Answer: Rs. 20,840

Problem 15: Effective Rate of Interest

A bank offers 10% per annum compounded semi-annually. What is the effective annual rate?

Solution:

Effective Rate = (1 + R/(n x 100))^n - 1
               = (1 + 10/200)^2 - 1
               = (1.05)^2 - 1
               = 1.1025 - 1
               = 0.1025

Effective Rate = 10.25%

Answer: 10.25%

Problem 16: Population with Different Growth Rates

The population of a city is 1,00,000. It increases by 10% in the first year, decreases by 5% in the second year, and increases by 8% in the third year. Find the population after 3 years.

Solution:

Population = 100000 (1 + 10/100)(1 - 5/100)(1 + 8/100)
           = 100000 x 1.10 x 0.95 x 1.08

Step by step:
  100000 x 1.10 = 110000
  110000 x 0.95 = 104500
  104500 x 1.08 = 112860

Population after 3 years = 1,12,860

Answer: 1,12,860

Advanced Level (Problems 17-24)

Problem 17: CI with Fractional Time

Find CI on Rs. 15,000 at 12% per annum for 2 years and 8 months.

Solution:

T = 2 years + 8 months = 2 + 8/12 = 2 + 2/3 years

A = P(1 + R/100)^2 x (1 + (2/3 x R)/100)
A = 15000 (1.12)^2 x (1 + (2/3 x 12)/100)
A = 15000 x 1.2544 x (1 + 8/100)
A = 15000 x 1.2544 x 1.08
A = 15000 x 1.354752
A = 20321.28

CI = 20321.28 - 15000 = Rs. 5321.28

Answer: Rs. 5321.28

Problem 18: Finding Principal from CI-SI Difference (3 Years)

The difference between CI and SI on a certain sum at 5% per annum for 3 years is Rs. 122. Find the sum.

Solution:

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

122 = P (5/100)^2 (3 + 5/100)
122 = P x (1/400) x (3.05)
122 = P x 3.05/400
122 = P x 0.007625

P = 122 / 0.007625

Simplify: 122 x 400 / 3.05
        = 48800 / 3.05
        = 16000

P = Rs. 16,000

Verification:
  SI = 16000 x 5 x 3 / 100 = 2400
  A = 16000(1.05)^3 = 16000 x 1.157625 = 18522
  CI = 18522 - 16000 = 2522
  CI - SI = 2522 - 2400 = 122  ✓

Answer: Rs. 16,000

Problem 19: SI and CI on Same Sum -- Comparison

The simple interest on a sum for 2 years at 12% per annum is Rs. 1500. Find the compound interest on the same sum at the same rate for the same period.

Solution:

Step 1: Find the principal from SI
  SI = PRT/100
  1500 = P x 12 x 2 / 100
  1500 = 24P/100
  P = 1500 x 100/24 = 6250

Step 2: Find CI
  CI = P[(1 + R/100)^2 - 1]
  CI = 6250 [(1.12)^2 - 1]
  CI = 6250 [1.2544 - 1]
  CI = 6250 x 0.2544
  CI = Rs. 1590

Alternative (faster):
  CI = SI + SI x R / (2 x 100)
  Wait -- use the difference formula:
  CI - SI = P(R/100)^2 = 6250 x (12/100)^2 = 6250 x 0.0144 = 90
  CI = 1500 + 90 = Rs. 1590

Answer: Rs. 1590

Problem 20: Depreciation -- Finding Original Value

The value of a machine depreciates at 20% per annum. After 2 years, its present value is Rs. 3,20,000. What was the original value?

Solution:

V_after = V_original (1 - R/100)^T
320000 = V_original (1 - 20/100)^2
320000 = V_original (0.8)^2
320000 = V_original x 0.64

V_original = 320000 / 0.64
V_original = Rs. 5,00,000

Answer: Rs. 5,00,000

Problem 21: Investment Comparison

Ram invests Rs. 10,000 at 10% SI, and Shyam invests Rs. 10,000 at 10% CI. After 3 years, how much more does Shyam earn?

Solution:

Ram's earning (SI):
  SI = 10000 x 10 x 3 / 100 = Rs. 3000

Shyam's earning (CI):
  CI = 10000[(1.1)^3 - 1]
  CI = 10000[1.331 - 1]
  CI = 10000 x 0.331
  CI = Rs. 3310

Difference = 3310 - 3000 = Rs. 310

Shortcut verification:
  CI - SI (3 years) = P(R/100)^2(3 + R/100)
                    = 10000 x 0.01 x 3.1
                    = Rs. 310  ✓

Answer: Shyam earns Rs. 310 more

Problem 22: Equal Annual Installments

A sum of Rs. 11,520 is borrowed at 25/3 % compound interest and is to be paid back in 2 equal annual installments. Find the value of each installment.

Solution:

Let each installment = X
Rate = 25/3 %  =>  R/100 = 25/300 = 1/12
Multiplier = 1 + 1/12 = 13/12

The present value of two installments equals the loan amount:

  11520 = X/(13/12) + X/(13/12)^2
  11520 = 12X/13 + 144X/169

  LCM of 13 and 169 = 169

  11520 = (12 x 13 x X + 144X) / 169
  11520 = (156X + 144X) / 169
  11520 = 300X / 169
  11520 x 169 = 300X
  1946880 = 300X
  X = 1946880 / 300
  X = Rs. 6489.60

Answer: Rs. 6489.60

Problem 23: Compound Interest -- Finding Rate When Amount Doubles

At what rate of compound interest will a sum double itself in 3 years? (Find the approximate rate.)

Solution:

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

Taking cube root:
  (1 + R/100) = 2^(1/3) = 1.2599 (approx.)
  R/100 = 0.2599
  R ≈ 26%

Using Rule of 72 for quick check:
  72/3 = 24%  (rough approximation)

The exact answer is approximately 25.99%, which rounds to 26%.

Answer: Approximately 26%

Problem 24: Complex Semi-Annual Problem

Rs. 40,000 is lent at 8% per annum compounded semi-annually. The borrower pays Rs. 10,000 at the end of the first year and Rs. 15,000 at the end of the second year. How much does the borrower still owe at the end of 3 years?

Solution:

Semi-annual rate = 4%, compounding every 6 months.

Amount at end of Year 1 (2 half-year periods):
  A1 = 40000 (1.04)^2 = 40000 x 1.0816 = 43264

After paying Rs. 10000:
  Balance = 43264 - 10000 = 33264

Amount at end of Year 2 (2 more half-year periods):
  A2 = 33264 (1.04)^2 = 33264 x 1.0816 = 35,977.53 (approx.)

After paying Rs. 15000:
  Balance = 35977.53 - 15000 = 20977.53

Amount at end of Year 3 (2 more half-year periods):
  A3 = 20977.53 (1.04)^2 = 20977.53 x 1.0816 = 22,693.50 (approx.)

Answer: Approximately Rs. 22,693.50

Problem 25: CI on Rs. 1 Lakh at Different Frequencies

Find the CI on Rs. 1,00,000 at 12% for 1 year when compounded (a) annually, (b) half-yearly, (c) quarterly.

Solution:

(a) Annually (n = 1):
  A = 100000 (1 + 12/100)^1 = 100000 x 1.12 = 112000
  CI = Rs. 12,000

(b) Half-yearly (n = 2):
  A = 100000 (1 + 6/100)^2 = 100000 x 1.1236 = 112360
  CI = Rs. 12,360

(c) Quarterly (n = 4):
  A = 100000 (1 + 3/100)^4 = 100000 x (1.03)^4
  (1.03)^2 = 1.0609
  (1.03)^4 = 1.0609^2 = 1.12550881
  A = 100000 x 1.12550881 = 112550.88
  CI = Rs. 12,550.88 (approx.)

Key Insight: More frequent compounding yields higher CI for the same nominal rate.

  Annual CI:     Rs. 12,000.00
  Semi-annual:   Rs. 12,360.00    (+360 more)
  Quarterly:     Rs. 12,550.88    (+550.88 more)

Answer: (a) Rs. 12,000 (b) Rs. 12,360 (c) Rs. 12,550.88

Summary of Problem Types

#Problem TypeKey Formula / Approach
1-3Basic CI calculationA = P(1+R/100)^T
4Finding P from AP = A / (1+R/100)^T
5CI-SI differenceP(R/100)^2 for 2 years
6Population growthP(1+R/100)^T
7DepreciationV(1-R/100)^T
8Fraction-based rateUse fraction multiplier
9-10Non-annual compoundingAdjust rate and periods
11Rate from successive CI(I2-I1)/I1 x 100
12Finding timeEquate and match powers
13CI-SI diff (3 years)P(R/100)^2(3+R/100)
14Variable ratesMultiply individual multipliers
15Effective rate(1+R/n)^n - 1
16Mixed growth/declineMultiply each year's factor
17Fractional timeCompound + SI for fraction
18P from CI-SI diff (3yr)Rearrange difference formula
19SI given, find CIFind P first, then CI
20Depreciation reverseV_original = V_after / (factor)
21SI vs CI comparisonUse CI-SI shortcut
22InstallmentsPresent value of annuity
23Doubling time / rateRule of 72 or direct solve
24Partial repaymentsStep-by-step balance tracking
25Frequency comparisonCompare annual/semi/quarterly