Essential Business Math and Financial Formulas

Posted on Oct 25, 2025 in Mathematics

1. Average Calculation Formula

Formula: Average = (Sum of all values) / (Number of values)

Example: Find the average of 10, 15, 20, 25, 30

  • Sum = 10 + 15 + 20 + 25 + 30 = 100
  • Number of values = 5
  • Average = 100 / 5 = 20

2. Ratio and Proportion Formulas

Ratio: a:b = a/b

Proportion: If a:b = c:d, then a/b = c/d or ad = bc

Example: If 3:4 = x:12, find x

  • 3/4 = x/12
  • 3 × 12 = 4 × x
  • 36 = 4x
  • x = 9

3. Percentage Formulas

Basic Percentage: Percentage = (Part / Whole) × 100

Percentage Increase/Decrease: ((New Value – Old Value) / Old Value) × 100

Example: What percentage is 25 out of 200?

  • Percentage = (25 / 200) × 100 = 12.5%

4. Profit and Loss Formulas

  • Cost Price (CP): The price at which an item is purchased.
  • Selling Price (SP): The price at which an item is sold.
  • Profit = SP – CP (when SP > CP)
  • Loss = CP – SP (when CP > SP)
  • Profit % = (Profit / CP) × 100
  • Loss % = (Loss / CP) × 100

Example: CP = ₹500, SP = ₹600

  • Profit = 600 – 500 = ₹100
  • Profit % = (100 / 500) × 100 = 20%

5. Commission Formula

Commission = (Commission Rate / 100) × Sales Amount

Example: A sales agent gets 5% commission on ₹50,000 sales

  • Commission = (5 / 100) × 50,000 = ₹2,500

6. Discount Formulas

  • Discount = Marked Price – Selling Price
  • Discount % = (Discount / Marked Price) × 100
  • Selling Price = Marked Price – Discount

Example: Marked Price = ₹1000, Discount = 20%

  • Discount Amount = (20 / 100) × 1000 = ₹200
  • Selling Price = 1000 – 200 = ₹800

7. Brokerage Formula

Brokerage = (Brokerage Rate / 100) × Transaction Value

Example: Calculate 2% brokerage on a ₹10,000 transaction

  • Brokerage = (2 / 100) × 10,000 = ₹200

8. Simple and Compound Interest Formulas

Simple Interest (SI)

SI = (P × R × T) / 100

Amount = Principal + SI

Where:

  • P = Principal amount
  • R = Rate of interest per annum
  • T = Time in years

Example: P = ₹1000, R = 10%, T = 2 years

  • SI = (1000 × 10 × 2) / 100 = ₹200
  • Amount = 1000 + 200 = ₹1200

Compound Interest (CI)

Amount = P(1 + R/100)T

CI = Amount – Principal

Example: P = ₹1000, R = 10%, T = 2 years

  • Amount = 1000(1 + 10/100)² = 1000(1.1)² = ₹1210
  • CI = 1210 – 1000 = ₹210

9. Valuation of Simple Loans and Debentures

Present Value Formula

PV = FV / (1 + r)n

Where:

  • PV = Present Value
  • FV = Future Value
  • r = Interest rate
  • n = Number of periods

Bond Valuation

Bond Price = (C × [1 – (1 + r)-n] / r) + (M / (1 + r)n)

Where:

  • C = Coupon payment
  • r = Required rate of return
  • n = Number of periods
  • M = Maturity value

10. Sinking Fund Problems

Sinking Fund Formula: A = PMT × [((1 + r)n – 1) / r]

Where:

  • A = Future value of sinking fund
  • PMT = Regular payment
  • r = Interest rate per period
  • n = Number of periods

Example: Monthly deposit of ₹5000 at 12% annual interest for 5 years

  • r = 12% / 12 = 1% per month = 0.01
  • n = 5 × 12 = 60 months
  • A = 5000 × [((1.01)60 – 1) / 0.01] = ₹4,09,435
  • t = ₹16

11. Indices and Logarithms

Index Laws

  • am × an = a(m+n)
  • am ÷ an = a(m-n)
  • (am)n = a(mn)
  • a0 = 1
  • a-n = 1/an

Logarithm Laws

  • log(ab) = log(a) + log(b)
  • log(a/b) = log(a) – log(b)
  • log(an) = n × log(a)
  • log₁₀(10) = 1
  • log₁₀(1) = 0

12. Arithmetic and Geometric Progression

Arithmetic Progression (AP)

General term: an = a + (n-1)d

Sum of n terms: Sn = n/2[2a + (n-1)d]

Where:

  • a = First term
  • d = Common difference
  • n = Number of terms

Example: Find the 10th term of AP: 2, 5, 8, 11…

  • a = 2, d = 3
  • a₁₀ = 2 + (10-1) × 3 = 2 + 27 = 29

Geometric Progression (GP)

General term: an = ar(n-1)

Sum of n terms: Sn = a(rn – 1) / (r – 1) (when r ≠ 1)

Where:

  • a = First term
  • r = Common ratio

Example: Find the 5th term of GP: 3, 6, 12, 24…

  • a = 3, r = 2
  • a₅ = 3 × 2(5-1) = 3 × 16 = 48

13. Sum of Squares and Cubes

Sum of First n Natural Numbers

Σn = n(n+1) / 2

Sum of Squares of First n Natural Numbers

Σn² = n(n+1)(2n+1) / 6

Sum of Cubes of First n Natural Numbers

Σn³ = [n(n+1) / 2]²

Example: Sum of cubes of first 5 natural numbers

  • Σn³ = [5(5+1) / 2]² = [5 × 6 / 2]² = 15² = 225
  • Verification: 1³ + 2³ + 3³ + 4³ + 5³ = 1 + 8 + 27 + 64 + 125 = 225 ✓

14. Linear Programming Concepts

Standard Form of LPP

Objective Function: Maximize/Minimize Z = c₁x₁ + c₂x₂ + … + cₙxₙ

Subject to constraints:

  • a₁₁x₁ + a₁₂x₂ + … + a₁ₙxₙ ≤/≥ b₁
  • a₂₁x₁ + a₂₂x₂ + … + a₂ₙxₙ ≤/≥ b₂
  • …
  • x₁, x₂, …, xₙ ≥ 0 (Non-negativity constraints)

Graphical Method Steps

  1. Convert inequalities to equations.
  2. Find intersection points.
  3. Identify the feasible region.
  4. Evaluate the objective function at corner points.
  5. Select the optimal solution.

Example Problem: Maximize Z = 3x + 4y Subject to: x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

Solution:

  • Corner points: (0,0), (0,4), (2,3), (4,0)
  • Z values: 0, 16, 18, 12
  • Maximum Z = 18 at (2,3)

15. Business Applications of Linear Programming

Common Business Applications:

  1. Production Planning

    • Maximize profit from limited resources.
    • Minimize production costs.

  2. Transportation Problems

    • Minimize shipping costs.
    • Optimize distribution routes.

  3. Assignment Problems

    • Assign workers to jobs optimally.
    • Minimize total assignment cost.

  4. Diet Problems

    • Minimize cost while meeting nutritional requirements.

  5. Investment Problems

    • Maximize returns subject to risk constraints.

Sample Business Problem:

A company produces two products A and B. Product A gives ₹3 profit per unit and B gives ₹4 profit per unit. Each unit of A requires 2 hours of labor and each unit of B requires 3 hours. Total available labor is 12 hours. How many units of each product should be produced to maximize profit?

Solution:

  • Let x = units of A, y = units of B
  • Maximize Z = 3x + 4y
  • Subject to: 2x + 3y ≤ 12, x ≥ 0, y ≥ 0
  • Optimal solution: x = 0, y = 4, resulting in Maximum profit.