Notes on Business Mathematics & General Mathematics

BUSINESS MATHEMATICS

LESSON 1: FUNDAMENTAL OPERATIONS ON: FRACTIONS DECIMALS AND PERCENTAGE

  • Focuses on operations and conversions among fractions, decimals, and percentages.
  • Key tasks include:
    • Performing operations on fractions (addition, subtraction, multiplication, division).
    • Performing operations on decimals.
    • Working with percentages (conversion and calculations).
    • Converting between fractions, decimals, and percentages.
  • Core formulas and rules:
    • Fraction operations (with common denominator where needed):
    • Addition: ab+cd=ad+bcbd\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}
    • Subtraction: ab−cd=ad−bcbd\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}
    • Multiplication: ab⋅cd=acbd\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}
    • Division: ab÷cd=adbc\frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc}
    • Converting fractions to decimals: compute the division ab=a÷b\frac{a}{b} = a \div b.
    • Decimal to fraction: write the decimal with a power of 10 in the denominator, then simplify to lowest terms.
    • Fraction/Decimal to Percent: multiply by 100%: ab×100%\frac{a}{b} \times 100\%; Decimal to Percent: d×100%d \times 100\%; Percent to Decimal: p%=p100p\% = \frac{p}{100}.
    • Converting between all three representations:
    • Fraction ⇄ Decimal ⇄ Percent are just different forms of the same quantity.
  • Illustrative examples:
    • Addition: 35+23=1915 .\frac{3}{5} + \frac{2}{3} = \frac{19}{15}\,.
    • Subtraction: 47−13=521 .\frac{4}{7} - \frac{1}{3} = \frac{5}{21}\,.
    • Multiplication: 23⋅34=12 .\frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}\,.
    • Division: 23÷45=56 .\frac{2}{3} \div \frac{4}{5} = \frac{5}{6}\,.
    • Decimal ⇄ Fraction: 34=0.75=75%\frac{3}{4} = 0.75 = 75\%.
    • Decimal to Fraction: 0.6=35 ;  0.125=180.6 = \frac{3}{5}\,;\;0.125 = \frac{1}{8}.
    • Percent to Decimal: 25%=0.25  ;  0.83=83%25\% = 0.25\; ;\; 0.83 = 83\%.
  • Conversions and relationships:
    • 1=100%  ⇒  x%=x100 (as decimal)1 = 100\%\;\Rightarrow\; x\% = \frac{x}{100}\text{ (as decimal)}
    • Decimal d=p100⇒p%=d×100%\text{Decimal } d = \frac{p}{100} \Rightarrow p\% = d\times 100\%
  • Practical tips:
    • Align denominators before adding/subtracting fractions.
    • Check simplifications after every operation.
    • Use conversions to verify answers (e.g., convert to a common form).

LESSON 2: RATIO & PROPORTION

  • Key concepts:
    • Ratio: a comparison expressed as a:b (or a/b).
    • Proportion: an equation that expresses two ratios as equal: a:b = c:d, with cross-multiplication ad = bc.
    • Base, Rate, and Percentage in ratio-context:
    • Base amount (the reference quantity).
    • Rate (the percentage or multiplier applied to the base).
    • Percentage (the rate expressed as a percent).
    • Practical applications include mixing ratios, scaling recipes, and financial calculations (percent changes).
  • Proportion and solving use cross-multiplication:
    • If a:b = c:d, then ad = bc.
    • Example: If the ratio of apples to oranges is 3:2 and there are 15 apples, oranges = 15 × (2/3) = 10. Solve via cross-multiplication: (\frac{3}{2} = \frac{15}{x} \Rightarrow x = 10).
  • Ratios and percent examples:
    • Percentage as a ratio: 5% corresponds to 5 parts per 100, i.e., 5/100.
    • Percentage of a total: if B is the base and r% is the rate, final value V = B(1 + r/100) for an increase, or V = B(1 - r/100) for a decrease.
    • Example: 5% of 200 = 200×5100=10200 \times \frac{5}{100} = 10.
    • Increasing a base by a percentage: V = B(1 + r/100).
  • Worked cross-checks:
    • If you know a target for one part of a ratio, you can compute the other part using ad = bc.
  • Summary equations:
    • Proportion: ab=cd⇒ad=bc\frac{a}{b} = \frac{c}{d} \quad\Rightarrow\quad ad = bc
    • Base–Rate–Percentage relation: V=B(1±r100),r in percentV = B\left(1 \pm \frac{r}{100}\right), \quad r\text{ in percent}

LESSON 3: BUYING AND SELLING

  • Core concepts:
    • Markup: the amount added to cost to reach selling price.
    • Margin: profit as a portion of selling price or cost (two common definitions).
    • Markdown: reduction from original price to a lower price.
    • Trade discounts: reductions given by suppliers to retailers for bulk purchases.
  • Formulas:
    • Cost (C), Selling Price (SP):
    • Markup (amount): M=SP−CM = SP - C
    • Markup percentage on cost: Markup% on cost=SP−CC×100%\text{Markup\% on cost} = \frac{SP - C}{C} \times 100\%
    • Margin (profit as a percentage):
    • Margin on selling price: Margin (on SP)=SP−CSP×100%\text{Margin (on SP)} = \frac{SP - C}{SP} \times 100\%
    • Margin on cost: Margin (on cost)=SP−CC×100%\text{Margin (on cost)} = \frac{SP - C}{C} \times 100\%
    • Markdown: Markdown=Original Price−New Price\text{Markdown} = \text{Original Price} - \text{New Price}
    • Markdown percentage: Markdown%=MarkdownOriginal Price×100%\text{Markdown\%} = \frac{\text{Markdown}}{\text{Original Price}} \times 100\%
    • Trade discounts (net price):
    • If List Price = $L$ and there is a discount $d\%$: Net Price=L(1−d100)\text{Net Price} = L\left(1 - \frac{d}{100}\right)
    • For multiple discounts, apply sequentially: Net Price=L(1−d<em>1100)(1−d</em>2100)⋯\text{Net Price} = L\left(1 - \frac{d<em>1}{100}\right)\left(1 - \frac{d</em>2}{100}\right)\cdots
  • Illustrative examples:
    • If cost C = 80 and selling price SP = 100, then markup M = 20; markup on cost = (\frac{20}{80} = 25\%).
    • Margin on selling price: (\frac{SP - C}{SP} = \frac{20}{100} = 20\%).
    • Markdown example: Original Price 100, New Price 85 → Markdown = 15; Markdown% = (\frac{15}{100} \times 100\% = 15\%).
    • Trade discount example: List Price 100, trade discount 15% → Net Price = 85.
  • Practical relevance:
    • Accurate calculation of costs, pricing, and profitability.
    • Understanding how markups and margins affect business decisions.
    • Use of discounts and markdowns in competitive pricing while maintaining profitability.

GENERAL MATHEMATICS

LESSON 1: FUNCTIONS AND RELATIONS

  • Distinction:
    • A relation is a set of ordered pairs (x, y) that defines how two sets relate.
    • A function is a special relation where every input x from the domain maps to exactly one output y.
  • Key terms:
    • Domain: all possible input values x for which the relation is defined.
    • Range: all possible output values y produced by the relation.
    • Notation: f: X → Y denotes a function mapping from domain X to codomain Y.
    • Representations: explicit form f(x), mapping diagram, and graph.
  • Examples:
    • Relation R: { (1,2), (2,3), (3,2) } is a function (each x has a single y).
    • Relation R: { (1,2), (1,3) } is not a function (x = 1 maps to two y-values).
  • Foundational ideas:
    • Functions are the fundamental way to model rules that assign outputs to inputs.
    • Domain and range determine where a function is defined and what outputs it can produce.

LESSON 2: EVALUATION OF FUNCTIONS

  • Purpose:
    • Evaluate a function at a specified input value: compute f(a).
  • Typical notation:
    • Given f(x), evaluate at x = a: f(a).
  • Example:
    • If f(x) = 3x - 7, then f(4) = 3(4) - 7 = 12 - 7 = 5.
  • Considerations:
    • Some inputs may be outside the domain; ensure a is in the domain before evaluating.
    • If the function has piecewise definitions, choose the piece that applies to the input a.

LESSON 3: OPERATION OF FUNCTIONS

  • Operations on functions:
    • Sum: (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x)
    • Difference: (f−g)(x)=f(x)−g(x)(f - g)(x) = f(x) - g(x)
    • Product: (f⋅g)(x)=f(x)⋅g(x)(f \cdot g)(x) = f(x) \cdot g(x)
    • Quotient: (f÷g)(x)=f(x)g(x)(f \div g)(x) = \frac{f(x)}{g(x)} with $g(x) \neq 0$.
  • Examples:
    • If $f(x) = x^2$ and $g(x) = 2x$, then
    • $(f + g)(x) = x^2 + 2x$
    • $(f - g)(x) = x^2 - 2x$
    • $(f \cdot g)(x) = x^2 \cdot 2x = 2x^3$
    • $(f \div g)(x) = \frac{x^2}{2x} = \frac{x}{2}$ for $x \neq 0$.
  • Domain considerations:
    • For division, avoid values of x where $g(x) = 0$.
    • Ensure that the resulting expression is defined for the chosen x.

LESSON 4: COMPOSITE FUNCTIONS

  • Definition:
    • The composite function (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)).
  • Domain considerations:
    • The domain of $f \circ g$ consists of all $x$ in the domain of $g$ such that $g(x)$ lies in the domain of $f$.
  • Example:
    • If $f(x) = x^2$ and $g(x) = x + 1$, then $(f \circ g)(x) = (x + 1)^2$.
  • Properties:
    • Associativity (where defined): (f∘g)∘h=f∘(g∘h)(f \circ g) \circ h = f \circ (g \circ h) provided the expressions are defined.
  • Visual intuition:
    • First apply the inner function to the input, then apply the outer function to the result.
  • Practical note:
    • Composite functions are central to modeling multi-step processes and transformations in mathematics and applied contexts.