paper no. ad23 accounting technician...

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PAPER NO. AD23

ACCOUNTING TECHNICIAN DIPLOMA

(ATD)

LEVEL II

BUSINESS

MATHEMATICS AND STATISTICS

STUDY NOTES

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KASNEB SYLLABUS

GENERAL OBJECTIVE

This paper is intended to equip the candidate with knowledge, skills and attitudes that will enable him/her

to apply mathematical and statistical skills in business transactions.

7.0 LEARNING OUTCOMES

A candidate who passes this paper should be able to:

Apply linear, quadratic and simultaneous equations to solve business problems

Compute simple and compound interests

Solve problems involving appreciation, depreciation, hire purchase and foreign exchange

Solve business problems using matrix algebra

Solve business problems involving commercial mathematics

Present data in form of tables, graphs and curves

Calculate measures of location, dispersion skewness and kurtosis

Compute simple, general and weighted index numbers.

CONTENT

7.1Equations

Linear equations; solving and graphs

Quadratic equations: solving and graphs

Differentiation

Simultaneous equations; solving

Break-even analysis

Total revenue, total cost and profit equations; application of errors

7.2 Matrices

Introduction; order, types

Addition, subtraction and multiplication

Determinants of 2x2 matrices

Inverses of 2x2 matrices

Application of matrices to business problems

7.3 Commercial mathematics

Buying and selling; discounts, profit and loss, margins and mark-ups

Commissions and salaries: piece and hourly rates, gross and net pay, PAYE

Bills calculations; water and electricity /

Simple and compound interest

Depreciation and appreciation of assets

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Hire purchase

Foreign exchange

7.4 Elementary statistics

Introduction: definitions and branches of statistics^

Methods of data collection: primary and secondary data./

Sampling techniques J

Presentation of data:

Tables

Diagrams: bar charts and pie charts

Graphs: Basic time series graphs, Z-charts, Lorenz curves and semi-log graphs

Frequency distribution tables

Histogram and frequency polygons'

Cumulative frequency curve (ogive) and its application

7.5 Descriptive statistics

Measures of central tendency:

Mean: arithmetic mean, weighted arithmetic mean, geometric mean and harmonic mean

Mode

Median

Measures of dispersion: range, quartile, deciles, percentiles, mean deviation, standard deviation

and coefficient of variation

Measures of skewness and kurtosis excluding computation of the coefficients

7.6 Set theory

Introduction to set theory

Types of sets: universal, empty/null. Sublets, finite and infinite

Operation of sets: unions, intersecting, complements and set difference

Venn diagrams

7.7 Basic probability theory

Introduction to probability: definitions, events, outcomes, sample space

Types of events: simple, compound, independent, mutually exclusive,

Mutually inclusive, dependent events

Rules of probability: additive and multiplicative rules

Baye‘s Theorem

Elementary probability trees

7.8 Index numbers

Construction of index numbers Purpose of index numbers

Simple index numbers; fixed base method and chain base method

Consumer Price Index (CPI)

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Weighted index numbers; Laspeyre‘s, Paasche's, Fisher‘s ideal and Marshall- Edgeworth‘s

methods (both price and quantity index numbers)

Limitations of index numbers

Emerging issues and trends

7.9 Emerging trends

CONTENT PAGE

Chapter 1: Equations………………………………………………………………………5

Chapter 2: Matrices………………………………………………………………….…...31

Chapter 3: Commercial mathematics…………………………………………………….52

Chapter 4: Elementary statistics……………………………………………………….....77

Chapter 5: Descriptive statistics………………………………………………………...109

Chapter 6: Set theory……………………………………………………………………146

Chapter 7: Index numbers……………………………………………………………….176

Chapter 8: Basic probability theory……………………………………………………..205

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CHAPTER 1

EQUATIONS

Linear and Quadratic Equations

Equations often express relationships between given quantities (the knowns) and quantities yet to

be determined (the unknowns). By convention, unknowns are denoted by letters at the end of the

alphabet, x, y, z, w, …, while knowns are denoted by letters at the beginning, a, b, c, d, … . The

process of expressing the unknowns in terms of the knowns is called solving the equation. In an

equation with a single unknown, a value of that unknown for which the equation is true is called

a solution or root of the equation. In a set simultaneous equations, or system of equations,

multiple equations are given with multiple unknowns. A solution to the system is an assignment

of values to all the unknowns so that all of the equations are true. Two kinds of equations are

linear and quadratic.

LINEAR EQUATIONS

It takes the form y = a + bx

Where x and y are variables while a and b are constants.

e.g y = 20 + 2x

y = 5x

y = 15-0.3x

In graphical presentation of a linear equation, the constant ‗a‘ represents y-intercept and ‗b‘

represents the gradient of the slope. Δ𝑦

Δ𝑥

The linear equation can be presented either as 2 by 2 simultaneous equation. In general 2 by 2

equations are given as follows:

2x2

𝑎1𝑥 + 𝑏1𝑦 = 𝑐1

𝑎2𝑥 + 𝑏2𝑦 = 𝑐2

3x3 is given by as,

𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1

𝑎2𝑥 + 𝑏2𝑦 + 𝑐2𝑧 = 𝑑2

𝑎3𝑥 + 𝑏3𝑦 + 𝑐3𝑧 = 𝑑3

FOR FULLNOTES CALL/TEXT:0713440925

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Solving and graphs

Solve for x in the following equation.

Example 1:

5 x - 6 = 3 x - 8

Subtract 3x from both sides of the equation:

Add 6 to both sides of the equation:

Divide both sides by 2:

The answer is x = - 1

Check the solution by substituting -1 in the original equation for x. If the left side of the equation

equals the right side of the equation after the substitution, you have found the correct answer.

Left side: 5(-1) - 6 = - 11

Right side: 3(-1) - 8 = - 11

Exercise

Solve each of the following equations.

(a)

(b)

(c)

(d)

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Solving Linear Systems Graphically

Solve this system of equations graphically.

4x - 6y = 12

2x + 2y = 6

Example using a graphical method:

Solve graphically: 4x - 6y = 12

2x + 2y = 6

To solve a system of equations graphically, graph both equations and see where

they intersect. The intersection point is the solution.

First, solve each equation for "y =".

4x - 6y = 12

slope =

y-intercept = -2

2x + 2y = 6

slope = -1

y-intercept = 3


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Graph the lines:

The slope intercept method of graphing was used in this example.

The point of intersection of the two lines, (3,0), is the solution to the system of equations.

This means that (3,0), when substituted into either equation, will make them both true. See the

check.

Check: Since the two lines cross at (3,0), the solution is x = 3 and y = 0. Checking these value

shows that this answer is correct. Plug these values into the ORIGINAL equations and get a true

result.

4x - 6y = 12

4(3) - 6(0) = 12

12 - 0 = 12

12 = 12 (check)

2x + 2y = 6

2(3) + 2(0) = 6

6 + 0 = 6

6 = 6 (check)

QUADRATIC EQUATIONS

The general formula is 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

When the equation is ploted, it yields either a valley or a mountain depending on constant a. if <

0 a mountain, if> 0 a valley.

In order to solve a linear equation, the equation is equated to zero.

The Quadratic Formula Explained

Often, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set

each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy,

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or it doesn't factor at all, or you just don't feel like factoring. While factoring may not always be

successful, the Quadratic Formula can always find the solution.

The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are

just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to

solve. The Quadratic Formula is derived from the process of completing the square, and is

formally stated as:

For ax2 + bx + c = 0, the value of x is given by:

For the Quadratic Formula to work, you must have your equation arranged in the form

"(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything

above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that

you are careful not to drop the square root or the "plus/minus" in the middle of your calculations,

or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself

up. Remember that "b2" means "the square of ALL of b, including its sign", so don't leave b

2

being negative, even if b is negative, because the square of a negative is a positive.

In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the

long run. Trust me on this!

Solving and using quadratic equations

Quadratic equations can be solved by factorising, completing the square and by formula.

Solving quadratic equations by factorising

To solve a quadratic equation, the first step is to write it in the form: ax2 + bx + c = 0. Then

factorise the equation as you have revised in the previous section.

If we have two numbers, A and B, and we know that A × B = 0, then it must follow that either A

= 0, or B = 0 (or both). When we multiply any number by 0, we get 0.

Example

Solve the equation: x2 - 9x + 20 = 0


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Solution

First, factorise the quadratic equation x2- 9x + 20 = 0

Find two numbers which add up to 9 and multiply to give 20. These numbers are 4 and 5.

(x - 4) (x - 5) = 0

Now find the value x so that when these brackets are multiplied together the answer is 0.

This means either (x - 4) = 0 or (x - 5) = 0

So x = 4 or x = 5.

You can check these answers by substitutuing 4 and 5 in to the equation:

x2- 9x + 20

Substituting 4 gives:

42 - 9 × 4 + 20 = 16 - 36 + 20 = 0

Substituting 5 gives:

52 - 9 × 5 + 20 = 25 - 45 + 20 = 0

Remember these 3 simple steps and you will be able to solve quadratic equations.

Completing the square

This is another way to solve a quadratic equation if the equation will not factorise.

It is often convenient to write an algebraic expression as a square plus another term. The other

term is found by dividing the coefficient of x by 2, and squaring it.

Any quadratic equation can be rearranged so that it can be solved in this way.

Have a look at this example.

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Example 1

Rewrite x2 + 6x as a square plus another term.

The coeffient of x is 6. Dividing 6 by 2 and squaring it gives 9.

x2 + 6x = (x

2 + 6x + 9) - 9

= (x + 3)2 - 9

Example 2

We have seen in the previous example that x2 + 6x = (x + 3)

2 - 9

So work out x2 + 6x - 2

x2 + 6x - 2 = ( x

2 + 6x + 9 ) - 9 - 2 = (x + 3)

2 - 11

Questions:

Question 1

Solve x2 + 6x - 2 = 0

Answer

From the previous examples, we know that x2 + 6x - 2 = 0 can be written as (x + 3)

2 - 11

= 0

So, to solve the equation, take the square root of both sides. So (x + 3)2 = 11

x + 3 = +

or x + 3 = -

x = - 3 +

or x = - 3 -

x = - 3 + 3.317 or x = - 3 - 3.317 ( is 3.317)

x = 0.317 (3 s.f) or x = - 6.317 (3 s.f)


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Example 3

Rewrite 2x2 + 20x + 3

Rewrite to get x2 on its own.

2( x2 + 10x ) + 3

The coefficient of x is 10. Divide 10 by 2, and square to get 25.

= 2 ( ( x + 5)2 - 25) + 3

= 2 (x + 5)2 - 50 + 3

= 2 (x + 5)2 - 47

Question

Now use the previous example to solve 2x2 + 20x + 3 = 0

Answer

From the previous example, we know that 2x2 + 20x + 3 can be rewritten as:

2 (x + 5)2 - 47

Therefore, we can rewrite the equation as:

2(x + 5 )2 - 47 = 0

2(x + 5 )2 = 47

(x + 5 )2 = 23.5 (dividing both sides by 2)

Take the square root of both sides.

x + 5 =

or x + 5 = -

x = - 5 +

or x = - 5 -

x = - 5 +

or x = - 5 -

x = - 0.152 (3 s.f) or x = - 9.85 (3 s.f)

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Using the quadratic formula

Using the technique of completing the square, you can make a formula which works for all

quadratic equations.

The most general way to write a quadratic equation is:

ax2 + bx + c = 0

Here a, b and c are numbers that vary for different equations.

So if the equation was:

2x2 + 7x + 11 = 0

then a = 2, b = 7, c = 11.

The formula for the solution is:

This formula will work for all equations that can be solved.

Always try to factorise first. If the equation factorises, this is the easier method. In an exam,

any question that asks for an answer to a quadratic equation correct to x decimal places should be

solved using this formula. Take a look at this example:

Solve 2x2 - 5x - 6 = 0

Here a = 2, b = -5, c = -6

Substituting these values in the formula, gives you:


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Solving a linear and quadratic simultaneous equation

Another way to solve a quadratic equation is to draw its graph.

This is the graph of: y = x2 - 9x + 20

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We can use it to solve: x2 - 9x + 20 = 0

The answers are along the x axis where the graph reaches y = 0 (where it crosses the x axis).

Using the graph the answers are x = 4 and x = 5. We can also solve this equation by factorising:

y = x2 - 9x + 20 y = (x - 4) (x - 5)

This shows that the solutions are x = 4 and x = 5 which matches the answers in the graph above.

A graphical method can be used to find the solution of a quadratic and linear equation. Suppose

we want to find the solutions for:

x2 - 9x + 20 = x - 1

One side of this equation is x2 - 9x + 20. We already know what this graph looks like.

The other side of the equation is x - 1. Drawing y = x - 1 on the same graph we get:


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The solutions to x2 - 9x + 20 = x - 1 can be found where the line and curve crosses. The solutions are found reading values from the x-axis. The solutions are x = 3 and x = 7.

We can check this with the following:

x = 3

1. 32 - 9 x 3 + 20 = 3 - 12. 9 - 27 + 20 = 23. 9 - 7 = 24. 2 = 2

x = 7

1. 72 - 9 x 7 + 20 = 7 - 12. 49 - 63 + 20 = 63. 49 - 43 = 64. 6 = 6

For the following questions, you may find it useful to look back at the graph sections Straight-

line graphs and Curved graphs.

Question

Solve: x2 + x + 2 = 5 – x

Answer

First draw the curve y = x2 + x + 2 and the line y = 5 – x.

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The solutions are where the curve and line cross. In this case x = -3 and x = 1.

The answers will not usually be whole numbers.

Question

Solve: x2 + 2x + 1 = x + 4

Answer

First draw the curve y = x2 + 2x + 1 and the line: y = x + 4.

The solutions are where the curve and line cross. In this case x = -2.3 and x =1.3.

Because these solutions are correct to one decimal place, a check will give numbers that

are close but not exactly the same.

x = -2.3

-2.32 + (2 x -2.3) + 1 = -2.3 + 4

5.29 + (-4.6) + 1 = 1.7

1.69 and 1.7 are very close

x = 1.3

-1.32 + (2 x 1.3) + 1 = 1.3 + 4

1.69 + 2.6 + 1 = 5.3

5.29 and 5.3 are very close


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Differentiation

It is concerned with rates of change e.g. profit with respect to output

i) Revenue with respect to output ii) Change of sales with respect to level of advertisementSavings with respect to income, interest rates Rates of changes and slope (gradients)

-It estimates a slope (gradient of graph) of a particular point. The derivative of a function Δ𝑦

Δ𝑥Gives the exact change of a point. It‘s the process of finding the derivative o f a function.

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Business purpose for differentiation For business purposes, it‘s used in the profit functions, cost functions and revenue functions . it will help the management in determining the levels of activities that will maximize profits or minimize cost for max/min points in a function,

Δ𝑦

Δ𝑥= 0

Example 1 A firm has analyzed the operating conditions, prices and cost and they have developed the following functions. Where Q is the no. of units sold.

𝑇𝑅 = 400𝑄 + 4𝑄2

𝑇𝐶 = 𝑄2 + 10𝑄 + 3𝑄

The firm wishes to maximize revenue/profit and wishes to know

i. The quantity to be sold.

ii. Price

iii. Amount of profit


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Solution: Profit is maximized when 𝑀𝑅 = 𝑀𝐶 = 0

i. The quantity to be sold.

𝑀𝐶 =Δ𝑇𝐶

Δ𝑄= 2𝑄 + 10 = 0

𝑀𝑅 =Δ𝑇𝑅

Δ𝑄= 400 + 8𝑄 = 0

𝑀𝑅 = 𝑀𝐶

400 + 8𝑄 = 2𝑄 + 10

6𝑄

6=

390

6

𝑄 = −65

𝑄 = 65

ii. Price

𝑃 =𝑇𝑅

𝑄=

400 65 + 4(65)2

65

= 60

iii. Amount of profit

𝑇𝑅 − 𝑇𝐶

(400 65 + 4(65)2 − 65𝑋10 + 30

= 37,995

SOLVING SIMULTANEOUS EQUATIONS:

Substitution Method

Example:

𝑥 = 86000 + 0.01𝑦

𝑦 = 44000 + 0.02𝑥

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Rewrite:

86000 + 0.01𝑦 − 𝑥 44000 + 0.02𝑥 − 𝑦

𝑌 = 44000 + 0.02(86000 + 0.01)

𝑌 = 44000 + 1720 + 0.0002𝑦

𝑦 − 0.0002𝑦 = 44000 + 1720

0.9998𝑦 = 45720

𝑦 =45720

0.998

= 45,729.15

𝑋 = 86000 + (0.01𝑥45.729)

= 86457.29

Elimination

𝑥 + 2𝑦 = 3 2𝑥 + 3𝑦 = 4

2(𝑥 + 2𝑦) = 3(2)

2𝑥 + 4𝑦 = 6

−2𝑥 + 3𝑦 = 4 𝑦 = 2

𝑥 = −3

BREAK-EVEN ANALYSIS

The break-even point (BEP) in economics, business, and specifically cost accounting, is the

point at which total cost and total revenue are equal: there is no net loss or gain, and one has

"broken even." A profit or a loss has not been made, although opportunity costs have been

"paid", and capital has received the risk-adjusted, expected return. In short, all costs that needs to

be paid are paid by the firm but the profit is equal to 0


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Overview

The break-even level or break-even point (BEP) represents the sales amount—in either unit or

revenue terms—that is required to cover total costs (both fixed and variable). Total profit at the

break-even point is zero. Break-even is only possible if a firm‘s prices are higher than its

variable costs per unit. If so, then each unit of the product sold will generate some ―contribution‖

toward covering fixed costs

For example, if a business sells fewer than 200 tables each month, it will make a loss; if it sells

more, it will make a profit. With this information, the business managers will then need to see if

they expect to be able to make and sell 200 tables per month.

If they think they cannot sell that many, to ensure viability they could:

1. Try to reduce the fixed costs (by renegotiating rent for example, or keeping better control

of telephone bills or other costs)

2. Try to reduce variable costs (the price it pays for the tables by finding a new supplier)

3. Increase the selling price of their tables.

Any of these would reduce the break-even point. In other words, the business would not need to

sell so many tables to make sure it could pay its fixed costs.

Purpose

The main purpose of break-even analysis is to determine the minimum output that must be

exceeded in order to make profit. It also is a rough indicator of the earnings impact of a

marketing activity.

The break-even point is one of the simplest yet least used analytical tools in management. It

helps to provide a dynamic view of the relationships between sales, costs, and profits. For

example, expressing break-even sales as a percentage of actual sales can give managers a

chance to understand when to expect to break even (by linking the percent to when in the

week/month this percent of sales might occur).

The break-even point is a special case of Target Income Sales, where Target Income is 0

(breaking even). This is very important for financial analysis.

Construction

In the linear Cost-Volume-Profit Analysis model (where marginal costs and marginal revenues

are constant, among other assumptions), the break-even point (BEP) (in terms of Unit Sales

(X)) can be directly computed in terms of Total Revenue (TR) and Total Costs (TC) as:

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where:

TFC is Total Fixed Costs,

P is Unit Sale Price, and

V is Unit Variable Cost.

The Break-Even Point can alternatively be computed as the point where Contribution equals

Fixed Costs.

The quantity, , is of interest in its own right, and is called the Unit Contribution

Margin (C): it is the marginal profit per unit, or alternatively the portion of each sale that

contributes to Fixed Costs. Thus the break-even point can be more simply computed as the point

where Total Contribution = Total Fixed Cost:

To calculate the break-even point in terms of revenue (a.k.a. currency units, a.k.a. sales

proceeds) instead of Unit Sales (X), the above calculation can be multiplied by Price, or,

equivalently, the Contribution Margin Ratio (Unit Contribution Margin over Price) can be

calculated:

R=C, Where R is revenue generated, C is cost incurred i.e. Fixed costs + Variable Costs or Q * P

(Price per unit) = TFC + Q * VC (Price per unit), Q * P - Q * VC = TFC, Q * (P - VC) = TFC,

or, Break Even Analysis Q = TFC/c/s ratio=Break Even


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TURNING POINTS

It‘s a point of local minima or maxima. These points are of great interest/importance to

businessmen because they represent points at which costs are minimized or revenue is

maximized. These points are of zero slopes.

This is done by getting the 2nd

derivative.

If the 2nd

derivative is –ve; then the turning point is at maximum If

the 2nd

derivative is +ve; the turning point is minimum.

𝑹 = 𝟒𝟎𝟎𝑸 − 𝟒𝑸𝟐

d𝑅

d𝑄= 400 + 8𝑄

d2𝑅

dQ 2= −8 This turning point is maximum

Point of maximum revenue is not necessary to be points of maximum profits. This is because

the revenue function does not include the cost function

TR=TC. It‘s that point where total revenue equates to total cost. It means that break-even point

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where profits are equal to zero. It‘s important for the management to identify this point so as to

ensure that the business is not running at a loss.

𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶

𝜋 − 𝑇𝑅 − 𝑇𝐶

𝜋 − 𝑇𝑅 − (𝐹𝐶 + 𝑉𝐶)

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 = 𝑇𝑅 − 𝑉𝐶

Thus,

𝜋 = 𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 − 𝐹𝐶

It means that contribution should be equal to FC therefore

𝑏𝑟𝑒𝑎𝑘𝑒𝑣𝑒𝑛 𝑝𝑜𝑖𝑛𝑡𝑠 𝐵. 𝐸. 𝑃 =𝐹𝐶

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝑁𝑜. 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑠𝑜𝑙𝑑

Example: A carpenter makes tables for sale. The carpenter needs materials and labour for sh.60 every table is sold for sh.120. The fixed cost is sh.1200

1. How many tables must be sold per month to breakeven?

𝐵. 𝐸. 𝑃 =𝐹𝐶

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛=

1200

60

2. If profit of sh.600 is derived, how many tables must be sold?

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 = 𝜋 + 𝐹𝐶

= 600 + 1200

= 1800

𝑁𝑜. 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑠𝑜𝑙𝑑 =𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡

=1800

(120 − 60)

= 30 𝑡𝑎𝑏𝑙𝑒𝑠


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3. Suppose fixed cost is sh.1500 and selling price sh.150 how many tables must be sold tobreakeven point.

𝐵. 𝐸. 𝑃 =𝐹𝐶

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡=

1200

(150 − 60)

= 17 𝑡𝑎𝑏𝑙𝑒𝑠

4. Suppose VC=90, how many tables should be sold at old selling price and at the old FCand profit.

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 = 600 + 1200

= 1800

𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝐶𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡==

1800

(120 − 90)

= 𝟔𝟎 𝒕𝒊𝒎𝒆𝒔

Assumptions of breakeven point

a) There is a linear relationship between cost and revenue

b) VC and MR are constant over the relevant range

c) FC is assumed to be constant

Application:

Example 1

A company A sells out to company B for sh.200 per unit. The cost of sales per week is given by

the following functions

𝐶 = 2𝑄2 + 40𝑄 + 80

Where Q is the value of weekly sales

Company B uses output of company A to manufacture a product a product whose demand is dependent on the sales price. The revenue per week of company B is given by,

𝑅 = 1000𝑄 − 16𝑄2

And the cost per week of the company B excluding the cost of the product both from company A are given by the function

𝐶 = 2𝑄2 + 80𝑄 + 400 Company A can restrict weekly supply of its product to company B but cannot raise the unit

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price above sh.200. the two companies are considering whether to merge together into a single company. Required:

i. At what weekly sales will company A maximize profit?

ii. What would be the profit/loss of company B if A was able to supply a profitmaximizing quantity of each product for each week.

Example 2

If the two companies are merged, what will be the profit maximization output per week and what will be the weekly profit


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Example 3

ABC ltd is a small estate developer. It has 7 permanent employees.

1- Managing Director 100,000 2- Manager, Development 60,000 3- Manager, marketing 45,000 4- Project manager 55,000 5- Financial manager 40,000 6- Office manager 30,000 7- Receptionist 20,000

ABC leases a building for sh.200,000 a month. The cost of supplies and utilities is sh.30,000 per month. ABC builds only one style house in the valley land for each house is sh.550,000 supplies per house is sh.280,000. Labour cost is sh.200,000 per house. One sale representative of ABC is paid a commission of sh.20,000 on sale of each house. The selling price of each house is sh.1150,000. Required:

I. Identify all the costs that denote MR and costs for each house?

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MR= selling price of each house-(cost of land+labour+supplies per house)

II. Determine the monthly cost function, revenue function and profit function

Cost function= (550000+280000+200000))Q + 400000 =1,030,000Q + 400000

Revenue function=(1150000-20000)Q =1,130,000Q

Profit function= TR-TC =1130000Q-(1030000Q+400000) =100,000Q-400,000

III. Determine breakeven point for monthly sales for the house.

100,000Q-400,000=C 100,000Q=400,000

400000 Q=

100000

Q=4

IV. Determine monthly profit if 12 houses are built and sold=100,000(12)-400,000 =1,200,000-400,000 =800,000

APPLICATION OF EQUATION OR FUNCTIONS IN COST-VOLUME-PROFIT

ANALYSIS

It can be applied in the following areas:

a) Demand function (price function)

b) Revenue function

c) Cost function

d) Profit function

Demand function

This is an expression that enables one to determine price of a given item. It takes the form of an equation of straight line

𝑝 = 𝑎 + 𝑏𝑥

Where p – price of an item

a & b –constants

x – quantity demanded


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Revenue function The total revenue equation is the function of a price

TR=P.Q

Where,

P –price of an item

Q-quantity sold

TR- total revenue

Total cost function It has two components namely fixed and variable cost per unit. TC is a linear equation

𝑇𝐶 = 𝑎 + 𝑏𝑥

Where, a is fixed cost b is variable cost Q is quantity produced

Profit function

It‘s expressed as 𝝅 = 𝑻𝑹 − 𝑻𝑪

TR - total revenue TC - total cost

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