Business Mathematics (BC 104) PDF Past Paper - University of Jammu

Directorate of Distance Education UNIVERSITY OF JAMMU JAMMU STUDY MATERIAL FOR B.COM, SEMESTER - I Course No. : BC 104 Unit I to V Business Mathematics...

Directorate of Distance Education UNIVERSITY OF JAMMU JAMMU STUDY MATERIAL FOR B.COM, SEMESTER - I Course No. : BC 104 Unit I to V Business Mathematics Lesson No. 1 to 25 Course Co-ordinator Rohini Gupta Suri Mob. No. 9419186716 http:/www.distanceeducationju.in Printed and Published on behalf of the Directorate of Distance Education, University of Jammu by the Director, DDE, University of Jammu, Jammu. ( 206 ) Course No. BC-104 Title : Business Mathematics Duration of Exam. : 3 Hrs. Total marks : 100 Theory Examination: 80 Internal Assessment : 20 OBJECTIVE : To impart knowledge about fundamental mathematics used in business. UNIT – I : COMMERCIAL ARITHMETIC Introduction of business mathematics; Scope and importance of quantitative techniques; Concept of equated monthly instalment (EMI), profit and loss, simple and compound interest including half yearly and quarterly calculations, bill of discounting-Business applications. UNIT – II: SET THEORY Concept of a set , operation of sets, Algebra of sets, Cartesian product of two sets and its application to business mathematics. UNIT – III : PROGRESSIONS Arithmetics progression, Finding the nth term, Sum of n terms, representation of an A.P. Geomatric progression , Finding the nth term , Sum of n terms and sum of infinity, representation of an G.P. Special cases , , UNIT – IV : MATRIX AND MEASURMENT Concept of a matrix , algebra of matrices, inverse of matrices, determinant of a square matrix, expansion rule, properties of determinant, solution of a system of linear equation upto 3 variable using 1. Cramer’s Rule 2. The method of matrix inverse. (i) SKILL DEVELOPMENT (SPECIMEN FOR CLASS ROOM TEACHING AND INTERNAL ASSESSMENT) A minimum of five exercises to be undertaken from above said courses selecting atleast from one unit. BOOKS RECOMMENDED: 1. Dr. A.K. Arte & : A Text book of Business-Mathematics R.V. Prabhakar 2. Dorai Raj : Business Mathematics 3. Sanchethi & Kapoor : Business Mathematics 4. Zamiruddin & khanna : Business Mathematics 5. Saha : Business Mathematics 6. Kavita Gupta : Business Mathematics 7. VK Kapoor : Linear Programming NOTE FOR PAPER SETTER: Equal weightage shall be given to all the units of the syllabus. The external paper shall be of the two sections viz, A&B. Section-A: This section will contain four short answer questions selecting one from each unit. Each question carries 5 marks. A candidate is required to attempt all the four questions. Total weightage to this section shall be 20 marks. Section- B: This section will contain eight long answer questions of 15 marks each. Two questions with internal choice will be set from each unit. A candidate has to attempt any four questions selecting one from each unit. Total weightage to this section shall be 60 marks (ii)  MODEL QUESTION PAPER BUSINESS MATHEMATICS Section – A (20 Marks) Attempt all questions. Each question carries five marks. 1. Explain the importance of quantative techniques in business? 2. Explain the concept of set with suitable examples? 3. Differentiate between arithmetic and geometric progression? 4. Explain the concept of expansion rule? Section – B (60 Marks) Attempt any four questions selecting one question from each unit. Each question carries 15 marks. 1. A sum of Rs. 800 amounts to Rs. 920 in 3 years at a simple inerest. If the interest rate is increased by 3% , it would amount to how much. OR What annual instalment will discharge a debt of Rs.1092 due in 3 years at 125 simple interest. 2. Which of the following sets are equal. A={0,L,W,F} B={Letters of the word follow} C={Letters of the word wolf} D={Letters of the word flow} OR Find true & False for the following set V={1,2,3,4,5} A={1,2,3} (iii)  B={3,5} C={2,4} 3. What is arithmetic progression. Explain with suitable example? OR What is the Geographic progression. Explain with example ? 4. Evaluate a. |5 4| b. |x-1 1| c. |X2 +xy+y2 X+Y | |-2 3| |X3 X2+X+1| | X2 - xy+y2 X-Y | OR Use matrix method to solve the following system of equations 5x-&y=2, 7x-5y=3 (iv) Business Mathematics- 104 Unit-I Commercial Arithmetic Lesson No. 1-5 STRUCTURE 1.1 Introduction 1.2 Objective 1.3 Scope of quantitative techniques 1.4 Importance of quantitative techniques 1.5 Concept of Equated Monthly Instalment (EMI) 1.6 Profit and Loss 1.7 Simple Interest 1.8 Compound Interest (half yearly and quarterly calculations) 1.9 Bill of Discounting-Business Applications 1.10 Summary 1.11 Self Assessment exercises 1.12 Suggested Reading 1.1 INTRODUCTION Business mathematics is mathematics used by commercial enterprises to record and manage business operations. Commercial organisations use mathematics in accounting, inventory management, marketing, sales forecasting, and financial analysis. Mathematics typically used in commerce includes elementary arithmetic, elementary algebra, statistics and probability. Business management an be done more effective in some cases by use of more advanced mathematics such as calculus, matrix algebra and linear programming. 1 Business mathematics, sometimes called commercial math or consumer math, is a group of practical subjects used in commerce and everyday life. In schools, these subjects are often taught to students who are not planning a university education. In the United States, they are typically offered in high schools and in schools that grant associate’s degrees. The emphasis in these courses is on computational skills and their practical application, with practical application being predominant. A U.S. business math course might include a review of elementary arithmetic, including fractions, decimals, and percentages. Elementary algebra is often included as well, in the context of solving practical business problems. The practical applications typically include checking accounts, price discounts payroll calculations, simple and compound interest, consumer and business credit, and mortgages and [revenues]. Business mathematics is mathematics used by commercial enterprises to record and manage business operations. Commercial organisations use mathematics in accounting, inventory management, marketing, sales forecasting, and financial analysis. Mathematics typically used in commerce includes elementary arithmetic, elementary algebra, statistics and probability. Business management can be made more effective in some cases by use of more advanced mathematics such as calculus, matrix algebra and linear programming. Business Mathematics is very important for modern business management. The forecasting and operating procedures are based primarily on business mathematics. Things such as simple interest and compound interest show a company what it will lose or get over the years if it invests in a particular asset. Business mathematics also helps in cost and price calculations which are the basis of cash inflows and outflows that all companies have to deal with. In academia, Business Mathematics includes mathematics courses taken at an undergraduate level by business students at University. These courses are slightly less difficult and do not always go into the same depth as other mathematics courses for people majoring in mathematics or science fields. The two most common math courses taken in this form are Business Calculus and 2 Business Statistics. Examples used for problems in these courses are usually real- life problems from the business world to help students gain a more detailed understanding. Thus, to conclude Business Mathematics in management system is more effective in some cases by the use of more advanced mathematics such as calculus, matrix algebra and linear programming. Business organisations use mathematics in accounting, inventory management, marketing, sales forecasting, financial analysis etc. 1.2 OBJECTIVE: After reading this unit you would be able: To understand importance of quantitative techniques. To know about applications of bills of discounting in business. To provide the student clear concept about profit and loss. To enhance the student in purchasing and selling activity. To make recognise the student with money value in marketing system. 1.3 SCOPE OF QUANTITATIVE TECHNIQUES Quantitative techniques may be defined as those techniques which provide the decision maker a systematic and powerful means of analysis, based on quantitative data. It is a scientific method employed for problem solving and decision making by the management. With the help of quantitative techniques, the decision maker is able to explore policies for attaining the pre-determined objectives. In short, quantitative techniques are inevitable in decision-making process. There are different types of quantitative techniques. We can classify them into three categories. They are: 1. Mathematical Quantitative Techniques 2. Statistical Quantitative Techniques 3. Programming Quantitative Techniques. 1.3.1 Mathematical Quantitative Techniques: 3 A technique in which quantitative data are used along with the principles of mathematics is known as mathematical quantitative techniques. Mathematical quantitative techniques involve: 1. Permutations and Combinations: Permutation means arrangement of objects in a definite order. The number of arrangements depends upon the total number of objects and the number of objects taken at a time for arrangement. Combination means selection or grouping objects without considering their order. 2. Set Theory: - Set theory is a modern mathematical device which solves various types of critical problems. Matrix Algebra: Matrix is an orderly arrangement of certain given numbers or symbols in rows and columns. It is a mathematical device of finding out the results of different types of algebraic operations on the basis of the relevant matrices. 4. Determinants: It is a powerful device developed over the matrix algebra. This device is used for finding out values of different variables connected with a number of simultaneous equations. 5. Differentiation: It is a mathematical process of finding out changes in the dependent variable with reference to a small change in the independent variable. 6. Integration: Integration is the reverse process of differentiation. 7. Differential Equation: It is a mathematical equation which involves the differential coefficients of the dependent variables. 1.3.2 Statistical Quantitative Techniques: Statistical techniques are those techniques which are used in conducting the statistical enquiry concerning to certain Phenomenon. They include all the statistical methods beginning from the collection of data till interpretation of those collected data. Statistical techniques involve: 1. Collection of data: One of the important statistical methods is collection of data. There are different methods for collecting primary and secondary data. 2. Measures of Central tendency, dispersion, skewness and Kurtosis Measures of Central tendency is a method used for finding he average of a series while measures of dispersion used for finding out the variability in a series. Measures of Skewness measures asymmetry of a distribution while measures of Kurtosis measures the flatness of peakedness in a distribution. 3. Correlation and Regression Analysis: Correlation is used to study the degree of relationship among two or more variables. On the other hand, 4 regression technique is used to estimate the value of one variable for a given value of another. 4. Index Numbers: Index numbers measure the fluctuations in various Phenomena like price, production etc over a period of time, they are described as economic barometres. 5. Time series Analysis: Analysis of time series helps us to know the effect of factors which are responsible for changes: 6. Interpolation and Extrapolation: Interpolation is the statistical technique of estimating under certain assumptions, the missing figures which may fall within the range of given figures. Extrapolation provides estimated figures outside the range of given data. 7. Statistical Quality Control Statistical quality control is used for ensuring the quality of items manufactured. The variations in quality because of assignable causes and chance causes can be known with the help of this tool. Different control charts are used in controlling the quality of products. 8. Ratio Analysis: Ratio analysis is used for analyzing financial statements of any business or industrial concerns which help to take appropriate decisions. 9. Probability Theory: Theory of probability provides numerical values of the likely hood of the occurrence of events. 10. Testing of Hypothesis: Testing of hypothesis is an important statistical tool to judge the reliability of inferences drawn on the basis of sample studies. 1.3.3 Programming Techniques: Programming techniques are also called operations research techniques. Programming techniques are model building techniques used by decision makers in modern times. Programming techniques involve: 1. Linear Programming: Linear programming technique is used in finding a solution for optimizing a given objective under certain constraints. 2. Queuing Theory: Queuing theory deals with mathematical study of queues. It aims at minimizing cost of both servicing and waiting. 3. Game Theory: Game theory is used to determine the optimum strategy in a competitive situation. 4. Decision Theory: This is concerned with making sound decisions under conditions of certainty, risk and uncertainty. 5. Inventory Theory: Inventory theory helps for optimizing the inventory levels. It focuses on minimizing cost associated with holding of inventories. 6. Net work programming: It is a technique of planning, scheduling, controlling, monitoring and co-ordinating large 5 and complex projects comprising of a number of activities and events. It serves as an instrument in resource allocation and adjustment of time and cost up to the optimum level. It includes CPM, PERT etc. 7. Simulation: It is a technique of testing a model which resembles a real life situations 8. Replacement Theory: It is concerned with the problems of replacement of machines, etc due to their deteriorating efficiency or breakdown. It helps to determine the most economic replacement policy. 9. Non Linear Programming: It is a programming technique which involves finding an optimum solution to a problem in which some or all variables are non-linear. 10. Sequencing: Sequencing tool is used to determine a sequence in which given jobs should be performed by minimizing the total efforts. 11. Quadratic Programming: Quadratic programming technique is designed to solve certain problems, the objective function of which takes the form of a quadratic equation. 12. Branch and Bound Technique It is a recently developed technique. This is designed to solve the combinational problems of decision making where there are large number of feasible solutions. Problems of plant location, problems of determining minimum cost of production etc. are examples of combinational problems. 1.3.4 Objectives of Quantitative Techniques 1. To facilitate the decision-making process. 2. To provide tools for scientific research. 3. To help in choosing an optimal strategy. 4. To enable in proper deployment of resources. 5. To help in minimising costs. 6. To help in minimising the total processing time required for performing a set of jobs. 6 1.4 Importance of Quantitative Techniques Quantitative techniques render valuable services in the field of business and industry. Today, all decisions in business and industry are made with the help of quantitative techniques. Some important uses of quantitative techniques in the field of business and industry are given below: 1. Quantitative techniques of linear programming is used for optimal allocation of scarce resources in the problem of determining product mix 2. Inventory control techniques are useful in dividing when and how much items are to be purchase so as to maintain a balance between the cost of holding and cost of ordering the inventory 3. Quantitative techniques of CPM and PERT helps in determining the earliest and the latest times for the events and activities of a project. This helps the management in proper deployment of resources. 4. Decision tree analysis and simulation technique help the management in taking the best possible course of action under the conditions of risks and uncertainty. 5. Queuing theory is used to minimise the cost of waiting and servicing of the customers in queues. 6. Replacement theory helps the management in determining the most economic replacement policy regarding replacement of equipment. 1.4.1 Applications of Quantitative Techniques in: i) Marketing: Analysis of marketing research information Statistical records for building and maintaining an extensive market Sales forecasting ii) Production Production planning, control and analysis Evaluation of machine performance 7  Quality control requirements Inventory control measures iii) Finance, Accounting and Investment: Financial forecast, budget preparation Financial investment decision Selection of securities Auditing function Credit, policies, credit risk and delinquent accounts iv) Personnel: Labour turnover rate Employment trends Performance appraisal Wage rates and incentive plans v) Economics Measurement of gross national product and input-output analysis Determination of business cycle, long-term growth and seasonal fluctuations Comparison of market prices, cost and profits of individual firms Analysis of population, land economics and economic geography Operational studies of public utilities Formulation of appropriate economic policies and evaluation of their effect vi) Research and Development Development of new product lines Optimal use of resources Evaluation of existing products 1.4.2 The following are the important limitations of quantitative techniques: Even though the quantitative techniques are inevitable in decision-making process, they are not free from short comings. 8  1. Quantitative techniques involves mathematical models, equations and other mathematical expressions 2. Quantitative techniques are based on number of assumptions. Therefore, due care must be ensured while using quantitative techniques, otherwise it will lead to wrong conclusions. 3. Quantitative techniques are very expensive. 4. Quantitative techniques do not take into consideration intangible facts like skill, attitude etc. 5. Quantitative techniques are only tools for analysis and decision-making. They are not decisions itself. 1.5 Equated Monthly Instalment (EMI) 1.5.1 Concept A fixed payment amount made by a borrower to a lender at a specified date each calendar month. Equated monthly instalments are used to pay off both interest and principal each month, so that over a specified number of years, the loan is paid off in full. With most common types of loans, such as real estate mortgages, the borrower makes fixed periodic payments to the lender over the course of several years with the goal of repaying the loan. EMIs differ from variable payment plans, in which the borrower is able to pay higher payment amounts at his or her discretion. In EMI plans, borrowers are usually allowed one fixed payment amount each month 1.5.2 Calculation of EMI (Equated Monthly Installments) When a person takes loan from a bank to buy a house for a fixed tenure, he has to pay EMI (Equated Monthly Installments). EMI is the term used for the monthly payment made by a borrower to the lender towards interest and principal money borrowed. EMI amount depends on the following factors: Amount of Loan: EMI depends primarily on the amount of loan you have taken. With an increase in the loan amount, the EMI to be paid also increases. 9 Tenure of the Loan: Next most important factor is the time for which you have taken the loan. The EMI decreases with the increase in the tenure of the loan. But one should understand the increase in tenure means that you will have to pay more interest to the bank. Since you will be having an outstanding amount against you for a longer time you will have to pay some extra for taking more time. EMI increases with the shorter tenure; in this case one should do the budgeting properly and make sure that the EMI can be paid on time. Complete your Loan as Soon as Possible: For a longer tenure of loan one has to pay more interest, so one should plan prudently and pay back the loan as soon as possible. Fluctuation in Interest Rates: Interest rate is a floating parameter. It keeps changing with Inflation and changing policies of the government. One has to pay the EMI according to the prevailing interest rates. RBI has recently cut repo rate by 50 bases. Many banks have already reduced the interest rates accordingly, in this case the buyer will have to pay the EMIs on the reduced rates and this will make a significant difference. Formula for finding EMI is given as under: The mathematical formula to calculate EMI is: EMI = P x r x (1 + r) n/ ((1 + r)n - 1) Where P= Loan amount, r = interest rate, n=tenure in number of months. Let’s try to understand it using an example. Suppose Ram has borrowed Rs. 5 lakhs from a bank on the interest rate of 12 per cent for 10 years. Do in this case: M (Loan period in months) : No of Years X 12 = 10 X 12 = 120 I (Interest rate per Annum / 12) : ( 12/100) / 12 =.01 10 L(LoanAmount : Rs. 5 ,00,000 EMI (Equated Monthly Instalments) So, in the example discussed above: EMI that Ram has to pay is Rs. 7147. Total payment made by Ram to the bank in 10 years (EMI X Total tenure in months (7174 X 120) is Rs 8, 60, 880. The total interest rate payable will be (Total payment – loan amount) Rs. 3,60,88 1.5.3 Reasons for Varied EMI Payments The other major factor which determines the EMI payments is the type of interest on the loan. In case of fixed rate loans, the EMI payments remain constant during the tenure. In case of floating rate loans, the interest rates vary based on the prevailing market rates. Hence, the EMI payments also vary whenever there is a change in the base rates. The other factor which effects the EMI payments is the pre-closure or partial payments made towards the loan. Any partial payments made towards the loan are deducted from the principal amount of the loan. This results in reduction of total interest that is to be paid. Generally an individual who is making a partial payment will be given an option to keep the tenure constant or keep the EMI constant. If one opts for keeping the tenure constant, the monthly EMI payments will be reduced. Similarly, if one opts for keeping the EMI constant, the tenure of the loan will be reduced. 1.5.4 Floating Rate EMI Calculation Calculate floating or variable rate EMI by taking into consideration two opposite scenarios, i.e., optimistic (deflationary) and pessimistic 11  (inflationary) scenario. Loan amount and loan tenure are the two components required to calculate the EMI i.e., you are going to decide how much loan you have to borrow and how long your loan tenure should be. But interest rate is decided by the banks & HFCs based on rates and policies set by RBI. As a borrower, you should consider the two extreme possibilities of increase and decrease in the rate of interest and calculate how much would be your EMI under these two conditions. Such calculation will help you decide how much EMI is affordable, how long your loan tenure should be and how much you should borrow. Optimistic (deflationary) scenario: Assume that the rate of interest comes down by 1% – 3% from the present rate. Consider this situation and calculate your EMI. In this situation, your EMI will come down or you may opt to shorten the loan tenure. Ex: If you avail home loan to purchase a house as an investment, then optimistic scenario enables you to compare this with other investment opportunities. Pessimistic (inflationary) scenario: In the same way, assume that the rate of interest is hiked by 1% – 3%. Is it possible for you to continue to pay the EMI without much struggle? Even a 2% increase in rate of interest can result in significant rise in your monthly payment for the entire loan tenure. Such calculation helps you to plan for such future possibilities. When you take a loan, you are making a financial commitment for next few months, years or decades. 1.6 PROFIT AND LOSS In our day to day life, we come across a number of situations wherein we use the concept of percent. Here, in this section we discuss the application of percentage in problems of profit and loss, & discount. Profit and loss is the branch of basic mathematics which deals with the study of profit and loss made in a business transaction. The profit and loss account is fundamentally a summary of the trading transactions of a business and shows whether it has made a profit or loss during a particular period of account. Indeed, by deducting the total expenditure 12 from total income the profit or loss of a business can be calculated. Along with the balance sheet, it is one of the key financial statements that make up a company's statutory accounts. Let us begin with the terms and formulae related to profit and loss. Cost Price (C.P.): The Price at which an article is purchased, is called its cost price. Selling Price (S.P.): The Price at which an article is sold, is called its selling price. Profit (Gain): When S.P. > C.P., then there is profit, and Profit = S.P. – C.P. Loss: When C.P. > S.P., then there is loss, and Loss = C.P. – S.P. Formulae Profit % = (Profit/CP * 100) % Loss% = (Loss/CP *100) % One thing always kept in mind that Gain % or loss % is always calculated on C.P. Example 1: A shopkeeper buys an article for Rs. 360 and sells it for Rs. 270. Find his gain or loss percent. Sol: Here C.P. = Rs. 360, and S.P. = Rs. 270 Since C.P. > S.P., there is a loss. Loss = C.P. – S.P. = Rs (360 – 270) = Rs. 90 Loss % = (loss/CP*100) % = 25% Example 2: Sudha purchased a house for Rs. 4, 52, 000 and spent Rs. 28,000 on its repairs. She had to sell it for Rs. 4, 92, 000. Find her gain or loss percent. Sol: Here C.P. = Cost price + Overhead charges = Rs. (452000 + 28000) = Rs. 4,80,000 S.P. = Rs. 4,92,000 13 Since, S.P. > C.P., Gain = Rs. (492000 – 480000) = Rs. 12000 Gain % = 2.5% Example 3: By selling a book for Rs. 258, a publisher gains 20%. For how much should he sell it to gain 30%? Sol: S.P. = Rs. 258 Profit = 20% C.P. = SP*100/100+ profit% = Rs. 215 Now, if Profit = 30% and C.P. = Rs. 215, then, S.P. = CP (100+profit %) / 100 = 215×130/ 100 = Rs. 279.50 Example 4: A man bought oranges at 25 for Rs. 100 and sold them at 20 for 100. Find his gain or loss percent. Sol: C.P. of 25 oranges = Rs. 100 C.P. of 1 orange = Rs. 100/ 25 = Rs. 4 and S.P. of 1 orange = Rs. 100/ 20 = Rs. 5 Profit on 1 orange = Rs (5 – 4) = Rs. 1 Profit % = 100 25% Example 5: A man sold two horses for Rs. 29700 each. On one he lost 10% while he gained 10% on the other. Find his total gain or loss percent in the transaction. Sol: S.P. of first horse = Rs. 29700 Loss = 10% C.P. = Rs. 29700×100/90 = Rs. 33,000 S.P. of 2nd horse = Rs. 29700, Profit = 10% C.P. = Rs. 29700×100/110 = Rs. 27,000 14 Total CP = Rs. (33000 + 27000) = Rs. 60,000 Total SP = Rs. (2 × 29700) = Rs. 59400 Net Loss = Rs. (60000 – 59400) = Rs. 600 Loss % = 600/60000*100 = 1% Example 6: The cost price of 15 articles is equal to the selling price of 12 articles. Find the gain percent. Sol: Let the C.P. of 15 articles be Rs. 15 then S.P. of 12 articles = Rs. 15 S.P. of 15 articles = Rs. 15/12*15 = 75/4 75 Gain = (75/4 – 15) = Rs. 15/4 Gain % = 15/4/15*100 = 25% Example 7: A watch was sold at a profit of 12%. Had it been sold for 33 more, the profit would have been 14%. Find the cost price of the watch. Sol: Let the cost price of the watch be ` x Then S.P = x*112/100 = 112x/100. If the watch is sold for Rs. 33 more then S.P. = (112x/100 +33) New profit 14% CP. = x= (112x/100+ 33)× 100/114 or 114x = 112 x + 3300 or 2x = 3300 x = 1650 C.P. = ` 1650 1.7 SIMPLE INTEREST Interest is money paid by a borrower to a lender for a credit or a similar liability. Important examples are bond yields, interest paid for bank loans, and returns on savings. Interest differs from profit in that it is paid to a lender, whereas profit is paid to an owner. In economics, the various forms of credit are also referred to as loanable funds. When money is borrowed, interest is typically calculated as a percentage of the principal, the amount owed to the lender. The percentage of the principal that is paid over a certain period of time (typically a year) is called 15 the interest rate. Interest rates are market prices which are determined by supply and demand. They are generally positive because loanable funds are scarce. Interest is often compounded, which means that interest is earned on prior interest in addition to the principal. The total amount of debt grows exponentially, and its mathematical study led to the discovery of the number e. In practice, interest is most often calculated on a daily, monthly, or yearly basis, and its impact is influenced greatly by its compounding rate. When a person has to borrow some money as a loan from his friends, relatives, bank etc. he promises to return it after a specified time period along with some extra money for using the money of the lender. The money borrowed is called the Principal, usually denoted by P, and the extra money paid is called the Interest, usually denoted by I. The total money paid back, that is, the sum of Principal and the Interest is called the Amount, and is usually denoted by A. Thus, A = P + I The interest is mostly expressed as a rate percent per year (per annum). Interest depends on, how much money (P) has been borrowed and the duraton of time (T) for which it is used. Interest is calculated according to a mutually agreed rate percent, per annum (R). [i.e. R = r % = r/100] Thus, Interest = (Principal) × (Rate % per annum) × time I=P×R×T Interest calculated as above, is called simple interest. Example 1: Find the simple interest in each of the following cases: (a) P= Rs. 8000 R= 5% T= 2 yrs (b) Rs. 20,000 R= 15% T= 1*1/2 years Sol: (a) I= 8000* 5* 2/100 = Rs. 800 (b) I= 20000* 15* 3/2*100= Rs. 4500 Example 2: Find at what rate of simple interest per annum will 5000 amount to Rs. 6050 in 3 years. 16 Sol: Here A = Rs. 6050, P = Rs. 5000, T = 3 yrs I = Rs. (6050 – 5000) = Rs. 1050 I = P × R × T or r% = I/ P*T R= 1050*100/5000*3 = 7% Example 3: A sum amounts to Rs. 4875 at 12× % simple interest per annum after 4 years. Find the sum. Sol: Here A = Rs. 4875, R = 12-1/2 %, T= 4 years I=P×R×T I = Rs. P* 25/200* 4 = Rs. P/2 A = P + P/2 =Rs. 3P/2 Thus, 3P/2 = 4875 => 3P= 9750=> P= Rs. 3250 Example 4: In how many years will a sum of Rs. 2000 yield an interest (Simple) of Rs. 560 at the rate of 14% per annum? Sol: Here P = Rs. 2000, I = Rs. 560 R = 14% I = P × R × T or 560 = 2000*14/100*T T = 560 *100/ 2000* 14 = 2 years Example 5: A certain sum of money at simple interest amounts to Rs. 1300 in 4 years and to Rs. 1525 in 7 years. Find the sum and rate percent. Sol: Amount after 4 years = Rs. 1300 Amount after 7 years = Rs. 1525 Interest for 3 years = Rs. [1525 – 1300] = Rs. 225 Interest for 1 year = Rs. 225/3 = 75 17  1300 = P + Interest for 4 yrs = P + 4 × 75 or P = Rs. (1300 – 300) = Rs. 1000 R = 75 ×100/ 1000 ×1 = 7.5% Example 6: A certain sum of money doubles itself in 10 years. In how many years will it become 2× 1/2 times at the same rate of simple interest. Sol: Let P = Rs. 100, T = 10 yrs, A = Rs. 200, I = Rs. 100 100 = 100× × 10 or R = 10% Now P = Rs. 100, R = 10% and A = Rs. 250 I = Rs. 150 150 = 100× 10 × = = 15 Thus, in 15 yrs, the sum will become 2 times Example 7: Out of ` 70,000 to invest for one year, a man invests Rs. 30,000 at 4% and Rs. 20,000 at 3% per annum simple interest. At what rate percent, should he lend the remaining money, so that he gets 5% interest on the total amount he has? Sol: Interest on total amount at 5% for one year = Rs. 70, 000× ×1 = Rs. 3500 Interest on Rs. 30,000 at 4% for 1 year = Rs. 30000 × ×1 = Rs. 1200 Interest on ` 20,000 at 3% for 1 year = 20000 × ×1 = Rs. 600 Interest on remaining Rs. 20,000 for 1 yr = Rs. [3500 – 1200 – 600] = Rs. 1700 The remaining amount should be invested at 8.5% per annum. 1.8 COMPOUND INTEREST To define an interest rate fully, allowing comparisons with other interest rates, both the interest rate and the compounding frequency must be disclosed. Since 18 most people prefer to think of rates as a yearly percentage, many governments require financial institutions to disclose the equivalent yearly compounded interest rate on deposits or advances. For instance, the yearly rate for a loan with 1% interest per month is approximately 12.68% per annum (1.01 12 1). This equivalent yearly rate may be referred to as annual percentage rate (APR), annual equivalent rate (AER), effective interest rate, effective annual rate, and other terms. When a fee is charged up front to obtain a loan, APR usually counts that cost as well as the compound interest in converting to the equivalent rate. These government requirements assist consumers in comparing the actual costs of borrowing more easily. For any given interest rate and compounding frequency, an equivalent rate for any different compounding frequency exists. Compound interest may be contrasted with simple interest, where interest is not added to the principal (there is no compounding). Compound interest is standard in finance and economics, and simple interest is used infrequently (although certain financial products may contain elements of simple interest). 1.8.1 Terminology: The effect of compounding depends on the frequency with which interest is compounded and the periodic interest rate which is applied. Therefore, to accurately define the amount to be paid under a legal contract with interest, the frequency of compounding (yearly, half- yearly, quarterly, monthly, daily, etc.) and the interest rate must be specified. Different conventions may be used from country to country, but in finance and economics the following usages are common: The periodic rate is the amount of interest that is charged (and subsequently compounded) for each period divided by the amount of the principal. The periodic rate is used primarily for calculations and is rarely used for comparison. The nominal annual rate or nominal interest rate is defined as the periodic rate multiplied by the number of compounding periods per year. For example, a monthly rate of 1% is equivalent to an annual nominal interest rate of 12%. 19  The effective annual rate is the total accumulated interest that would be payable up to the end of one year divided by the principal.Economists generally prefer to use effective annual rates to simplify comparisons, but in finance and commerce the nominal annual rate may be quoted. When quoted together with the compounding frequency, a loan with a given nominal annual rate is fully specified (the amount of interest for a given loan scenario can be precisely determined), but the nominal rate cannot be directly compared with that of loans that have a different compounding frequency.Loans and financing may have charges other than interest, and the terms above do not attempt to capture these differences. Other terms such as annual percentage rate andannual percentage yield may have specific legal definitions and may or may not be comparable, depending on the jurisdiction. When interest is calculated on the Principal for the entire period of loan, the interest is called simple interest and is given by: I= P*R*T/100 But if this interest is due (not paid) after the decided time period, then it becomes a part of the principal and so is added to the principal for the next time period, and the interest is calculated for the next time period on this new principal. Interest calculated, this way is called compound interest. The time period after which the interest is added to the principal for the next time period is called the Conversion Period. The conversion period may be one year, six months or three months and the interest is said to compounded, annually, semi-annually or quarterly, respectively. In other words, if you walk into a bank and open up a savings account you will earn interest on the money you deposit in the bank. If the interest is calculated once a year then the interest is called “simple interest”. If the interest is calculated more than once per year, then it is called “compound interest”. The mathematical formula for calculating compound interest depends on several factors. These 20 factors include the amount of money deposited called the principal, the annual interest rate (in decimal form), the number of times the money is compounded per year, and the number of years the money is left in the bank. Example1: Find the compound interest on a sum of Rs. 2000, for two years when the interest is compounded annually at 10% per annum. Sol: Here P = Rs. 2000 and R = 10% Interest for the first conversion time period (i.e. first year) = 2000*10*1/100 = Rs. 200 Principal for the second year (or 2nd conversion period) = Rs. (2000 + 200) = Rs. 2200 Interest for the 2nd time period = Rs. 2200*10*1/100 = Rs. 220 Amount payable at the end of two years = Rs. (2200 + 220) = Rs. 2420 Total interest paid at the end of two years = Rs. (2420 – 2000) = Rs. 420 or [Rs. (200 + 220) = Rs. 420] Compound interest = Rs. 420 Thus, for calculating the compound interest, the interest due after every conversion period is added to the principal and then interest is calculated for the next period. Formula for Compound Interest Let a sum P be borrowed for n years at the rate of r% per annum, then Interest for the first year = P*R*1/100 = PR/100 Amount after one year = Principal for 2nd year = P + PR/100 21  P (1+R/100) Interest for 2nd year = P (1+R/100) * R*1/100 = PR/100(1+R/100) Amount after 2 years = P + PR/100 + PR/100(1+R/100) = P(1+R/100)2 and so on. Amount after n years = P (1+r/100)n Thus, if A represents the amount and R represents r% or r/100 A = P(1 + R)n and compound interest = A – P = P (1 + R)n – P = P[(1 + R)n –1] or P(1+r/100) - P Simple interest and compound interest are equal for first year (first conversion period). To solve compound interest problems, we need to take the given information at plug the information into the compound interest formula and solve for the missing variable. The method used to solve the problem will depend on what we are trying to find. If we are solving for the time, t, then we will need to use logarithms because the compound interest formula is an exponential equation and solving exponential equations with different bases requires the use of logarithms. Example 2: Calculate the compound interest on Rs. 20,000 for 3 years at 5% per annum, when the interest is compounded annually. Sol: Here P = Rs. 20,000, R = 5% and n =3 CI = P [(1 + R)n –1] = Rs. 20000[(1+ 5/100) – 1] = Rs. 20000[(21/20) – 1] = Rs. 3152.50 Example 3: Calculate the compound interest on Rs. 20,000 for 1-1/2 years at the rate of 10% per annum, when the interest is compounded semi-annually. 22 Sol: Here P = Rs. 20,000, R = 10% per annum = 5% per half year and n = 3 half years CI = P[(1 + R)n –1] = Rs. 20000[(1+5/100) – 1] = 20000 [(9261/8000 – 1)] = Rs. 3125.50 Example 4: Calculate the compound interest on Rs. 20,000 for 9 months at the rate of 4% per annum, when the interest is compounded quarterly. Sol: Here P = Rs. 20,000, R = 4% per annum = 1% per quarter of year and n = 3/4 yrs = 3 quarters CI = P[(1 + R)n –1] = 20000[(1+1/100) – 1] = 20000 * 30301/ 100*100*100 = = Rs. 606.02 Example 5: calculate the amount and compound interest on Rs. 12000 for 1- 1/2years at the rate of 10% per annum compounded annually. Sol: Here P = Rs. 12000, R = 10% and n = 1-1/2 years Since interest is compounded, annually, so, amount at the end of 1 year is given by: A= P(1+ r/100)n A= 12000(1+ 10/100)1 = Rs. 13200 Principal for next 6 months = Rs. 13200. and Rate R = 10%/2 = 5% A = 13200( 1+ 5/100)1 =13200*21/20 = ` 13860 Amount after 1-1/2 years = Rs. 13860 And Compound interest = Rs. [13860 – 12000] = Rs. 1860 23 Example 6: At what rate per cent per annum will Ron lends a sum of Rs. 2000 to Ben. Ben returned after 2 years Rs. 2205, compounded annually? Sol : Let the required rate be R% per annum. Here, A = Rs. 2205, P = Rs. 2000 and n = 2 years. Using the formula A = P(1 + R/100)n, 2205 = 2000 × ( 1 + R/100)2 (1 + R/100)2 = 2205/2000 = 441/400 = (21/20)2 ( 1 + R/100) = 21/20 R/100 = (21/20 – 1) = 1/20 R = (100 × 1/20) = 5 Hence, the required rate of interest is 5% per annum. Example 7: A man deposited Rs. 1000 in a bank. In return he got Rs. 1331. Bank gave interest 10% per annum. How long did he kept the money in the bank? Sol: Let the required time be n years. Then, amount = Rs. {1000 × (1 + 10/100)n} = Rs. {1000 × (11/10)n} Therefore, 1000 × (11/10)n = 1331 [since, amount = Rs. 1331 (given)] (11/10)n = 1331/1000 = 11 × 11 × 11/ 10 × 10 × 10 = (11/10)3 24  (11/10)n = (11/10)3 n = 3. Thus, n = 3. Hence, the required time is 3 years. 1.9 BILL OF DISCOUNTING-BUSINESS APPLICATIONS Bill discounting, as a fund-based activity, emerged as a profitable business in the early nineties for finance companies and represented a diversification in their activities in tune with the emerging financial scene in India. In the post-1992 (scam) period its importance has substantially declined primarily due to restrictions imposed by the Reserve Bank of India. 1.9.1 Concept According to the Indian Negotiable Instruments Act, 1881: “The bill of exchange is an instrument in writing containing an unconditional order, signed by the maker, directing a certain person to pay a certain sum of money only to, or to the order of, a certain person, or to the bearer of that instrument.” The bill of exchange (B/E) is used for financing a transaction in goods which means that it is essentially a trade-related instrument. 1.9.2 Definition and Meaning When a buyer buys goods from the seller, the payment is usually made through letter of credit. The credit period may vary from 30 days to 120 days. Depending upon the credit worthiness of the buyer, the bank discounts the amount that needs to be paid at the end of credit period. It means that the bank will charge the interest amount for the credit period as an advance from the buyer’s account. After that, the bill amount is paid as per the end of the time span with respect to the agreed upon document between the buyer and seller. In other words, If the drawer of the bill does not want to wait till the due date of the bill and is in need of money, he may sell his bill to a bank at a certain rate of discount. The bill will be endorsed by the drawer with a signed and dated order to 25 pay the bank. The bank will become the holder and the owner of the bill. After getting the bill, the bank will pay cash to the drawer equal to the face value less interest or discount at an agreed rate for the number of days it has to run. This process is known as discounting of a bill of exchange. In capitalist countries, the purchase of bills of exchange by banks before maturity. The bank pays less thannominal value of the bill, deducting a certain percentage for interest. At maturity, the bank collects the full nominal value from the drawee. The discounting of bills expands commercial credit and speeds up the circulation of capital. At the same time, the discounting of commercialpaper and especially of finance bills encourages speculation, increases the anarchy of capitalism, and has tens the onset of economic crises.In the age of the general crisis of capitalism, the proportion of bank assets represented by longterm loans and investments in secur ities ofvarious types has increased, while that represented by discounted bills has decreased. At the same time, the structure of discounting haschanged, with the discounting of treasury bills becoming increasingly important. The proceeds from the sale of these bills are used bygovernments for such purposes as financing military expenditures and covering budget deficits. 1.9.2 Types of Bills There are various types of bills. They can be classified on the basis of when they are due for payment, whether the documents of title of goods accompany such bills or not, the type of activity they finance, etc. Some of these bills are: a. Demand Bill -This is payable immediately “at sight” or “on presentment” to the drawee. A bill on which no time of payment or “due date” is specified is also termed as a demand bill. b. Usance Bill -This is also called time bill. The term usance refers to the time period recognized by custom or usage for payment of bills. c. Documentary Bills - These are the B/Es that are accompanied by documents that confirm that a trade has taken place between the buyer and the seller of goods. These documents include the invoices and other documents of title such as railway receipts, lorry receipts and bills of 26 lading issued by custom officials. Documentary bills can be further classified as: (i) Documents against acceptance (D/A) bills and (ii) Documents against payment (DIP) bills. D/ A Bills In this case, the documentary evidence accompanying the bill of exchange is deliverable against acceptance by the drawee. This means the documentary bill becomes a clean bill after delivery of the documents. Clean Bills are not accompanied by any documents that show that a trade has taken place between the buyer and the seller. Because of this, the interest rate charged on such bills is higher than the rate charged on documentary bills. Creation of a B/E Suppose a seller sells goods or merchandise to a buyer. In most cases, the seller would like to be paid immediately but the buyer would like to pay only after some time, that is, the buyer would wish to purchase on credit. To solve this problem, the seller draws a B/E of a given maturity on the buyer. The seller has now assumed the role of a creditor; and is called the drawer of the bill. The buyer, who is the debtor, is called the drawee. The seller then sends the bill to the buyer who acknowledges his responsibility for the payment of the amount on the terms mentioned on the bill by writing his acceptance on the bill. The acceptor could be the buyer himself or any third party willing to take on the credit risk of the buyer. Discounting of a B/E The seller, who is the holder of an accepted B/E has two options: 1. Hold on to the B/E till maturity and then take the payment from the buyer. 2. Discount the B/E with a discounting agency. Option (2) is by far more attractive to the seller. The seller can take over the accepted B/E to a discounting agency bank, NBFC, company, high net worth individual] and obtain ready cash. The act of handing over an endorsed B/E for ready money is called discounting the B/E. The margin between the ready money paid and the face value of the bill is called the discount and is calculated at a rate percentage per annum on the maturity value. The maturity a B/E is defined as the date on which payment will fall due. Normal maturity periods are 30, 60, 90 or 120 days but bills maturing within 90 days seem to be the most popular. 27 1.9.3 Advantages: The advantages of bill discounting to investors and banks and finance companies are as follows: To Investors 1. Short-term sources of finance; 2. Bills discounting being in the nature of a transaction is outside the purview of Section 370 of the Indian Companies Act 1956, that restricts the amount of loans that can be given by group companies; 3. Since it is not a lending, no tax at source is deducted while making the payment charges which is very convenient, not only from cash flow point of view, but also from the point of view of companies that do not envisage tax liabilities; 4. Rates of discount are better than those available on ICDs; 5. Flexibility, not only in the quantum of investments but also in the duration of investments. To Banks 1. Safety of Funds- The greatest security for a banker is that a B/E is a negotiable instrument bearing signatures of two parties considered good for the amount of bill; so he can enforce his claim easily. 2. Certainty of Payment -A B/E is a self-liquidating asset with the banker knowing in advance the date of its maturity. Thus, bill finance obviates the need for maintaining large, unutilised, ideal cash balances as under the cash credit system. It also provides banks greater control over their drawls. 3. Profitability- Since the discount on a bill is front ended, the yield is much higher than in other loans and advances, where interest is paid quarterly or half yearly. 4. Evens out Inter-Bank Liquidity Problems -The development of healthy parallel bill discounting market would have stabilized the violent fluctuations in the call money market as banks could buy and sell bills to even out their liquidity mismatches. 28 Discount Rate and Effective Rate of Interest Banks and finance companies discounting bills prefer to discount LlC (letter of credit)-backed bills compared to clean bills. The rate of discount applicable to clean bills is usually higher than the rate applicable to LlC-based bills. The bills are generally discounted up-front, that is, the discount is payable in advance. As a consequence, the effective rate of interest is higher than the quoted rate (discount). The discount rate varies from time to time depending upon the short-term interest rate. The computation of the effective rate of interest on bills discounting is shown in following illustration Example: The Hypothetical Finance ltd. discounts the bills of its clients at the rate specified below: L/C - backed bills, 22 per cent per annum Clean bill, 24 per cent per annum Required: Compute the effective rate of interest implicit in the two types of bills assuming usance period of (a) 90 days for the L/C - based bill and (b) 60 days for the clean bill and value of the bill, Rs 10,000. Sol. Effective Rate of Interest on L/C - based Bill: Value of the bill, Rs 10,000 Discount charge Rs. 550 (Rs 10,000x 0.22 x 90/360) Amount received by the client, = Rs 9,450 (Rs 10,000 - Rs 550) Quarterly effective interest rate = 5.82% [Rs 90 x 100/Rs. 9450] Annualised effective rate of interest, [(1.0582)4- 1] x 100 = 25.39% Effective Rate of Interest on Clean Bill: Value of bill Rs 10,000 Discount Charge, = Rs 400 (Rs 10,000 x 0.24 x 60/360) Amount received by the client = Rs 9,600 (Rs 10,000 - Rs 400) Quarterly rate of interest = 4.17% (Rs. 400/Rs 9,600) x100 Effective rate of interest per annum, = 17.75%. 1.9.4 Present Position of Bills Discounting 29 Financial services companies had been acting till the early nineties as bill-brokers for sellers and buyers of bills arising out of business transactions. They were acting as link between banks and business firms. At times they used to take up bills on their own account, using own funds or taking short-term accommodation from banks working as acceptance/discount houses. They had been handling business approximating Rs 5,000 crores annually. Bill discounting, as a fund- based service, made available funds at rates I per cent lower than on ash credit finance and bill finance constituted about one-fourth of bank finance. However, the bill re-discounting facility was misused by banks as well as the bill-brokers. The Jankiraman Committee appointed by the RBI which examined the factors responsible for the securities-scam identified the following misuse of the scheme: Banks have been providing bill finance outside the consortium without informing the consortium bankers; 1.9.5 Bills of discounting is of two types 1. Purchase bills discounting and 2. Sales bill discounting. A purchase bill discounting means that the investor discounts the purchase bill of the company and pays the company, who in turn pay their supplier. The investor gets his money back from the company at the end of the discounting period. A sales bills discounting means the investor discounts the sales bill of the company and pays directly to the company. The investor gets his return from the company at the end of the discounting period. Funds -The funds generally required for this type of transaction start from Rs300, 000 to upto Rs2 mn. The tenure, generally, ranges from 60 days to upto 180 days. Procedure- The procedure is that a broker will contact you with proposals to discount bills of different companies at different rates of discounting. The better companies command discounting rates of 13% to 15%, while the lesser known, by size and by safety, have to pay discounting rates of 17% to as high as 28%. It is later explained what factors determine the discount rates. When an investor and the company agree to a particular bill discounting transaction, the following is what the company gives to the investor: The original copies of bills to be 30 discounted; A hundi / promissory note; Post dated cheque. The investor simply has to issue a cheque. The amount of cheque is arrived at after deducting the discount rate. The post dated cheque that the company gives is of the full amount of the transaction. This can be better explained with an example as follows. Company A wants to discount its purchase bill of Rs200, 000 for a period of three months. Investor P agrees to do so at a discount rate of 21%. The deal is mutually agreed. Now, the investor will issue a cheque of Rs189, 500. This figure is arrived at as follows: = 200,000 x 21% x 3/12 Thus the investor gets his/her interest before the end of the period on discounting. The company on its part will issue a post-dated cheque of Rs200, 000 for three months period. Here we see that the investor benefits in two ways: He gets the interest element at the first day of issuing the cheque. i.e. he does not include that part in his cheque amount. Thus he can earn interest on this interest for a three-month period. Even a simple bank fixed deposit on it will earn @5% p.a. By investing Rs189, 500 for three months, the investor earns Rs10, 500 on it. A return of 22. 16%. Discount Rates- The rates depend on the following factors: The Broker: The broker has a good influence on the rates offered by companies. His relations with the company and the investor do make a difference of a couple of percentage point in discounting rates. Market Liquidity: Liquidity crunch in the market tends to hike up the rates even in the best of the companies. Since this instrument is a short tenure one, short-term changes in the market liquidity greatly affect the discount rates. Volume/Value of Discounting: When the volume/value of discounting done by the investor is high, he is looking at security more than returns. The company on its part is looking at savings by way of reduced legal paper work and a higher amount of dedicated funds for a said period and hence on the whole reduced costs to the company. Frequency: An investor who is regular bills discounter for the company may get upto 1% to 1.5% points higher interest rates than a new investor. As for the investor he is trying it out with a new company and will agree to a lesser rate to ensure safety. Company’s finance resources: This is one of the biggest factors that decide the discount rates. A 31 Public limited company generally tends to have a cheaper source of finance as against any other form of company. Working capital financing of companies to a large extent manipulates the rates the companies are willing to discount their bills at. Caveat- The following points need to be remembered when dealing in this instrument. One must have a thorough knowledge of the company whose bills are discounted. Their industry, competition, people at the helm and their reputation in the market. This is necessary as it is going to be the company that is going to pay you from its earnings. There is no legal fall back option in case of default by the company. The company does sign a promissory note, but legal respite using this will take years to happen. The investor is not a secured creditor for the company nor does he get any preference on winding up of the company. Brokers need to be people who are well known to you. Since most of the deals happen through them, you should know the broker well enough to trust him and his deals. Spurious brokers are plenty out there in the market and a watchful eye must be kept. Even after investing in the company a regular watch on the company’s fortune and a constant touch with the broker would be warranted. 1.10 SUMMARY Business mathematics is mathematics used by commercial enterprises to record and manage business operations. Commercial organisations use mathematics in accounting, inventory management, marketing, sales forecasting, and financial analysis. Mathematics typically used in commerce include elementary arithmetic, elementary algebra, statistics and probability. Business management can be made more effective in some cases by use of more advanced mathematics such as calculus, matrix algebra and linear programming. Business Mathematics is very important for modern business management. The forecasting and operating procedures are based primarily on business mathematics. Things such as simple interest and compound interest show a company what it will lose or get over the years if it invests in a particular 32 asset. Business mathematics also helps in cost and price calculations which are the basis of cash inflows and outflows that all companies have to deal with. Equated Monthly Instalment (EMI) refers to the monthly payment a borrower makes on his loan. Though it is a combination of interest payment and principal repayment, the total monthly amount is calculated in such a way that it remains constant all through the repayment tenure. The below mentioned example is given to explain the methodology. Consider that a loan of Rs.1,00, 000 is to be repaid over 25 years in equal monthly installments. If the annual interest rate is 7%, the monthly EMI is calculated as follows: (Since the repayments are monthly). So, The EMI will be fixed as Rs.710 in this case by rounding off to the nearest 10 rupees. To determine the specific percent of a number or quantity, we change the percent to a fraction or a decimal and then multiply. When the selling price is more than the cost price of the goods, there is a profit (or gain). When the selling price is less than the cost price of the goods, there is a loss. The simple interest (I.) on a principal (P) at the rate of R% for a time T years, is calculated, using the formula: I= P × R × T Discount is a reduction in the list price of goods. Discount is always calculated on the marked price of the goods. (Marked price – discount), gives the price, which a customer has to pay while buying an article. A discount series can be reduced to a single discount. Sales tax is charged on the sale price of goods. An instalment plan enables a person to buy costlier goods. In the case of compound interest. Amount (A) = P (1 + R)n, where P is the Principal, R = rate% and n = time. Compound interest is greater than simple interest, except for the first conversion period. The interest on loans and mortgages that are amortized-that is, have a smooth monthly payment until the loan has been paid off-is often compounded monthly. 1.11 SELF ASSESSMENT EXERCISES 1. Bobby Cash deposited Rs 10,000 at 8% compounded quarterly. Two years after she makes the first deposit, he adds another Rs. 20,000, 33  also at 8% rate compounded quarterly. What total amount will he have 4 years after his first deposit? 2. For one year, a student loan of Rs. 52,000 at 9% compounded semi- annually resulted in a maturity value of Rs. 5,934.06. 3. Billy Jean King deposited Rs. 6500 in an account paying 7.5 % compounded quarterly. After 3 years the rate drops to 4% compounded semi-annually. Find the amount in her account at the end of 7 years. 4. John and Jill have Rs. 20,000 cash for the down payment of a house and they can afford a 15-year mortgage payment of Rs. 2,500/month. If the best mortgage rate that they can get is 7.5% then what will be the most affordable home that they can buy by their current budget plan? 5. If money can be borrowed at 8 % compound monthly, which one is larger: Rs. 10,000 now or Rs. 15,000 in 5 years? Use present value to decide. 6. Mention the advantages of quantitative approach to management. 7. What factors in modern society contribute to the increasing importance of quantitative approach to management? 8. A retailer buys a cooler for Rs. 3800 but had to spend Rs. 200 on its transport and repair. If he sells the cooler for Rs. 4400, determine, his profit percent. 9. A vendor buys lemons at the rate of 5 for Rs. 7 and sells them at Rs. 1.75 per lemon. Find his gain percent. 34 10. A man purchased a certain number of oranges at the rate of 2 for Rs. 5 and sold them at the rate of 3 for Rs. 8. In the process, he gained Rs. 20. Find the number of oranges he bought. 11. By selling a bi-cycle for Rs. 2024, the shopkeeper loses 12%. If he wishes to make a gain of 12% what should be the selling price of the bi-cycle? 12. By selling 45 oranges for Rs. 160, a woman loses 20%. How many oranges should she sell for Rs. 112 to gain 20% on the whole? 13. A dealer sold two machines at Rs. 2400 each. On selling one machine, he gained 20% and on selling the other, he lost 20%. Find the dealer’s net gain or loss percent. 14. Harish bought a table for Rs. 960 and sold it to Raman at a profit of 5%. Raman sold it to Mukul at a profit of 10%. Find the money paid by Mukul for the table. 15. A man buys bananas at 6 for Rs. 5 and an equal number at Rs. 15 per dozen. He mixes the two lots and sells them at Rs. 14 per dozen. Find his gain or loss percent, in the transaction. 16. If the selling price of 20 articles is equal to the cost price of 23 articles, find the loss or gain percent. 17. Rama borrowed Rs. 14000 from her friend at 8% per annum simple interest. She returned the money after 2 years. How much did she pay back altogether? 35 18. Ramesh deposited Rs. 15600 in a financial company, which pays simple interest at 8% per annum. Find the interest he will receive at the end of 3 years. 19. Naveen lent Rs. 25000 to his two friends. He gave Rs. 10,000 at 10% per annum to one of his friend and the remaining to other at 12% per annum. How much interest did he receive after 2 years. 20. Shalini deposited Rs. 29000 in a finance company for 3 years and received Rs. 38570 in all. What was the rate of simple interest per annum? 21. At what rate of interest will simple interest be half the principal in 5 Years. 22. A sum of money amounts to Rs. 1265 in 3 years and to Rs. 1430 in 6 years, at simple interest. Find the sum and the rate percent. 23. Out of Rs. 75000 to invest for one year, a man invested Rs. 30000 at 5% per annum and Rs. 24000 at 4% per annum. At what percent per annum, should he invest the remaining money to get 6% interest on the whole money? 24. A certain sum of money doubles itself in 8 years. In how much time will it become 4 times of itself at the same rate of interest? 25. Calculate the compound interest on Rs. 16000 for 9 months at 20% per annum, compounded quarterly. 26. Find the sum of money which will amount to Rs.27783 in 3 years at 5% per annum, the interest being compounded annually. 36  27. Find the difference between simple interest and compound interest for 3 years at 10% per annum, when the interest is compounded annually on Rs. 30,000. 28. A sum of money is invested at compound interest for 9 months at 20% per annum, when the interest is compounded half yearly. If the interest were compounded quarterly, it would have fetched Rs. 210 more than in the previous case. Find the sum. 29. A sum of Rs. 15625 amounts to Rs. 17576 at 8% per annum, compounded semi-annually. Find the time. 30. Find the rate at which Rs. 4000 will give Rs. 630.50 as compound interest in 9 months, interest being compounded quarterly. 31. A sum of money becomes Rs. 17640 in two years and Rs. 18522 in 3 years at the same rate of interest, compounded annually. Find the sum and the rate of interest per annum. 1.12 SUGGESTED READINGS Statistical Methods. Gupta S.P. Business Mathematics and Statistics. P.R. Vital Business mathematics. Kavita Choudhary. 37 Business Mathematics-104 Unit-II Set Theory Lessons 6-10 STRUCTURE 2.1 Introduction 2.2 Objectives 2.3 Concept of Sets 2.3.1 How to Describe or Specify a Set 2.3.2 Types of Set 2.3.3 Some Results on Subsets 2.4 Operation of Sets 2.5 Algebra of Sets 2.6 Cartesian Product of Two Sets 2.7 Application of Sets to Business Mathematics 2.8 Summary 2.9 Self assessment exercises 2.10 Suggested Readings 2.1 INTRODUCTION The set concept, fundamental in modern mathematics, was introduced by Georg Cantor in the nineteenth century. Cantor’s concept profoundly influenced mathematical thinking, but most of his fellow mathematicians initially rejected his ideas. Everything about the concept was new and shocking. It contradicted the accepted way of thinking, especially with regard to infinity. However, after the mathematics community became familiar with the Cantorian concept, flaws were discovered in his set theory. Probably the best-known 38 description of those flaws is found in the Russell paradox. Bertrand Russell made a successful attempt to solve the flaws within the Cantorian set theory by elaborating a new approach. (With regard to the Russell set theory and the Russell paradox, see Wilder, 1965.) Nowadays the concept of set is an integral part of mathematics, and is one of its central, seminal components. In our mathematical language, everything in this universe, whether living or non-living, is called an object. If we consider a collection of objects given in such a way that it is possible to tell beyond doubt whether a given object is in the collection under consideration or not, then such a collection of objects is called a well defined collection of objects. The term ‘set’ is introduced in mathematics without an initial definition. It is rather described as a collection. Nevertheless, a number of properties are established especially in order to distinguish the mathematical concept of set from the term ‘collection’ from which it is derived. 2.2 OBJECTIVES This unit explains basic concepts in Set Theory, describing sets, elements, Venn diagrams and the union and intersection of sets. The specific objectives are: To know about the concept of sets. To understand the application of sets in business mathematics. 2.3 CONCEPT OF SETS In school mathematics, the concept of set is used in various contexts but, generally, in an inconsistent manner. A major difficulty for students is the fact that the set concept is accepted in mathematics as a primitive, undefined concept (like, point, straight line, etc.) in contrast to the formal definition one starts with in an intuitive model, the idea of a collection of objects. But the mathematical set concept has a number of formal properties and aspects which contradict the initial collection model. SET: A well defined collection o f objects is called a set. The objects in a set are called its members or elements. 39 We usually denote sets by capital letters and their elements by small letters. If is an element o f a set A, we write x A, which means that, 'x belongs to A' or that 'x is an element o f A'. If x does not belong to A, we write, x A. Ex.1. The collection o f vowels in English alphabet is a set containing five elements, namely a, e, i, o, u. Ex.2. The collection o f first four prime numbers is a set containing the elements 2, 3, 5, 7. Ex.3. The collection o f all beautiful girls o f India, is not a set, since the term ‘beautiful’ is vague and it is not well defined. Similarly, ‘rich persons’; ‘honest persons’; ‘good players’ etc. do not form sets. Ex.4. We denote the sets o f all natural numbers, all integers; all rational numbers and all real numbers by N , I , Q and R respectively. 2.3.1 How to Describe or Specify a Set There are two methods for describing a set. (i) Tabulation method or Roster form (ii) Rule method or Set builder form. I. ROSTER FORM: Under this method; we make a list of the elements of the set and put it within braces {}. Ex.1. I f A is the set o f vowels in English alphabet, then A = [ a , e, i, o , u }. Ex.2. I f B is the set o f prime numbers less than 15, then B= { 2 , 3, 5, 7, 11,13}. Ex.3. I f C is the set o f all even numbers lying between 10 and 500, then C = {12, 14, 16, 18,..., 494, 496, 498 }. II. SET-BUILDER FORM: Under this method, we list the property or properties satisfied by the elements of the set. We write, ( x : x satisfies properties P}. This means, 40 ‘the set of all those x such that each x satisfies the properties P’. Ex.1. I f A = {1, 2, 3, 4, 5 ), then we can write, A = { x e N : x < 6). Ex.2. I f B = { 1 , 2 , 3, 4, 6, 8, 12, 24 }, then we can write, B = [ x : x is a factor o f 2 4 }. 2.3.2 Types of Set 1. EMPTY SET: A set consisting o f no element at all is called an empty set or a null set or a void set and it is denoted by (J). In Roster form, is denoted by {}. Ex.1. { x : x N , 2 < x < 3 } =. Ex.2. { x: x R, x2= -1} =. A set which has at least one element is called a non-empty set. 2. SINGLETON SET: A set consisting o f a single element is called a singleton set. e.g. [ 0 } is a singleton set, whose only member is 0 3. FINITE AND INFINITE SET: A set in which the process o f counting o f elements surely comes to an end, is called a finite set, otherwise it is called an infinite set. Ex.1. Each one o f the sets given below is a finite set: (i) set of all persons on the earth ; (ii) { x : x N , x < 5 crores}; (iii) {x : x I x is a factor of 1000). Ex.2. Each one o f the sets given below is an infinite set: (i) set of all points on an arc of a circle; (ii) set of all concentric circles with a given centre ; (iii) [ X I : x < l } ; ( iv ) { x Q : 0 < x < l }. 41 Cardinal number of a finite set: The number o f distinct elements contained in a finite set A is called its cardinal number, to be denoted by n (A). Ex. If A = (2, 3, 5, 7, 11}, then n (A) = 5. 4. EQUAL SETS: Two sets A and B are said to be equal, written as A = B , i f every element o f A is in B and every element o f B is in A. Remarks, (i) The elements of a set may be listed in any order. Thus, (1, 2, 3} = { 2, 1, 3 } = { 3, 2, 1 } etc. (ii) The repetition of elements in a set is meaningless. Ex.1. { x : x is a letter in the word, ‘follow’} = {f, o, I , w }. Ex.2. Show that , (0) and 0 are all different. Sol. Since is a set containing no element at all; {0} is a set containing one element, namely 0; And, 0 is a number, not a set. Hence, {0} 0. 5. EQUIVALENT SETS: Two finite sets A and B are said to be equivalent, i f n (A) = n (B). Clearly, equal sets are equivalent but equivalent sets need not be equal. Ex. A = {1, 3, 5 } and B = { 2, 4, 6 } are equivalent but not equal 6. SUB SETS: I f A and B are two sets such that every element of A is in B, we say that A is a subset o f B and write, A B. C If A B and A * B , then A is called a proper subset of B, written as A B. Remarks, (i) If there exists even a single element in A, which is not in B, then A is not a subset of B and we write, A B. (ii) has no proper subset. Ex.1. { 3 } C { 2 , 3, 5 ). Type equation here. Ex.2. { 1 , 2 ) ( 2 , 3 , 5 }. 42  Ex.3. N W I Q B. But, W N a n d { 0 } N. 7. UNIVERSAL SET: If there are some sets under consideration, then there happens to be a set which is a superset o f each one o f the given sets. Such a set is known as the universal set, to be denoted by U. Ex.1. I f A = { 1, 2, 3, 4 ) , B = { 2, 3, 5, 7 } and C = { 2, 4, 6, 8 }, then U = {1, 2, 3, 4, 5, 6, 7, 8 } is the universal set. Ex.2. When we discuss sets o f lines, triangles or circles in two dimen- sional geometry, the plane in which these lines, triangles or circles lie, is the universal set. 8. POWER SET: The collection o f all possible subsets o f a given set A is called the power set o f A, to be denoted by P (A). Ex.1. I f A = { 1 , 2 , 3 } , then P(A) = { , {1}, { 2 }, { 3 }, {1, 2 }, {l , 3 }, { 2, 3 }, { 1, 2, 3 } }. Ex.2. I f A = {1, { 2 } }, we may write, A = {1, B } , where B = { 2 }..-.P(A) = { , { 1 } , { B } , { 1 , B } } = {«>,{1},{{2}},{1,{2}}}. Ex.3. If A = {1, {2, 3 }}, we may write, A = {1, B }, where B = { 2, 3 }..-. P (A) = { , { 1 }, {B }, { 1, B } } = { , , { 1 }, { { 2, 3 } }, {1, { 2 , 3 } } }. Ex.4. Write down the power sets o f : (i) (ii) { , { } }. Sol. ( i) P ( ) = { }. (iii) L e t A = { , { } } = { , B } , where B={ }. P (A) = { , { } , { B } , { , B} } or or P(A) = {. { } , { B } , { , ( ) } }. Ex.5. Show that n {P [ P (P ( ) ) ] } = 4. Sol. P ( ) ={ } P (P ( ) ) = { , { } } =P [ { P ( ))] ={ , ( }}} n { P [ P ( P ( ) ) ] } = 4. 43 2.3.3 SOME RESULTS ON SUBSETS Theorem 1. Every set is a subset o f itself. Proof. Let A be any set. Then, each element of A is clearly, there in A Hence, A A. Theorem 2. The empty set is a subset o f every set. Proof. Let A be any set. In order to show that A, we must show that there is no element of $ which is not contained in A. And, since contains no element at all, no such element can be found out. Hence, A. Theorem 3. The total number o f all possible subsets o f a given set containing n elements is 2 n. Proof. Let A be any set containing n elements. Then, one of its subsets is the empty set. Apart from this, the number of singleton subsets of A = n = nC1; the number of subsets of A, each containing 2 elements = nC2 ; the number of subsets of A, each containing 3 elements = nC3 ; the number of subsets of A, each containing n elements = nCn. Total number of all possible subsets of A = { l + nC1 + nC2 + nC3 +...+ nCn) = (1 + 1)n = 2n. [Using binomial theorem] 2.4 OPERATION OF SETS In ordinary arithmetic and algebra there are four common operations that can be performed; namely, addition, subtraction, multiplication and division. With sets, however, just two operations are defined. These are set union and set intersection. Both of these operations are described, with examples, in the following sections. The Union of two sets A and B is written as AUB and defined as that set which contains all the elements lying within either A or B or both. For example, if A = (c,d,f,h,j) and B = (d,m,c,f,n,p), then the union of A and B is AUB = (c,d,f,h,j,m,n,p), these being the elements that lie in either A 44  or B. So that any element of A must be an element of AUB; similarly any element of B must also be an element of AUB. Set union for three or more sets is defined in an obvious way. 2.4.1 UNION OF SETS: The union o f two sets A and B, denoted by AUB is the set of all those elements, each one o f which is either in A or in B or in both A and B, Thus, AU B = { x : x A or x B }. Clearly, x A U B => x A or x B. “And, x € A U B = > x € A a nd x € B. Ex.1. I f A = {1, 2, 3, 4} and B = {2, 4, 6, 8 }, then AUB= {1,2,3,4,6,8}. Ex.2. I f A = {x: x is a positive integer} and B = {x: x is a negative integer} then, A U B = [ x : x is an integer, x 0}. 2.4.2 INTERSECTION OF SETS: The intersection of two sets A and B, denoted by A B is the set of all elements, common to both A and B. Thus, A B = {x: x A and x B}. Clearly, x A B = x A and X B. And, x € A B = x € A or x € B. Ex.1. Let A = {1, 2, 3, 4}, and B = { 2, 4, 6 }. Then, A B = {2, 4}. Ex.2. Let A = {x : x = 3n, n N} and B = {x : x = 4n, n N). Then, A B= {x : x = 12n, n N}. Disjoint Set: Two sets A and B are said to be disjoint, i f A B=. If A B = then A and B are said to be intersecting sets or overlaping sets. Ex. if A = {1, 3, 5, 7, 9}, B = { 2 , 4 , 6, 8 } and C = { 2, 3, 5, 7, 11 }, then A and B are disjoint sets, while A and C are intersecting. 2.4.3 DIFFERENCE OF SETS: I f A and B are two sets, then their difference A-B is defined by; 45  A- B = [ x : x A and x B). Thus, x A- B => x A a nd x B. Also, the symmetric difference A B is defined by. A B = (A-B) ( B - A). Ex. Let A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7, 11}. T he n , A- B = { 1 , 9 } ; B - A = {2,11}. And, A B = ( A - B) ( B - A) = { l, 2 , 9 , l l}. 2.4.4 COMPLEMENT OF A SET: Let U be the universal set and let A U. Then, the complement o f A, denoted by A' or U - A is defined as A'= ( x U : x A}. Clearly, i. e A ' x A. Ex.1. I f U = {1, 2, 3, 4, 5 } and A = { 2, 4 }, then A' = {1, 3, 5 }. Ex.2. Let U = (x : x is a letter in English, alphabet) and A = x : x is a vowel} , then A' = { x : x i s a consonant}. SOME RESULTS ON COMPLEMENTATION: (i) U '= { x U: x U} = (ii) = {x U : x } = U (iii) (A')' = [ x U : x A'} = [ x U : x A} = A ; (iv) A U A ' = { x U: x A}U{x U: x A) = U; (v) A A '= { x U:x A} {x U: x A} = 2.5 ALGEBRA OF SETS Intuitively, a set is a “collection” of objects known as “elements.” But in the early 1900’s, a radical transformation occurred in mathematicians’ understanding of sets when the British philosopher Bertrand Russell identified a fundamental paradox inherent in this intuitive notion of a set. Consequently, 46  in a formal set theory course, a set is defined as a mathematical object satisfying certain axioms. These axioms detail properties of sets and are used to develop an elegant and sophisticated theory of sets. Definition: denotes the empty set { }, which does not contain any elements. N denotes the set of natural numbers {1, 2, 3,... }. Z denotes the set of integers {.. , 3, 2, 1, 0, 1, 2, 3,... }. Q denotes the set of rational numbers {p/q : p, q 2 Z with q 6= 0 }. R denotes the set of real numbers consisting of directed distances from adesignated point zero on the continuum of the real line C denotes the set of complex numbers { a + bi : a, b 2 R with i = p 1 }. In this definition, various names are used for the same collection of numbers. For example, the natural numbers are referred to by the mathematical symbol “N,” the English words “the natural numbers,” and the set-theoretic notation “{1, 2, 3,...}.” Mathematicians move freely among these different ways of referring to the same number system as the situation warrants. In addition, the mathematical symbols for these sets are “decorated” with the superscripts “¤,” “+,” and “ ” to designate the corresponding sub-collections of nonzero, positive, and negative numbers, respectively. For example, applying this symbolism to the integers Z = {... , 3, 2, 1, 0, 1, 2, 3,...}, we have Z = {... , 3, 2, 1, 1, 2, 3,...}, Z+ = {1, 2, 3,...}, Z = { 1, 2, 3,...}. I. Idempotent Laws: (i) A U A= A (ii) A A = A. Proof. (i) AU A = { x: x A or x A} = {x : x A} = A. (ii) A A = x : x e A and x A} = [ x : x A } =A. II. Identity Laws: (i) A U = A (ii)A U =A. 47  Proof (i) A U = { x: x A or x } == {x: x A} = A ['.' has no element] (ii) A U= {x: x A and x U ] = { x : x A } = A. III. Commutative; Laws: (i) A U B = B U A ( i i ) A B = B A. Proof (i) A B = { x : x A or x B } (ii) A B = { x : x A and x B }. = { x : x B and x A} = B A. IV. Associative Laws: ( i) ( A U B ) U C = A U( B U C ) (ii) ( A B ) C = A ( B C) Proof. Let x be an arbitrary element of (A U B) U C Then, x (A U B) U C => x (AU B) or x C. => (x A or x B) or x e C => x A or (x B or x C) = > x A or x (B C) = > x A ( B C). (A B) C A U ( B UC ). Similarly, A U (B UC) c: (A U B) U C. Hence, ( A U B ) U C = AU ( B U C ). V. Distributive Laws: (i) A U ( B C ) = ( A U B ) (A C). (ii) A ( B U C ) = ( A B ) U ( A C ). Proof. We prove (ii) and leave (i) as an exercise, (ii) x A ( B C ) = > x A and x (B C ) => x A and (x B or x C) => (x A & x B) or ( x A & x C) 48  [ 'and' distributes 'or’] => x (A B) or x (A C) = »x ( A B ) U ( A C ). A (B C ) ( A B ) U ( A C). Similarly, ( A B ) (A C) A (B C). Hence, A (B C) = (A B) O (A C). VI.De-Morgan’s Laws: (i) ( A U B ) ' = A ' B ’ (ii) ( A B ) ' = A ' B'. Proof. (i) x (A U B)' = x x A a n d x B = > x A' and x B' = > x A ' B'. ( A B) ' ( A ' n B ' ). Similarly, (A' B) (A B)'. Hence, (A u B)'= (A' B'). ii) Hint, x A B => x A or x B. SOME MORE RESULTS ON OPERATIONS ON SETS: Theorem 1: For any sets A and B, prove that: (i) A B A and A B B ; (ii) A - B = A B ' ; (iii) ( A - B ) B = A B ; (iv) ( A - B ) B = (v) ( A - B ) ( B - A ) = ( A B ) - ( A B ). Proof: (i) x A B x A and x B =>x A (surely) A B A. Similarly, A B B. 49 (ii) x (A - B) => x A and x B => x A and x B' =>x ( A B ' ). :. ( A - B ) ( A B ' ). Similarly, (A B') (A - B). Hence, A - B = A B '. (iii) x (A - B) U B => x (A - B) or x B _ =>(x A&x B)or x B => (x A or x B) & (x B or x B) = > x ( AU B ). ( A- B )U B ( AU B )...(I) Again, y (A UB) => y A or y B => (y A o r y B) & (y B o r y B ) [Note] (y A & y B) o r y B = > y ( A- B ) or y B = > y ( A- B) U B. AU B ( A- B) U B...(II) Hence, from (I) & (II), we have: (A - B) U B = (A U B). , (iv) If possible, let ( A - B ) B and let x (A-B) B. T he n , x ( A - B ) B= > x ( A - B ) a nd x B = > (x A and x B) and x B => x A and ( x B and x B). But, x B and x B both can never hold simultaneously. Thus, we arrive at a contradiction. Since the contradiction arises by assuming that ( A - B ) B. , so (A - B) B= (v) x ( A- B) U ( B - A) =>(x A & x B) or ( x B & x A) 50  => (x A or x B) and (x B or x A) => x ( A U B ) and x ( A B ) => x ( A U B ) - ( A B ). ( A - B ) U ( B - A) ( A U B ) - ( A B ). Similarly, ( A U B ) - ( A B ) ( A- B) (B - A). Hence, ( A - B ) U ( B - A ) = (AUB)-( A B ) Theorem 2. For any sets A and B, prove that: (i) A B < = > B ' A ' ; (ii) A - B = A & A B = (iii) A U B = A B < = > A = B Proof, (i) To prove: A B B' A'. We shall prove it in two parts, namely: ( A B = > B ' A') and (B' A' = > A B ) (ii) To prove: A-B = A A = We prove it into two parts, namely: [ A - B = A = > A B = ] and [A B = A-B =A ]. 51 To prove: (iii) ( A U B ) = ( A B ) A = B. We prove it in two parts, namely: [A B=A B A = B ] and [A = B => A B = A B ]. First part Second part G i v e n : A U B = A B. Given : A = B To prove: A- B. To prove: AU B = A B. Proof. Let x A. Then, Proof: A B = B U B = B X A = >X AU B [ A AU B] [ A = B] ==>x A B And, A B = B B = B. [ AUB=A B] [ A = B ] => x A and x B AU B = A B [each equal to B ] => x B (surely).: A B. Thus, A = B AUB = A B. Similarly, B A. 52  A = B. Thus, A U B = A B =>A = B. Hence, A U B = A B A = B. Theorem 3. For any sets A, B, C prove that: (i) A - (B Q = (A - B) (A - C); (ii) A - (B U O = (A - B) (A - C); (iii) A ( B - C ) = ( A B ) - ( A C ) ; (iv) A ( B C ) = ( A B ) ( A C ). Proof, (i) Let x A - (B C). Then, x A - ( B C ) = > x A and x (B n C) = > x A and (x B or x C) (x A and x B) or (x A and x C ) = > X (A - B) or x (A-C) = > X ( A- B )U ( A - C ). A- ( B C) ( A- B) U( A- C ). Similarly, (A- B ) U (A - C) A - ( B C). Hence, A - (B C) = (A - B) U (A - C). Similarly, (ii) may be proved, (iii) L e t x A ( B - C ).Then, x A ( B - C ) => x A and x ( B - C ) => x A and (x B and i. e C) => (x A & x B ) & (x A & x => C) => x => (A B) & x (A C) = > X ( A B ) - ( A C). => A ( B -C ) c( A B ) - ( A n C ). Similarly, ( A B )- ( A C ) A ( B - C ). Hence, A ( B - C ) = ( A B ) - ( A C ). 53  (iv) A ( B C ) = A [ ( B -C ) U ( C - B) ] = [ A B - C ) ] U [ A ( C - B ) ] [distributive law ] = [(A B)-( A C)] U[( A C)-(A B)] = (A B) ( A C). 2.5.1 EULER-VENN DIAGRAMS To express the relationship among sets in a perspective way, we represent them pictorially by means of diagrams, known as Venn-diagrams. The universal set is usually represented by a rectangular region a n d its subsets by closed bounded regions inside this rectangular region. 2.5.1.1 Venn diagrams in different situations: I. Given: A U , where U is the universal set. Let U = ( 1 , 2 , 3 , 4 , 5 , 6 , 7 }. ' ?... And, A= ( 1 , 3 , 5 , 7 } Then, the rectangular region shown in Fig. 1 represents the universal set a n d the region enclosed by a closed curve inside the rectangular region represents the set A. Clearly, the shaded region represents A'. So, A' = (2, 4, 6 }. II. Given: Two intersecting subsets o f a universal set. Let U = { 1, 2, 3, 4, 5, 6, 7, 8 } be the universal set and Let A = {1, 3, 4, 5} and B = { 2, 4, 5, 6 } be its two subsets. As shown in Fig. 2, the rectangular region represents the universal set. 54  Since the sets A and B are intersecting, we draw two intersecting circles (or bounded figures), representing A and B respectively. Then (i) The common region between the two circles represents A B. T hu s , A B = {4, 5}. (ii) The total region bounded by both A and B represents A U B. Thus, A U B = {1, 2, 3, 4, 5, 6 }. (iii) Excluding the portion o f B from the region of A, we get the region representing (A - B). Thus, A - B = {1, 3 }. (iv) Excluding the portion of A from the region o f B , we get the region representing (B - A). T hu s , B- A = { 2 , 6 }. 7 U 1 4 2 8 3 5 6 A B III. Given : Two disjoint subsets o f a universal set. Let U = 1, 2, 3, 4, 5, 6, 7 } be the universal set and let A = { 1, 3, 5 } & B = {2, 4} be two of its subsets. Clearly, A and B are disjoint: Since the sets A and B have no common element, we represent them by two disjoint circles, as shown in Fig. 3. Clearly, A B = , A- B = A , B - A = B & AU B ={1,2,3,4,5) IV. Given: A B U , where U is the universal set. Let U = 55  {1, 2, 3, 4, 5, 6, 7, 8} be the universal set and let A = {1, 3 } and B = {1, 3, 4, 5 } be two of its subsets. Since A B, the region representing A must therefore lie wholly inside the region representing B, as shown in Fig. 4. Clearly, A B = A; A U B = B; A- B = and B- A = { 4 , 5 }. 2.5.1.2 Important Results from Venn Diagrams Let A and B be two nonempty intersecting sets. In counting the elements of AU B , the elements of A B are counted twice, once in the counting of elements of A and second time in the counting of elements of B. Thus, n (A U B ) = n (A) + n ( B ) - n (A B). If A B = , The n ( AUB) = n (A) + n( B ). It is also clear from the given Venn-diagram that: (i) n( A - B ) + n( A B ) = n( A ) ; (ii) n ( B- A) + n( A B ) = n ( B ) ; (iii) N (A U B ) = N (A-B) + n (A B) + n (B- A). 56 Ex.1. A survey shows that 74% o f the Indians like apples, whereas 68% like oranges. What percentage o f the Indians like both apples and oranges? Sol. Let A and B denote the sets of Indians who like apples and oranges respectively. Then,n ( A ) = 74; n ( B ) = 68 and n ( A U B) = 100. n ( A B ) = n ( A ) + n ( B ) - n ( A U B ). = (74 + 68-100) = 42. Hence, 42% of the Indians like both apples and orange. Ex. 2. In a group o f 850 persons, 600 can speak Hindi and 340 can speak Tamil. Find :(i) How many can speak both Hindi and Tamil; (ii) How many can speak Hindi only ; (iii) How many can speak Tamil only. Sol. Let A and B denote the sets of persons who can speak Hindi and Tamil respectively. Then, n ( A ) = 600; n ( B ) = 340 & n (A U B ) = 850: (i) n ( A B ) = n (A) + n ( B ) - n ( A U B ) = (600 + 340 - 850) = 90. Thus, 90 persons can speak both Hindi a nd Tamil. (ii) n ( A - B ) + n ( A n B ) = n ( A ). n ( A - B ) = n( A ) - n ( A n B ) = (600 - 90) = 510. Thus, 510 persons can speak Hindi only. (iii) n ( B - A ) + n ( A B ) = n ( B ). (iv) n ( B - A ) = n ( B ) - n ( A B ) = (340 - 90) = 250. Thus, 250 persons can speak Tamil only. 57 Ex.3. In a group o f 52 persons, 16 drink tea but not coffee and 33 drink tea. Find: (i) how many drink tea and coffee both; (ii) how many drink coffee but not tea. Sol. Let A and B be sets of persons who drink tea and coffee respectively. Then (A - B) is the set of persons who drink tea but not coffee. And, (B -A) is the set of persons who drink coffee but not tea. n (A) = 33,n ( A- B) = 16 and n ( A U B ) = 52. Now, (i) n (A-B) + n (A B) = n (A) n( A B ) = n( A ) - n( A - B) = (33 - 16) = 17. Thus, 17 persons drink tea and coffee both, ( i i i ) n ( A U B ) = n (A) + n (B)-n ( A B ). n(B) = n( A u B) -n (A) + n ( A B ) So, n (B) = (52 - 33 + 17) = 36. Now, n ( B - A ) + n ( A B ) = n(B). n ( B - A ) = n ( B ) - n ( A B ) = (36 - 17) = 19. Hence, 250 persons drink coffee but not tea. Ex.4. For any three sets A, B, C prove that : n ( A U B U C ) = n(A) + n (B)+ n (C) + n (A B C) ] -[ n( A B ) + n( B C) + n( A C )]. Proof, n (A U B U C) = n[ ( A U B ) U C ] = n (A U B) + n (C) - n [(A U B) C ] = re (A U B) + n (C) - n [(A C) U (B C)] = n(A) + n( B) -n( A B)+ n(C) 58  -n(A C ) + n( B C) - n( A C B C) ] = [ n(A) + n(B) + n(C) + n( A B C)] - [ n ( A B ) + n ( A C ) + n ( B C) ]. Ex.5, In a group of athletic teams in a school, 21 are in the basket ball team; 26 in the hockey team and 29 in the football team. I f 14 play hockey and basket ball; 12 play football and basket ball; 15 play hockey and football and 8 play all the three games. Find: (i) how many players are there in all; (ii) how many play football only. Sol. Let A, B, C be the sets of players forming basket ball team; hockey team and football team respectively. Then, n (A) = 21; n (B) = 26; n (C) = 29; n (A B ) = 14 ; n (A C) = 12 ; n (B C) = 15 & n (A B C) = 8. n (A U B U C) = [n(A) + n (B) + n (C) + n (A B C) ] - [ n (A B) + n (A C) + n(B C) ] A , = [ (21 + 26 + 29 + 8 ) - (14 + 12 + 15) ] = 43. Thus, there are 43 players in the group. With the given data, we may draw the Venn diagram as shown herewith. Clearly, the number of players playing football only = [29 - (7 + 8 + 4) ] = 10. Ex.6. A class has 175 students. The following is the description showing the number o f students studying one or more o f the following subjects in this class. Mathematics 100; Physics 70; Chemistry 46; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. 59 Find (i) How many students are enrolled in Mathematics alone, Physics alone and Chemistry alone ; (ii) the number of students who have not offered any of these three subjects. Sol. Let A, B, C denote the sets of students enrolled in Mathematics, Physics and Chemistry respectively. Let us denote the number of elements contained in each bounded region by small letters a, b, c, d , e, f, g as shown in_the figure. Using the given data, we have a + b + c + d = 100; b + c+ e+ f=70