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    Math In Business boa Math In Business boa Presentation Transcript

    • Math in Business Presents……….
    • Contents……
      • Chapter I. Basic operation
      • Chapter 2. Operation with Decimals
      • Chapter 3. Fractions
      • Chapter 4 Percent in Business
      • Chapter 5: Employee’s Compensation
      • Chapter 6: BUYING AND SELLING
      • Chapter 7:SIMPLE INTEREST
      • Chapter 8: COMPOUND INTEREST
      • Chapter 9: DEPRECIATION
      • Chapter 10: INCOME STATEMENT
    • Chapter I. Checking answers of the Basic Operations
      • Addition, subtraction, multiplication, and division are the basic operations in mathematic. Users confronted with different problems in business require thorough knowledge of these basic operations. With the invention of calculators, a person can solve a mathematical problem with greater speed. Since calculators are mere instrument in performing the operations, accuracy is largely dependent on mastery of the operations by the user and on his skills in checking answers .
    • Basic Math Operations
      • Addition
      • Commutative Property of Addition · Associative Property of Addition · Distributive Property · Additive Identity: Zero
      • We'll assume the reader can add digits, so that
      • 2 + 2 = 4
      • is not a surprise. In addition, we'll also assume that the fundamentals of "carrying" are not that big of a problem, and so most readers will immediately know how we did
      • We're more interested here in the general properties of addition that impact algebra. With that in mind, let a, b and c be three real numbers. Then the following properties of addition turn out to be usable and important:
      • Commutative Property of Addition
      • (Addition is the same regardless of the order one adds the numbers, i.e., forwards addition is the same as backwards addition).
      • Examples
      • Note that negative numbers are sometimes enclosed in parentheses to avoid confusion between the sign of the number and the addition operation. This is merely a matter of style – many textbook writers use spacing to set off the difference instead.
      • Associative Property of Addition
      • (Addition of a list of numbers is the same regardless of which are added together first, i.e., grouping does not matter)
      • Examples
      • Multiplication
      • Commutative Property of Multiplication · Associative Property of Multiplication · Distributive Property · Multiplicative Identity
      • Again, we’ll assume that the basics of multiplication are well known, so that 2x2=2· 2=2*2=4. There are obviously several different notations in use, depending when one learns it and what context one learns it. We will use all of these notations, as well as another: when using variables, multiplication is assumed when symbols are merely written next to each other.
      • Multiplication follows the same two laws just described for addition.
      • Commutative Property of Multiplication
      • Examples
      • We are following the same convention as with addition. Note that just writing numbers next to each other is a poor idea because, for example, 32 can be confused with 3· 2. Thus, we need some sort of symbol to make the two distinct. There are several correct ways to do this – in other articles, we’ll make a lot of use of parentheses, so that 3· 2 will be written 3(2) or (3)(2). Experience with the notation will help make it very clear what is meant.
      • Associative Property of Multiplication
      • Examples
      • Please note that we do not have to limit ourselves to parentheses; the last computation could have been written
      • This looks a bit neater. Again, such things are a matter of style, and the reader is encouraged to use whatever bracketing makes the most sense and allows clear, proper ordering of calculations.
      • Distributive Property
      • When both addition and multiplication appear in a single mathematical expression, this distributive law controls the operation. This is probably one of the most important laws in mathematics; getting it wrong guarantees bad calculations! The "reverse" is NOT CORRECT:
      • Examples
      • Subtraction
      • We’re going to define subtraction in terms of addition of the negative :
      • This means that subtraction is a shortcut or an abbreviation of the above addition operation. The "triple equals" sign used here means definition , and is meant to signify an operation that is always true . It is a stronger statement than a simple equals sign.
      • Examples
      • Subtraction is not in general commutative:
      • Example
      • Nor is it in general associative:
      • Example
      • Division
      • We’re going to define division in terms of multiplication of the inverse . Please see the discussion of multiplicative inverse below for more info. For now, suppose that a > 1 and that
      • The number b has a very important function – it is called the multiplicative inverse of a, also known as the inverse of a. Naively, we can write
      • but it should be pointed out that this is not really a definition. In order to properly define division, we would have to discuss rational numbers and how they work. Instead of doing that now, we’ll simply take it as given that the reader intuitively understands fractions. With this definition of inverse, we can tackle division:
      • Example
      • Suppose a = 2. Then
      • Note that the decimal expansion 0.5000… is the result of long division, a subject we are avoiding here (for a discussion, see the division article). We can "prove" that 0.5000… is the inverse of 2 by multiplying:
      • To the extent that this is rather unsatisfying, we must ask the reader to suspend disbelief. The mechanics of division are complicated, and deserve a separate article.
      • Now we’ll use this notion to do a division.
      • Example
      • Note that we expanded 6 into 2· 3 and canceled the 3’s. This is a bit sloppy, but again, our definition of division is intuitive rather than precise.
    • Chapter 2. Operation with Decimals
      • Our money representation is based on a decimal system. The peso is composed of 100 centavos. Since we rarely deal with amounts less than a centavo, the lowest value we use is P.01. Thus, in any mathematical computation, the final answer should be exact to the nearest centavo. An answer of P 135.256 should be written as P 135.26. Usually, it serves no purpose to express an answer id peso value with fractional part of a centavo. For discussion purpose, the money value P 135.26 is read as “one hundred thirty five pesos and twenty six centavos.” Note that the decimal point is read as “and” to separate peso value from the centavo. Sometimes, it is necessary to indicate fractional parts of a centavo in such cases as statistical figures, indices, real estate tax, and surveys.
      • In performing addition and subtraction, the numbers containing decimal components are arranged according to their place value. In multiplication and division, the numbers with decimal components are treated as whole numbers. Care must be taken in placing the decimal points. The basic operations will be discussed in detail in the following sections.
    • Addition with Decimals
      • Addends with decimal components should be arranged according to their proper order (column). The decimal points should fall in one vertical column in order to align the figures according to their values. Hence, units of equal value will fall under the same column such as tens, hundreds, thousands, millions, etc.
      • Example 1. Add 3.085; 12.314 and 94.65
      • 3. 085
      • 12. 314
      • 94. 65
      • 110. 0492
      • Same column
    • Subtraction with Decimals
      • The rule used in addition applies equally to subtraction. Decimal points of the minuend and subtrahend are placed in the same vertical column.
      • Example 2.
      • Subtract 125.2684 from 436.1052
      • 436. 1052
      • - 125. 2684
      • 310. 8368
      • Same column
    • Multiplication with Decimals
      • The number of the decimal places of the product is equal to the total number of decimal places (total digit numbers to the right of the decimal points) of the factors.
      • Example 3.
      • Determine the product of factors 3.248 and 1.26
      • Note: For convenience, use the shorted factor ( lesser digit numbers) as multiplier (1.26)
      • 3.248 multiplicand (3 decimal places)
      • 1.26 multiplier (2 decimal places)
      • 19488
      • 64976
      • 3248
      • 4.09248 product (3+2 = 5 decimal places)
    • Division with decimals
      • When the dividend contains decimal fractions and the divisor is a whole number, the decimal point of the quotient is aligned vertically above (same vertical column) the decimal point of the dividend.
      • Example 4. Divide 125.265 by 25
      • Same column
      • 5. 0106 quotient
      • Divisor 25/ 125. 265
      • 125
      • 2
      • 0
      • 26
      • 25
      • 15
      • 0
      • 150
      • 150
      • 0 remainder
    • Rounding off Numbers
      • Generally, numbers are rounded off in making estimates or forecasts. In monetary computations, it is the practice to round off the final answer to the nearest centavo because the third decimal place (thousandths) is often immaterial.
    • Chapter 3. Fraction
      • While our exposure to the numeration system dealt mostly with whole numbe5rs, we also come across figures wherein the use of fractions is inescapable. Such terms as “halves,” “fourths” or “quarters” are used with reference to measurement like “quarter to twelve,” “half a mile” “three fourths full” and so on. Like negative numbers, fractions `and computation. If a unit is divided equally into two or more parts, each equal part is the fractional part of the unit which is usually expresses in a fraction. For example, a parent bought a whole pizza pie and would want to share the pie equally among four (4) children. The parent would cut the pie into four equal parts and one part is given to each which is one-fourth of the whole pie. As illustrated below
      • 1/4
      • 1/4 1/4
      1/4
      • ¼ means one part of the whole which was divided into 4 equal parts.
      • The number written above the line (numerator indicates the number of parts and the number written below the line (denominator) indicates the total units or parts the cake was divided into. A fraction may be used to represent a ratio.
      • Fractions are classified into:
      • proper fractions , ½, ¾, 5/6, 7/8, etc., are fractions
      • whose numerators are less than the denominators. These fractions indicate values less than one (1).
      • Improper fractions . 3/2, 4/3, 5/4, 7/6, etc., are fractions whose numerators are greater than or equal to the denominator. These fractions indicate values equal to or greater than (1).
      • Mixed numbers . 1 ½ 2 ¼ 3 1/3 4 ¼ etc., are fractions written as the sum of integer or whole number and a proper fraction. An improper fraction can also be expressed as a mixed number or vice versa.
    • Fundamental Rules in Dealing with Fractions
      • One should understand the fundamental rules in dealing with fractions. These rules are very important in the basic operations of addition, subtraction, multiplication and division of fractions. The rules are the following:
      • The value of a fraction is not changed when the numerator and the denominator are both multiplied by the same number other than zero.
      • Example 1. 1/3 multiplied both by 3
      • 1/3 x 3/3 = 1 x 3
      • 3 x 1 = 3/3
      • Note: 3/3 is the same as 1.
      • 2. The value of a fraction is not changed if we divide both the numerator and the denominator by the same number other than zero. The result is like dividing the fraction by 1.
    • Changing Improper Fraction to a Whole or Mixed Number
      • Mathematically, the line between the numerator and denominator indicates division. In solving problems, it is more convenient or easier to work with an improper fraction than mixed number.
      • To change an improper fraction to a whole or mixed number, divide the numerator by the denominator. Examples are:
      • Example 3:
      • Change the improper faction 2/2 to a whole or mixed number.
      • Solution:
      • 2/2 = 2 divided by 2 = 1
    • Changing Mixed Numbers to Improper Fractions
      • This conversion is very convenient when we perform most of the basic operations with fractions. Below are the ways in converting a mixed number to an improper fraction:
      • Multiply the whole number and the denominator;
      • Add the numerator to the product ; and
      • Make the result of (2) as the numerator of the new fraction and use the same denominator as that of the original mixed number.
      • Example 5 . Change 4 2 to improper fraction
      • 3
      • Solution: follow the steps given above
      • 4 x 3 (whole number x denominator) = 12
      • 12 + 2 (add numerator to product) = 14
      • 14/3 (improper fraction with same denominator as the original mixed number).
    • Reduction of Fractions
      • The process of converting fractions to other equivalent forms of either higher or lower terms without changing the value of the fraction is called reduction. As in division, multiplying or dividing both the numerator and denominator by the same number other than zero does not affect the quotient.
      • Examples are:
      • Reduce ¾ to higher terms.
      • 3 = 3 x 3 = 9 ; or 3 = 3 x 5 = 15
      • 4 4 x 3 12 4 4 x 5 20
      • Reduce 25/625 to lower terms.
      • 25 = 25 ÷ 5 = 5 or 1
      • 625 625 ÷ 5 125 25
      • When the numerator and the denominator have no more common divisor except 1 the fraction has been reduced to its lowest term. To reduce a fraction to its lowest term, both the numerator and the denominator should be divided by their greatest possible common divisor.
      • For example,
      • in the proper fraction 46 23 is the greatest possible 161
      • common divisor of 46 and 161, hence,
      • 46 = 46 ÷ 23 = 2
      • 161 161 23 7
    • Finding the greatest Common Divisor (g.c.d)
      • The greatest common divisor is called highest common factor . If the greatest common divisor is not readily apparent, the following steps are recommended to determine the desirable divisor:
      • Divide the larger number of the numerator and denominator by the smaller number.
      • If there is a remainder in step 1, divide the smaller number by the remainder.
      • If there is still a remainder in step 2, divide the remainder in step 1 by the remainder in step 2.
      • Continue dividing each remainder by its succeeding remainder until the remainder is 0. The last divisor is the greater common divisor.
      • Example: Find the g.c.d. of 69
      • 184
      • 2
      • 69/184 step 1. 69 is less than 184
      • 138
      • 46 1
      • 46 /69 step 2.
      • 46
      • 23
      • 2
      • 23/46 step 3 and 4
      • 46
      • 0 remainder
    • Changing Fractions to Decimal
      • There are two ways of expressing parts of a whole, namely, common fractions and decimal fractions. Most of the problems in business require converting fractions into decimals. For example, if a mathematical operation with fractions is performed on a pocket calculator, all fractions must be converted to decimals before entering them in the calculator.
      • Fraction indicates division, which means that the numerator is to be divided by the denominator. An example is ¼ which is same as 1 divided by 4.
      • Example:
      • 0.4
      • 2 = 2 ÷ 5 = 5/ 2.0
      • 2.0
      • 0
      • The answer is 0.4
    • Changing Decimal to Fractions
      • When a decimal fraction is written in a common fractions, the decimal fraction without the decimal point is used as the numerator, and the denominator is 1 with as many zeros annexed as there are decimal places in the original decimal fraction. Then the common fraction is reduced to its lowest term.
      • Example: express .5 to fraction
      • .5 = 5 = 1
      • 10 2 lowest term
    • Addition of fractions
      • Two different things cannot be added. The same is true with numbers and fractions.
      • To add fractions, all denominators must be converted to a common number without changing the values of the fraction.
      • When two or more fractions with different denominators are to be added, we must change the fractions to their equivalent fractions with the same denominator. The procedure involves the convertion of the fractions to its higher or lower terms as discussed in the preceding topics.
      • Example: 1 & 1
        • 2 4
      • 1 = 1 x 2 = 2
      • 2 2 x 2 4
      • After converting 1/2 to ¼, lets add this to ¼ w/c has same denominator
      • 2 + 1 = 3
      • 4 4 4
      • Solution:
      • 1st step – Determine the “lowest possible” number that can be divided by the different denominators (this number is called least common denominator). In the example, the least common denominator is 12.
      • 2nd step – Determine the equivalent fractions of each fraction with the same denominator as 12.
      • 1 = 1 x 6 = 6
      • 2 2 x 6 12
      • 3rd step – Add the equivalent fractions with the same denominators.
      • 1 + 1 + 1 = 6 + 4 + 3
      • 2 3 4 12 12 12
      • = 6 + 4 + 3
      • 12
      • = 13/12 or 1 ½
    • Addition of Mixed Numbers
      • When mixed numbers are added, it is unnecessary to convert the numbers to their equivalent improper fractions. Add the whole integers and the fractions separately. The fractional part of the answer should reduce to its lowest term.
    • Subtraction of Fractions
      • Fractions subtracted must first be converted to their equivalents with the least common denominator (l.c.d.). then the numerators of the converted fractions are subtracted. The answer will have the numerator difference as the numerator and the least common denominator as the denominator.
    • Subtracting a Mixed Number from a Whole Number
      • By way of an example, the following are the steps in subtracting a mixed from a whole number:
      • Step 1. Convert one unit of the minuend into an improper fraction with the same denominator as the fraction in the subtrahend. Reduce the whole number by one so as not to change the value of the minuend.
      • 12 = 11 8/8
      • Step 2. Subtract
      • 11 8/8 – 5 1/8 = (11-5)+(8/8 – 1/8)
      • = 6 + 7/8
      • = 6 7/8
    • Subtracting Mixed Numbers
      • When necessary, convert the fractional parts of both the minuend and the subtrahend so that they have a common denominator. If the fraction in the subtrahend is smaller than the fraction in the minuend, you can subtract, but if the fraction is greater than the fraction in the minuend, follow the steps below.
      • Convert one unit of the minuend into an improper fraction with “correct’ denominator, and add this unit to the existing fraction in the minuend.
      • Reduce the whole number in the minuend by one (the unit which is now a fraction).
      • You can now subtract the fraction.
    • Multiplication of Fractions
      • There is no need to determine the common denominator of the fractions in multiplication. The basic principle in multiplication of fraction is to determine the product of both numerators and both denominators and the products become the numerator and denominator of the answer.
      • In multiplying a whole number by a mixed number, change the mixed number to an improper fraction and determine the product.
      • If both factors are mixed numbers, convert each mixed number to an improper fraction before multiplying.
      • Example: Multiply 1/2 by 2/3
      • 1/2 x 2/3 = 1 x 2 = 2 or 1 reduce to lowest term
      • 2 x 3 = 6 3
    • Divisions of Fractions
      • The simplest way to divide fractions is to multiply the dividend by the reciprocal (inverted form) of the divisor. The numbers 3 and 1/3 are called reciprocal to each other. Likewise, 4/5 is the reciprocal of 5/4.
      • When dividing a mixed number by another mixed number, change each mixed number to an improper fraction and then multiply after inverting the divisor.
      • Example:
      • Divide: 5/8 by 2/3
      • 5 ÷ 2 = 5 x 3
      • 8 3 8 2
      • = 5 x 3 = 15
      • 8 2 16
    • Chapter 4 Percent in Business
      • The use of percent is gaining wide acceptance in all spheres of activity. When relating the parts to a whole, relationship is expressed in percent.
      • Percent is also frequently used in presenting accounting and statistical data relationship.
    • Percent to Fraction
      • Percent numbers are not really “new” since they are in reality fractions.
      • To express 80% as a fraction in its lowest term, convert 80% in its fractional form first and perform the operation of division of proper fractions as discussed in Chapter III.
      • 80% = 80
      • 100
      • Then = 80 ÷ 20
      • 100÷20
      • = 4
      • 5
      • thus, 80% is 4/5 in fraction
    • Percent to Decimal
      • In any mathematical computation, percent is converted to decimal before multiplying of dividing it with other quantifying numbers. There are several percent values encountered in business transactions that do not convert into fractions making such percent values as decimals rather than as fractions.
      • The basic principle in converting percent to decimal is to move the decimal point two places to the left and drop the percent sign.
      • Example 4. Change 15 % to a decimal
      • Solution:
      • 15% = 15 = .15
      • 100
    • Representing Decimals as Percents
      • The fastest way to convert decimals to percent is to the decimal point two places to the right. This method is actually the reverse order of converting percent to its decimal equivalent as discussed in the preceding sections.
      • Example:
      • 0.25 = 0.25 x 100 = 25 %
    • Representing Fractions as Percents
      • In converting a fraction to percent, change the fraction to its decimal equivalent as discussed in the previous chapter. Then multiply the decimal equivalent by 100 or simply move the decimal point two places to the right and suffix the percent symbol.
      • Example: Convert 1 7/8 to its equivalent percent form,
      • Solution:
      • 1 7 = (8 x 1) + 7 = 15
      • 8 8 8
      • = 1.875 x 100
      • = 187.5 %
    • Fraction-Decimal-Percent Equivalents
      • The equivalents discussed in the previous chapters can be tabulated into what are known as aliquot parts of 100. An aliquot part is a portion of a number by which the number may be divided leaving no remainder
      • Example:
      • Multiply 60 by 33 1 %
      • 3
      • 60 by 33 1 % = 60 x
      • 3
      • = 20
    • Determining the Percentage
      • In the statement, 40 is 40/700 0f 700 the denominator of the fraction which is the basis of comparison is called the base. The rate is the percent indicating the number or quantity for every 100. The percentage refers to the number of items in the desired situation or condition.
      • Based on the above concept, the principle is written.
      • P= b x r Equation 4.1
      • When: P = percentage
      • B = base
      • r = rate, usually expresses in %
      • Stated as a principle, percentage is the product of the base times the rate. In any mathematical computation, before we add, subtract, multiply or divide, we must convert the percent to its decimal or fractional equivalent whichever is convenient to use.
      • Example: An employee who earned P 1,500 spent 30 % of his earnings to buy a wrist watch. How much is the wrist watch?
      • Given: B = 1,500 (base)
      • r = 30 %
      • Solution: P = B x r
      • = 1,500 x 30 % = 1,500 x .3
      • = 450.00
    • Finding the Rate
      • One of the most commonly used representations of a number in business is the rate. Although representation of number based on rate involves different interpretations in business or surveys.
      • Based on the preceding discussion, a formula can be derived from Equation 4.1 as follows:
      • P = B x r Equation 4.1
      • r = P ÷ B
      • r= P/b Equation 4.2
    • Finding the Base
      • Basically, the base quantity can be determined by applying the same concept in Equation 4.2. the formula is written below.
      • B = P ÷ r = P/r Equation 4.3
    • Ratio and Proportion
      • As mentioned earlier, percent is another way of comparing a number with the whole, while its equivalent in decimal is comparing with one unit.
      • Ratio is a relation between two numbers expressed in terms of a quotient.
      • The result of reducing ratios to lowest terms is called equivalent reduced ratio.
      • The equality of two ratios is called proportion .
    • Chapter 5: Employee’s Compensation
      • Among other expenses, a business enterprise pays the salaries to its personnel. Salary is the renumeration received by an employees for services rendered. The labor code of the Philippines defines wages or salaries as “ the renumeration or earning, however designated, capable of being expressed in terms of money, whether fixed or ascertained on a time, task, piece or commissions basis or other method of calculating the same which is payable by an employer to an employee and includes the fair and reasonable value of board, lodging or other facilities customarily furnished by the employer to the employee.”
      • Salaries may be based on period of time, production, or percentage on an agreed basis. Employees may be classified according to the bases of their salary payments. There are the fixed wags earners, the piece-workers, the commission earners and the salary-plus-commission earners.
    • Computing Wages Based on Time (period)
      • Fixed wage earners are paid based on the number of hours worked. For work done beyond the regular eight (8) working hours a day, the employee is given additional pay called overtime pay.
      • The Labor Code fixes the minimum rates of overtime pay, night shift differential, and work on regular and special holidays. The rates range from 25 %, 30% to 100% over and above the regular pay rate. For purposes of discussion, overtime pay as herein used will be regular pay plus 50% of said pay which means that for every one hour overtime, the pay will be equivalent to 1.5 of the regular hourly rate.
    • Computing Wages on Piecework Basis
      • The minimum rates set by the Labor Code do not apply to employees whose salaries are not based on the number of working hours. Employees may be paid by results like “pakyaw” (pakiao), “takay” and by the pieces of product.
      • Workers in selected industries such as those engaged in manufacturing, packing, and handicrafts are paid on straight piecework basis. The wage of a straight pieceworker depends solely on number of units completed each period of time.
    • Computing Wages on Commission Basis
      • Most of the commission wage earners are staff in the sales of a business. They are paid on the basis of number of units sold. Since sales is the bloodline of the company, the sales people are doubly motivated to increase sales output. The three most commonly used compensation schemes in sales are the straight commission, commission and bonus, and salary plus commission.
      • The regular sales force of a company is usually paid a fixed monthly rate plus commission earned from sales.
      • To compute for the commission, multiply the sales (in number of units or peso value) by the commission rate.
    • Incentive or Bonuses
      • Incentive pay and bonuses in money or in kind are given to employees to motivate them. The incentive pay may be based on the number of units completed or sold in excess of a required minimum.
    • Deductions
      • The earnings of an employee are subject to deductions which are required by law of by agreement between the employer and the employee. Deductions required by law are income taxes, social security taxes, medicare contributions, contributions to pension plans and contributions to pag-ibig.
    • Chapter 6: Buying and Selling
      • Buying and selling are the essential functions of a trading concern. Manufacturers or producers buy raw materials or parts to produce finished products. A trading concern tries to buy merchandise at the lowest possible cost in order to maximize its profits. A manufacturer endeavors to acquire materials and/or finished parts in order to minimize its cost of production which is a way of increasing the income upon the sale of the finished products.
      • In a trading business, the trader of the buyer should determine the goods needed by the customers and the price the customers are willing to pay for the goods. He should also know where to obtain the goods at the lowest price to minimize cost and maximize profit.
      • In general, a manufacturer sells its products in bulk to a wholesaler. In turn, the wholesaler sells in smaller lots to retailers who sell the product to the end users or consumers.
    • Discounts
      • Discounts are of two types, namely:
      • Trade discount . It is a discount offered by a seller to induce trading. This is usually offered by a manufacturer or wholesaler. This type is usually encountered in catalogs where a list price is printed together with the trade discount thereon. The list price less the trade discount is the suggested selling price upon release of the product.
      • Cash discount . It is a reduction on the selling price offered to a buyer to induce him to pay promptly.
    • Finding the Net Price
      • There are two methods in determining the net price to the buyer when trade discount is given. These are the Discount Rate Method and the Net Price Rate Method.
      • Method A. Discount Rate Method. The formula for discount is the same for percentage where the terms percentage (P), base (b) and rate (r) in Equation 4.1 on page 56 are substituted with the terms discount (D), selling or list price (L.P) and discount rate (d) respectively. Thus,
      • D = L.P. x d Equation 6.1
      • L.P. = Selling price or list price
      • D = Discount rate usually expressed in percent
      • The net price then is the difference between the selling price and the discount which is shown below.
      • N.P. = L.P. − D Equation 6.2
      • N.P. = L.P. x (net price rate)
      • = L.P. x (100% - d in%) Equation 6.3
      • Example 1. The list price of a cassette tape recorder at AVESCO is P1, 850. The Trade discount rate is 12 %. How much will the buyer pay for the recorder? Use Discount Rate Method and the Net Price Rate Method.
      • Given: L.P. = P1,850 d = 12%
      • Solutions:
      • Method A. Discount Rate Method
      • PhP 1,850 (L.P.)
      • X .12 (d)
      • P 222 (D)
      • Determining the net price:
      • PhP 1,850 (L.P.)
      • ─ 222 (D)
      • Php1,628 (N.P)
      • =======
      • Method B. Using Equation 6.3
      • N.P. = L.P. x N.P.R
      • = Php 1,850 x (100%-12%)
      • = Php 1,850 x 88% = 1, 850 x .88
      • = Php 1, 628.00
    • Finding the Discount Rate
      • A supplier may list only the net price and list price of the products being sold. The buyer should know how to determine the discount rate in order to evaluate the reasonableness of the discount offered.
      • Example 2. What is the discount rate if the list price is P1,500 and the discount is P450.00?
      • Given: L.P. = P1,500, D = P450
      • Solution:
      • Substituting the values in Equation 6.1
      • D = L.P. x d
      • 450 Php = 1,500 Php x d
      • d = 450 = .3 or 30%
      • 1,500
      • Example 3. What is the discount rate, if the net price is 246 and the discount is P83.00?
      • Given: N.P.= P246 D = P83
      • Solution:
      • N.P. = L.P. ─ D
      • N.P. = is net price
      • L.P. = N.P. + D
      • = P246 + P83
      • = P 329.00
      • Substituting the list price in Equation 6.1 :
      • D = L.P. x d
      • d = D/L.P.
      • = 83/329
      • = .252 or 25.2%
    • Discounts in Series
      • To dispose their goods quickly, wholesalers may offer successive trade discount rates which we call discounts in series. If a manufacturing concern would give 15% and 10% trade discounts on its product, this does not mean that the total discount rate is 25%. As a discount series, it simply means that the first discount of 15% is applied to the original list price and the second discount of 10% to the balance of the original list price and the first discount price.
      • Example 4. A home appliance dealer was offered television set with a list price of P5,680 less 15% and 10%. What is the net price?
      • Given: List Price (L.P) = P5,680
      • First discount rate (d1) = 15%
      • Second discount rate(d2 ) = 10%
      • Solution:
      • Method A. Discount Rate Method
      • 1rst Step . Using Equation 6.1
      • D1 = L.P.1 x d1
      • = P5,680 x .15
      • = P852.00
      • 2nd Step. To get the balance (L.P2) on which the second discount rate (d2) will be applied, subtract the first discount rate from the list price–
      • PhP 5,680 (L.P1)
      • ─ 852 (d1)
      • Php 4,828 (L.P2)
      • =======
      • 3rd Step. Solve for D2 using Equation 1
      • D2 = L.P2 x d2
      • = 4,828 x .10
      • = P482.80
      • 4rth Step. Solve for the net price (N.P.) by subtracting the second discount (D2) from L.P2
      • Php 4,828 (L.P2)
      • 482 (D2)
      • Php 4,345.20 (N.P.)
      • =======
    • Method B. Net Price Rate Method
      • 1rst Step . The net price rate is the difference between 100% and the discount rate (See Equation 6.3). The same formula is also applied when two or more discounts are given. The final net price is the product of all the net price rates found by using Equation 6.3.
      • Final net price rate = (100% ─ d1) x (100% - d2) = (100% ─ 15%) (100% - 10%)
      • = 85% x 90%
      • = .85 x .90
      • = .765
      • 2nd Step . The net price is the product of the list price and the final net price rate.
      • Net price = Php 5,680 x .765
      • = Php 4,345.20
      • Obviously, the net price rate method is relatively shorter and simpler than the discount method, more so when there are more than two series of discounts. In such cases, the formula to use in determining the net price is :
      • N.P. = L.P. x (100% ─ d1) (100% ─ d3) (100% ─ dn) Equation 6.4
      • Where dn indicates the last discount rate
      • Note: It is immaterial as to which order the discounts are taken.
    • Single-Discount Rate Equivalent
      • There are two practical ways to evaluate discount series offered by two or more sellers. Let us assume that in terms of services, neither firm held any advantage over the other. Hence, the buyer’s only concern is to determine which firm’s net selling price was smaller. As earlier discussed, the discount series of, for example, 10% and 15% are not equivalent to the total discount rate of 25%.
      • One way of evaluating discount series is to determine the final net price rate which was illustrated in Example 4, Method B. From the buyer’s point of view, the smaller the final net price rate, the lesser is the net price.
      • An easier way to evaluate the discount rates in a discount series is the equivalent single discount rate . From the buyer’s point of view, the bigger the single discount rate equivalent, the lesser is the net price.
      • Example 5 . Charing’s Beauty Parlor was offered beauty product with trade discounts of 15% and 10%. Find the single discount rate equivalent.
      • Given: d1 = 15%
      • d2 = 10%
      • Solution:
      • First Step . Derive the net price rate (NPR) of each discount rate.
      • Since NPR = 100% ─ d
      • a. NPR1 = 100% ─ d1
      • = 100% ─ 15%
      • = 85%
      • b. NPR2 = 100% ─ 10%
      • = 90%
      • Second Step. Determine the single discount rate equivalent (SDRE).
      • SDRE = 100% ─ (NPR1 x NPR2)
      • = 100% ─ (85% x 90|%)
      • = 100% ─ (.85 x .90)
      • = 100% ─ (.765)
      • = 23.5%
      • Example 6. A product is being offered in the market by Supplier A and Supplier B. The trade discounts being offered by Supplier A are 20% and 10% while that of Supplier B are 15% and 15%. If you were the buyer, which is the better offer?
      • Given:
      • Supplier A = 20% and 10% (d1 and d2)
      • Supplier B = 15% and 15%. (d1 and d2)
      • Solution:
      • Supplier A. Solving for the single discount rate equivalent of Supplier A, determine first the net price rate applicable
      • NPR1 = 100% - 20% = 80%
      • NPR2 = 100% - 10% = 90%
      • Therefore,
      • SDRE = 100% - (80% x 90%)
      • = 100%- (.80 x .90)
      • = 100% - (.72)
      • = 100% - 72%
      • = 28%
      • Supplier B. Solve for the net price applicable to the single discount rate equivalent of Supplier B.
      • NPR1 = 100% - 15% = 85%
      • NPR2 = 100% - 15% = 85%
      • Therefore,
      • SDRE = 100% - (85% x 85%)
      • = 100%- (.85 x .85)
      • = 100% - (.7225)
      • = 100% - 72.25%
      • = 27.75%
      • Based on the net price rates being offered, the offer of a lower net price rate (72%) by Supplier A is the better offer. But, based on the single discount rate equivalent, the offer of a higher discount (28%) by Supplier A is also the better offer.
    • Cash Discounts
      • To encourage wholesalers or retailers to buy is one thing and to induce the buyers to pay their purchases is another thing. While sellers give buyers credit lines in that the buyer are given several days after delivery within which to pay their purchases, sellers encourage buyers to pay promptly their purchases by offering cash discounts. A cash discount should not be confused with a trade discount. The former is a reduction in the net price or invoice price upon payment on a specified period of time while the latter is a deduction from the list price.
      • The cash discount rates offered are clearly written on the invoice. An example of a cash discount is” 5, n ”
      • 10 30
      • which means that a cash discount of 5% will be deducted
      • from the net price if the invoice paid within 10 days from the date of the invoice. The "n/30" means that after 10 days, no discount is given and that the invoice price should be paid not later than 30 days of the invoice date.
      • If the retailer or buyer would avail of the cash discount offered, the price to be paid is the invoice price less the amount of cash discount.
      • Example 7. An invoice with a list price of P18,200 dated October 12, has trade discounts of 15% and 10% and the terms are 4/10 ,2/30 , n/60. How much must be paid if the invoice is settled on (a) October 22? (b) November 11?
      • Given:
      • List price = P 18,200
      • Trade Discount series = 15% and 10%
      • Cash discounts = 4 , 2 , and n
      • 10 30 60
      • Solution :
      • (a.) If the amount is settled on October 22
      • 1rst Step. Solve first the net price (N.P.)
      • N.P. = L.P. x (NPR1) (NPR2)
      • = P 18,200 x (100% - 15%) (100% - 10%)
      • = 18,200 x (85%) (90%)
      • = 18,200 x (.85) (.90)
      • = P 13,923.00
      • 2nd Step. Having solve for the N.P., which is P13,923.00, we can now proceed to solved the cash discount (Dc). If the purchaser pays on October 22 which is 10 days from the invoice date, the cash discount term is 4/10. Thus, the cash discount rate is 4%
      • Dc = P13, 923 x 4%
      • = 13, 923 x .04
      • = P556.92
      • 3rd Step . Determine the net invoice price (NIP) by simply subtracting cash discount from the net price.
      • NIP = N.P. - Dc
      • = P13, 923 - P556.92
      • = P13,366.08
      • Alternatively, the computation may be as follows:
      • NIP = N.P. x (100% - dc)
      • = (100% - 4%)
      • = 13,923 x (96%)
      • = 13,923 x (.96)
      • = P 13, 666.08
      • If the invoice is paid on November 11.
      • First Step . Same as solution (a.).
      • N.P. = P 13,923.00
      • 2nd Step. If the Purchaser pays on November 11 which is 30 days from the invoice date, the cash discount rate is 2%.
      • Dc = 13,923 x 2% = P278.46
      • NIP = 13,923 – P278.46
      • = P 13,644.54
      • Alternatively, the computation may be as follows:
      • NIP = N.P. x (100% - dc)
      • = P 13, 923 x (100% - 2%)
      • = 13, 923 x (98%)
      • = 13, 923 x .98
      • = P 13,644.54
    • Mathematics of Pricing
      • The price of a commodity is determined by many factors obtaining in the market. It is greatly affected by the need of the consumers for the commodity, its availability, its quality level and the number of different lines with similar products. If there is a great demand for the commodity and the stock is limited, the seller could dictate his own selling price. But if the price demanded is prohibitive such that the average consumer could not afford it, the consumers may shift to other available similar products and the seller may stand to lose.
      • The business should understand pricing terms such as cost, mark-up, margin and the like to enable himself to price his commodities reasonably. Cost refers to the amount the purchaser acquired the good. Margin (gross profit) is the excess of selling price over cost from viewpoint of the seller. If an original selling price is adjusted upward, the addition is called Mark-up . The reverse is called Mark-down.
      • To illustrate clearly the above terminologies, let us take an example. A car dealer bought a car for P14,500.00 and sold it for P17,200.00.
      • P14,500 is the cost of the article, assuming he did not spend any amount for recondition the car before selling it.
      • P 17,200 is the selling price, tag price, or retail price of the article. (P17,200 – P14,500) = P2,700 is the gross profit or margin.
      • In the above illustration, the margin P2,700 can be expressed in percent based in either the selling price or cost. This is called percent margin. If the selling price is used as the base, the percent margin on selling price is P2,700 = .15697 = 15.7%.
      • If the cost is P17,200 used as the base, the percent margin on cost is P2,700 = .1862 =1 =18.62%
      • P14,500
      • Formula-wise, the equations for the above are the following:
      • Cost + Margin = Selling price
      • Or
      • Margin = Selling price – Cost
      • Percent Margin(%) = Margin x 100, based on selling price
      • Selling price
      • Percent Margin(%) = Margin x 100, based on cost
      • Cost
      • Example 8. The merchandise that cost a dealer P4,150 was sold for P5,200. Determine the percent margin (a) based on selling price, (b) based on cost.
      • Given:
      • Selling price = P5,200
      • Cost = P4,150
      • Solution: Immediately subtract P4,150 from P5,200 to get the margin.
      • Margin = Selling price - Cost
      • = P5,200 -= P4,150
      • = P 1,050
      • If the percent margin is required based on selling price divide the margin by the selling price.
      • Percent Margin (%) = P1,050 x 100
      • P5,200
      • = .202 x 100
      • = 20.2%
      • If percent margin is required based on cost, divide the margin by the cost.
      • Percent Margin (%) = P1,050 x 100
      • P4,150
      • = .253 x 100
      • = 25.3%
      • Note : Another term used for gross margin is mark on.
    • Chapter 7: Simple Interest
      • Money borrowed usually bears a cost called interest. The amount borrowed is called principal.
      • Sources of loans are the banks, investment houses, savings and loan associations, cooperatives, credit unions, and other financing companies.
      • Determining the Interest
      • The interest or interest charge is usually expressed in percent, such as 6%,which means a charge of P6.00 for every P100.00 for a definite period of time. Unless otherwise stated, quoted interest rate is for one year.
      • The formula for computing interest or interest charge is:
      • I = P x r x t Equation 7.1
      • Where:
      • I = interest expressed in monetary value.
      • P = principal or the amount borrowed.
      • R = interest rate on the principal, usually stated in percent.
      • T = time or duration of the loan, expressed in number of years, months,
      • days, etc.
      • To illustrate, if Mr. X borrowed P2,600.00 from the bank at an interest of 5%, what is the interest in 3 years?
      • Given:
      • P = P2,600.00
      • r = 5% (per year)
      • t = 3 years
      • I = to find
      • Solution: Substituting the values in Equation 7.1
      • I = P x r x t
      • = P 2,600.00 x 5% x 3
      • = 2,600 x .05 x 3
      • = P390.00
      • The amount to be paid by the borrower upon maturity of the loan is –
      • Where :
      • A = sum or total amount to be paid.
      • P = principal
      • I = interest
      • Thus,
      • A = P 2,600 + P390.00
      • = P2,990.00
      • Example 1. Pedro borrowed from Jose P5,200 with an interest of 6%. How much should Pedro pay after 16 months?
      • Given:
      • P = P5,200
      • r = 6% (per year)
      • t = 16 months
      • Solution:
      • Method A. Convert 16 months to number of years.
      • I = P x r x t
      • = P 5,200 x 6% x 16
      • 12
      • = 5,200 x .06 x 4
      • 3
      • = P 416.00
      • Method B. Convert interest rate per month.
      • I = P x r x t
      • = P 5,200 x 6%/12n x 16 months
      • 12
      • = 5,200 x .5% x 16
      • = 5,200 x.005 x16
      • = P 416.00
      • The amount to be paid after 16 months is –
      • A= P + I
      • = P5,200 + P416
      • = P5,616.00
    • Determining the Rate of Interest (r)
      • This particular problem is encountered in evaluating two or more alternative choices. A lender wants a higher rate of interest for his money, while a borrower prefers a lesser rate of interest on money borrowed.
      • Assume that Mr. Y would like to borrow P5,000 payable after one (1) year from a bank whose interest charge is 18% per year. On the other hand, Mr. Z is willing to lend him the same principal and interest charge is P950 per year. Since the rate of interest on the bank loan is already know, Mr. Y must know the rate of interest on Mr. Z’s to enable him to determine which is the better offer.
      • Solution:
      • r = I Equation 7.4
      • Pt
      • = 950 = 0.19
      • 5,000
      • = 0.19 x 100%
      • = 19%
      • Thus, as compared to Mr. Y’s proposal of a 19% interest rate per year, the bank loan with 18% interest rate per year is definitely the better offer.
    • Determining the Time (t)
      • As derived from Equation 7.1
      • t = I Equation 7.5
      • Pr
      • Example 3. How long will it take for a deposit of P1,500.00 to earn P186.00 invested at the rate of 7- 1/2%?
      • Given:
      • I = P186.00
      • P = P1,500.00
      • r = 7- 1/2%
      • Solution: Solve for t, using Equation 7.5
      • t = I_ = Php186.00
      • Pr Php1,500 x 7 1 %
      • 2
      • = Php186.00
      • Php1,500 x .075
      • = 1.65 years
      • The time may be converted to its appropriate units of time measure like months or days
      • (Months) t = 1.65 years x 12 months per year
      • = 19.8 months
      • Or
      • (Days) t = 1.65 years x 360 days per year
      • = 594 days
      • If it is required to express the time (t) in combination of two or more units such as in years; months, and days the procedure is as follows:
      • In the above example, to convert t = 1.65 years to years, months and days.
      • 1rst Step. The period of 1.65 indicates one(1) year and a fraction of year expressed in decimal part as 0.65, thus number of years is 1.
      • 2nd Step . The fractional part of a year a (0.65) is multiplied by the conversion factor of 12 months per year.
      • Thus, number of months is 7.
      • 3rd Step . The fractional part of month (0.8) is multiplied by 30 days per month (approximate).
      • Therefore, the final answer is:
      • t = 1 year, 7 months and 24 days.
    • Determining the Principal, Given (A) (r) (t)
      • This particular problem involves Equations 7.1 and 7.2. In any mathematical computation, solving two unknown variables using one equation or formula is not possible. However, two equations with at least one common variable can be combined to derive a new equation.
      • If A = P + I (Equations 7.2) and I = Prt (Equation 7.1), by substituting (I) using its equivalent (Prt) in Equation 7.2, in the following manner.
      • A = P + I
      • A = P + Prt
      • And factoring out P, we have
      • A = P (I + rt) Equation 7.6
        • P = A Equation 7.7 I + rt
      • Example 4. If a bank offers 8 ½% interest rate and a depositor wants to have P3,878.20 after 5 years, how much should he deposit now?
      • Given:
      • r = 8 1 %
      • 2
      • A = P3,878.20
      • t = 5 years
      • Solution: Using Equation P = A___
      • 1 + r x t
      • P = P3,878.20
      • 1 + 8 1 % x 5 years
      • = P3,878.20
      • 1 + .085 x 5
      • = P3,878.20
      • 1 + .425
      • = P3,878.20
      • 1.425
      • = P 2,721.54
    • Determining the Sum (A)
      • Example 5. The bank offers an annual rate of 10 1 % interest, for 3 ½ years, 2
      • how much will the accumulated sum be if the initial deposit is P 1,350.00?
      • Given:
      • P = P 1, 350
      • r = 10 1 %
      • 2
      • t = 3 ½ years
      • In this particular problem, two methods may be used to determine the amount (A).
      • 1rst Method. Determine the interest by using
      • Equation 7.1, I = Prt and then, add the interest and the principal to determine the amount A, using 7.2
      • 2nd Method .
      • By A = P(1 + r x t)
      • A = P 1,350 (1 + 10 1 % x 3 1 )
      • 2 2
      • = 1,350 (1 + .105 x 3.5)
      • = 1,350 (1.3675)
      • = P 1, 846.13
    • Promissory Notes
      • A promissory note may be used for raising money. The borrower , in exchange for a loan, may issue his own promissory note or an immature note received by him form another person. A promissory note is a written promise to pay to a person a certain sum of money on a specified date. The person who signs the promissory note is the maker. The person to whom his payment is promised is the payee. The period, or term of the note is the duration or length of time from the date of the note to the date of maturity. Maturity date is also referred to as the expiration date or due date of the note. The maturity value is the total amount to be paid on the maturity or due date. The face value of a promissory note is the amount specifically mentioned in the note. If the interest is mentioned in the note, it is an interest- bearing note . If no interest is mentioned, then it is a non-interest bearing note .
      • To illustrate, let us assume that
        • Mr. Darwin borrowed P10,000 from his friend Carlos on November 8, Mr. Darwin issued a promissory note (shown below) due on February 8 worth P10,000 plus 10% interest.
        • The following is an example of promissory note:
          • P 10,000 November 8
        • Ninety days after I promise to pay to the order of Mr. Carlos the sum of Ten Thousand Pesos and no/100 For value received with interest at 10%.Due
        • Date: February 8
        • Mr. Darwin Signed
      • Where:
      • Date of the note is November 8.
      • Term of the note is 90 days
      • Maker is Mr. Darwin]
      • Face value o the note is P10,000
      • Interest rate is 10% per year
      • Maturity date is February 8.
      • To compute for the Maturity Value (a) of the promissory note in the above example –
      • 1rst Step. Determine the interest by using the formula I = P x r x t
      • I = P10,000 x 10% x 90
      • 360
      • = 10,000 x .1 x .25
      • = P 250.00
      • 2nd Step. Determine the amount or maturity value by using formula A = P + I
      • A = P10,000 + P250
      • = P10,250
      • Alternatively, the amount or maturity value can be computed as follows using formula A = P(1 + r x t)
      • A = P10,000 (1 +10% x 90/100)
      • = 10,000 (1 + .1 x.25)
      • = 10,000 (1.025)
      • = P10,250.00
      • On the other hand, if no interest is mentioned in the note, the face value is the maturity value. To illustrate, convert above interest-bearing note to a non-interest bearing note as shown below:
        • P 10,250.00 November 8
        • Ninety days after I promise to pay to the order of Mr. Carlos the sum of Ten Thousand Two hundred fifty Pesos and no/100 For value received with interest at 10%.
        • Due Date: February 8 Mr. Darwin Signed
      • The omission of a stated interest rate does not necessarily mean that the original debt bears no interest. The interest, if any, might have been added to the original debt when the face value was determined. As illustrated, the original debt was P 10,000 but the face value indicated in the note was P10, 250.00 which is also the maturity value.
    • Discounting of Notes
      • Discounting a note is based on the maturity value while simple interest is based on the principal. Nonetheless, the basic principles of discounting a note or any unmatured drafts or negotiable instruments at a bank are the same as those for obtaining a loan from a bank from which the interest is deducted in advance. For example, a businessman was granted a one (1) year loan by a bank worth P10,000. If the bank charges 10% interest the amount the businessman will receive is P10,000 less interest in advance worth (I= P x r x t = 10,000 x .1 x 1) P1,000. Therefore, the businessman will receive only the discounted amount of P9,000 but the maturity value after one year is still P10,000.
      • The discount rate (d) is usually expressed in percent or its equivalent decimal and is generally quoted on a yearly basis. Its is the ration of the discount for the period to the maturity value. If the discount rate is not mentioned in any given problem, it is assumed that the interest rate is also the discount rate. The discount ( D) is the amount to be deducted to the maturity value. The proceeds (Pd) is the amount a creditor is willing to pay a note before its maturity or due date.
      • Term of the discount (td), is the number of days from the maturity date to the discount date (date the note was sold), Otherwise specified, the Banker’s Rule will be used in computing the term of the discount. Hence, to determine the discount (D), the formula is
      • D = A x d x td Equation 7.8
      • Such formula is also referred to as the bank discount. The formula to determine the proceeds is
      • Pd = A - D Equation 7.9
      • Combining Equations 7.8 and 7.9 and factoring out A, the resulting equation is
      • Pd = A (1 – d x td ) Equation 7.9
    • Discounting an Interest-Bearing Note
      • Example 8. Mr. Dennis had a note for P8,000 with an interest rate of 6%. The note was dated November 15, 1981, and the maturity date is after 90 days. On November 30, 1981, he took the note to his bank, which discounted it at the rate of 7%. How much did he receive from the bank?
      • Given:
      • P = Php 8,000 face value
      • r = 6% (per year)
      • Date of the note = November 15, 1981
      • Term of the note = 90 days = t
      • Discount date = November 30, 1981
      • Discount rate (d) = 7% (per year)
      • P = to find
      • Solution:
      • 1rst Step . Find the maturity value according to the face value, the rate of interest and the time stipulated on the note. Substituting the values in the formula: A = P (1 + r x t)
      • A = P 8,000 ( 1 + 6% x 90 )
      • 360
      • = 8,000 (1 + .015)
      • = 8,000 (1.015)
      • = P 8, 120.00
      • 2nd Step . Determine the discount by using formula D = A x d x td).
      • D = A x d x td
      • = 8,120 x 7% x 75
      • 360
      • = P 118.42
      • Therefore, the proceeds is
      • Pd = A – D
      • = P 8, 120 – P 118.42
      • = P 8, 001.58
      • Or, using Equation 7.10 after determining the maturity value and the term of the discount.
      • Pd = A (1 – d x td)
      • = 8,120 (1 – 7% x 75/360)
      • = 8,120 (1 - .0145833)
      • = P8,001.58
      • Example 9 . Cruz pays his account with Santos on October 1, 1985 with a P8,000 60-day, 5% note. On October 26, 1985, Santos offered to discount Cruz’ note with the bank at 6%.
      • Given:
      • Date of the note = October 1, 985
      • Face value of note = P 8,000
      • Term of the note = 60 days
      • Interest rate = 5% (per year)
      • Discount date = October 26, 1985
      • Discount rate = 6% (per year)
      • Solution:
      • Analyzing the problem, the note of Mr. Cruz in the possession of Santos has already earned an interest on October 26. On the other hand, the bank will be discounting at 6% the same note whose maturity value will be 35 days from October 26 to November 30, 1985.
      • October 1, 1985 October 26 November 30, 1985__
      • Date of the note Date of the Maturity date discount
      • Discounted for P 8, 000
      • 35 days
      • Maturity value is
      • A = 8,000 ( 1 + 5% x 60 )
      • 360
      • = 8,000 (1 + .008333)
      • = 8,000 (1 .008333)
      • = P 8,066.66
      • Therefore, the proceeds is
      • Pd = A (1 – d x td)
      • = P 8,066.66 ( 1 - 6% x 35)
      • 360
      • = P 8,066.66 ( 1 - .0058333)
      • = P 8,019.60
    • Chapter 8: Compound Interest
      • In loan and saving transactions, interest arrangements may be simple or compounded .
      • In the former the periodic interest is based only on the original principal . In the later, the periodic uncollected interest are added to the principal and their sum is the basis of the interest for the succeeding interest period.
      • The process of accumulating a principal to obtain a compound amount is called compound accumulation. Compound interest is the difference between the original principal and the compound amount (principal plus total interest). The period for computing interest, usually at regularly stated intervals such as annually, semi-annually, quarterly, or monthly, is called the conversion period. The stated interest rate is called nominal rate. Therefore, if the nominal rate of interest is 8% compound semi-annually, the interest rate is expressed as 8% compound semi-annually, the interest rate per conversion period is 8 % or 4%. 2
      • Thus,“6% compounded annually” means that the interest per year is 6% and the interest is paid annually. Therefore, the conversion period is 1 and the interest per conversion periods is 6% also.
      • “ 6% compounded semi-annually” means that interest per year is 6% but the interest is paid every six(6) months or semi-annually. The interest per conversion is 6 % or3%.
      • 2
      • “ 6% compounded quarterly” means that the interest per year is 6% but the interest is paid every 3 months or quarterly. To determine the interest per quarter divide 6% by the conversion factor of 4 (four 3-months for every year), thus, interest per conversion period is 6/4% or 1 1 %.
      • To illustrate, let us assume that a loan of P10,000 for 5 years was taken on January 1, 1982. If the arrangement is for simple interest, for every year, the annual interest is
      • P 1,000.00 ( P1,000 x .10). If the interest is compounded, the annual interest for 1983 is −
      • (P 10,000 x.10) −P1,000; 1984 − (P 10,000 + P1,000) x .10) − P1,000; 1985 − (P12,000 x .10) −P1,210; 1986 − (P13,310 x .10) −P1,321; and 1987− (P14, 461.10) −P1,464.
    • Simple Interest Formula
      • Example 1. Adorada deposited P1,000 with the Pacific Bank and allowed it to remain for 4 years. If the bank paid its depositors 5% on their savings once a year, how much money did she have in the bank at the end of 4 years?
      • Solution: Using simple interest formula.
      • 1rst year interest on P 1,000.00
      • I = P x r x t
      • = P 1,000.00 x 5% x 1
      • = P 1,000.00 x .05 x 1
      • = P 50.00
      • At the end of the first year, Adorada has accumulated P 1,000.00 and P50.00 or P1,050.00
      • 2nd year interest on P1,050.00
      • I = P x r x t
      • = P 1,050.00 x 5% x 1
      • = P 1,050.00 x .05 x 1
      • = P 52.50
      • When this idea is extend, the compound amount at the end of the nth period may be expressed as:
      • A = P (+i)n Equation 8.1
      • Where:
      • A = Compound amount usually in P
      • P = Original value of the principal in P
      • i = Rate of interest per conversion period
      • n = Number of conversion periods
      • Thus, in the above example, if P = P 1,000, r = I = 5% per year, n = 4 years, using the equation
      • A= P ( 1 + 1) n
      • = 1,000 (1 + 5%)4
      • = 1,000 (1.21550625)
      • = A,215.50625 or P 1,215.51
      • The table on the Compound Amount of 1 in the index shows the compound interest accumulation at different rates and periods. The formula using this Table for the computation of the 3rd year interest on P 1,102.50
      • I = P x r x t
      • = P 1,102.50.00 x 5% x1
      • = P 1,102.50.00 x .05 x 1
      • = P 55.13
      • At the end of the 3rd year, Adorada have accumulated P 1,102.50 and P55.13 or P 1,157.63.
      • 4rth year interest on P1,157.63
      • I = P x r x t
      • = P 1,157.63 x 5% x 1
      • = P 1,157.63 x .05 x 1
      • = P 57.88
      • Therefore, at the end of the 4th year, Adorada has accumulated P1,157.63 and P 57.88 or P1,215.51.
      • Summarizing the computations:
      • At the end of Interest Principal
      • 1rst year P 50.00 Php1, 050.00
      • 2nd year 52.50 1,102,50
      • 3rd year 55.13 1,157.63
      • 4rth year 57.881 1,215.51 (final answer)
      • As shown in the preceding example, the computation using simple interest method is quite tedious and time consuming.
      • Compound amount is
      • A = Principal x equivalent in the index
      • The index A provides eight (8) decimal places for each equivalent. It would be a waste of time if we use 8 decimal places in multiplying with a small amount. All interest rates used on Index A are in percent per conversion period. Therefore, if the computed or given of interest is expressed in decimal form, compute its equivalent in percent before using the index. Then, a column indicates the number of conversion periods.
      • Example 3. If the rate of interest is 5% compounded annually and the period is 5 years, what is its equivalent in index A?
      • Solution:
      • Since the rate of interest is in the percent per year and period is 5 years, determine its equivalent in the Index A by using their original values (n=5) and (1 =5%). On the “n” column, locate 5 and move horizontally to the right until reaching “I” column with 5%. Thus the Index equivalent = 1.2728156
      • Example 4 . Using example 3, get the Index equivalent if the rate of interest is 5% compounded semi-annually.
      • Solution: If the rate of interest is 5% compounded semi-annually and the interest per conversion period is every 6 months, interest per conversion period is 5/2% = 2 ½ %. The period given in 5 years which must be converted into 6-month periods. 5 x 2 = 10(6-month period). Using now is = 2 ½% and n = 10. Index equivalent = 1.28008454
      • Example 3. Using example 3, if the rate of interest is 5% compounded quarterly.
      • i = 5%/4 = 1 1 %
      • 4
      • n = 5 x 4 = 20 (interest Period)
      • Index equivalent = 1.28203723
    • Chapter 9: Depreciation
      • The long-lived properties of business are of two types, the non-depreciable and depreciable . Non-depreciable assets have an almost perpetual life. An example is land. Depreciable assets have limited economic (useful) lives. Examples are building, machines, equipment, furniture, and the like.
      • The use of long-lived assets is indispensable in the earning of revenues. If this is so, then the value of this use is a cost of earning the revenue.. as part of the costs of operations, the costs of depreciable assets are allocated over their useful years. This is allocated cost is called depreciation expense or simply depreciation.
      • There are different methods of calculating periodic depreciation but only the Averages and the Reducing Charge Methods will be discussed.
      • Average Methods
        • Straight line Method
        • Service Hour Method
    • Average Methods
      • In the straight line method , the depreciation expense is distributed in equal amounts over the estimated useful life of the asset. The formula is:
      • R = C − S Equation 9.1
      • n
      • Where:
      • R = depreciation charge per year
      • n = number of period in years, months and the like
      • C = original cost of the asset
      • S = estimated scrap value or salvage value
      • C – S = Total depreciable cost
      • Example 1 . A car purchased for P69,000 has an estimated useful life of three (3) years and a scrap value of P9,000. Use the straight line method to find the depreciation expense per year and construct a depreciation schedule.
      • Solution:
      • Given: C = P69,000
      • S = P 9,000
      • n = 3 years
      • Substituting the values in Equation 9.1
      • R = C − S
      • n
      • = P 69,000 – P9,0000
      • 3
      • = 60,000 = P 20,000
      • 3
      • Depreciation Schedule − Straight Line Method
      • End of Year Annual Depreciation Accumulated Book Value of Assets
      • Expense Depreciation
      • 0 − − − − 0 P 90,000
      • 1 P 20,000 P 20,000 49,000
      • 2 20,000 40,000 29,000
      • 3 20,000 60,000 9,000
      • Total Php 60,000
      • The Service hours Method estimates depreciation on the productive capacity of the asset per service hour. It assumes that the number of productive hours decreases as the property becomes older.
      • Example 2. A sewing machine purchased for P4,500 has an estimated life of four years and a scrap value of P 300.00. Use the straight line method equation to find the depreciation. Assume that the useful life of the machine is estimated to be 15, 000 hours and the actual number of hours spent in production each year is as follows:
      • 1rst year: 4,500 service hours
      • 2nd year: 4,100 service hours
      • 3rd year: 3,500 service hours
      • 4rth year: 2,900 service hours
      • Use the service hours method to find the depreciation expenses for each year and construct a depreciation schedule.
      • Given:
      • C = P 4,500
      • S = P 300
      • N = 15,000 service hours
      • Solution:
      • Substituting the value in equation 9.1
      • R = P 4,500 – P 300
      • 15,000
      • = P 0.28 per hour
      • The annual depreciation as computed are:
      • 1rst year: 4,500 x P 0.28 = P P 1,260.00
      • 2nd year: 4,100 x P 0.28 = P 1,148.00
      • 3rd year: 3,500 x P 0.28 = P980.00
      • 4rth year: 2,900 x P 0.28 = P812.00
      • Depreciation Schedule − Straight Line Method
      • End of Year Annual Depreciation Accumulated Book Value of Assets
      • Expense Depreciation
      • 0 − − − − − − − − P 4,500.00
      • 1 P 1,260.00 P1,260.00 3,240.00
      • 2 1,148.00 2,408.00 2,092.0
      • 3 980.00 3,388.00 1,112.00
      • 4 812.00 4,200.00 300.00
      • Total Php 60,000
    • Chapter 10: Income statement
      • Businessmen invest their money in business primarily to make profit. Profit is realized (earned) when revenues exceeds costs and expenses (outgoing property). Say for example, a car bought for P30,000 was sold for cash of P35,000. The revenue of P 35,000 exceeds cost of P 30,000. A profit of P 5,000 is realized.
      • In order to record intelligently the various activities of a business one should understand the important terminologies used for bookkeeping.
      • Sales. The revenue from the selling of goods or services is called sales. A trading concern sells goods that he buys in the same form. Example is a bookstore .
      • Sales Returns and Allowances. Goods sold and returned by the buyer are called sales returns . Reduction in the selling price of goods sold due to low quality or defect is called sales allowance.
      • Net Sales. This is the total selling price of the goods sold less the amount of returns and allowances on such sale.
      • Purchases. Goods bought for sale are called purchases. The cost of goods purchased includes not only the invoice price but also the incidental costs relating to merchandise acquisition, preparation, and placement for sale. Example of incidental costs are transportation charges, taxes etc.
      • Inventory. Goods bought and remaining unsold are called inventory.
      • Cost of goods sold. The cost of the goods bought and sold is called cost of goods sold. This is computed by adding inventory at the beginning plus net purchases less inventory ending.
      • Example 1. At the beginning of the year, the inventory of the Erwin Gas Station was P1,250,125.55. Purchases during the year amounted to P698.254.05. At the year end, the manager verified that the station had P968,056.58 remaining inventory. Determine cost of goods sold.
      • Cost of Goods Sold = P1,250,125.55 + P698, 254.05─ P968,056.58
      • 7. Gross Profit. It is the excess of net sales over the cost of goods sold. It is also called gross margin on sales. The formula to determine the gross profit is as follows.
      • Example 2. An Auto Supply bought spare parts for P26,245. After a month of operation all the spare parts were sold of P27,895.45 net. Determine the gross profit.
      • G.P. = Net Sales – Cost of Goods Sold
      • G.P. = Php27,895.45- Php26,245
      • = Php1,650.45
      • Selling Expenses. Expenses in connection with the selling function of the business are called Selling expenses. Examples are advertising expenses, traveling expenses of Salesmen and the like.
      • Administrative expenses. These are items of expenses incurred in the administrative operations of the business which are not connected with the selling function. Examples are insurance expense, postage dues, property taxes, office supplies and the like.
      • Net Profit. The excess of the gross profit over the total operating expenses (selling and administrative expenses is the net income (net profit) from operations before income taxes.
      • Example 5. The gross profit of Nancy’s Grocery Store last year was P346,985.50. Below is the illustration to determine net profit.
      • Net profit= Gross Profit – Operating expenses
      • Gross Profit ………………………… ……….P346,985.50
      • Less: Operating Expenses
      • Selling Expenses P 56,925.00
      • Light and water 10,850.00
      • Salaries and wages 25,345.70
      • Repair and Maintenance 61,250.46 156,371.16
      • Net Profit Php 192,614.34
      • The life or existence of a business concern usually covers a long period of time involving many years. During this period, the businessman must be guided by some reliable business data in order that he may be informed of the trend of his business. In other words, even with estimates only, he must be of whether he is making a profit or suffering a loss.
      • Hence, the life of a business is divided into periods of equal length at the end of each of which summary reports are submitted to him. Usually, each of these periods cover one year or 12 months. These summary reports are the following:
      • 1. Balance sheet or statement of assets and liabilities - This report or statement lists the assets (properly), liabilities (obligations), and the owner’s equity as of a given date.
      • 2. Income statement or statement of operations – This statement shows the detail of the revenues, costs expenses and losses for the period which is usually a year.
      • This present chapter will deal with the income statement. Below is an illustration of the income statement of a trading concern.
    • TRADING COMPANY INCOME STATEMENT YEAR ENDED DECEMBER 31,19_
      • Sales PXX
      • Less: Sales Returns, discounts, and allowances XX
      • Net Sales PXX
      • Cost of goods sold
      • Merchandise, beginning PXX
      • Add – Net Purchases XX
      • Available for sale PXX
      • Merchandise, ending XX XX
      • Gross Margin on Sales PXX
      • Less: Operating Expenses PXX
      • Selling Expenses PXX
      • Administrative expenses XX
      • Net Operating Profit PXX
      • Less: Income Taxes XX
      • Net Income for the Year PXX
      • =====
    • Thank you…….
    • Jasmin A. Abiva & Marie Carl Pasacsac BOA IV-1