1. Math in BusinessMath in Business
Presents……….Presents……….

2. Contents……Contents……
 Chapter I.Chapter I. Basic operationBasic operation
 Chapter 2. Operation withChapter 2. Operation with
DecimalsDecimals
 Chapter 3. FractionsChapter 3. Fractions
 Chapter 4 Percent in BusinessChapter 4 Percent in Business
 Chapter 5: Employee’s CompensationChapter 5: Employee’s Compensation

3.  Chapter 6: BUYING AND SELLINGChapter 6: BUYING AND SELLING
 Chapter 7:SIMPLE INTERESTChapter 7:SIMPLE INTEREST
 Chapter 8: COMPOUND INTERESTChapter 8: COMPOUND INTEREST
 Chapter 9: DEPRECIATIONChapter 9: DEPRECIATION
 Chapter 10: INCOME STATEMENTChapter 10: INCOME STATEMENT

4. Chapter I.Chapter I.
CheckingChecking
answers ofanswers of
the Basicthe Basic
OperationsOperations

5.  Addition, subtraction, multiplication,Addition, subtraction, multiplication,
and division are the basic operations inand division are the basic operations in
mathematic. Users confronted withmathematic. Users confronted with
different problems in business requiredifferent problems in business require
thorough knowledge of these basicthorough knowledge of these basic
operations. With the invention ofoperations. With the invention of
calculators, a person can solve acalculators, a person can solve a
mathematical problem with greatermathematical problem with greater
speed. Since calculators are merespeed. Since calculators are mere
instrument in performing theinstrument in performing the
operations, accuracy is largelyoperations, accuracy is largely
dependent on mastery of the operationsdependent on mastery of the operations
by the user and on his skills in checkingby the user and on his skills in checking
answersanswers..

6. Basic Math OperationsBasic Math Operations
 AdditionAddition
 Commutative Property of Addition ·Commutative Property of Addition ·
Associative Property of Addition ·Associative Property of Addition ·
Distributive Property · AdditiveDistributive Property · Additive
Identity: ZeroIdentity: Zero
 We'll assume the reader can add digits, so thatWe'll assume the reader can add digits, so that
 2 + 2 = 42 + 2 = 4
 is not a surprise. In addition, we'll also assume that theis not a surprise. In addition, we'll also assume that the
fundamentals of "carrying" are not that big of a problem,fundamentals of "carrying" are not that big of a problem,
and so most readers will immediately know how we didand so most readers will immediately know how we did
 We're more interested here in the general properties ofWe're more interested here in the general properties of
addition that impact algebra. With that in mind, let a, b andaddition that impact algebra. With that in mind, let a, b and
c be three real numbers. Then the following properties ofc be three real numbers. Then the following properties of
addition turn out to be usable and important:addition turn out to be usable and important:

7.  Commutative Property of AdditionCommutative Property of Addition
 (Addition is the same regardless of the order(Addition is the same regardless of the order
one adds the numbers, i.e., forwards addition isone adds the numbers, i.e., forwards addition is
the same as backwards addition).the same as backwards addition).
 ExamplesExamples
 Note that negative numbers are sometimesNote that negative numbers are sometimes
enclosed in parentheses to avoid confusionenclosed in parentheses to avoid confusion
between the sign of the number and the additionbetween the sign of the number and the addition
operation. This is merely a matter of style –operation. This is merely a matter of style –
many textbook writers use spacing to set off themany textbook writers use spacing to set off the
difference instead.difference instead.

8.  Associative Property of AdditionAssociative Property of Addition
 (Addition of a list of numbers is the same regardless(Addition of a list of numbers is the same regardless
of which are added together first, i.e., grouping doesof which are added together first, i.e., grouping does
not matter)not matter)
 ExamplesExamples

9.  MultiplicationMultiplication
 Commutative Property of Multiplication ·Commutative Property of Multiplication ·
Associative Property of Multiplication ·Associative Property of Multiplication ·
Distributive Property · MultiplicativeDistributive Property · Multiplicative
IdentityIdentity
 Again, we’ll assume that the basics ofAgain, we’ll assume that the basics of
multiplication are well known, so that 2x2=2·multiplication are well known, so that 2x2=2·
2=2*2=4. There are obviously several different2=2*2=4. There are obviously several different
notations in use, depending when one learns itnotations in use, depending when one learns it
and what context one learns it. We will use all ofand what context one learns it. We will use all of
these notations, as well as another: when usingthese notations, as well as another: when using
variables, multiplication is assumed whenvariables, multiplication is assumed when
symbols are merely written next to each other.symbols are merely written next to each other.
 Multiplication follows the same two laws justMultiplication follows the same two laws just
described for addition.described for addition.

10.  Commutative Property of MultiplicationCommutative Property of Multiplication
 ExamplesExamples
 We are following the same convention as withWe are following the same convention as with
addition. Note that just writing numbers next toaddition. Note that just writing numbers next to
each other is a poor idea because, for example,each other is a poor idea because, for example,
32 can be confused with 3· 2. Thus, we need32 can be confused with 3· 2. Thus, we need
some sort of symbol to make the two distinct.some sort of symbol to make the two distinct.
There are several correct ways to do this – inThere are several correct ways to do this – in
other articles, we’ll make a lot of use ofother articles, we’ll make a lot of use of
parentheses, so that 3· 2 will be written 3(2) orparentheses, so that 3· 2 will be written 3(2) or
(3)(2). Experience with the notation will help(3)(2). Experience with the notation will help
make it very clear what is meant.make it very clear what is meant.

11. 
Associative Property ofAssociative Property of MultiplicationMultiplication
 ExamplesExamples
 Please note that we do not have to limit ourselvesPlease note that we do not have to limit ourselves
to parentheses; the last computation could haveto parentheses; the last computation could have
been writtenbeen written
 This looks a bit neater. Again, such things are aThis looks a bit neater. Again, such things are a
matter of style, and the reader is encouraged to usematter of style, and the reader is encouraged to use
whatever bracketing makes the most sense andwhatever bracketing makes the most sense and
allows clear, proper ordering of calculations.allows clear, proper ordering of calculations.

12.  Distributive PropertyDistributive Property
 When both addition and multiplication appear in aWhen both addition and multiplication appear in a
single mathematical expression, this distributive lawsingle mathematical expression, this distributive law
controls the operation. This is probably one of thecontrols the operation. This is probably one of the
most important laws in mathematics; getting itmost important laws in mathematics; getting it
wrong guarantees bad calculations! The "reverse" iswrong guarantees bad calculations! The "reverse" is
NOT CORRECT:NOT CORRECT:
 ExamplesExamples

13.  SubtractionSubtraction
 We’re going to define subtraction in terms ofWe’re going to define subtraction in terms of
addition of the negativeaddition of the negative::
 This means that subtraction is a shortcut or anThis means that subtraction is a shortcut or an
abbreviation of the above addition operation. Theabbreviation of the above addition operation. The
"triple equals" sign used here means"triple equals" sign used here means definitiondefinition,,
and is meant to signify anand is meant to signify an operation that is alwaysoperation that is always
truetrue. It is a stronger statement than a simple equals. It is a stronger statement than a simple equals
sign.sign.
 ExamplesExamples
 Subtraction is notSubtraction is not in generalin general commutative:commutative:

14.  ExampleExample
 Nor is itNor is it in generalin general associative:associative:
 ExampleExample

15.  DivisionDivision
 We’re going to define division in terms ofWe’re going to define division in terms of multiplicationmultiplication
of the inverseof the inverse. Please see the discussion of multiplicative. Please see the discussion of multiplicative
inverse below for more info. For now, suppose that a > 1inverse below for more info. For now, suppose that a > 1
and thatand that
 The number b has a very important function – it is calledThe number b has a very important function – it is called
the multiplicative inverse of a, also known as the inversethe multiplicative inverse of a, also known as the inverse
of a. Naively, we can writeof a. Naively, we can write
 but it should be pointed out that this is not really abut it should be pointed out that this is not really a
definition. In order to properly define division, we woulddefinition. In order to properly define division, we would
have to discusshave to discuss rational numbersrational numbers and how they work.and how they work.
Instead of doing that now, we’ll simply take it as given thatInstead of doing that now, we’ll simply take it as given that
the reader intuitively understands fractions. With thisthe reader intuitively understands fractions. With this
definition of inverse, we can tackle division:definition of inverse, we can tackle division:

16.  ExampleExample
 Suppose a = 2. ThenSuppose a = 2. Then
 Note that the decimal expansion 0.5000… is theNote that the decimal expansion 0.5000… is the
result of long division, a subject we are avoidingresult of long division, a subject we are avoiding
here (for a discussion, see the division article). Wehere (for a discussion, see the division article). We
can "prove" that 0.5000… is the inverse of 2 bycan "prove" that 0.5000… is the inverse of 2 by
multiplying:multiplying:
 To the extent that this is rather unsatisfying, weTo the extent that this is rather unsatisfying, we
must ask the reader to suspend disbelief. Themust ask the reader to suspend disbelief. The
mechanics of division are complicated, and deservemechanics of division are complicated, and deserve
a separate article.a separate article.
 Now we’ll use this notion to do a division.Now we’ll use this notion to do a division.

17.  ExampleExample
 Note that we expanded 6 into 2· 3 and canceled theNote that we expanded 6 into 2· 3 and canceled the
3’s. This is a bit sloppy, but again, our definition of3’s. This is a bit sloppy, but again, our definition of
division is intuitive rather than precise.division is intuitive rather than precise.

18. Chapter 2.Chapter 2.
OperationOperation
withwith
DecimalsDecimals

19.  Our money representation is based on a decimalOur money representation is based on a decimal
system. The peso is composed of 100 centavos.system. The peso is composed of 100 centavos.
Since we rarely deal with amounts less than aSince we rarely deal with amounts less than a
centavo, the lowest value we use is P.01. Thus, in anycentavo, the lowest value we use is P.01. Thus, in any
mathematical computation, the final answer shouldmathematical computation, the final answer should
be exact to the nearest centavo. An answer of Pbe exact to the nearest centavo. An answer of P
135.256 should be written as P 135.26. Usually, it135.256 should be written as P 135.26. Usually, it
serves no purpose to express an answer id pesoserves no purpose to express an answer id peso
value with fractional part of a centavo. For discussionvalue with fractional part of a centavo. For discussion
purpose, the money value P 135.26 is read as “onepurpose, the money value P 135.26 is read as “one
hundred thirty five pesos and twenty six centavos.”hundred thirty five pesos and twenty six centavos.”
Note that the decimal point is read as “and” toNote that the decimal point is read as “and” to
separate peso value from the centavo. Sometimes, itseparate peso value from the centavo. Sometimes, it
is necessary to indicate fractional parts of a centavois necessary to indicate fractional parts of a centavo
in such cases as statistical figures, indices, realin such cases as statistical figures, indices, real
estate tax, and surveys.estate tax, and surveys.
 In performing addition and subtraction, the numbersIn performing addition and subtraction, the numbers
containing decimal components are arrangedcontaining decimal components are arranged
according to their place value. In multiplication andaccording to their place value. In multiplication and
division, the numbers with decimal components aredivision, the numbers with decimal components are
treated as whole numbers. Care must be taken intreated as whole numbers. Care must be taken in
placing the decimal points. The basic operations willplacing the decimal points. The basic operations will
be discussed in detail in the following sections.be discussed in detail in the following sections.

20. Addition with DecimalsAddition with Decimals
 Addends with decimal components should be arrangedAddends with decimal components should be arranged
according to their proper order (column). The decimalaccording to their proper order (column). The decimal
points should fall in one vertical column in order to alignpoints should fall in one vertical column in order to align
the figures according to their values. Hence, units of equalthe figures according to their values. Hence, units of equal
value will fall under the same column such as tens,value will fall under the same column such as tens,
hundreds, thousands, millions, etc.hundreds, thousands, millions, etc.
 Example 1.Example 1. Add 3.085; 12.314 and 94.65Add 3.085; 12.314 and 94.65
 3. 0853. 085
 12. 31412. 314
 94. 6594. 65
 110. 0492110. 0492
 Same columnSame column

21. Subtraction with DecimalsSubtraction with Decimals
 The rule used in addition applies equally toThe rule used in addition applies equally to
subtraction. Decimal points of the minuend andsubtraction. Decimal points of the minuend and
subtrahend are placed in the same vertical column.subtrahend are placed in the same vertical column.
 Example 2.Example 2.
 Subtract 125.2684 from 436.1052Subtract 125.2684 from 436.1052
436. 1052436. 1052
 -- 125. 2684125. 2684
 310. 8368310. 8368
 Same columnSame column

22. Multiplication with DecimalsMultiplication with Decimals
 The number of the decimal places of the product isThe number of the decimal places of the product is
equal to the total number of decimal places (total digitequal to the total number of decimal places (total digit
numbers to the right of the decimal points) of thenumbers to the right of the decimal points) of the
factors.factors.
 Example 3.Example 3.
 Determine the product of factors 3.248 and 1.26Determine the product of factors 3.248 and 1.26
Note: For convenience, use the shorted factor ( lesser digitNote: For convenience, use the shorted factor ( lesser digit
numbers) as multiplier (1.26)numbers) as multiplier (1.26)
 3.248 multiplicand3.248 multiplicand (3 decimal places)(3 decimal places)
 1.26 multiplier1.26 multiplier (2 decimal places)(2 decimal places)
 1948819488
 6497664976
 32483248
 4.09248 product4.09248 product (3+2 = 5 decimal places)(3+2 = 5 decimal places)

23. Division with decimalsDivision with decimals
 When the dividend contains decimal fractions and theWhen the dividend contains decimal fractions and the
divisor is a whole number, the decimal point of thedivisor is a whole number, the decimal point of the
quotient is aligned vertically above (same vertical column)quotient is aligned vertically above (same vertical column)
the decimal point of the dividend.the decimal point of the dividend.
 Example 4. Divide 125.265 by 25Example 4. Divide 125.265 by 25
 Same columnSame column
 5. 0106 quotient5. 0106 quotient
 Divisor 25/ 125. 265Divisor 25/ 125. 265
 125125
 22
 00
 2626
 2525
 1515
 00
 150150
 150150
 0 remainder0 remainder

24. Rounding off NumbersRounding off Numbers
 Generally, numbers are rounded offGenerally, numbers are rounded off
in making estimates or forecasts. Inin making estimates or forecasts. In
monetary computations, it is themonetary computations, it is the
practice to round off the final answerpractice to round off the final answer
to the nearest centavo because theto the nearest centavo because the
third decimal place (thousandths) isthird decimal place (thousandths) is
often immaterial.often immaterial.

25. Chapter 3.Chapter 3.
FractionFraction

26. 1/41/4
 While our exposure to the numeration system dealtWhile our exposure to the numeration system dealt
mostly with whole numbe5rs, we also come acrossmostly with whole numbe5rs, we also come across
figures wherein the use of fractions is inescapable.figures wherein the use of fractions is inescapable.
Such terms as “halves,” “fourths” or “quarters” areSuch terms as “halves,” “fourths” or “quarters” are
used with reference to measurement like “quarter toused with reference to measurement like “quarter to
twelve,” “half a mile” “three fourths full” and so on.twelve,” “half a mile” “three fourths full” and so on.
Like negative numbers, fractions `and computation.Like negative numbers, fractions `and computation.
If a unit is divided equally into two or more parts,If a unit is divided equally into two or more parts,
each equal part is the fractional part of the uniteach equal part is the fractional part of the unit
which is usually expresses in a fraction. Forwhich is usually expresses in a fraction. For
example, a parent bought a whole pizza pie andexample, a parent bought a whole pizza pie and
would want to share the pie equally among four (4)would want to share the pie equally among four (4)
children. The parent would cut the pie into fourchildren. The parent would cut the pie into four
equal parts and one part is given to each which isequal parts and one part is given to each which is
one-fourth of the whole pie. As illustrated belowone-fourth of the whole pie. As illustrated below

 1/41/4

 1/4 1/41/4 1/4

27.  ¼ means¼ means oneone part of the whole which was divided into 4part of the whole which was divided into 4
equal parts.equal parts.
 The number written above the line (numeratorThe number written above the line (numerator
indicates the number of parts and the number writtenindicates the number of parts and the number written
below the line (denominator) indicates the total units orbelow the line (denominator) indicates the total units or
parts the cake was divided into. A fraction may be used toparts the cake was divided into. A fraction may be used to
represent a ratio.represent a ratio.
 Fractions are classified into:Fractions are classified into:
 proper fractionsproper fractions, ½, ¾, 5/6, 7/8, etc., are fractions, ½, ¾, 5/6, 7/8, etc., are fractions
 whose numerators are less than the denominators. Thesewhose numerators are less than the denominators. These
fractions indicate values less than one (1).fractions indicate values less than one (1).
 Improper fractionsImproper fractions. 3/2, 4/3, 5/4, 7/6, etc., are fractions. 3/2, 4/3, 5/4, 7/6, etc., are fractions
whose numerators are greater than or equal to thewhose numerators are greater than or equal to the
denominator. These fractions indicate values equal to ordenominator. These fractions indicate values equal to or
greater than (1).greater than (1).
 Mixed numbersMixed numbers. 1 ½ 2 ¼ 3 1/3 4 ¼ etc., are fractions. 1 ½ 2 ¼ 3 1/3 4 ¼ etc., are fractions
written as the sum of integer or whole number and awritten as the sum of integer or whole number and a
proper fraction. An improper fraction can also beproper fraction. An improper fraction can also be
expressed as a mixed number or vice versa.expressed as a mixed number or vice versa.

28. Fundamental Rules in Dealing withFundamental Rules in Dealing with
FractionsFractions
 One should understand the fundamental rules inOne should understand the fundamental rules in
dealing with fractions. These rules are very importantdealing with fractions. These rules are very important
in the basic operations of addition, subtraction,in the basic operations of addition, subtraction,
multiplication and division of fractions. The rules aremultiplication and division of fractions. The rules are
the following:the following:
 The value of a fraction is not changed when theThe value of a fraction is not changed when the
numerator and the denominator are both multipliednumerator and the denominator are both multiplied
by the same number other than zero.by the same number other than zero.
 Example 1. 1/3 multiplied both by 3Example 1. 1/3 multiplied both by 3
 1/3 x 3/3 =1/3 x 3/3 = 1 x 31 x 3
 3 x 1 = 3/33 x 1 = 3/3
 Note: 3/3 is the same as 1.Note: 3/3 is the same as 1.
 2. The value of a fraction is not changed if we2. The value of a fraction is not changed if we
divide both the numerator and the denominator bydivide both the numerator and the denominator by
the same number other than zero. The result is likethe same number other than zero. The result is like
dividing the fraction by 1.dividing the fraction by 1.

29. Changing Improper Fraction to aChanging Improper Fraction to a
Whole or Mixed NumberWhole or Mixed Number
 Mathematically, the line between the numerator andMathematically, the line between the numerator and
denominator indicates division. In solving problems, it isdenominator indicates division. In solving problems, it is
more convenient or easier to work with an impropermore convenient or easier to work with an improper
fraction than mixed number.fraction than mixed number.
 To change an improper fraction to a whole or mixedTo change an improper fraction to a whole or mixed
number, divide the numerator by the denominator.number, divide the numerator by the denominator.
Examples are:Examples are:
 Example 3:Example 3:
 Change the improper faction 2/2 to a whole or mixedChange the improper faction 2/2 to a whole or mixed
number.number.
 Solution:Solution:
 2/2 = 2 divided by 2 = 12/2 = 2 divided by 2 = 1

30. Changing Mixed Numbers toChanging Mixed Numbers to
Improper FractionsImproper Fractions This conversion is very convenient when we perform mostThis conversion is very convenient when we perform most
of the basic operations with fractions. Below are the waysof the basic operations with fractions. Below are the ways
in converting a mixed number to an improper fraction:in converting a mixed number to an improper fraction:
 Multiply the whole number and the denominator;Multiply the whole number and the denominator;
 Add the numerator to the product ; andAdd the numerator to the product ; and
 Make the result of (2) as the numerator of the new fractionMake the result of (2) as the numerator of the new fraction
and use the same denominator as that of the originaland use the same denominator as that of the original
mixed number.mixed number.
 Example 5Example 5. Change 4. Change 4 22 to improper fractionto improper fraction
33
 Solution:Solution: follow the steps given abovefollow the steps given above
 4 x 3 (whole number x denominator) = 124 x 3 (whole number x denominator) = 12
 12 + 2 (add numerator to product) = 1412 + 2 (add numerator to product) = 14
 14/3 (improper fraction with same denominator as the14/3 (improper fraction with same denominator as the
original mixed number).original mixed number).

31. Reduction of FractionsReduction of Fractions
 The process of converting fractions to otherThe process of converting fractions to other
equivalent forms of either higher or lower termsequivalent forms of either higher or lower terms
without changing the value of the fraction is calledwithout changing the value of the fraction is called
reduction. As in division, multiplying or dividing bothreduction. As in division, multiplying or dividing both
the numerator and denominator by the samethe numerator and denominator by the same
number other than zero does not affect the quotient.number other than zero does not affect the quotient.
 Examples are:Examples are:
 Reduce ¾ to higher terms.Reduce ¾ to higher terms.
 33 == 3 x 33 x 3 == 99; or; or 33 == 3 x 53 x 5 == 1515
 4 4 x 3 124 4 x 3 12 4 4 x 5 204 4 x 5 20

 Reduce 25/625 to lower terms.Reduce 25/625 to lower terms.
 2525 == 2525 ÷÷ 55 == 55 oror 11
625625 625 ÷ 5 125 25625 ÷ 5 125 25

32.  When the numerator and the denominator have noWhen the numerator and the denominator have no
more common divisor except 1 the fraction hasmore common divisor except 1 the fraction has
been reduced to its lowest term. To reduce abeen reduced to its lowest term. To reduce a
fraction to its lowest term, both the numerator andfraction to its lowest term, both the numerator and
the denominator should be divided by their greatestthe denominator should be divided by their greatest
possible common divisor.possible common divisor.
 For example,For example,
 in the proper fractionin the proper fraction 4646 23 is the greatest possible23 is the greatest possible
161161
 common divisor of 46 and 161, hence,common divisor of 46 and 161, hence,
 4646 == 4646 ÷÷ 2323 == 22
 161 161161 161 23 723 7

33. Finding the greatest CommonFinding the greatest Common
Divisor (g.c.d)Divisor (g.c.d)
 The greatest common divisor is calledThe greatest common divisor is called highesthighest
commoncommon factorfactor. If the greatest common divisor is not. If the greatest common divisor is not
readily apparent, the following steps arereadily apparent, the following steps are
recommended to determine the desirable divisor:recommended to determine the desirable divisor:
 Divide the larger number of the numerator andDivide the larger number of the numerator and
denominator by the smaller number.denominator by the smaller number.
 If there is a remainder in step 1, divide the smallerIf there is a remainder in step 1, divide the smaller
number by the remainder.number by the remainder.
 If there is still a remainder in step 2, divide theIf there is still a remainder in step 2, divide the
remainder in step 1 by the remainder in step 2.remainder in step 1 by the remainder in step 2.
 Continue dividing each remainder by its succeedingContinue dividing each remainder by its succeeding
remainder until the remainder is 0. The last divisor isremainder until the remainder is 0. The last divisor is
the greater common divisor.the greater common divisor.

34.  Example: Find the g.c.d. ofExample: Find the g.c.d. of 6969
 184184
 22
 69/184 step 1.69/184 step 1. 69 is less than 18469 is less than 184
 138138
 4646 11
 46 /69 step 2.46 /69 step 2.
 4646
 2323
 22
 23/46 step 3 and 423/46 step 3 and 4
 4646
 0 remainder0 remainder

35. Changing Fractions to DecimalChanging Fractions to Decimal
 There are two ways of expressing parts of a whole,There are two ways of expressing parts of a whole,
namely, common fractions and decimal fractions.namely, common fractions and decimal fractions.
Most of the problems in business require convertingMost of the problems in business require converting
fractions into decimals. For example, if afractions into decimals. For example, if a
mathematical operation with fractions is performedmathematical operation with fractions is performed
on a pocket calculator, all fractions must beon a pocket calculator, all fractions must be
converted to decimals before entering them in theconverted to decimals before entering them in the
calculator.calculator.
 Fraction indicates division, which means thatFraction indicates division, which means that
the numerator is to be divided by the denominator.the numerator is to be divided by the denominator.
An example is ¼ which is same as 1 divided by 4.An example is ¼ which is same as 1 divided by 4.
 Example:Example:
 0.40.4
 2 = 2 ÷ 5 = 5/ 2.02 = 2 ÷ 5 = 5/ 2.0
 2.02.0
 00
 The answer is 0.4The answer is 0.4

36. Changing Decimal to FractionsChanging Decimal to Fractions
 When a decimal fraction is written in a commonWhen a decimal fraction is written in a common
fractions, the decimal fraction without the decimalfractions, the decimal fraction without the decimal
point is used as the numerator, and thepoint is used as the numerator, and the
denominator is 1 with as many zeros annexed asdenominator is 1 with as many zeros annexed as
there are decimal places in the original decimalthere are decimal places in the original decimal
fraction. Then the common fraction is reduced to itsfraction. Then the common fraction is reduced to its
lowest term.lowest term.
 Example:Example: express .5 to fractionexpress .5 to fraction
 .5 =.5 = 55 == 11
 10 210 2 lowest termlowest term

37. Addition of fractionsAddition of fractions
 Two different things cannot be added. The same is true withTwo different things cannot be added. The same is true with
numbers and fractions.numbers and fractions.
 To add fractions, all denominators must be converted to aTo add fractions, all denominators must be converted to a
common number without changing the values of thecommon number without changing the values of the
fraction.fraction.
 When two or more fractions with different denominators areWhen two or more fractions with different denominators are
to be added, we must change the fractions to theirto be added, we must change the fractions to their
equivalent fractions with the same denominator. Theequivalent fractions with the same denominator. The
procedure involves the convertion of the fractions to itsprocedure involves the convertion of the fractions to its
higher or lower terms as discussed in the preceding topics.higher or lower terms as discussed in the preceding topics.
 Example:Example: 11 && 11
2 42 4
 11 == 1 x 21 x 2 == 22
 2 2 x 2 42 2 x 2 4
 After converting 1/2 to ¼, lets add this to ¼ w/c has sameAfter converting 1/2 to ¼, lets add this to ¼ w/c has same
denominatordenominator
 22 ++ 11 == 33
 4 4 44 4 4


38.  Solution:Solution:
 1st step – Determine the1st step – Determine the “lowest possible”“lowest possible” numbernumber
that can be divided by the different denominatorsthat can be divided by the different denominators
(this number is called least common denominator). In(this number is called least common denominator). In
the example, the least common denominator is 12.the example, the least common denominator is 12.
 2nd step – Determine the equivalent fractions of each2nd step – Determine the equivalent fractions of each
fraction with the same denominator as 12.fraction with the same denominator as 12.
 11 == 1 x 61 x 6 == 66
 22 2 x 62 x 6 1212
 3rd step – Add the equivalent fractions with the same3rd step – Add the equivalent fractions with the same
denominators.denominators.
 11 ++ 11 ++ 11 == 66 ++ 44 ++ 33
 2 32 3 4 12 12 124 12 12 12
 == 6 + 4 + 36 + 4 + 3
 1212
 = 13/12 or 1 ½= 13/12 or 1 ½

39. Addition of Mixed NumbersAddition of Mixed Numbers
 When mixed numbers are added,When mixed numbers are added,
it is unnecessary to convert theit is unnecessary to convert the
numbers to their equivalentnumbers to their equivalent
improper fractions. Add the wholeimproper fractions. Add the whole
integers and the fractionsintegers and the fractions
separately. The fractional part ofseparately. The fractional part of
the answer should reduce to itsthe answer should reduce to its
lowest term.lowest term.

40. Subtraction of FractionsSubtraction of Fractions
 Fractions subtracted must firstFractions subtracted must first
be converted to theirbe converted to their
equivalents with theequivalents with the leastleast
commoncommon denominatordenominator (l.c.d.).(l.c.d.).
then the numerators of thethen the numerators of the
converted fractions areconverted fractions are
subtracted. The answer willsubtracted. The answer will
have the numerator differencehave the numerator difference
as the numerator and the leastas the numerator and the least
common denominator as thecommon denominator as the
denominator.denominator.

41. Subtracting a Mixed Number fromSubtracting a Mixed Number from
a Whole Numbera Whole Number
 By way of an example, theBy way of an example, the
following are the steps infollowing are the steps in
subtracting a mixed from asubtracting a mixed from a
whole number:whole number:

42.  Step 1. Convert one unit of the minuend into anStep 1. Convert one unit of the minuend into an
improper fraction with the same denominator as theimproper fraction with the same denominator as the
fraction in the subtrahend. Reduce the wholefraction in the subtrahend. Reduce the whole
number by one so as not to change the value of thenumber by one so as not to change the value of the
minuend.minuend.
 12 = 11 8/812 = 11 8/8
 Step 2. SubtractStep 2. Subtract
 11 8/8 – 5 1/8 = (11-5)+(8/8 – 1/8)11 8/8 – 5 1/8 = (11-5)+(8/8 – 1/8)
= 6 + 7/8= 6 + 7/8
= 6 7/8= 6 7/8

43. Subtracting Mixed NumbersSubtracting Mixed Numbers
 When necessary, convert the fractional parts ofWhen necessary, convert the fractional parts of
both the minuend and the subtrahend so that theyboth the minuend and the subtrahend so that they
have a common denominator. If the fraction in thehave a common denominator. If the fraction in the
subtrahend is smaller than the fraction in thesubtrahend is smaller than the fraction in the
minuend, you can subtract, but if the fraction isminuend, you can subtract, but if the fraction is
greater than the fraction in the minuend, follow thegreater than the fraction in the minuend, follow the
steps below.steps below.
 Convert one unit of the minuend into an improperConvert one unit of the minuend into an improper
fraction with “correct’ denominator, and add this unitfraction with “correct’ denominator, and add this unit
to the existing fraction in the minuend.to the existing fraction in the minuend.
 Reduce the whole number in the minuend by oneReduce the whole number in the minuend by one
(the unit which is now a fraction).(the unit which is now a fraction).
 You can now subtract the fraction.You can now subtract the fraction.

44. Multiplication of FractionsMultiplication of Fractions
 There is no need to determine the commonThere is no need to determine the common
denominator of the fractions in multiplication.denominator of the fractions in multiplication.
The basic principle in multiplication of fractionThe basic principle in multiplication of fraction
is to determine the product of both numeratorsis to determine the product of both numerators
and both denominators and the productsand both denominators and the products
become the numerator and denominator of thebecome the numerator and denominator of the
answer.answer.
 In multiplying a whole number by a mixedIn multiplying a whole number by a mixed
number, change the mixed number to annumber, change the mixed number to an
improper fraction and determine the product.improper fraction and determine the product.
 If both factors are mixed numbers, convertIf both factors are mixed numbers, convert
each mixed number to an improper fractioneach mixed number to an improper fraction
before multiplying.before multiplying.

45.  Example: Multiply 1/2 by 2/3Example: Multiply 1/2 by 2/3
1/2 x 2/3 =1/2 x 2/3 = 1 x 21 x 2 == 22 oror 11 reduce to lowest termreduce to lowest term
2 x 3 = 6 32 x 3 = 6 3

46. Divisions of FractionsDivisions of Fractions
 The simplest way to divide fractions is toThe simplest way to divide fractions is to
multiply the dividend by the reciprocalmultiply the dividend by the reciprocal
(inverted form) of the divisor. The(inverted form) of the divisor. The
numbers 3 and 1/3 are called reciprocalnumbers 3 and 1/3 are called reciprocal
to each other. Likewise, 4/5 is theto each other. Likewise, 4/5 is the
reciprocal of 5/4.reciprocal of 5/4.
 When dividing a mixed number byWhen dividing a mixed number by
another mixed number, change eachanother mixed number, change each
mixed number to an improper fractionmixed number to an improper fraction
and then multiply after inverting theand then multiply after inverting the
divisor.divisor.

47.  Example:Example:
 Divide: 5/8 by 2/3Divide: 5/8 by 2/3
 55 ÷÷ 22 == 55 xx 33
 88 3 8 23 8 2
== 55 xx 33 == 1515
8 2 168 2 16

48. Chapter 4Chapter 4
Percent inPercent in
BusinessBusiness

49.  The use of percent is gainingThe use of percent is gaining
wide acceptance in all sphereswide acceptance in all spheres
of activity. When relating theof activity. When relating the
parts to a whole, relationship isparts to a whole, relationship is
expressed in percent.expressed in percent.
 Percent is also frequentlyPercent is also frequently
used in presenting accountingused in presenting accounting
and statistical data relationship.and statistical data relationship.

50. Percent to FractionPercent to Fraction
 Percent numbers are not really “new” sincePercent numbers are not really “new” since
they are in reality fractions.they are in reality fractions.
 To express 80% as a fraction in its lowestTo express 80% as a fraction in its lowest
term, convert 80% in its fractional form firstterm, convert 80% in its fractional form first
and perform the operation of division ofand perform the operation of division of
proper fractions as discussed in Chapterproper fractions as discussed in Chapter
III.III.
 80% =80% = 8080
 100100

 Then =Then = 80 ÷ 2080 ÷ 20
 100÷20100÷20
 == 44
 55
 thus, 80% is 4/5 in fractionthus, 80% is 4/5 in fraction

51. Percent to DecimalPercent to Decimal
 In any mathematical computation, percent isIn any mathematical computation, percent is
converted to decimal before multiplying of dividing itconverted to decimal before multiplying of dividing it
with other quantifying numbers. There are severalwith other quantifying numbers. There are several
percent values encountered in businesspercent values encountered in business
transactions that do not convert into fractionstransactions that do not convert into fractions
making such percent values as decimals rather thanmaking such percent values as decimals rather than
as fractions.as fractions.
 The basic principle in converting percent toThe basic principle in converting percent to
decimal is to move the decimal point two places todecimal is to move the decimal point two places to
the left and drop the percent sign.the left and drop the percent sign.
 Example 4. Change 15 % to a decimalExample 4. Change 15 % to a decimal
 Solution:Solution:
 15%15% == 1515 = .15= .15
 100100

52. Representing Decimals as PercentsRepresenting Decimals as Percents
 The fastest way to convert decimals to percent is toThe fastest way to convert decimals to percent is to
the decimal point two places to the right. Thisthe decimal point two places to the right. This
method is actually the reverse order of convertingmethod is actually the reverse order of converting
percent to its decimal equivalent as discussed in thepercent to its decimal equivalent as discussed in the
preceding sections.preceding sections.
 Example:Example:
 0.25 =0.25 = 0.25 x 100 = 25 %0.25 x 100 = 25 %

53. Representing Fractions asRepresenting Fractions as
PercentsPercents
 In converting a fraction to percent, changeIn converting a fraction to percent, change
the fraction to its decimal equivalent asthe fraction to its decimal equivalent as
discussed in the previous chapter. Thendiscussed in the previous chapter. Then
multiply the decimal equivalent by 100 ormultiply the decimal equivalent by 100 or
simply move the decimal point two placessimply move the decimal point two places
to the right and suffix the percent symbol.to the right and suffix the percent symbol.
 Example: Convert 1 7/8 to its equivalent percentExample: Convert 1 7/8 to its equivalent percent
form,form,
 Solution:Solution:
 11 77 == (8 x 1) + 7(8 x 1) + 7 == 1515
 8 8 88 8 8
 = 1.875 x 100= 1.875 x 100
 == 187.5 %187.5 %

54. Fraction-Decimal-PercentFraction-Decimal-Percent
EquivalentsEquivalents
 The equivalents discussed in the previous chaptersThe equivalents discussed in the previous chapters
can be tabulated into what are known as aliquotcan be tabulated into what are known as aliquot
parts of 100. An aliquot part is a portion of a numberparts of 100. An aliquot part is a portion of a number
by which the number may be divided leaving noby which the number may be divided leaving no
remainderremainder
 Example:Example:
 Multiply 60 by 33Multiply 60 by 33 11 %%
 33

 60 by 3360 by 33 11 %% = 60 x= 60 x
 33
= 20= 20

55. Determining the PercentageDetermining the Percentage
 In the statement, 40 is 40/700 0f 700 theIn the statement, 40 is 40/700 0f 700 the
denominator of the fraction which is thedenominator of the fraction which is the
basis of comparison is called the base. Thebasis of comparison is called the base. The
rate is the percent indicating the number orrate is the percent indicating the number or
quantity for every 100. The percentagequantity for every 100. The percentage
refers to the number of items in the desiredrefers to the number of items in the desired
situation or condition.situation or condition.
 Based on the above concept, the principleBased on the above concept, the principle
is written.is written.
 P= b x r Equation 4.1P= b x r Equation 4.1
 When:When: P = percentageP = percentage
 B = baseB = base
 r = rate, usually expresses in %r = rate, usually expresses in %

56.  Stated as a principle, percentage is the product ofStated as a principle, percentage is the product of
the base times the rate. In any mathematicalthe base times the rate. In any mathematical
computation, before we add, subtract, multiply orcomputation, before we add, subtract, multiply or
divide, we must convert the percent to its decimal ordivide, we must convert the percent to its decimal or
fractional equivalent whichever is convenient to use.fractional equivalent whichever is convenient to use.
 Example: An employee who earned P 1,500 spentExample: An employee who earned P 1,500 spent
30 % of his earnings to buy a wrist watch. How30 % of his earnings to buy a wrist watch. How
much is the wrist watch?much is the wrist watch?
 Given:Given: B = 1,500 (base)B = 1,500 (base)
 r = 30 %r = 30 %
 Solution: P = B x rSolution: P = B x r

 = 1,500 x 30 %= 1,500 x 30 % = 1,500 x .3= 1,500 x .3
 = 450.00= 450.00

57. Finding the RateFinding the Rate
 One of the most commonly used representations ofOne of the most commonly used representations of
a number in business is the rate. Althougha number in business is the rate. Although
representation of number based on rate involvesrepresentation of number based on rate involves
different interpretations in business or surveys.different interpretations in business or surveys.
 Based on the preceding discussion, a formulaBased on the preceding discussion, a formula
can be derived from Equation 4.1 as follows:can be derived from Equation 4.1 as follows:
 P = B x r Equation 4.1P = B x r Equation 4.1
 r = P ÷ Br = P ÷ B
 r= P/b Equation 4.2r= P/b Equation 4.2

58. Finding the BaseFinding the Base
 Basically, the base quantity can be determined byBasically, the base quantity can be determined by
applying the same concept in Equation 4.2. theapplying the same concept in Equation 4.2. the
formula is written below.formula is written below.
 B = P ÷ r = P/r Equation 4.3B = P ÷ r = P/r Equation 4.3

59. Ratio and ProportionRatio and Proportion
 As mentioned earlier, percent is another way ofAs mentioned earlier, percent is another way of
comparing a number with the whole, while itscomparing a number with the whole, while its
equivalent in decimal is comparing with one unit.equivalent in decimal is comparing with one unit.
 RatioRatio is a relation between two numbers expressedis a relation between two numbers expressed
in terms of a quotient.in terms of a quotient.
 The result of reducing ratios to lowest terms isThe result of reducing ratios to lowest terms is
called equivalent reduced ratio.called equivalent reduced ratio.
 The equality of two ratios is calledThe equality of two ratios is called proportionproportion..

60. Chapter 5:Chapter 5:
Employee’sEmployee’s
CompensatioCompensatio
nn

61.  Among other expenses, a business enterprise paysAmong other expenses, a business enterprise pays
the salaries to its personnel. Salary is thethe salaries to its personnel. Salary is the
renumeration received by an employees forrenumeration received by an employees for
services rendered. The labor code of the Philippinesservices rendered. The labor code of the Philippines
defines wages or salaries as “ the renumeration ordefines wages or salaries as “ the renumeration or
earning, however designated, capable of beingearning, however designated, capable of being
expressed in terms of money, whether fixed orexpressed in terms of money, whether fixed or
ascertained on a time, task, piece or commissionsascertained on a time, task, piece or commissions
basis or other method of calculating the same whichbasis or other method of calculating the same which
is payable by an employer to an employee andis payable by an employer to an employee and
includes the fair and reasonable value of board,includes the fair and reasonable value of board,
lodging or other facilities customarily furnished bylodging or other facilities customarily furnished by
the employer to the employee.”the employer to the employee.”
 SalariesSalaries may be based on period of time,may be based on period of time,
production, or percentage on an agreed basis.production, or percentage on an agreed basis.
Employees may be classified according to theEmployees may be classified according to the
bases of their salary payments. There are the fixedbases of their salary payments. There are the fixed
wags earners, the piece-workers, the commissionwags earners, the piece-workers, the commission
earners and the salary-plus-commission earners.earners and the salary-plus-commission earners.

62. Computing Wages Based on TimeComputing Wages Based on Time
(period)(period)
 Fixed wage earnersFixed wage earners are paid based on the numberare paid based on the number
of hours worked. For work done beyond the regularof hours worked. For work done beyond the regular
eight (8) working hours a day, the employee iseight (8) working hours a day, the employee is
given additional pay calledgiven additional pay called overtime pay.overtime pay.
 The Labor Code fixes the minimum rates ofThe Labor Code fixes the minimum rates of
overtime pay, night shift differential, and work onovertime pay, night shift differential, and work on
regular and special holidays. The rates range fromregular and special holidays. The rates range from
25 %, 30% to 100% over and above the regular pay25 %, 30% to 100% over and above the regular pay
rate. For purposes of discussion, overtime pay asrate. For purposes of discussion, overtime pay as
herein used will be regular pay plus 50% of said payherein used will be regular pay plus 50% of said pay
which means that for every one hour overtime, thewhich means that for every one hour overtime, the
pay will be equivalent to 1.5 of the regular hourlypay will be equivalent to 1.5 of the regular hourly
rate.rate.

63. Computing Wages on PieceworkComputing Wages on Piecework
BasisBasis
 The minimum rates set by the Labor CodeThe minimum rates set by the Labor Code
do not apply to employees whose salariesdo not apply to employees whose salaries
are not based on the number of workingare not based on the number of working
hours. Employees may be paid by resultshours. Employees may be paid by results
likelike “pakyaw” (pakiao),“pakyaw” (pakiao), “takay”“takay” and byand by
the pieces of product.the pieces of product.
 Workers in selected industries such asWorkers in selected industries such as
those engaged in manufacturing, packing,those engaged in manufacturing, packing,
and handicrafts are paid onand handicrafts are paid on straightstraight
pieceworkpiecework basis. The wage of a straightbasis. The wage of a straight
pieceworker depends solely on number ofpieceworker depends solely on number of
units completed each period of time.units completed each period of time.

64. Computing Wages on CommissionComputing Wages on Commission
BasisBasis
 Most of the commission wage earners are staff inMost of the commission wage earners are staff in
the sales of a business. They are paid on the basisthe sales of a business. They are paid on the basis
of number of units sold. Since sales is the bloodlineof number of units sold. Since sales is the bloodline
of the company, the sales people are doublyof the company, the sales people are doubly
motivated to increase sales output. The three mostmotivated to increase sales output. The three most
commonly used compensation schemes in salescommonly used compensation schemes in sales
are the straight commission, commission andare the straight commission, commission and
bonus, and salary plus commission.bonus, and salary plus commission.
 The regular sales force of a company isThe regular sales force of a company is
usually paid a fixed monthly rate plus commissionusually paid a fixed monthly rate plus commission
earned from sales.earned from sales.
 To compute for the commission, multiply theTo compute for the commission, multiply the
sales (in number of units or peso value) by thesales (in number of units or peso value) by the
commission rate.commission rate.

65. Incentive or BonusesIncentive or Bonuses
 Incentive pay and bonuses in money or inIncentive pay and bonuses in money or in
kind are given to employees to motivatekind are given to employees to motivate
them. The incentive pay may be based onthem. The incentive pay may be based on
the number of units completed or sold inthe number of units completed or sold in
excess of a required minimum.excess of a required minimum.

66. DeductionsDeductions
 The earnings of an employee are subject toThe earnings of an employee are subject to
deductions which are required by law of bydeductions which are required by law of by
agreement between the employer and theagreement between the employer and the
employee. Deductions required by law areemployee. Deductions required by law are
income taxes, social security taxes,income taxes, social security taxes,
medicare contributions, contributions tomedicare contributions, contributions to
pension plans and contributions to pag-ibig.pension plans and contributions to pag-ibig.

67. Chapter 6:Chapter 6:
Buying andBuying and
SellingSelling

68.  Buying and selling are the essentialBuying and selling are the essential
functions of a trading concern. Manufacturersfunctions of a trading concern. Manufacturers
or producers buy raw materials or parts toor producers buy raw materials or parts to
produce finished products. A trading concernproduce finished products. A trading concern
tries to buy merchandise at the lowest possibletries to buy merchandise at the lowest possible
cost in order to maximize its profits. Acost in order to maximize its profits. A
manufacturer endeavors to acquire materialsmanufacturer endeavors to acquire materials
and/or finished parts in order to minimize itsand/or finished parts in order to minimize its
cost of production which is a way of increasingcost of production which is a way of increasing
the income upon the sale of the finishedthe income upon the sale of the finished
products.products.
 In a trading business, the trader of theIn a trading business, the trader of the
buyer should determine the goods needed bybuyer should determine the goods needed by
the customers and the price the customers arethe customers and the price the customers are
willing to pay for the goods. He should alsowilling to pay for the goods. He should also
know where to obtain the goods at the lowestknow where to obtain the goods at the lowest
price to minimize cost and maximize profit.price to minimize cost and maximize profit.
 In general, a manufacturer sells its productsIn general, a manufacturer sells its products
in bulk to a wholesaler. In turn, the wholesalerin bulk to a wholesaler. In turn, the wholesaler
sells in smaller lots to retailers who sell thesells in smaller lots to retailers who sell the
product to the end users or consumers.product to the end users or consumers.

69. DiscountsDiscounts
 Discounts are of two types, namely:Discounts are of two types, namely:
 Trade discountTrade discount. It is a discount offered by a seller. It is a discount offered by a seller
to induce trading. This is usually offered by ato induce trading. This is usually offered by a
manufacturer or wholesaler. This type is usuallymanufacturer or wholesaler. This type is usually
encountered in catalogs where a list price is printedencountered in catalogs where a list price is printed
together with the trade discount thereon. The listtogether with the trade discount thereon. The list
price less the trade discount is the suggestedprice less the trade discount is the suggested
selling price upon release of the product.selling price upon release of the product.
 Cash discountCash discount.. It is a reduction on the sellingIt is a reduction on the selling
price offered to a buyer to induce him to payprice offered to a buyer to induce him to pay
promptly.promptly.

70. Finding the Net PriceFinding the Net Price
 There are two methods in determining the net priceThere are two methods in determining the net price
to the buyer when trade discount is given. Theseto the buyer when trade discount is given. These
are theare the Discount Rate MethodDiscount Rate Method and theand the Net PriceNet Price
Rate Method.Rate Method.
 Method A.Method A. Discount Rate Method.Discount Rate Method. The formulaThe formula
for discount is the same for percentage where thefor discount is the same for percentage where the
terms percentage (P), base (b) and rate (r) interms percentage (P), base (b) and rate (r) in
Equation 4.1 on page 56 are substituted with theEquation 4.1 on page 56 are substituted with the
terms discount (D), selling or list price (L.P) andterms discount (D), selling or list price (L.P) and
discount rate (d) respectively. Thus,discount rate (d) respectively. Thus,
 D = L.P. x d Equation 6.1D = L.P. x d Equation 6.1


 L.P.L.P. = Selling price or list price= Selling price or list price
 DD = Discount rate usually expressed in percent= Discount rate usually expressed in percent

71.  The net price then is the difference betweenThe net price then is the difference between
the selling price and the discount which is shownthe selling price and the discount which is shown
below.below.


 N.P. = L.P. − D Equation 6.2N.P. = L.P. − D Equation 6.2

N.P. = L.P. x (net price rate)N.P. = L.P. x (net price rate)
= L.P. x (100% - d in%) Equation 6.3= L.P. x (100% - d in%) Equation 6.3

Example 1. The list price of a cassette tapeExample 1. The list price of a cassette tape
recorder at AVESCO is P1, 850. The Traderecorder at AVESCO is P1, 850. The Trade
discount rate is 12 %. How much will the buyer paydiscount rate is 12 %. How much will the buyer pay
for the recorder? Usefor the recorder? Use Discount Rate MethodDiscount Rate Method andand
thethe Net Price Rate Method.Net Price Rate Method.
 Given:Given: L.P. = P1,850L.P. = P1,850 d = 12%d = 12%

72.  Solutions:Solutions:

 Method A.Method A. Discount Rate MethodDiscount Rate Method
 PhPPhP1,850 (L.P.)1,850 (L.P.)
 X .12X .12 (d)(d)
 P 222 (D)P 222 (D)

 Determining the net price:Determining the net price:
 PhPPhP1,850 (L.P.)1,850 (L.P.)
 ── 222222 (D)(D)
 Php1,628 (N.P)Php1,628 (N.P)
 ==============

 Method B.Method B. Using Equation 6.3Using Equation 6.3
 N.P. = L.P. x N.P.RN.P. = L.P. x N.P.R
 == PhpPhp1,850 x (100%-12%)1,850 x (100%-12%)
 == PhpPhp1,850 x 88% = 1, 850 x .881,850 x 88% = 1, 850 x .88
 == PhpPhp1, 628.001, 628.00

73. Finding the Discount RateFinding the Discount Rate
 A supplier may list only the net price and list price ofA supplier may list only the net price and list price of
the products being sold. The buyer should knowthe products being sold. The buyer should know
how to determine the discount rate in order tohow to determine the discount rate in order to
evaluate the reasonableness of the discountevaluate the reasonableness of the discount
offered.offered.
 Example 2. What is the discount rate if the listExample 2. What is the discount rate if the list
price is P1,500 and the discount is P450.00?price is P1,500 and the discount is P450.00?
 Given: L.P. = P1,500, D = P450Given: L.P. = P1,500, D = P450
 Solution:Solution:
 Substituting the values in Equation 6.1Substituting the values in Equation 6.1

 D = L.P. x dD = L.P. x d
 450450PhpPhp = 1,500= 1,500PhpPhp x dx d
d =d = 450450 = .3 or 30%= .3 or 30%
 1,5001,500

74.  Example 3.Example 3. What is the discount rate, if theWhat is the discount rate, if the
net price is 246 and the discount is P83.00?net price is 246 and the discount is P83.00?
 Given: N.P.= P246Given: N.P.= P246 D = P83D = P83
 Solution:Solution:

 N.P. = L.P. ─ DN.P. = L.P. ─ D
 N.P. = is net priceN.P. = is net price
 L.P. = N.P. + DL.P. = N.P. + D
 = P246 + P83= P246 + P83
 = P 329.00= P 329.00

 Substituting the list price in Equation 6.1Substituting the list price in Equation 6.1::
 D = L.P. x dD = L.P. x d
 d = D/L.P.d = D/L.P.
 = 83/329= 83/329
 = .252 or 25.2%= .252 or 25.2%

75. Discounts in SeriesDiscounts in Series
 To dispose their goods quickly, wholesalers mayTo dispose their goods quickly, wholesalers may
offer successive trade discount rates which we calloffer successive trade discount rates which we call
discounts in series.discounts in series. If a manufacturing concernIf a manufacturing concern
would give 15% and 10% trade discounts on itswould give 15% and 10% trade discounts on its
product, this does not mean that the total discountproduct, this does not mean that the total discount
rate is 25%. As a discount series, it simply meansrate is 25%. As a discount series, it simply means
that the first discount of 15% is applied to the originalthat the first discount of 15% is applied to the original
list price and the second discount of 10% to thelist price and the second discount of 10% to the
balance of the original list price and the first discountbalance of the original list price and the first discount
price.price.
 Example 4.Example 4. A home appliance dealer was offeredA home appliance dealer was offered
television set with a list price of P5,680 less 15% andtelevision set with a list price of P5,680 less 15% and
10%. What is the net price?10%. What is the net price?
 Given:Given: List Price (L.P)List Price (L.P) = P5,680= P5,680
 First discount rate (d1)First discount rate (d1) = 15%= 15%
 Second discount rate(d2Second discount rate(d2) = 10%) = 10%
 Solution:Solution:
 Method A.Method A. Discount Rate MethodDiscount Rate Method
 1rst Step1rst Step. Using Equation 6.1. Using Equation 6.1
 D1 = L.P.1 x d1D1 = L.P.1 x d1
 = P5,680 x .15= P5,680 x .15
 = P852.00= P852.00

76.  2nd Step. To get the balance (L.P2) on which the2nd Step. To get the balance (L.P2) on which the
second discount rate (d2) will be applied, subtractsecond discount rate (d2) will be applied, subtract
the first discount rate from the list price–the first discount rate from the list price–
 PhPPhP 5,680 (L.P1)5,680 (L.P1)
 ── 852852 (d1)(d1)
 PhpPhp 4,828 (L.P2)4,828 (L.P2)
 ==============
 3rd Step. Solve for D2 using Equation 13rd Step. Solve for D2 using Equation 1
 D2 = L.P2 x d2D2 = L.P2 x d2
 = 4,828 x .10= 4,828 x .10
= P482.80= P482.80
 4rth Step. Solve for the net price (N.P.) by4rth Step. Solve for the net price (N.P.) by
subtracting the second discount (D2) from L.P2subtracting the second discount (D2) from L.P2

 PhpPhp 4,828 (L.P2)4,828 (L.P2)
 482482 (D2)(D2)
 PhpPhp 4,345.20 (N.P.)4,345.20 (N.P.)
 ==============

77. Method B.Method B. Net Price RateNet Price Rate
MethodMethod
 1rst Step1rst Step. The net price rate is the difference. The net price rate is the difference
between 100% and the discount rate (See Equationbetween 100% and the discount rate (See Equation
6.3). The same formula is also applied when two or6.3). The same formula is also applied when two or
more discounts are given. The final net price is themore discounts are given. The final net price is the
product of all the net price rates found by usingproduct of all the net price rates found by using
Equation 6.3.Equation 6.3.
Final net price rate = (100% ─ d1) x (100% - d2)Final net price rate = (100% ─ d1) x (100% - d2)
= (100% ─ 15%) (100% -= (100% ─ 15%) (100% -
10%)10%)
= 85% x 90%= 85% x 90%
= .85 x .90= .85 x .90
= .765= .765
2nd Step2nd Step. The net price is the product of the list price. The net price is the product of the list price
and the final net price rate.and the final net price rate.

 Net price = Php 5,680 x .765Net price = Php 5,680 x .765
 = Php 4,345.20= Php 4,345.20

78.  Obviously, the net price rate method isObviously, the net price rate method is
relatively shorter and simpler than the discountrelatively shorter and simpler than the discount
method, more so when there are more than twomethod, more so when there are more than two
series of discounts. In such cases, the formula toseries of discounts. In such cases, the formula to
use in determining the net price isuse in determining the net price is::

 N.P. = L.P. x (100% ─ d1) (100% ─ d3)N.P. = L.P. x (100% ─ d1) (100% ─ d3)
(100% ─ dn) Equation 6.4(100% ─ dn) Equation 6.4

 Where dn indicates the last discount rateWhere dn indicates the last discount rate
 Note:Note: It is immaterial as to which order theIt is immaterial as to which order the
discounts are taken.discounts are taken.

79. Single-Discount RateSingle-Discount Rate
EquivalentEquivalent
 There are two practical ways to evaluate discountThere are two practical ways to evaluate discount
series offered by two or more sellers. Let usseries offered by two or more sellers. Let us
assume that in terms of services, neither firm heldassume that in terms of services, neither firm held
any advantage over the other. Hence, the buyer’sany advantage over the other. Hence, the buyer’s
only concern is to determine which firm’s net sellingonly concern is to determine which firm’s net selling
price was smaller. As earlier discussed, theprice was smaller. As earlier discussed, the
discount series of, for example, 10% and 15% arediscount series of, for example, 10% and 15% are
not equivalent to the total discount rate of 25%.not equivalent to the total discount rate of 25%.
 One way of evaluating discount series is toOne way of evaluating discount series is to
determine the final net price rate which wasdetermine the final net price rate which was
illustrated in Example 4, Method B. From theillustrated in Example 4, Method B. From the
buyer’s point of view, the smaller the final net pricebuyer’s point of view, the smaller the final net price
rate, the lesser is the net price.rate, the lesser is the net price.
 An easier way to evaluate the discount rates inAn easier way to evaluate the discount rates in
a discount series is thea discount series is the equivalent single discountequivalent single discount
raterate.. From the buyer’s point of view, the biggerFrom the buyer’s point of view, the bigger
the single discount rate equivalent, the lesser is thethe single discount rate equivalent, the lesser is the
net price.net price.

80.  Example 5Example 5. Charing’s Beauty Parlor was offered beauty. Charing’s Beauty Parlor was offered beauty
product with trade discounts of 15% and 10%. Find theproduct with trade discounts of 15% and 10%. Find the
single discount rate equivalent.single discount rate equivalent.
 Given: d1 = 15%Given: d1 = 15%
 d2 = 10%d2 = 10%

 Solution:Solution:
 First StepFirst Step. Derive the net price rate (NPR) of each. Derive the net price rate (NPR) of each
discount rate.discount rate.
 Since NPR = 100% ─ dSince NPR = 100% ─ d
 a. NPR1a. NPR1 = 100% ─ d1= 100% ─ d1
 = 100% ─ 15%= 100% ─ 15%
 = 85%= 85%
 b. NPR2 = 100% ─ 10%b. NPR2 = 100% ─ 10%
 = 90%= 90%
 Second Step.Second Step. Determine the single discount rateDetermine the single discount rate
equivalent (SDRE).equivalent (SDRE).
 SDRE = 100% ─ (NPR1 x NPR2)SDRE = 100% ─ (NPR1 x NPR2)
 = 100% ─ (85% x 90|%)= 100% ─ (85% x 90|%)
 = 100% ─ (.85 x .90)= 100% ─ (.85 x .90)
 = 100% ─ (.765)= 100% ─ (.765)
 = 23.5%= 23.5%

81.  Example 6.Example 6. A product is being offered in the market byA product is being offered in the market by
Supplier A and Supplier B. The trade discounts beingSupplier A and Supplier B. The trade discounts being
offered by Supplier A are 20% and 10% while that ofoffered by Supplier A are 20% and 10% while that of
Supplier B are 15% and 15%. If you were the buyer,Supplier B are 15% and 15%. If you were the buyer,
which is the better offer?which is the better offer?
 Given:Given:
 Supplier A =Supplier A = 20% and 10% (d1 and d2)20% and 10% (d1 and d2)
 Supplier B =Supplier B = 15% and 15%. (d1 and d2)15% and 15%. (d1 and d2)
 Solution:Solution:

 Supplier A. Solving for the single discount rate equivalentSupplier A. Solving for the single discount rate equivalent
of Supplier A, determine first the net price rate applicableof Supplier A, determine first the net price rate applicable

 NPR1 = 100% - 20% = 80%NPR1 = 100% - 20% = 80%
 NPR2 = 100% - 10% = 90%NPR2 = 100% - 10% = 90%
 Therefore,Therefore,
 SDRE = 100% - (80% x 90%)SDRE = 100% - (80% x 90%)
 = 100%- (.80 x .90)= 100%- (.80 x .90)
 = 100% - (.72)= 100% - (.72)
 = 100% - 72%= 100% - 72%
 = 28%= 28%

82.  Supplier B. Solve for the net price applicable to theSupplier B. Solve for the net price applicable to the
single discount rate equivalent of Supplier B.single discount rate equivalent of Supplier B.
 NPR1NPR1 = 100% - 15% = 85%= 100% - 15% = 85%
 NPR2 = 100% - 15% = 85%NPR2 = 100% - 15% = 85%

 Therefore,Therefore,
 SDRE = 100% - (85% x 85%)SDRE = 100% - (85% x 85%)
 = 100%- (.85 x .85)= 100%- (.85 x .85)
 = 100% - (.7225)= 100% - (.7225)
 = 100% - 72.25%= 100% - 72.25%
 = 27.75%= 27.75%

Based on the net price rates being offered, theBased on the net price rates being offered, the
offer of a lower net price rate (72%) by Supplier A isoffer of a lower net price rate (72%) by Supplier A is
the better offer. But, based on the single discountthe better offer. But, based on the single discount
rate equivalent, the offer of a higher discount (28%)rate equivalent, the offer of a higher discount (28%)
by Supplier A is also the better offer.by Supplier A is also the better offer.

83. Cash DiscountsCash Discounts
 To encourage wholesalers or retailers to buy is one thingTo encourage wholesalers or retailers to buy is one thing
and to induce the buyers to pay their purchases is anotherand to induce the buyers to pay their purchases is another
thing. While sellers give buyers credit lines in that thething. While sellers give buyers credit lines in that the
buyer are given several days after delivery within which tobuyer are given several days after delivery within which to
pay their purchases, sellers encourage buyers to paypay their purchases, sellers encourage buyers to pay
promptly their purchases by offering cash discounts.promptly their purchases by offering cash discounts. AA
cash discountcash discount should not be confused with a tradeshould not be confused with a trade
discount. The former is a reduction in thediscount. The former is a reduction in the net pricenet price oror
invoice priceinvoice price upon payment on a specified period of timeupon payment on a specified period of time
while the latter is a deduction from the list price.while the latter is a deduction from the list price.
 The cash discount rates offered are clearly writtenThe cash discount rates offered are clearly written
on the invoice. An example of a cash discount is”on the invoice. An example of a cash discount is”5,5, n ”n ”
 10 3010 30
 which means that a cash discount of 5% will be deductedwhich means that a cash discount of 5% will be deducted
 from the net price if the invoice paid within 10 daysfrom the net price if the invoice paid within 10 days
from the date of the invoice. The "n/30" means that afterfrom the date of the invoice. The "n/30" means that after
10 days, no discount is given and that the invoice price10 days, no discount is given and that the invoice price
should be paid not later than 30 days of the invoice date.should be paid not later than 30 days of the invoice date.
 If the retailer or buyer would avail of the cashIf the retailer or buyer would avail of the cash
discount offered, the price to be paid is the invoice pricediscount offered, the price to be paid is the invoice price
less the amount of cash discount.less the amount of cash discount.

84.  Example 7. An invoice with a list price of P18,200Example 7. An invoice with a list price of P18,200
dated October 12, has trade discounts of 15% anddated October 12, has trade discounts of 15% and
10% and the terms are 4/10 ,2/30 ,10% and the terms are 4/10 ,2/30 , n/60. Hown/60. How
much must be paid if the invoice is settled on (a)much must be paid if the invoice is settled on (a)
October 22? (b) November 11?October 22? (b) November 11?

 Given:Given:
 List price = P 18,200List price = P 18,200
 Trade Discount series = 15% and 10%Trade Discount series = 15% and 10%
 Cash discounts =Cash discounts = 44 ,, 22 , and, and nn
 10 30 6010 30 60
 SolutionSolution::
 (a.) If the amount is settled on October 22(a.) If the amount is settled on October 22
 1rst Step. Solve first the net price (N.P.)1rst Step. Solve first the net price (N.P.)
 N.P. = L.P. x (NPR1) (NPR2)N.P. = L.P. x (NPR1) (NPR2)
 = P 18,200 x (100% - 15%)= P 18,200 x (100% - 15%)
(100% - 10%)(100% - 10%)
 = 18,200 x (85%) (90%)= 18,200 x (85%) (90%)
 = 18,200 x (.85) (.90)= 18,200 x (.85) (.90)
 = P 13,923.00= P 13,923.00

85.  2nd Step.2nd Step. Having solve for the N.P., which isHaving solve for the N.P., which is
P13,923.00, we can now proceed to solved the cashP13,923.00, we can now proceed to solved the cash
discount (Dc). If the purchaser pays on October 22 whichdiscount (Dc). If the purchaser pays on October 22 which
is 10 days from the invoice date, the cash discount term isis 10 days from the invoice date, the cash discount term is
4/10. Thus, the cash discount rate is 4%4/10. Thus, the cash discount rate is 4%
 Dc = P13, 923 x 4%Dc = P13, 923 x 4%
 = 13, 923 x .04= 13, 923 x .04
 = P556.92= P556.92
 3rd Step3rd Step. Determine the net invoice price (NIP) by simply. Determine the net invoice price (NIP) by simply
subtracting cash discount from the net price.subtracting cash discount from the net price.
 NIP = N.P. - DcNIP = N.P. - Dc
 = P13, 923 - P556.92= P13, 923 - P556.92
 = P13,366.08= P13,366.08

 Alternatively, the computation may be as follows:Alternatively, the computation may be as follows:
 NIP = N.P. x (100% - dc)NIP = N.P. x (100% - dc)
 = (100% - 4%)= (100% - 4%)
 = 13,923 x (96%)= 13,923 x (96%)
 = 13,923 x (.96)= 13,923 x (.96)
 = P 13, 666.08= P 13, 666.08
 If the invoice is paid on November 11.If the invoice is paid on November 11.
 First StepFirst Step. Same as solution (a.).. Same as solution (a.).
 N.P. = P 13,923.00N.P. = P 13,923.00

86.  2nd Step.2nd Step. If the Purchaser pays on November 11If the Purchaser pays on November 11
which is 30 days from the invoice date, the cashwhich is 30 days from the invoice date, the cash
discount rate is 2%.discount rate is 2%.
 Dc = 13,923 x 2% = P278.46Dc = 13,923 x 2% = P278.46
 NIP = 13,923 – P278.46NIP = 13,923 – P278.46
 = P 13,644.54= P 13,644.54
 Alternatively, the computation may be asAlternatively, the computation may be as
follows:follows:
 NIP = N.P. x (100% - dc)NIP = N.P. x (100% - dc)
 = P 13, 923 x (100% - 2%)= P 13, 923 x (100% - 2%)
 = 13, 923 x (98%)= 13, 923 x (98%)
 = 13, 923 x .98= 13, 923 x .98
 = P 13,644.54= P 13,644.54

87. Mathematics of PricingMathematics of Pricing
 The price of a commodity is determined by manyThe price of a commodity is determined by many
factors obtaining in the market. It is greatly affectedfactors obtaining in the market. It is greatly affected
by the need of the consumers for the commodity, itsby the need of the consumers for the commodity, its
availability, its quality level and the number ofavailability, its quality level and the number of
different lines with similar products. If there is adifferent lines with similar products. If there is a
great demand for the commodity and the stock isgreat demand for the commodity and the stock is
limited, the seller could dictate his own selling price.limited, the seller could dictate his own selling price.
But if the price demanded is prohibitive such thatBut if the price demanded is prohibitive such that
the average consumer could not afford it, thethe average consumer could not afford it, the
consumers may shift to other available similarconsumers may shift to other available similar
products and the seller may stand to lose.products and the seller may stand to lose.
 The business should understand pricing termsThe business should understand pricing terms
such as cost, mark-up, margin and the like tosuch as cost, mark-up, margin and the like to
enable himself to price his commodities reasonably.enable himself to price his commodities reasonably.
CostCost refers to the amount the purchaser acquiredrefers to the amount the purchaser acquired
the good.the good. MarginMargin (gross profit) is the excess of(gross profit) is the excess of
selling price over cost from viewpoint of the seller.selling price over cost from viewpoint of the seller.
If an original selling price is adjusted upward, theIf an original selling price is adjusted upward, the
addition is calledaddition is called Mark-upMark-up.. The reverse is calledThe reverse is called
Mark-down.Mark-down.

88.  To illustrate clearly the above terminologies, letTo illustrate clearly the above terminologies, let
us take an example. A car dealer bought a carus take an example. A car dealer bought a car
for P14,500.00 and sold it for P17,200.00.for P14,500.00 and sold it for P17,200.00.
 P14,500 is the cost of the article, assumingP14,500 is the cost of the article, assuming
he did not spend any amount for reconditionhe did not spend any amount for recondition
the car before selling it.the car before selling it.
 P 17,200 is the selling price, tag price, orP 17,200 is the selling price, tag price, or
retail price of the article. (P17,200 – P14,500) =retail price of the article. (P17,200 – P14,500) =
P2,700 is the gross profit or margin.P2,700 is the gross profit or margin.
 In the above illustration, the margin P2,700In the above illustration, the margin P2,700
can be expressed in percent based in either thecan be expressed in percent based in either the
selling price or cost. This is calledselling price or cost. This is called percentpercent
margin.margin. If the selling price is used as the base,If the selling price is used as the base,
thethe percent margin on selling pricepercent margin on selling price isis P2,700P2,700
= .15697 = 15.7%.= .15697 = 15.7%.

89.  If the cost is P17,200 used as the base, theIf the cost is P17,200 used as the base, the percentpercent
marginmargin on cost ison cost is P2,700P2,700 = .1862 =1 =18.62%= .1862 =1 =18.62%

P14,500P14,500

 Formula-wise, the equations for the above are theFormula-wise, the equations for the above are the
following:following:
 Cost + Margin = Selling priceCost + Margin = Selling price
 OrOr
 Margin = Selling price – CostMargin = Selling price – Cost
 Percent Margin(%) =Percent Margin(%) = MarginMargin x 100, based onx 100, based on
selling priceselling price
 Selling priceSelling price
 Percent Margin(%) =Percent Margin(%) = MarginMargin x 100, based on costx 100, based on cost
 CostCost

90.  Example 8.Example 8. The merchandise that cost a dealerThe merchandise that cost a dealer
P4,150 was sold for P5,200. Determine the percentP4,150 was sold for P5,200. Determine the percent
margin (a) based on selling price, (b) based on cost.margin (a) based on selling price, (b) based on cost.
 Given:Given:
 Selling price = P5,200Selling price = P5,200
 Cost = P4,150Cost = P4,150
 Solution:Solution: Immediately subtract P4,150 fromImmediately subtract P4,150 from
P5,200 to get the margin.P5,200 to get the margin.
 MarginMargin = Selling price - Cost= Selling price - Cost
 = P5,200 -= P4,150= P5,200 -= P4,150
 = P 1,050= P 1,050
 If the percent margin is required based on sellingIf the percent margin is required based on selling
price divide the margin by the selling price.price divide the margin by the selling price.
 Percent Margin (%)Percent Margin (%) == P1,050P1,050 x 100x 100
 P5,200P5,200
 = .202 x 100= .202 x 100
 = 20.2%= 20.2%

91.  If percent margin is required based on cost,If percent margin is required based on cost,
divide the margin by the cost.divide the margin by the cost.
 Percent Margin (%)Percent Margin (%) == P1,050P1,050 x 100x 100
 P4,150P4,150
 = .253 x 100= .253 x 100
 = 25.3%= 25.3%
 NoteNote:: Another term used for gross margin isAnother term used for gross margin is
mark on.mark on.

92. Chapter 7:Chapter 7:
SimpleSimple
InterestInterest

93.  Money borrowed usually bears a cost calledMoney borrowed usually bears a cost called interest.interest.
The amount borrowed is calledThe amount borrowed is called principal.principal.
 Sources of loans are the banks, investmentSources of loans are the banks, investment
houses, savings and loan associations, cooperatives,houses, savings and loan associations, cooperatives,
credit unions, and other financing companies.credit unions, and other financing companies.
 Determining the InterestDetermining the Interest
 The interest or interest charge is usually expressed inThe interest or interest charge is usually expressed in
percent, such as 6%,which means a charge of P6.00percent, such as 6%,which means a charge of P6.00
for every P100.00 for a definite period of time. Unlessfor every P100.00 for a definite period of time. Unless
otherwise stated, quoted interest rate is for one year.otherwise stated, quoted interest rate is for one year.
 The formula for computing interest or interest chargeThe formula for computing interest or interest charge
is:is:
 I = P x r x t Equation 7.1I = P x r x t Equation 7.1

 Where:Where:
 I = interest expressed in monetary value.I = interest expressed in monetary value.
 P = principal or the amount borrowed.P = principal or the amount borrowed.
 R = interest rate on the principal, usuallyR = interest rate on the principal, usually
stated in percent.stated in percent.
 T = time or duration of the loan, expressedT = time or duration of the loan, expressed
in number of years, months,in number of years, months,
 days, etc.days, etc.

94.  To illustrate, if Mr. X borrowed P2,600.00 from theTo illustrate, if Mr. X borrowed P2,600.00 from the
bank at an interest of 5%, what is the interest in 3bank at an interest of 5%, what is the interest in 3
years?years?
 Given:Given:
 P = P2,600.00P = P2,600.00
 r = 5% (per year)r = 5% (per year)
 t = 3 yearst = 3 years
 I = to findI = to find
 Solution:Solution: Substituting the values in Equation 7.1Substituting the values in Equation 7.1

I = P x r x tI = P x r x t
 = P 2,600.00 x 5% x 3= P 2,600.00 x 5% x 3
 = 2,600 x .05 x 3= 2,600 x .05 x 3
 = P390.00= P390.00

95.  The amount to be paid by the borrower upon maturity of the loanThe amount to be paid by the borrower upon maturity of the loan
is –is –

 WhereWhere::
 A = sum or total amount to be paid.A = sum or total amount to be paid.
 P = principalP = principal
 I = interestI = interest
 Thus,Thus,
 A = P 2,600 + P390.00A = P 2,600 + P390.00
 = P2,990.00= P2,990.00
 Example 1. Pedro borrowed from Jose P5,200 with anExample 1. Pedro borrowed from Jose P5,200 with an
interest of 6%. How much should Pedro pay after 16interest of 6%. How much should Pedro pay after 16
months?months?
 Given:Given:
 P = P5,200P = P5,200
 r = 6% (per year)r = 6% (per year)
 t = 16 monthst = 16 months
 Solution:Solution:
 Method A. Convert 16 months to number of years.Method A. Convert 16 months to number of years.
 I = P x r x tI = P x r x t
 = P 5,200 x 6% x= P 5,200 x 6% x 1616
 1212
 = 5,200 x .06 x= 5,200 x .06 x 44
 33
 = P 416.00= P 416.00

96.  Method B. Convert interest rate per month.Method B. Convert interest rate per month.

 I = P x r x tI = P x r x t
 = P 5,200 x 6%/12n x 16 months= P 5,200 x 6%/12n x 16 months
 1212
 = 5,200 x .5% x 16= 5,200 x .5% x 16
 = 5,200 x.005 x16= 5,200 x.005 x16
 = P 416.00= P 416.00

The amount to be paid after 16 months is –The amount to be paid after 16 months is –
 A= P + IA= P + I
 = P5,200 + P416= P5,200 + P416
 = P5,616.00= P5,616.00

97. Determining the Rate ofDetermining the Rate of
Interest (r)Interest (r) This particular problem is encountered in evaluating two orThis particular problem is encountered in evaluating two or
more alternative choices. A lender wants a higher rate ofmore alternative choices. A lender wants a higher rate of
interest for his money, while a borrower prefers a lesser rate ofinterest for his money, while a borrower prefers a lesser rate of
interest on money borrowed.interest on money borrowed.
 Assume that Mr. Y would like to borrow P5,000 payable afterAssume that Mr. Y would like to borrow P5,000 payable after
one (1) year from a bank whose interest charge is 18% perone (1) year from a bank whose interest charge is 18% per
year. On the other hand, Mr. Z is willing to lend him the sameyear. On the other hand, Mr. Z is willing to lend him the same
principal and interest charge is P950 per year. Since the rateprincipal and interest charge is P950 per year. Since the rate
of interest on the bank loan is already know, Mr. Y must knowof 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 whichthe rate of interest on Mr. Z’s to enable him to determine which
is the better offer.is the better offer.
 Solution:Solution:
 r =r = II Equation 7.4Equation 7.4
 PtPt

 == 950950 = 0.19= 0.19
 5,0005,000
 = 0.19 x 100%= 0.19 x 100%
 = 19%= 19%
 Thus, as compared to Mr. Y’s proposal of a 19% interestThus, as compared to Mr. Y’s proposal of a 19% interest
rate per year, the bank loan with 18% interest rate perrate per year, the bank loan with 18% interest rate per
year is definitely the better offer.year is definitely the better offer.

98. Determining the Time (t)Determining the Time (t)
 As derived from Equation 7.1As derived from Equation 7.1
 t =t = II Equation 7.5Equation 7.5
 PrPr
 Example 3. How long will it take for a deposit ofExample 3. How long will it take for a deposit of
P1,500.00 to earn P186.00 invested at the rate of 7-P1,500.00 to earn P186.00 invested at the rate of 7-
1/2%?1/2%?
 Given:Given:
 I = P186.00I = P186.00
 P = P1,500.00P = P1,500.00
 r = 7- 1/2%r = 7- 1/2%
 Solution: Solve for t, using Equation 7.5Solution: Solve for t, using Equation 7.5
 t =t = I_I_ == Php186.00Php186.00
 PrPr Php1,500 x 7Php1,500 x 7 11%%
 22
 == Php186.00Php186.00
 Php1,500 x .075Php1,500 x .075
 == 1.65 years1.65 years

99.  The time may be converted to its appropriate unitsThe time may be converted to its appropriate units
of time measure like months or daysof time measure like months or days
 (Months) t(Months) t == 1.65 years x 121.65 years x 12
months per yearmonths per year
 == 19.8 months19.8 months
 OrOr
 (Days) t(Days) t == 1.65 years x 360 days1.65 years x 360 days
per yearper year
 == 594 days594 days
 If it is required to express the time (t) inIf it is required to express the time (t) in
combination of two or more units such as in years;combination of two or more units such as in years;
months, and days the procedure is as follows:months, and days the procedure is as follows:
 In the above example, to convert t = 1.65In the above example, to convert t = 1.65
years to years, months and days.years to years, months and days.
 1rst Step.1rst Step. The period of 1.65 indicates one(1) yearThe period of 1.65 indicates one(1) year
and a fraction of year expressed in decimal part asand a fraction of year expressed in decimal part as
0.65, thus number of years is 1.0.65, thus number of years is 1.
 2nd Step2nd Step. The fractional part of a year a (0.65) is. The fractional part of a year a (0.65) is
multiplied by the conversion factor of 12 months permultiplied by the conversion factor of 12 months per
year.year.
 Thus, number of months is 7.Thus, number of months is 7.

100.  3rd Step3rd Step. The fractional part of month (0.8) is. The fractional part of month (0.8) is
multiplied by 30 days per month (approximate).multiplied by 30 days per month (approximate).
 Therefore, the final answer is:Therefore, the final answer is:

 t =t = 1 year, 7 months and 24 days.1 year, 7 months and 24 days.

101. Determining the Principal,Determining the Principal,
Given (A) (r) (t)Given (A) (r) (t)
 This particular problem involves Equations 7.1 andThis particular problem involves Equations 7.1 and
7.2. In any mathematical computation, solving two7.2. In any mathematical computation, solving two
unknown variables using one equation or formula isunknown variables using one equation or formula is
not possible. However, two equations with at leastnot possible. However, two equations with at least
one common variable can be combined to derive aone common variable can be combined to derive a
new equation.new equation.
 If A = P + I (Equations 7.2) and I = PrtIf A = P + I (Equations 7.2) and I = Prt
(Equation 7.1), by substituting (I) using its(Equation 7.1), by substituting (I) using its
equivalent (Prt) in Equation 7.2, in the followingequivalent (Prt) in Equation 7.2, in the following
manner.manner.
 A = P + IA = P + I
 A = P + PrtA = P + Prt
 And factoring out P, we haveAnd factoring out P, we have
 A = P (I + rt) Equation 7.6A = P (I + rt) Equation 7.6
P =P = AA Equation 7.7Equation 7.7
I + rtI + rt


102.  Example 4.Example 4. If a bank offers 8 ½% interest rate andIf a bank offers 8 ½% interest rate and
a depositor wants to have P3,878.20 after 5 years,a depositor wants to have P3,878.20 after 5 years,
how much should he deposit now?how much should he deposit now?
 Given:Given:
 r = 8r = 8 11%%
 22
 A = P3,878.20A = P3,878.20
 t = 5 yearst = 5 years
 Solution:Solution: Using EquationUsing Equation P =P = A___A___
 1 + r x t1 + r x t
 PP == P3,878.20P3,878.20
 1 + 81 + 8 11% x 5 years% x 5 years

 == P3,878.20P3,878.20
 1 + .085 x 51 + .085 x 5
 == P3,878.20P3,878.20
 1 + .4251 + .425
 == P3,878.20P3,878.20
 1.4251.425
 = P 2,721.54= P 2,721.54

103. Determining the Sum (A)Determining the Sum (A)
 Example 5. The bank offers an annualExample 5. The bank offers an annual
rate of 10rate of 10 11 % interest, for 3 ½ years,% interest, for 3 ½ years,
22
 how much will the accumulated sum be ifhow much will the accumulated sum be if
the initial deposit is P 1,350.00?the initial deposit is P 1,350.00?
 Given:Given:
 P = P 1, 350P = P 1, 350
 r = 10r = 10 11 %%
 22
 t = 3 ½ yearst = 3 ½ years

 In this particular problem, two methodsIn this particular problem, two methods
may be used to determine the amount (A).may be used to determine the amount (A).

104.  1rst Method.1rst Method. Determine the interest by usingDetermine the interest by using
 Equation 7.1, I = Prt and then, add the interestEquation 7.1, I = Prt and then, add the interest
and the principal to determine the amount A, usingand the principal to determine the amount A, using
7.27.2
 2nd Method2nd Method..
 By A = P(1 + r x t)By A = P(1 + r x t)
 A = P 1,350 (1 + 10A = P 1,350 (1 + 10 11 % x 3% x 3 11))
 2 22 2
 = 1,350 (1 + .105 x 3.5)= 1,350 (1 + .105 x 3.5)
 = 1,350 (1.3675)= 1,350 (1.3675)
 = P 1, 846.13= P 1, 846.13

105. Promissory NotesPromissory Notes
 A promissory note may be used for raising money. TheA promissory note may be used for raising money. The
borrower , in exchange for a loan, may issue his ownborrower , in exchange for a loan, may issue his own
promissory note or an immature note received by himpromissory note or an immature note received by him
form another person. Aform another person. A promissorypromissory notenote is a writtenis a written
promise to pay to a person a certain sum of moneypromise to pay to a person a certain sum of money
on a specified date.on a specified date. The person who signs theThe person who signs the
promissory note is thepromissory note is the maker.maker. The person to whom hisThe person to whom his
payment is promised is thepayment is promised is the payee.payee. TheThe period,period, oror termterm ofof
the note is the duration or length of time from the date ofthe note is the duration or length of time from the date of
the note to the date of maturity.the note to the date of maturity. Maturity dateMaturity date is alsois also
referred to as the expiration date or due date of the note.referred to as the expiration date or due date of the note.
TheThe maturity valuematurity value is the total amount to be paid on theis the total amount to be paid on the
maturity or due date. Thematurity or due date. The face valueface value of a promissoryof a promissory
note is the amount specifically mentioned in the note. Ifnote is the amount specifically mentioned in the note. If
the interest is mentioned in the note, it is anthe interest is mentioned in the note, it is an interest-interest-
bearing notebearing note.. If no interest is mentioned, then it is aIf no interest is mentioned, then it is a
non-interest bearing notenon-interest bearing note..
 To illustrate, let us assume thatTo illustrate, let us assume that
 Mr. Darwin borrowed P10,000 from his friend Carlos onMr. Darwin borrowed P10,000 from his friend Carlos on
November 8, Mr. Darwin issued a promissory noteNovember 8, Mr. Darwin issued a promissory note
(shown below) due on February 8 worth P10,000 plus(shown below) due on February 8 worth P10,000 plus
10% interest.10% interest.
The following is an example of promissory note:The following is an example of promissory note:

106. 
 P 10,000P 10,000 November 8November 8
 Ninety daysNinety days afterafter II promise to pay to the order ofpromise to pay to the order of
Mr. CarlosMr. Carlos the sum ofthe sum of Ten Thousand Pesos andTen Thousand Pesos and
no/100no/100For value received with interest at 10%.DueFor value received with interest at 10%.Due
Date:Date: February 8February 8
Mr. DarwinMr. Darwin
SignedSigned

 Where:Where:
 Date of the note is November 8.Date of the note is November 8.
 Term of the note is 90 daysTerm of the note is 90 days
 Maker is Mr. Darwin]Maker is Mr. Darwin]
 Face value o the note is P10,000Face value o the note is P10,000
 Interest rate is 10% per yearInterest rate is 10% per year
 Maturity date is February 8.Maturity date is February 8.
 To compute for the Maturity Value (a) of theTo compute for the Maturity Value (a) of the
promissory note in the above example –promissory note in the above example –

107.  1rst Step. Determine the interest by using the1rst Step. Determine the interest by using the
formula I = P x r x tformula I = P x r x t
 II == P10,000 x 10% xP10,000 x 10% x 9090
 360360
 = 10,000 x .1 x .25= 10,000 x .1 x .25
 = P 250.00= P 250.00
 2nd Step. Determine the amount or maturity value2nd Step. Determine the amount or maturity value
by using formula A = P + Iby using formula A = P + I
 AA == P10,000 + P250P10,000 + P250
 == P10,250P10,250
 Alternatively, the amount or maturity value can beAlternatively, the amount or maturity value can be
computed as follows using formula A = P(1 + r x t)computed as follows using formula A = P(1 + r x t)
 AA == P10,000 (1 +10% x 90/100)P10,000 (1 +10% x 90/100)
 == 10,000 (1 + .1 x.25)10,000 (1 + .1 x.25)
 == 10,000 (1.025)10,000 (1.025)
 == P10,250.00P10,250.00

108.  On the other hand, if no interest is mentioned in theOn the other hand, if no interest is mentioned in the
note, the face value is the maturity value. Tonote, the face value is the maturity value. To
illustrate, convert above interest-bearing note to aillustrate, convert above interest-bearing note to a
non-interest bearing note as shown below:non-interest bearing note as shown below:

 P 10,250.00P 10,250.00 November 8November 8
 Ninety daysNinety days afterafter II promise to pay to the order ofpromise to pay to the order of
Mr. CarlosMr. Carlos the sum ofthe sum of Ten Thousand Two hundredTen Thousand Two hundred
fifty Pesos and no/100fifty Pesos and no/100For value received with interestFor value received with interest
at 10%.at 10%.
 Due Date:Due Date: February 8February 8 Mr. DarwinMr. Darwin
SignedSigned

 The omission of a stated interest rate does notThe omission of a stated interest rate does not
necessarily mean that the original debt bears nonecessarily mean that the original debt bears no
interest. The interest, if any, might have beeninterest. The interest, if any, might have been
added to the original debt when the face value wasadded to the original debt when the face value was
determined. As illustrated, the original debt was Pdetermined. As illustrated, the original debt was P
10,000 but the face value indicated in the note was10,000 but the face value indicated in the note was
P10, 250.00 which is also the maturity value.P10, 250.00 which is also the maturity value.

109. Discounting of NotesDiscounting of Notes
 Discounting a noteDiscounting a note is based on the maturity value whileis based on the maturity value while
simple interest is based on the principal. Nonetheless,simple interest is based on the principal. Nonetheless,
the basic principles of discounting a note or anythe basic principles of discounting a note or any
unmatured drafts or negotiable instruments at a bank areunmatured drafts or negotiable instruments at a bank are
the same as those for obtaining a loan from a bank fromthe same as those for obtaining a loan from a bank from
which the interest is deducted in advance. For example,which the interest is deducted in advance. For example,
a businessman was granted a one (1) year loan by a banka businessman was granted a one (1) year loan by a bank
worth P10,000. If the bank charges 10% interest theworth P10,000. If the bank charges 10% interest the
amount the businessman will receive is P10,000 lessamount the businessman will receive is P10,000 less
interest in advance worth (I= P x r x t = 10,000 x .1 x 1)interest in advance worth (I= P x r x t = 10,000 x .1 x 1)
P1,000. Therefore, the businessman will receive only theP1,000. Therefore, the businessman will receive only the
discounted amount of P9,000 but the maturity value afterdiscounted amount of P9,000 but the maturity value after
one year is still P10,000.one year is still P10,000.
 TheThe discount rate (d)discount rate (d) is usually expressed inis usually expressed in
percent or its equivalent decimal and is generally quotedpercent or its equivalent decimal and is generally quoted
on a yearly basis. Its is the ration of the discount for theon a yearly basis. Its is the ration of the discount for the
period to the maturity value. If the discount rate is notperiod to the maturity value. If the discount rate is not
mentioned in any given problem, it is assumed that thementioned in any given problem, it is assumed that the
interest rate is also the discount rate. Theinterest rate is also the discount rate. The discount (discount (D)D)
is the amount to be deducted to theis the amount to be deducted to the maturity value. Thematurity value. The
proceedsproceeds (Pd)(Pd) is the amount a creditor is willing to pay ais the amount a creditor is willing to pay a
note before its maturity or due date.note before its maturity or due date.