Module 5 Using Mathematical Techniques

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MODULE 5

USING

MATHEMATICAL

CONCEPTS AND

TECHNIQUES

ASIAN ACADEMY OF BUSINESS AND COMPUTERS

MAT 111

BUSINESS MATHEMATICS

COMPETENCY-BASED LEARNING MODULE ON


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MODULE 5

.

USING

MATHEMATICAL

CONCEPTS AND

TECHNIQUES Information Sheet 1 SIMPLE INTEREST

Information Sheet 2 SIMPLE DISCOUNT

Information Sheet 3 COMPOUND INTEREST

Information Sheet 4 SIMPLE ANNUITIES

ASIAN ACADEMY OF BUSIMESS AND COMPUTERS


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MODULE TITLEUSING MATHEMATICAL CONCEPTS AND

TECHNIQUES

DESCRIPTION: This module covers the knowledge, skills and attitudes required in

the application of mathematical concepts and techniques.

COURSE OBJECTIVE: Students completing this module will be able to: Identifymathematical tools and techniques to solve problem; pply

mathematical procedure!solution; and naly"e results.

CONTENTS: Information Sheet #.# $ Simple Interest

Self%&heck #.#

Information Sheet #.' $ Solving the (rincipal

Self%&heck #.'Information Sheet #.) $ Solving the Interest *ate

Self%&heck #.)Information Sheet #.+ $ Solving the Time

Self%&heck #.+

Information Sheet #. $ Total mountSelf%&heck #.

Information Sheet #.- $ (resent alue

Self%&heck #.-

Information Sheet #./ $ 01act Interest and ppro1imate TimeSelf%&heck #./

Self%&heck #.2Information Sheet '.# $ Simple 3iscountInformation Sheet '.' $ Solve 3iscount *ate

Self%&heck '

Information Sheet '.) $ (romissory 4oteSelf%&heck '.#

Information Sheet ).# $ &ompound Interest

Self%&heck )Self%&heck ).#

Self%&heck ).'

Information Sheet +.# $ Sample nnuity

Self%&heck +Self%&heck +.#



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LEARNING GUIDE

Learning Steps Resources

1. Read Information Sheet 1 on SimpleInterest

2. Answer Self-Check

3. Read Information Sheet 2 on SimpleDIscount


5. Read Information Sheet 3 onCompound interest


7. Read Information Sheet 4 on Simple

Annuity8. Answer Self-Check

Information Sheet 1

Self-Check 1

Information Sheet 2

Self-Check 2

Information Sheet 3

Self-Check 3

Information Sheet 4

Self-Check 4

USING MATHEMATICAL CONCEPTS AND TECHNIQUES

MODULE 5



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1.1 Simple Interest

Simple interest is the most basic type of interest. In order to understand howvarious types of transactions work, it helps to have a complete understanding ofsimple interest.

For example, you may pay interest on a loan, and it is important to understandhow interest works. Better yet, your bank may be paying you interest on yourdeposits and you can maximize your earnings by knowing more aboutinterest.

Simple Interest Overview

Simple interest is just the amount of money paid on a loan. It is the easiest typeof interest to calculate and understand

Simple Interest Formula

If you want to calculate simple interest, use this formula:

I=P r t

In other wordsInterest(I) is calculated by multiplying Principal(p) times theRate(r) times the number of Time(t) periods.

For example, if I invest $100 (the Principal) at a 5% annual rate for 1 year thesimple interest calculation is:

I=P r t

$5 = $100 x 5 % x 1 yr

Simple Interest Limitations

Simple interest is a very basic way of looking at interest. In fact, your interest whether youre paying it or earning it is usually calculated using differentmethods. However, simple interest is a good start that gives us a general idea ofwhat a loan will cost or what an investment will give us.

The main limitation that you should keep in mind is that simple interest doesnot take compounding into account.

INFORMATION SHEET 1.1

SIMPLE INTEREST

Using Mathematical Concepts and Techni


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Example 1

A student purchases a computer by obtaining a simple interest loan. The

computer costs $1500, and the interest rate on the loan is 12%. If the loan is to

be paid back in weekly instalments over 2 years, find the amount of interest

paid.

Given :

P = $1500

R = 12% = 0.12,T= 2 years

I = ?

Solution:

I = PRT

= 1500 0.12 2

= $360

Example 2

A man borrowed P 5,000 at the rate of 6% per annum payable at the end of 2

years. Find the interest.

Given:

P = P5,000

R = 6%

T = 2 years

I = ?



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Solution:

I = PRT

= P5,000 x 6% x 2 years

= P600

Example 3.

Mrs. Garcia borrowed P8,000 at 8% simple interest for 2 years and 3 months.

Find the simple interest.

Given:

P = P8,000

R = 8%

T = 2 years and 3 months

Solution:

I = PRT

= P8,800 x 0.8 x 2 3/12

= P1,584



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A. Compute for the interest.

I P R T

1.____________ P 2,400 6% 2 years

2.____________ P 1,670 4 % 10months

3.____________ P 24,000 8 % 2 years and 3mos.

4.

____________ P 16,000 5.25% 8 months

5.____________ P 14,500 11% 5 years

B. Solve the following problems:

1. Mrs. Loida Aquino invested P8,250 in a bank which yields 6 % simple

interest for 27 months.

2. Find the total interest earned from P27,000 investment, if invested at 9% interest for 18 months.

3. A man borrowed P125,000 to start a business. Find the interest if moneyis invested at 8% for 8 months.

4.

For a period of 3 years, money worth P 148,000 is invested at 9 %.Find the interest.

5. Mrs. Guzman invested P 4,850 in a bank with an interest of 5%, howmuch will be the interest if invested for 7 years.

SELF-CHECK 1.1



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1.2 Solving the Principal

The principal (P) is computed by dividing the interest by the product of the rate

of interest (R) and the time (T). Hence,

P = I

RT

Example 1.

Find the principal if the interest from a certain investment is P600 for 4 years atthe rate of 6% simple interest.

Given:

I = P600R = 6%

T = 4 years

P = ?

Solution

P = I

RT

= P600

(.06)(4years)



= P2,500


SOLVING THE PRINCIPAL

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Example 2.

How much should Mr. Cruz invest so that his money earns P1,408 borrowed at8% for two years and 9 months.

Given:

I = P1,408

R = 8%


P = ?

Solution:

P = I

RT

= P1408

(.08)(2 years and 9 months) or (11/4)

= P6,400



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A. Compute for principal.

I P R T

1. P 2,400 ____________ 6% 2 years

2. P11,000 ____________ 4 % 10months

3. P 4,580 ____________ 8 % 2 years and 3mos.

4. P 26,000 ____________ 5.25% 8 months

5. P 4,500 ____________ 11% 5 years


1.

How much should Mr. Cruz invest so that his money earns P8,980 borrowed at 4% for 7 years 4 months.

2. Find the principal if money earned P700 3% interest for 18months.

3. For a period of 3 years, a certain principal earned P335 at 8%simple interest. Determine the value of the principal.

4. Find the principal if money is invested for 5 years at 7% simpleinterest earning P300 .

5. Find the principal if money is invested for 15 years at 3% simpleinterest earning P5750

SELF-CHECK 1.2



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1.3 Solving the Interest Rate (R)

To solve for the rate of interest (R), divide the interest (I) by the product of the

principal (P) and the time (T). Hence,

R = I

PT

Example 1.

Mrs. Latar invested P3,500. For a period of 2 years, her money earned P665.

Find the rate of interest.

Given:I = P665

T = 2 years

P = P3,500

R = ?

Solution:

R = I

PT= P665

(P3,500) (2)



= .095 or 9.5%


SOLVING THE INTEREST RATE

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Example 2.

At what rate should one invest in a bank so that P6000 earns P1,537.50 for aperiod of 3 years and 5 months?

Given:

I = P1,537.50


P = P6,000

R = ?

Solution:

R = I

PT

= P1,537.50

(6000) (.075) (3 5/12)

= 7.5%



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A. Compute for rate

I P R T

1. P 210 P 1,680 ___________ 2 years

2.

P 500 P 4,100 ___________ 9 months3. P 3,000 P 16,000 ___________ 2 years and 6

mos.

4. P 500 P 4,000 ___________ 5 years

5. P 252 P 3,500 ___________ 4 years


1.

Mrs. Guzman invested P4,850 in a bank which gave her an interestamounting to P121.25 in 6 months. Find the rate of interest.

2. A man borrowed P12,000 to start a business which gave her an interestamounting to P530 after 5 months. Find the rate of interest

3. At what interest will P 4,850 be invested if it earned P425 in 1 year and 2months?

4. Mr. Cruz earned P2,500 in a borrowed money amounting to P25,000 in 2years. What is the interest rate?

5. Find the rate of interest after 5 years and 6 months, if money is worthP8,500 with an earning of P560.

SELF-CHECK 1.3



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1.4 Solving the Time (T)

To solve for the time, divide the interest (I) by the product of the principal (P)and the rate of interest. (R). Thus,

T = I

PR

Example 1.

How long will it take P2,500 to earn P200 if the money is invested at 8% simpleinterest?

Given:

I = P200

P = P2,500

R = 8%

T = ?

Solution:

T = I

PR= P200

(P2,500) (8%)



= 3 years


SOLVING THE TIME

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Example 2.

At what length of time must P4,500 be invested at 91/2% simple interest to earnP496.25?

Given:

I = P496.25

P = P4,500

R = 9 1/2%

T = ?

Solution:

T = I

PR= P496.25

(P4,500) (9 1/2%)

= 3.5 years



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A. Compute for rate

I P R T

1. P 960 P 4,000 12% ___________

2.

P 9,116.25 P 25,500 6 1/2% ___________3. P 25,937.50 P 150,000 9 1/2% ___________

4. P 800 P 8,500 6% ___________

5. P 580 P 10,000 8% ___________


1. At what length of time must P 4,850 be invested at 12 % simple interestto earn P2,425?

2. How long will it take a given principal to double itself if invested at 8%simple interest?

3. Mrs. Garcia borrowed P2,500 at 8% simple interest. Find the length oftime if the interest is P200.

4. At what length of time must P8,000 at 8% simple interest if moneyearned P1,584?

5. Find the length of time if money amounting to P2,500 earned P600 at 6%simple interest.

SELF-CHECK 1.4



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1.5 Total Amount

The total amount Sat the end of 5 years represents the principal plus thetotal interest earned for the given number of years. We say that the principalPhas accumulated to the total amount Sat the end of tyears. By definition,

S = P + I

S = P + Prt

S = P (1 + rt)

Example 1:

Mary Juan borrowd from her sister the amount of P 3,650 and promised topay at the end of 2 years and 3 months. If money is worth 12%, how muchwill she pay back at the end of the term?

Solution:

a. I = PRT

= P3,650 (.12) (2 )

= P 985.50

S = P + I

= P3,650 + 985.50

= P4,635.50

b. Using the formula

S = P (1 + rt)

= P3,650 ( 1 + .13 x 2.25)

= P 4,650.50




TOTAL AMOUNT

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Solve the following problems

1. Accumulate P3,590 at 7 % for 2 years.

2. At what rate must P8,450 be invested in order to accumulate to P8,957 in9 months.

3. An investment of P1,500 accumulates to P1,850 in 10 months. Find therate at which the money was invested?

4. Accumulate P4,200 in 3 years if it was invested at 6% for the first 15months and then at 81/2% for the remaining period.

5. An investment of P5,400 earned interest amounting to P420.75 for 11months. Find the rate of interest.

6. Mrs. Ledesma loaned P24,500 to open a beauty parlor shop. If money isworth 14%, how much will she have to pay back at the end of a year and8 months?

7. How long will it take for a given principal to earn half of itself if investedat 8%?

8. How long will it take P11,500 to accumulate to P14,087.50 if the amountis invested at 9%?

9. Mario borrowed P8,500 with a promise to pay the amount including theinterest at 12% for one year. What simple payment should she have paidat the end of the year?

SELF-CHECK 1.5



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1.6 Present Value

The amount to be invested at simple interest (r) for a given period at time (t)is called the present value (P) of the total amount (S). To compute the

present value (P) at simple interest, we have,

S = P (1 + rt)

P = S

I + rt

Example:

Find the present value of P2,317.50 due in 180 days if it is invested at 6%

Solution:

P = S

1 + rt

= P 2,317.50

1 + .06 (180/360)

= P 2,250




PRESENT VALUE

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A. Supply the blanks with the correct answers.

I P R T S

1.___________ P 2,500 8 1/2% 2 years ___________

2.___________ ___________ 7% 240 days P4,364.60

3.

P1,651.65 ___________ 6 % 3 months ___________4.___________ P12,500 ___________ 4 years P17,562.50

5.___________ ___________ 12% 45 days P984.55


1. Find the present value of P245.18 at 8% from March 23, 1986 to April 12,1987.

2. At 5% simple interest, find the present value of P325 due in 120 days.

3. How much must a man invest today at 4 1/2 % to pay an account wothP3,426 due in 1 year and 8 months?

4.

Mr. Carlos will need P50,000 in order to buy new machines 4 years fromnow. How must he invest today in the bank which gives 12% simpleinterest?

5. Mr. Macoy invested a certain amount which earns 10 1/2 % simple interest.How much did he invest if the money will mature after 18 months with anaccumulated value of P6,405?

SELF-CHECK 1.6



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1.7 Exact and Ordinary Interest, Exact and Approximate Time

The exact interest is based on using as fraction whose numerator is thenumber of days in the term of the transaction and whose denominator is theexact number of days in a year. If 360 days is used, then the interest used issaid to be ordinary.

Exact time is the actual number of days in the term of the transaction.

Approximate time arbitrarily considers each month as having 30 days.

For convention, we shall use the exact number of days between two dates.But, we shall consider the time as a whole number of months when we findthe time from a certain date in another month. For example, from June 8 toOctober 8 is 4 months. Hence, 4 times 30 equals 120. From October 8 toOctober 12 is 4 days, therefore the approximate time is 120 plus 4 equals124 days.

Another method of finding the approximate time is by subtracting the firstdate from the second date. Thus,

Year Months Date

October 12 10 12

June 8 6 8

4 mo. 4 days

(4 x 30) + 4 = 124 days




EXACT INTEREST AND APPROXIMATE TIME

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To determine the exact time between two dates, there are two methods thatwe may use. On method is simply determining the number of days frommonth to moth from the first date to the second date of the term. Anothermethod is by using Table 1 (found at the end of the module), the number ofeach day of the year.

Example 1.

Determine the exact number of days from July 2, 1986 to December 20,1986

Solution:

a. Determine the number of days from month to month

July = 29

August = 31

September = 30

October = 31

November = 30

December = 20 (last day of the transaction)

Total No. = 171 days.

b. Using Table 1

December 20 = 345thday of the year

Less: July 2 = 290thday of the year

Exact No. of days = 171 days.



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Example 2:

Find the different types of interest on P5,600 at 5% fromOctober 17, 1987 toJune 26, 1988.

Solution

a. Find the exact time using the Table:

Number of day from October 17, 1987 to December 31, 1987

December 31 - 365thday of the year

October 17 - 290th

day of the year

Exam time = 75 days

There are 177 days from January 1, 1988 to June 26, 1988. Hence, the exactnumber of days from October 17, 1987 to June 26, 1988 is 252. That is 75 days

plus 177 days.

b. Find the approximate time:

From October 17, 1987 to June 17, 1988 is 8 months. This is equivalentto 240 days. From June 17 to June 26 is 9 days. So the approximate number ofdays from October 17,1987 to June 26, 1988 is 240 days plus 9 days equals 249days

Another method is done by subtracting the first date from the second date.

Year Month Date

Second date 1987 18 26

First date 1987 10 17

Approximate time = 8 mos. 9 days



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Approximate number of days = (8 x 30) + 9 = 249 days. Note that 10 cannot besubtracted from 6, so borrow 1 year from 1988 convert to months and add to thecolumn of month, hence 6 plus 12 equals 18.

To find the simple interest between two dates, gives us four types of interest asfollows:

1. Ordinary interest on exact number of days

I = PRT

= P5,600 (.05) (252/360)= P196

2. Ordinary interest on approximate number of days

I = PRT= P5,600 (.05) (249/360)

= P193.67

3.

Exact interest on exact number of days

I = PRT= P5,600 (.05) (252/365)

= P193.32

4. Exact interest on approximate number of days

I = PRT= P5,600 (.05) (249/365)

= P191.01

Among the four types of interest, the ordinary interest on exact number of daysyields the highest interest. This type of interest is otherwise known as theBankers Rule which is widely used in business



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A. Using the table, find the exact number of days between the following dates:1) May 17, 1986 to December 3, 1986

2)November 15,1986 to August 9, 1987

3) January 22, 1984 to March 3, 1985

4) June 12, 1983 to July 15, 1986

B. Find the approximate number of days1) April 16, 1982 to October 4, 1982

2) February 11, 1986 to September 30, 1986

3)November 28, 1985 to May 25, 1986

4) October 15, 1983 to July 24, 1987

C. Using the exact time, find the ordinary interest on

1) P 3,265 at 7% from August 14, 1986 to November 28, 1987

2) P 25,500 at 91/2% from April 20,1983 to September 26, 1986

D. Find the exact interest using approximate time:

1) P7,200 at 6% from February 10,1980 to April 12, 1982

2) P10,500 at 6 % from August 24,1983 to May 14, 1985

E. Find the ordinary interest using approximate number of days.

1) P825 at 8 1/2% from March 30, 1984 to October 3, 1984

2) P3,450 at 10% from September 14, 1985 to April 24, 1986.

SELF-CHECK 1.7



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F. Find the exact interest using the exact number of days:1) P5,300 at 12 % from May 5, 1984 to November 18, 1984

2) P16,800 at 9 % from December 3, 1985 to August 18, 1987

G. Solve the following problems:

1) On November 18, 1986, Mrs. Acero borrowed P4,500 at 8% simpleinterest. If the loan was paid on May 27, 1987, how much interest shouldshe pay?

2) Mrs. Paras, who needed an additional amount of P8,260 for her business,borrowed from Mrs. Santos the said amount at 9% for 264 days. Howmuch interest did Mrs. Paras pay Mrs. Santos?

3) On December 15, 1986, Mrs Lee loaned P5,000 to pay the hospital bills.She promised to pay the amount plus the interest on April 9, 1987. how

much was the interest, if the amount is worth 9 % simple interest?

4) At the start of business, Mr. Liptona loaned P50,500 at ABC Bank whichcharges 12% simple interest. He promised to pay the amount plus theinterest at the end of one year and 275 days. How much was the interest

paid?



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d)

I = PRT= (P1,200) (.06) (6000/360)

= (P1,200) (1)

= P 1,200

Take note that the interest is simply found by multiplying the given principal by.001 for 6 days, .01 for 60 days, .1 for 600 days and by 1 for 6,000 days.

Example 2.

Find the interest earned by P5,250 at 12% for

a. 6 days b. 6000 days

Solution:

a.

I = (P5,250) (.001)

= P5.25 (interest at 6% for 6 days)

Since, 12% = 2 (6%), then the interest at 12% for 6 days is equals to theinterest at 6% for 6 days multiplied by 2. Hence,

I = (P5.25) (2)

= P10.50 (interest at 12% for 6 days



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b.

I = (P5,250) (1)= P5250 (interest at 6% for 6 days)

Since, 12% = 2 (6%), then the interest at 12% for 6 days is equals to theinterest at 6% for 6 days multiplied by 2. Hence,

I = (P5,250) (2)

= P10,500 (interest at 12% for 6 days



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A. Using the 6%-6 day method, discount each of the following amounts

1. P620 at 6% for 15 days

2. P3,847 at 6% for 130 days

3. P192.15 at 3% for 240 days

4.

P427.85 at 12% for 380 days

5. P33,712.20 at 15% for 72 days

SELF-CHECK 1.8



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2.1 Simple Discount

Normally, when a person applies for loans from banks or credits institutions, theinterest is collected in advanced which we call discount. To discount an amount isto find its value at a period earlier than its maturity date. Let S be the sum ofmoney to be discounted, d the rate of discount per annum and t the term ofdiscount in years. Then , the amount of discount represented by D is.

D = SDT

Example:

Discount the amount, P2,400 at 10% discount rate for 150 days.

Solution:

D = SDT= P2,400 (.10) (150/360)= P100

After the amount is discounted by the bank, the amount which the borrower recivesis thepresent value(P) of the amount (S). To find the present value P.

P = S D

So,P = P2,400- 100

= P2,300


SIMPLE DISCOUNT



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2.2Solve for the Discount rate (d)

To solve for the discount rate (r), divide the discount D by the product of the totalamount (S) and the term of discount. (t). Hence,

d = DS t

Example:

Find the discount rate if P820 yields a discount of P147.60 for 3 years.

Solution:

d = DS t

= P147.60P820

2.3 Solve for the Term of Discount (t)

To solve for the term of discount (t), divide the discount (D) by the product of thetotal amount (S) and the discount rate d. Thus,

t = DS d




SOLVE DISCOUNT RATE

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Example:

How long will it take P1,300 to earn P55.25 if the amount is discounted at 8 %?

Solution:

t = DS d

t = P55.25P1,300 (.085)

t = .5 years or 6 months

2.4 Solve for the Total Amount (S)

To solve for the total amount (s), divide the discount D by the product of thediscount rate (d) and the term of the discount (t). So,

S = D

d t

Example:

A certain amount is discounted at 8% for 120 days, and yields a discount of P24.Find the total amount.

Solution:

S = Dd t

= P24(.08) (120/360)

= P900



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A. Fill the space provided with the correct answer.

D S d t P

1.___________ P 820 6% 75 days ___________

2.___________ P 6,300 8% 720 days ___________

3. P33.60 ___________ 12% 120 days ___________

4. P200 ___________ 10% __________ P3,400

5. P612 P8,640 ________ 300 days __________

B. Find the present value of the following maturity value:

1.

P245.18 at 8% simple interest rate from March 29, 1987 to November 12, 1987.

2. P950 at 12 % discount rate from October 21, 1986 to August 16,1987

3. P4,200 due in 225 days at 9 % discount rate.

4. P15,500 due in 390 days at 12% simple interest rate

5. P2,350 at 8 % discount rate for 1 year and 15 days.

SELF-CHECK 2



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2.4 Promissory Notes

Whenever, a person borrows money from a creditor, that person whom who callthe debtor signs as agreement which is a promise to pay the amount at some futuretime. The agreement is called the promissory note. The date when the note wasdrawn is called the date of notewhile the date when the note matures is called thematurity dateor the due date. The amount written in the promissory note is calledface of the note. While the amount which includes the interest plus an agreed rateof interest on the face is called the maturity value. The person who signs the note isthe maker of the note.

There are two types of promissory notes, one is the interest-bearing note and theother is the non-interest bearing note. A non-interest bearing note includes theinterest in the maturity value while an interest-bearing note states the interest rate.Any note, for it to be negotiable or valid, must be dated, signed, with a fixedamount, unconditional, either demand or with a definite due date and withspecified interest rate, if any.

Example of interest bearing note:

Example of non- interest bearing note:

Diliman, Quezon CityApril 8, 1987

Five months from date, I promise to pay to the order of Juan dela Cruz twothousand six hundred pesos (P2,600) with interest at the rate of 10% per annum.

(Signed) Francisco Merced

Diliman, Quezon CityApril 8, 1987

Five months from date, I promise to pay to the order of Juan dela Cruz twothousand six hundred pesos (P2,600)

(Signed) Francisco Merced




PROMISSORY NOTE

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A person who holds a note may cash the note before the due date as the need ofcash arises. That is, he may sell the note to the bank and the bank is turn will take acertain percentage on the maturity value at a discount rate. This percentage will bededucted from the maturity value and the balance will be deducted from thematurity value and the balance will be given to the seller of the promissory note.

On the other hand, the bank then collects the maturity value from the maker of thenote on the due date.

Example:

Mr. Juan dela Cruz made a promissory note, promising to pay Mr. Garcia P2,400with an interest at 9% after 9 months. Three months after the date of note, Mr.Garcia sold the note to a bank which charges 10% discount rate. How much didMr. Garcia receive from the bank?

Solution:

b. Substitute from the formula,

P = S (1 dt)= P2,563 (1 - .10 x 6/12)= P2,562 (1 - .05)= P2,433.90

a. The maturity value S of the note is,

S = P (1 + rt)= P2,400 (1 + .09 x 9/12)= P2,562

From the time the note was sold to the timeit matures, the bank discounts it for 6months. Thus,

D = Sdt

D= P2,562 x .10 x 6/12= P128.10

P = S D= P2,562 128.10= P2,433.90



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Solve the following problems.

1. Mr. Juan promises to pay to the order of Mr. Lee (P6,500 with 12% simpleinterest after 6 months. Three months after, Mr. Lee sold the note to a bankwhich discounts it at 5% simple discount. How much will Mr. Lee receive fromthe bank?

2. On February 16, 1988 Mrs. Lope borrowed P50,000 from Mrs. Cunate. Shethen made the following note:

a. Find the maturity date.b. Find the maturity value.c. On July 10, 1988 Mrs. Cunate sold the note to ACE Bank which discounts it

at 6% simple discount. How much will she receive from the purchase of thenote?

3. From problem no. 2, ACE Bank in turn sold the note to ABC Bank whichrediscounts it at 3% discount rate. How much will ACE Bank receive?

Sampaloc, ManilaFeb. 16, 1988

Two hundred ninety five days after date, I promise to pay Mrs. Cunate fiftythousand pesos only (P50,000) with 10% simple interest per annum.

Signed: Mrs. Lope



SELF-CHECK 2.1

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Business and banking transactions become more attractive to businessmen anddepositors with the use of compound interest. Compound interest methods ofcomputing gains and profits give more benefits to lenders, depositors and thoseengaged in business.

3.1 What is COMPOUND INTEREST?

When interest is regularly added to the principal and this new sumbecomes theprincipal for the following time period and the process is repeated periodically, the

final amount si called compound amount.

Compound interest is usually paid on loans, mortgages, bank deposits, insurancepremiums and in calculating sinking funds.

The regular or periodic interval of time per year is called conversion period orinterst period. This is denoted by the symbol m. This is usally quarterly, whereinterest is computed for every three months or four times a year. Quarterly issymbolized by m = 4. This may also be semi-annually m = 2 or monthy m= 12. If no

period is stated, it is understood that the interest period is annually (m=1).

The formula is the principal (P) will be multiplied by n factors (1 + i) nHence, the compound amount at the end of n period is P (1 + i ) n

S = P (1 + i ) n

Where:

S = final compound amountP = Principal

i = interest rate per conversion periodsn = the total number of conversion periods for entire term of the t intervals.

Note: The total number of conversion period (n) for the entire term of the transactionis the product of the conversion period per year (m) and the entire term expressed interms of years (t)


COMPOUND INTEREST



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If interest is compounded more than once a year, it is conventional to quote thereate on annual basis. Such a rate is known as nominal rate, denoted by i. Forexample, a rate of 6% compounded sem-annually means actually 3% is the rate perconversion period, denoted by i. The rate per conversion period can be found bydividing the nominal rate by the conversion period per year, (i = j/m)

Example 1:

To what amount will P1,000 accumulate in one year if invested at 8% compoundedquarterly?

Given:

P= P1,000j = 8%m = 4 (quarterly)t = 1 year

Solution:

a.) i = j = 8% = 2%m 4

b.) n = mt = 4(1) = 4c.) S = P (1 + i ) n = P1,000 (1 + 2%)4

= P1,000 (1.08243216)= P1,082.43

Example 2:

Accumulate P8,500 at 12% m=12 for 10 years



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Given:P= P8,500

j = 12%m = 12t = 10 year

Solution:

a.) i = j = 12% = 1%m 12

b.) n = mt = 12(10) =120c.) S = P (1 + i ) n = P8,500 (1 + 1%)120

= P8,500 (3.30038689)= P28,053.29

Example 3:

Find the compound amount if 1 years if P500 is invested at 4% compounded

monthly.

Given:

P= P500j = 4%m = 12t = 1 years

Solution:

a.) i = j = 4% = 1/3%m 12

b.) n = mt = 12(1 ) = 18c.) S = P (1 + i ) n = P500 (1 + 1/3%)18

= P500 (1..061731)= P530.87



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A. Find the total number of conversion periods (n)quarterly for 8 years

A.semi-annually for 36 monthsB.weekly for 2 yearsC.quarterly for 15 years and 9 monthsD.monthly ofr 13 years and 5 months

B. Find the interest rate per conversion period (i)

15% compounded quarterly

18% compounded monthly

3% compounded semi-annually

12% m=6

12.46% m=3

Solve for the accumulated value of the following amounts..1. P12,000 for 6 yearsat 8% m=122. 4,500 for 12 years at 12% m = 43. P9,000 for 2 years at 8% compounded

a. monthly b. Quarterly c. Semi-annually d. Annually

4.

P25,000 for 4 years at 7% compoundeda. monthly b. Quarterly c. Semi-annually d. Annually

A.PP5,600 for 8 years and 3 months at 7% compounded semi-annually

D. Solve the following problems.1. Ms. Paulin Nicole invested P700,000 at 9% compounded semi-annually for

6 years and 3 months. Find the amount2. Erwin invested P650,000 for years and 10 months at 7% comounded

quarterly. Find the amount

3.

Mr. Ric Pangan borrowed P120,000 on May 31, 2006 at 12% compoundedquarterly . What is the amount of the loan on July 31, 2008?4. What amount of money is required to repay a loan of P36,000 on July 1,

2007, if the loan was made on Oct. 1, 2002 at the interest rate of 10%compounded semi-annually.

5. A note today has a face value of P7,500 bears interest at 8 1/2%compounded semi-annually. Find the amount.

SELF-CHECK 3



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3.2 Present Value of an Amount

the present value P is that principal which is invested on the given date at thegiven interest rate will accumualate to a specific amount S. At some later date.

To compute for the value of P we use the formula for accumulation .S = P (1 + i )

nand placed P as the unknown, thus,

P = S(1 + i ) n

To avoid long division , we write this expression using negative exponentwhich gives the formula

P = S (1 + i ) -n

Example 1.

If money is worth 6% m = 2 find the present value of P2,000 whic is due in 5years.

Given:j = 6%m = 2S = P2,000t = 5 years

Solution:

a.) i = j = 6% = 3%m 2

b.) n = mt = 2(5 ) = 10c.) P = S (1 + i ) -n = P2,000 (1 + 3%)-10

= P2,000 (.744094)= P1,488.19



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Example 2.

A non-interest bearing note whose face value is P6,250 is sold to a bank 19 montsbefore maturity. If money is worth 8% m=4 how much will the bank pay for thenote

Given:j = 8%m = 4S = P6,250t = 5 years

Solution:

a.) i = j = 8% = 2%m 4

b.) n = mt = 4(19/12 ) = 6 1/3c.) P = S (1 + i ) -n = P6,250 (1 + 2%)-6

(1 + 2%) 1/3

= P6,250 (.88791)1.006623

= P5,513.30



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A. Find the present value of each of the following amounts

Amount Nominal Rate Frequency of Conversion Term

1. P 600 8% quarterly 6 years2. P4,820 12% monthly 10 years3. P517.50 9% quarterly 4 years4. P3,629 8 % semi-annually 18 months5. P55,000 12% every 2 months 3 years & 8 mos.


1. What amount must be invested today at 12% compounded monthly in order toaccumulate to P6,000 in 5 years?

2. How much must a corporation deposit on a bank which credits interest at 18% m = 12. tocome up with P30,000 in 6 years needed for its expansion program?

3. Mr. dela Rosa wishes to invest a sum on his sons 7thbirthday in a trust fund which gives

9% interest compounded quarterly. How much must he invest if he wants the money toamount to P20,000 by the time his son reaches his 18thbirthday?

SELF-CHECK 3.1



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3.3 Finding the Unknown Rate

There are some investment problems that will ask for the value of the nominal rateor quoted rate (j) and the value of the interest per period (i). These problemsusually look for these rates for purposes of information and comparison. If P, S,and n are given it is possible to solve for the values of i and j.

To compute for the nominal rate, divide accumulated amount (S) to principal (P)then look for the nth root (n), subtract the result to 1. To get the nominal ratemultiply it to periodic conversion (m). Below is the formula

i = n S - 1P

j = i (m)

Example 1:

At what nominal rate compounded quarterly will P500 accumulate to P760 in 5years?

Given:P = P500S = P760m = 4n = 5 years

Solution:

i = n S - 1P

= 20 P760 - 1P500



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= 20 1.52 - 1

= 1.021156 1= .021156 x 4= 8.46%



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A. Find the value of i and j in each of the following:

P S m Term

6. P 400 P 620 quarterly 3 years7. P6,280 P8,400 semi-annually 5 years8.

P5,500 P 7520.36 monthly 2 years9. P731.24 P1024.12 annually 10 years10.P1085 P2247.76 quarterly 27 years


1. At what nominal rate compounded monthly will P4720.16 accumulate to P6,285.45 in 2years and 8 months?

2. At what rate, converted semi-annually is P896 the present value of P1104 which is due in8 years?

3. One investment pays 8% annual interest. Another investment accumulates P4000 toPP6200 in 7 years. Which of the two gives higher annual income?

4. If an investment increases from P68000 to P83000 in 4 years what is the nominal rate ofinterest compounded quarterly?

5. Find the annual rate of interest if an investment increases 25% in 6 years?



SELF-CHECK 3.2

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Everyday living entails monthly rentals for persons dwelling installment paymentsfor a car or some household appliances. The common housewife is usuallyconfronted with the question of how to budget the family income in such a waythat the basic needs of the family would be attended to.

Aside from the basic financial needs which must be attended to, a certain amountmust be set aside for some future use. A regular monthly, bi-monthly (or whatever)savings ensure a person of a modest amount in the future.

4.1 Simple Annuities

An annuity is a series of periodic payments (usually of equal amount) made atregular intervals of time. Installment payments for appliances, housing loans,memorial plans and monthly rentals are common examples of annuities.

A simple annuity is one wherein the payment interval and the interest conversion

period coincide.

The payment interval is the period of time between consecutive payments. It maybe of any convenient length like monthly, quarterly, semiannually and annually.

The term of an annuity extends from one payment interval before the first paymentup to the day of the last payment. Unless otherwise specified, the term of anyannuity is presumed to begin immediately.


SIMPLE ANNUITY



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TYPES OF ANNUITIES

1.Annuity certain is an annuity payable for a definite length of time notdependent on some outside contingency. It means that payment beginsand ends on definite or fixed dates. Examples are monthly payment on ahousing loan and monthly payment on a car loan since the payment startson a fixed date and continues until the required number of payments has

been made.

2. Contingent annuity is one whose payment extend over a period of timewhose length cannot be foretold accurately. Examples are pensions andlife insurance policies.

4.2 Solving for the Amount (S) of an annuity

The amount of an annuity (S) is the total value of all payments at the end of theterm. It can also be called as the accumulated value of the annuity. Use this

formula,

Example

Mrs. Cruz regularly deposits P1,000 at the end of each six months for 3 years. Thebank gives interest at 8%, m = 2. How much will be in her account at the end of thethird year?

S = R ( 1 + i) n-1i

n i



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Given:

R = P1000n = 6i = 4%

Solution

S = R ( 1 + i)n-1

in i

= P1000 ( 1 + .04)6-1

.04

= P1000 (1.265319 1).04

= P6,632.98



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A. Solve for the amount (S) of the following annuities:

1. P2300 paid at the end of each 6 months for 4 years with interest at 11%, m =2

2. P5500 regular monthly deposit in a bank which credits interest at 12%,m = 6 for 5 years.

3. P8000 due at the beginning of each 3 months for 3 years. Money is worth6%,m = 4

4. An annual payment of P200 every three months for 2 years. Money is worth4%, m = 4

5. A payment of P 2000 every three months for 2 years. Money is worth 4%,

B. Solve the following problems completely:

1.

Mrs. Reyes bought a sala set and paid P5000 as downpayment and promisesto pay P250 at the end of each three months for 2 years. If money is worth5% compounded quarterly, how much is the total payment for the sala set?

2. The contract for the sale of farm machineries calls for a downpayment ofP5,600 plus a quarterly payment of P2,250 for 5 years. If money is worth10%, m=4, how much is the total payment for the machineries?

3. A man made semiannual deposits of P2000 into a savings fund that paysinterest at 9% compounded twice a year. How much will be his savings in 8years? If after the eight year, no deposit and withdrawal have been made,how much will be in his account at the end of the twelfth year?

4. The buyer of a house and lot agrees to pay P2,830 at the beginning of eachmonth for 8 years. If money is worth 18% compounded monthly, find thetotal cost of the house and lot.

SELF-CHECK 4



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4.3 Solving for the Present Value (A) of an annuity

The present value (A) of an annuity is the total value of all payments computed atthe beginning of the term. It is the sum of the discounted values of the payments ofthe annuity at the beginning of the term.

Use this formula,

Example:

The contract for the sale of a lot calls for a monthly payment of P500 for 5 years. Ifmoney is worth 12% compounded monthly, what is the cash price of the lot

Solution:R = P500 n = 60 i = 1%

b. A = R 1 -( 1 + i)- n

i

n i

= P500 1 ( 1 + .01)- 60.01

= P500 1 0.550450.01

= P500(44.955048)= P22,477.52

A = R 1 -( 1 + i)- n

i

n i

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A. Solve for the present value (A) of the following annuities.

1. A quarterly payment of P1500 for 15 years if money is worth 6%, m = 2

2. P1000 paid at the end of every year for 25 years with interest at 4%, m=12

3. P7500 deposited monthly for 20 years for 1 year if money is worth 12%compounded monthly

4. A monthly payment of P2000 for 2 years if money is worth 6%, m =12

5. A semiannual deposit of P1,500 for 7 years if money is worth 5% m=2

B. Solve the following problems completely

1. The buyer of a farm pays P42000 downpayment and promises to pay P8000at the end of each 6 months for 7 years. If money is worth 8% compoundedsemiannually, find the equivalent cash price.

2. A sala set is bought for P6000 downpayment and P240 a month for 8

months. What is the equivalent cash price of the sala set if money is worth24% compounded monthly.

3. A man agrees to pay P2000 at the end of each month for 20 years inpurchasing a house. Find the present value of this agreement if money isworth 3% compounded monthly.

4. Upon retirement, Mrs. Reyes finds that her company pension calls forpayments of P2000 to her (or to her estate if she dies) at the beginning ofeach month for 25 years. Find the present value of this pension if money isworth 4% compounded monthly.

5. A stereo set is offered for sale for P4000 downpayment and P1,200 everythree months for the balance for 5 years. If interest is to be computed at 6%,m=4, what is the cash price of the set.

SELF-CHECK 4.1

Module 5. Using Mathematical Concepts and Techni



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TABLE 1.Module 5. Using Mathematical Concepts and Techni



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Module 5 - Using Mathematical Concepts and Techniques

Recording Sheet For Oral Questioning / Interie!

Student name"

Module Title/#o"

Quali$ication"

Oral/interie! questions Satis$actor% response

&es #o

1.

2.

3.

4.

5.

The student's underpinning knowledge was:

Satisfactory Not satisfactory

Student's Signature" (ate

Trainor's signature" (ate"

)ccepta*le ans!ers are"




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Rating SheetModule 5 - Using Mathematical Concepts and Techniques

Performance

Feedback

Remarks

S NS C NYC

1. Self-Check 1.1

2. Self-Check 1.2

3. Self-Check 1.3

4. Self-Check 1.4

5. Self-Check 1.5

6. Self-Check 1.6

7. Self-Check 1.7

8. Self-Check 1.8

9. Self-Check 2

10. Self-Check 2.1

11. Self-Check 3

12. Self-Check 3.1

13. Self-Check 3.2

14. Self-Check 3.3

15. Self-Check 4

16. Self-Check 4.1

S - SatisfactoryNS Not Satisfactory

C - Comlete!

N"C Not "et Comlete!

odule is

!o"pleted Not #et !o"pleted

$e"arks:

Student's Signature (ate


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Module 5 Using Mathematical Techniques