Module 5 Using Mathematical Techniques
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ASIAN ACADEMY OF BUSINESS AND COMPUTERS COMPETENCYBASED LEARNING MODULE ON MAT 111 BUSINESS MATHEMATICS
MODULE 5 USING MATHEMATICAL CONCEPTS AND TECHNIQUES
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
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MODULE TITLE
USING MATHEMATICAL CONCEPTS AND TECHNIQUES
DESCRIPTION: the
This module covers the knowledge, skills and attitudes required in application of mathematical concepts and techniques.
COURSE OBJECTIVE:
Students completing this module will be able to: Identify mathematical tools and techniques to solve problem; Apply mathematical procedure/solution; and Analyze results.
CONTENTS:
Information Sheet 1.1 – Simple Interest SelfCheck 1.1 Information Sheet 1.2 – Solving the Principal SelfCheck 1.2 Information Sheet 1.3 – Solving the Interest Rate SelfCheck 1.3 Information Sheet 1.4 – Solving the Time SelfCheck 1.4 Information Sheet 1.5 – Total Amount SelfCheck 1.5 Information Sheet 1.6 – Present Value SelfCheck 1.6 Information Sheet 1.7 – Exact Interest and Approximate Time SelfCheck 1.7 SelfCheck 1.8 Information Sheet 2.1 – Simple Discount Information Sheet 2.2 – Solve Discount Rate SelfCheck 2 Information Sheet 2.3 – Promissory Note SelfCheck 2.1 Information Sheet 3.1 – Compound Interest SelfCheck 3 SelfCheck 3.1 SelfCheck 3.2 Information Sheet 4.1 – Sample Annuity SelfCheck 4 SelfCheck 4.1
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MODULE 5 USING MATHEMATICAL CONCEPTS AND TECHNIQUES
LEARNING GUIDE Learning Steps
1. Read Information Sheet 1 on Simple Interest 2. Answer SelfCheck
Resources
Information Sheet 1 SelfCheck 1
3. Read Information Sheet 2 on Simple DIscount
Information Sheet 2
4. Answer SelfCheck
SelfCheck 2
5. Read Information Sheet 3 on Compound interest
Information Sheet 3
6. Answer SelfCheck
SelfCheck 3
7. Read Information Sheet 4 on Simple Annuity
Information Sheet 4
8. Answer SelfCheck
SelfCheck 4
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 1.1 SIMPLE INTEREST
1.1 Simple Interest Simple interest is the most basic type of interest. In order to understand how various types of transactions work, it helps to have a complete understanding of simple interest. For example, you may pay interest on a loan, and it is important to understand how interest works. Better yet, your bank may be paying you interest on your deposits – and you can maximize your earnings by knowing more about interest. Simple Interest Overview Simple interest is just the amount of money paid on a loan. It is the easiest type of interest to calculate and understand Simple Interest Formula If you want to calculate simple interest, use this formula: I=P r t In other words Interest (I) is calculated by multiplying Principal (p) times the Rate (r) times the number of Time (t) periods. For example, if I invest $100 (the Principal) at a 5% annual rate for 1 year the simple 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 you’re paying it or earning it – is usually calculated using different methods. However, simple interest is a good start that gives us a general idea of what a loan will cost or what an investment will give us. The main limitation that you should keep in mind is that simple interest does not take compounding into account. ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
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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Module 5 Using Mathematical Concepts and Techniques
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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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.1
A. Compute for the interest. I
P
R
T
1. ____________
P 2,400
6%
2 years
2. ____________ months
P 1,670
4 ½%
10
3. ____________ mos.
P 24,000
8 ¼%
4. ____________
P 16,000
5.25%
8 months
5. ____________
P 14,500
11%
5 years
2 years and 3
B. Solve the following problems: 1. Mrs. Loida Aquino invested P8,250 in a bank which yields 6 ½% simple interest for 27 months. 2. Find the total interest earned from P27,000 investment, if invested at 9 ½% interest for 18 months. 3. A man borrowed P125,000 to start a business. Find the interest if money is invested at 8% for 8 months. 4. For a period of 3 years, money worth P 148,000 is invested at 9 ¾%. Find the interest. 5. Mrs. Guzman invested P 4,850 in a bank with an interest of 5%, how much will be the interest if invested for 7 years.
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 1.2
SOLVING THE PRINCIPAL
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 at the rate of 6% simple interest. Given: I = P600 R = 6% T = 4 years P=? Solution P=
I RT
=
P600 (.06)(4years)
= P2,500
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Module 5 Using Mathematical Concepts and Techniques
Example 2. How much should Mr. Cruz invest so that his money earns P1,408 borrowed at 8% for two years and 9 months. Given: I = P1,408 R = 8% T = 2 years and 9 months P=? Solution: P=
I RT
=
P1408 (.08)(2 years and 9 months) or (11/4)
= P6,400
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.2
A. Compute for principal. I
P
R
T
1. P 2,400
____________
6%
2 years
2. P11,000 months
____________
4 ½%
10
3. P 4,580 mos.
____________
8 ¼%
4. P 26,000
____________
5.25%
8 months
5. P 4,500
____________
11%
5 years
2 years and 3
B. Solve the following problems: 1. How much should Mr. Cruz invest so that his money earns P 8,980 borrowed at 4% for 7 years 4 months. 2. Find the principal if money earned P700 3% interest for 18 months. 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% simple interest earning P300 . 5. Find the principal if money is invested for 15 years at 3% simple interest earning P5750
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 1.3
SOLVING THE INTEREST RATE
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%
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Module 5 Using Mathematical Concepts and Techniques
Example 2. At what rate should one invest in a bank so that P6000 earns P1,537.50 for a period of 3 years and 5 months? Given: I = P1,537.50 T = 3 years and 5 months P = P6,000 R=? Solution: R=
I PT
=
P1,537.50 (6000) (.075) (3 5/12)
= 7.5%
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.3
A. Compute for rate I
P
R
T
1. P 210
P 1,680
___________
2 ½ years
2. P 500
P 4,100
___________
9 months
3. P 3,000 mos.
P 16,000
___________
2 years and 6
4. P 500
P 4,000
___________
5 years
5. P 252
P 3,500
___________
4 years
B. Solve the following problems: 1. Mrs. Guzman invested P4,850 in a bank which gave her an interest amounting 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 interest amounting 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 2 months? 4. Mr. Cruz earned P2,500 in a borrowed money amounting to P25,000 in 2 years. What is the interest rate? 5. Find the rate of interest after 5 years and 6 months, if money is worth P8,500 with an earning of P560. ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 1.4
SOLVING THE TIME
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% simple interest? Given: I = P200 P = P2,500 R = 8% T=? Solution: T= =
I PR P200 (P2,500) (8%)
= 3 years ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
Example 2. At what length of time must P4,500 be invested at 91/2% simple interest to earn P496.25? 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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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.4
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%
___________
B. Solve the following problems: 1. At what length of time must P 4,850 be invested at 12 ½% simple interest to 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 of time if the interest is P200. 4. At what length of time must P8,000 at 8% simple interest if money earned P1,584? 5. Find the length of time if money amounting to P2,500 earned P600 at 6% simple interest.
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 1.5
TOTAL AMOUNT
1.5 Total Amount The total amount S at the end of 5 years represents the principal plus the total interest earned for the given number of years. We say that the principal P has accumulated to the total amount S at the end of t years. 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 to pay at the end of 2 years and 3 months. If money is worth 12%, how much will she pay back at the end of the term? Solution: a. I = PRT = P3,650 (.12) (2 ¼) = 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
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.5
Solve the following problems 1. Accumulate P3,590 at 7 ½% for 2 ½ years. 2. At what rate must P8,450 be invested in order to accumulate to P8,957 in 9 months. 3. An investment of P1,500 accumulates to P1,850 in 10 months. Find the rate at which the money was invested? 4. Accumulate P4,200 in 3 years if it was invested at 6% for the first 15 months and then at 81/2% for the remaining period. 5. An investment of P5,400 earned interest amounting to P420.75 for 11 months. Find the rate of interest. 6. Mrs. Ledesma loaned P24,500 to open a beauty parlor shop. If money is worth 14%, how much will she have to pay back at the end of a year and 8 months? 7. How long will it take for a given principal to earn half of itself if invested at 8%? 8. How long will it take P11,500 to accumulate to P14,087.50 if the amount is invested at 9%? 9. Mario borrowed P8,500 with a promise to pay the amount including the interest at 12% for one year. What simple payment should she have paid at the end of the year?
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 1.6
PRESENT VALUE
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
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.6
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 ½%
3 ½ months ___________
4. ___________
P12,500
___________ 4 ½ years
P17,562.50
12%
P984.55
5. ___________ ___________
45 days
B. Solve the following problems: 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 woth P3,426 due in 1 year and 8 months? 4. Mr. Carlos will need P50,000 in order to buy new machines 4 ½ years from now. How must he invest today in the bank which gives 12% simple interest? 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 an accumulated value of P6,405? ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 1.7
EXACT INTEREST AND APPROXIMATE TIME
1.7 Exact and Ordinary Interest, Exact and Approximate Time The exact interest is based on using as fraction whose numerator is the number of days in the term of the transaction and whose denominator is the exact number of days in a year. If 360 days is used, then the interest used is said 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 find the time from a certain date in another month. For example, from June 8 to October 8 is 4 months. Hence, 4 times 30 equals 120. From October 8 to October 12 is 4 days, therefore the approximate time is 120 plus 4 equals 124 days. Another method of finding the approximate time is by subtracting the first date from the second date. Thus, Year October 12 June 8
Months
Date
10
12
6
8
4 mo.
4 days
(4 x 30) + 4 = 124 days
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Module 5 Using Mathematical Concepts and Techniques
To determine the exact time between two dates, there are two methods that we may use. On method is simply determining the number of days from month to moth from the first date to the second date of the term. Another method is by using Table 1 (found at the end of the module), the number of each 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 Less:
December 20
=
345th day of the year
July 2
=
290th day of the year
Exact No. of days =
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171 days.
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Module 5 Using Mathematical Concepts and Techniques
Example 2: Find the different types of interest on P5,600 at 5% fromOctober 17, 1987 to June 26, 1988. Solution a.
Find the exact time using the Table: Number of day from October 17, 1987 to December 31, 1987 December 31

365th day 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 exact number 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 equivalent to 240 days. From June 17 to June 26 is 9 days. So the approximate number of days from October 17,1987 to June 26, 1988 is 240 days plus 9 days equals 249 days 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
8 mos.
9 days
Approximate time = ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
Approximate number of days = (8 x 30) + 9 = 249 days. Note that 10 cannot be subtracted from 6, so borrow 1 year from 1988 convert to months and add to the column of month, hence 6 plus 12 equals 18. To find the simple interest between two dates, gives us four types of interest as follows: 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 days yields the highest interest. This type of interest is otherwise known as the Banker’s Rule which is widely used in business
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.7
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 days 1) 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 ½% 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.
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Module 5 Using Mathematical Concepts and Techniques
F. Find the exact interest using the exact number of days: 1) P5,300 at 12 ½% from May 5, 1984 to November 18, 1984 2) P16,800 at 9 ½% 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% simple interest. If the loan was paid on May 27, 1987, how much interest should she 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. How much 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 ½% simple interest? 4) At the start of business, Mr. Liptona loaned P50,500 at ABC Bank which charges 12% simple interest. He promised to pay the amount plus the interest at the end of one year and 275 days. How much was the interest paid?
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Module 5 Using Mathematical Concepts and Techniques
1.8
6%, 6Day Mehod for Computing Simple Interest
The use of the 6%, 6day method of computing interest makes computation easier. This method has varied application as illustrated by the following examples: Example 1:
Find the interest earned from P1,200 at 6% for
a. 6 days
c. 600 days
b. 60 days
d. 6,000 days
Solution: a) I = PRT = (P1,200) (.06) (6/360) = (P1,200) (.001) = P 1.20 b) I = PRT = (P1,200) (.06) (60/360) = (P1,200) (.01) = P 12 c) I = PRT = (P1,200) (.06) (600/360) = (P1,200) (.1) = P 120
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Module 5 Using Mathematical Concepts and Techniques
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 the interest 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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Module 5 Using Mathematical Concepts and Techniques
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 the interest 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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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 1.8
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
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 2.1 SIMPLE DISCOUNT
2.1 Simple Discount Normally, when a person applies for loans from banks or credits institutions, the interest is collected in advanced which we call discount. To discount an amount is to find its value at a period earlier than its maturity date. Let S be the sum of money to be discounted, d the rate of discount per annum and t the term of discount 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 recives is the present value (P) of the amount (S). To find the present value P. P=S–D So, P = P2,400 100 = P2,300
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 2.2 SOLVE DISCOUNT RATE
2.2 Solve for the Discount rate (d) To solve for the discount rate (r), divide the discount D by the product of the total amount (S) and the term of discount. (t). Hence, d= D St
Example: Find the discount rate if P820 yields a discount of P147.60 for 3 years. Solution: d=
D St = P147.60 P820
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 the total amount (S) and the discount rate d. Thus, t=
D Sd
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Module 5 Using Mathematical Concepts and Techniques
Example: How long will it take P1,300 to earn P55.25 if the amount is discounted at 8 ½%? Solution: t=
D Sd t = P55.25 P1,300 (.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 the discount rate (d) and the term of the discount (t). So, S=
D dt
Example: A certain amount is discounted at 8% for 120 days, and yields a discount of P24. Find the total amount. Solution: S=
D dt = P24 (.08) (120/360) = P900
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 2
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 ½% discount rate from October 21, 1986 to August 16,1987 3. P4,200 due in 225 days at 9 ½% discount rate. 4. P15,500 due in 390 days at 12% simple interest rate 5. P2,350 at 8 ½% discount rate for 1 year and 15 days.
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 2.3 PROMISSORY NOTE
2.4 Promissory Notes Whenever, a person borrows money from a creditor, that person whom who call the debtor signs as agreement which is a promise to pay the amount at some future time. The agreement is called the promissory note. The date when the note was drawn is called the date of note while the date when the note matures is called the maturity date or the due date. The amount written in the promissory note is called face of the note. While the amount which includes the interest plus an agreed rate of interest on the face is called the maturity value. The person who signs the note is the maker of the note. There are two types of promissory notes, one is the interestbearing note and the other is the noninterest bearing note. A noninterest bearing note includes the interest in the maturity value while an interestbearing note states the interest rate. Any note, for it to be negotiable or valid, must be dated, signed, with a fixed amount, unconditional, either demand or with a definite due date and with specified interest rate, if any. Example of interest bearing note: Diliman, Quezon City April 8, 1987 Five months from date, I promise to pay to the order of Juan dela Cruz two thousand six hundred pesos (P2,600) with interest at the rate of 10% per annum. (Signed) Francisco Merced Example of non interest bearing note: Diliman, Quezon City April 8, 1987 Five months from date, I promise to pay to the order of Juan dela Cruz two thousand six hundred pesos (P2,600) (Signed) Francisco Merced ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
A person who holds a note may cash the note before the due date as the need of cash arises. That is, he may sell the note to the bank and the bank is turn will take a certain percentage on the maturity value at a discount rate. This percentage will be deducted from the maturity value and the balance will be deducted from the maturity 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 the note on the due date. Example: Mr. Juan dela Cruz made a promissory note, promising to pay Mr. Garcia P2,400 with 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 did Mr. Garcia receive from the bank? Solution: 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 time it matures, the bank discounts it for 6 months. Thus,
b. Substitute from the formula, P = S (1 –dt) = P2,563 (1  .10 x 6/12) = P2,562 (1  .05) = P2,433.90
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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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 2.1
Solve the following problems. 1. Mr. Juan promises to pay to the order of Mr. Lee (P6,500 with 12% simple interest after 6 months. Three months after, Mr. Lee sold the note to a bank which discounts it at 5% simple discount. How much will Mr. Lee receive from the bank? 2. On February 16, 1988 Mrs. Lope borrowed P50,000 from Mrs. Cunate. She then made the following note: Sampaloc, Manila Feb. 16, 1988 Two hundred ninety five days after date, I promise to pay Mrs. Cunate fifty thousand pesos only (P50,000) with 10% simple interest per annum. Signed: Mrs. Lope
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 the note? 3. From problem no. 2, ACE Bank in turn sold the note to ABC Bank which rediscounts it at 3% discount rate. How much will ACE Bank receive?
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 3.1 COMPOUND INTEREST
Business and banking transactions become more attractive to businessmen and depositors with the use of compound interest. Compound interest methods of computing gains and profits give more benefits to lenders, depositors and those engaged in business. 3.1 What is COMPOUND INTEREST? ● When interest is regularly added to the principal and this new sumbecomes the principal 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, insurance premiums and in calculating sinking funds. ● The regular or periodic interval of time per year is called conversion period or interst period. This is denoted by the symbol m. This is usally quarterly, where interest is computed for every three months or four times a year. Quarterly is symbolized by m = 4. This may also be semiannually 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) n Hence, the compound amount at the end of n period is P (1 + i ) n S = P (1 + i ) n Where: S = final compound amount P = Principal i = interest rate per conversion periods n = 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 transaction is the product of the conversion period per year (m) and the entire term expressed in terms of years (t)
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Module 5 Using Mathematical Concepts and Techniques
If interest is compounded more than once a year, it is conventional to quote the reate on annual basis. Such a rate is known as nominal rate, denoted by i. For example, a rate of 6% compounded semannually means actually 3% is the rate per conversion period, denoted by i. The rate per conversion period can be found by dividing 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% compounded quarterly? Given: P= P1,000 j = 8% m = 4 (quarterly) t = 1 year Solution: a.) i = j = 8% = 2% m 4 b.) n = mt = 4(1) = 4 c.) 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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Module 5 Using Mathematical Concepts and Techniques
Given: P= P8,500 j = 12% m = 12 t = 10 year Solution: a.) i = j m
= 12% = 1% 12
b.) n = mt = 12(10) =120 c.) 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 ½ years if P500 is invested at 4% compounded monthly. Given: P= P500 j = 4% m = 12 t = 1 ½ years Solution: a.) i = j = 4% = 1/3% m 12 b.) n = mt = 12(1 ½ ) = 18 c.) S = P (1 + i ) n = P500 (1 + 1/3%)18 = P500 (1..061731) = P530.87 ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 3
A. Find the total number of conversion periods (n)quarterly for 8 years A. semiannually for 36 months B. weekly for 2 years C. quarterly for 15 years and 9 months D. monthly ofr 13 years and 5 months B. Find the interest rate per conversion period (i) • 15% compounded quarterly • 18% compounded monthly • 3% compounded semiannually • 12% m=6 • 12.46% m=3 • 1. 2. 3.
Solve for the accumulated value of the following amounts.. P12,000 for 6 yearsat 8% m=12 4,500 for 12 years at 12% m = 4 P9,000 for 2 years at 8% compounded a. monthly b. Quarterly c. Semiannually d. Annually 4. P25,000 for 4 ½ years at 7% compounded a. monthly b. Quarterly c. Semiannually d. Annually A. PP5,600 for 8 years and 3 months at 7% compounded semiannually D. Solve the following problems. 1. Ms. Paulin Nicole invested P700,000 at 9% compounded semiannually for 6 years and 3 months. Find the amount 2. 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% compounded quarterly . 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 semiannually. 5. A note today has a face value of P7,500 bears interest at 8 1/2% compounded semiannually. Find the amount.
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Module 5 Using Mathematical Concepts and Techniques
3.2 Present Value of an Amount the present value P is that principal which is invested on the given date at the given 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 ) n and placed P as the unknown, thus, P =
S (1 + i ) n
To avoid long division , we write this expression using negative exponent which 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 5 years. Given: j = 6% m=2 S = P2,000 t = 5 years Solution: a.) i = j = 6% = 3% m 2 b.) n = mt = 2(5 ) = 10 c.) P = S (1 + i ) n = P2,000 (1 + 3%)10 = P2,000 (.744094) = P1,488.19 ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
Example 2. A noninterest bearing note whose face value is P6,250 is sold to a bank 19 monts before maturity. If money is worth 8% m=4 how much will the bank pay for the note Given: j = 8% m=4 S = P6,250 t = 5 years Solution: a.) i = j = 8% = 2% m 4 b.) n = mt = 4(19/12 ) = 6 1/3 c.) 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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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 3.1
A. Find the present value of each of the following amounts
1. 2. 3. 4. 5.
Amount
Nominal Rate
Frequency of Conversion
P 600 P4,820 P517.50 P3,629 P55,000
8% 12% 9% 8 ½% 12%
quarterly monthly quarterly semiannually every 2 months
Term 6 years 10 years 4 ½ years 18 months 3 years & 8 mos.
B. Solve the following problems: 1. What amount must be invested today at 12% compounded monthly in order to accumulate to P6,000 in 5 years? 2. How much must a corporation deposit on a bank which credits interest at 18% m = 12. to come up with P30,000 in 6 years needed for its expansion program? 3. Mr. dela Rosa wishes to invest a sum on his son’s 7th birthday in a trust fund which gives 9% interest compounded quarterly. How much must he invest if he wants the money to amount to P20,000 by the time his son reaches his 18th birthday?
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Module 5 Using Mathematical Concepts and Techniques
3.3 Finding the Unknown Rate There are some investment problems that will ask for the value of the nominal rate or quoted rate (j) and the value of the interest per period (i). These problems usually 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 rate multiply it to periodic conversion (m). Below is the formula i = n
S
 1
P j = i (m) Example 1: At what nominal rate compounded quarterly will P500 accumulate to P760 in 5 years? Given: P = P500 S = P760 m=4 n = 5 years Solution: i = n
S
 1
P = 20 P760  1 P500 ASIAN ACADEMY OF BUSINESS AND COMPUTERS
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Module 5 Using Mathematical Concepts and Techniques
= 20
1.52
 1
= 1.021156 – 1 = .021156 x 4 = 8.46%
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 3.2
A. Find the value of i and j in each of the following: P 6. P 400 7. P6,280 8. P5,500 9. P731.24 10. P1085
S
m
Term
P 620 P8,400 P 7520.36 P1024.12 P2247.76
quarterly semiannually monthly annually quarterly
3 years 5 years 2 ½ years 10 years 27 years
B. Solve the following problems: 1. At what nominal rate compounded monthly will P4720.16 accumulate to P6,285.45 in 2 years and 8 months? 2. At what rate, converted semiannually is P896 the present value of P1104 which is due in 8 years? 3. One investment pays 8% annual interest. Another investment accumulates P4000 to PP6200 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 of interest compounded quarterly? 5. Find the annual rate of interest if an investment increases 25% in 6 years?
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Module 5 Using Mathematical Concepts and Techniques
INFORMATION SHEET 4.1 SIMPLE ANNUITY
Everyday living entails monthly rentals for persons’ dwelling installment payments for a car or some household appliances. The common housewife is usually confronted with the question of how to budget the family income in such a way that the basic needs of the family would be attended to. Aside from the basic financial needs which must be attended to, a certain amount must be set aside for some future use. A regular monthly, bimonthly (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 at regular 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 may be of any convenient length like monthly, quarterly, semiannually and annually. The term of an annuity extends from one payment interval before the first payment up to the day of the last payment. Unless otherwise specified, the term of any annuity is presumed to begin immediately.
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Module 5 Using Mathematical Concepts and Techniques
TYPES OF ANNUITIES 1. Annuity certain is an annuity payable for a definite length of time not dependent on some outside contingency. It means that payment begins and ends on definite or fixed dates. Examples are monthly payment on a housing loan and monthly payment on a car loan since the payment starts on 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 time whose length cannot be foretold accurately. Examples are pensions and life 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 the term. It can also be called as the accumulated value of the annuity. Use this formula, S
=
R ( 1 + i) n 1 i
n i
Example Mrs. Cruz regularly deposits P1,000 at the end of each six months for 3 years. The bank gives interest at 8%, m = 2. How much will be in her account at the end of the third year?
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Module 5 Using Mathematical Concepts and Techniques
Given: R = P1000 n=6 i = 4% Solution = R ( 1 + i) n 1 i
S n i
= P1000 ( 1 + .04)6 1 .04 = P1000 (1.265319 – 1) .04 = P6,632.98
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Module 5 Using Mathematical Concepts and Techniques
SELFCHECK 4
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 ½ years. 3. P8000 due at the beginning of each 3 months for 3 years. Money is worth 6%,m = 4 4. An annual payment of P200 every three months for 2 years. Money is worth 4%, 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 promises to pay P250 at the end of each three months for 2 years. If money is worth 5% 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 of P5,600 plus a quarterly payment of P2,250 for 5 years. If money is worth 10%, m=4, how much is the total payment for the machineries? 3. A man made semiannual deposits of P2000 into a savings fund that pays interest at 9% compounded twice a year. How much will be his savings in 8 years? 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 each month for 8 years. If money is worth 18% compounded monthly, find the total cost of the house and lot.
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Module 5. Using Mathematical Concepts and Techniques
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 at the beginning of the term. It is the sum of the discounted values of the payments of the annuity at the beginning of the term.
Use this formula, A
R 1 ( 1 + i)  n i
=
n i
Example: The contract for the sale of a lot calls for a monthly payment of P500 for 5 years. If money is worth 12% compounded monthly, what is the cash price of the lot Solution: R = P500 b. A
n = 60
i = 1%
= 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
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Module 5. Using Mathematical Concepts and Techniques
SELFCHECK 4.1
A. Solve for the present value (A) of the following annuities. 1. A quarterly payment of P1500 for 15 ½ 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 P8000 at the end of each 6 months for 7 years. If money is worth 8% compounded semiannually, 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 worth 24% compounded monthly. 3. A man agrees to pay P2000 at the end of each month for 20 years in purchasing a house. Find the present value of this agreement if money is worth 3% compounded monthly. 4. Upon retirement, Mrs. Reyes finds that her company pension calls for payments of P2000 to her (or to her estate if she dies) at the beginning of each month for 25 years. Find the present value of this pension if money is worth 4% compounded monthly. 5. A stereo set is offered for sale for P4000 downpayment and P1,200 every three 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.
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Module 5. Using Mathematical Concepts and Techniques
TABLE 1.
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Module 5  Using Mathematical Concepts and Techniques Recording Sheet For Oral Questioning / Interview Student name: Module Title/No: Qualification: Oral/interview questions
Satisfactory response Yes
No
1.
2.
3.
4.
5.
The student's underpinning knowledge was: Satisfactory
Not satisfactory
Student's Signature:
Date
Trainor's signature:
Date:
Acceptable answers are:
Trainor's signature:
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Date:
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Rating Sheet Module 5  Using Mathematical Concepts and Techniques
Performance Feedback
Remarks
S
C
NS
NYC
1. SelfCheck 1.1 2. SelfCheck 1.2 3. SelfCheck 1.3 4. SelfCheck 1.4 5. SelfCheck 1.5 6. SelfCheck 1.6 7. SelfCheck 1.7 8. SelfCheck 1.8 9. SelfCheck 2 10. SelfCheck 2.1 11. SelfCheck 3 12. SelfCheck 3.1 13. SelfCheck 3.2 14. SelfCheck 3.3 15. SelfCheck 4 16. SelfCheck 4.1 S  Satisfactory NS – Not Satisfactory C  Completed NYC – Not Yet Completed Module is Completed Remarks:
Not Yet Completed
Student's Signature
Date
Trainor's signature:
Date:
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