General Mathematics - Grade 11 - Simple and Compound Interest PDF

GENERAL MATHEMATICS – GRADE 11 SIMPLE AND COMPOUND INTEREST I. Introduction This learner’s packet will help you illustrate and distinguish simple and compound interest as this may serve as your guide in making decisions in the planning of investing or borrowing mone...

GENERAL MATHEMATICS – GRADE 11 SIMPLE AND COMPOUND INTEREST I. Introduction This learner’s packet will help you illustrate and distinguish simple and compound interest as this may serve as your guide in making decisions in the planning of investing or borrowing money in the future. This also provides tasks and exercises that you will perform for you to be equipped with the necessary skills and knowledge needed. II. Learning Objectives At the end of this learner’s packet, you will be able to (a) illustrate simple and compound interest (M11GM-IIa-1); (b) distinguish simple and compound interest (M11GM-IIa-2) (c) compute interest, maturity value, future value, and present value in a simple interest environment, and solve problems involving simple interest. (M11GM-IIa-b-1) (d) compute interest, maturity value, future value, and present value in a compound interest environment and solve problems involving compound interest. (M11GM-IIa-b- 1) III. Learning Activities Lesson 1: Illustrating and Distinguishing Simple and Compound Interest If you borrow money or make a loan, you return an amount that is higher than what you have borrowed. If you save or invest your money in a bank or any financial institution, the amount returned will be higher than your savings or investment. The difference between the amount you paid or received and the amount you borrowed or invested is called the interest. Most of us view interest as “the amount paid by the lender for the use of money”. There are two types of interest, simple interest, and compound interest. The most common measure of interest is simple interest. RO_General Mathematics_Grade 11_Q2_LP 1 1  While both types of interest talk about growing your money, we will be focusing first on illustrating simple and compound in this lesson. Here is an example. Example 1: Mr. Cruz is planning to invest his PhP 10 000. Two banks offer him a 3% interest rate. Determine which of the two banks is offering him a better deal. Bank A Bank B Bank A offers a 3% simple Bank B offers a 3% interest interest payable for 5 years. compounded annually payable for 5 years. Bank A: Principal Interest(Is) Maturity Value Time (t) (P) Solution Answer (A) 1 (10,000)(0.03)(1) 300 Php10,300 2 (10,000)(0.03)(2) 600 Php10,600 3 Php10 000 (10,000)(0.03)(3) 900 Php10,900 4 (10,000)(0.03)(4) 1200 Php11,200 5 (10,000)(0.03)(5) 1500 Php11,500 Bank B: Maturity Value Time Principal Interest(Ic) (A) (t) (P) Solution Answer 1 Php10 000.00 (10,000)(0.03)(1) 300 Php10,300 (10,000)(0.03)(1) + 2 Php10,300.00 300 + 309 = 609 Php10,609 (10,300)(0.03)(1) (10,000)(0.03)(1) + 300 + 309 + 318.27 3 Php10,609.00 (10,300)(0.03)(1) + Php10,927.27 = 927.27 (10,609)(0.03)(1) (10,000)(0.03)(1) + 300 + 309 + 318.27 (10,300)(0.03)(1) + 4 Php10,927.27 + 327.82 = Php11,255.09 (10,609)(0.03)(1) + 1,255.09 (10,927.27)(0.03)(1) (10,000)(0.03)(1) + (10,300)(0.03)(1) 300 + 309 + 318.27 5 Php11,255.09 +(10,609)(0.03)(1) + + 327.82 + 337.65 Php11,592.74 (10,927.27)(0.03)(1) = 1,592.74 +(11,255.09)(0.03)(1) RO_General Mathematics_Grade 11_Q2_LP 1 2  In the illustration above, bank A offers PhP 900.00 simple interest (PhP 10 900 – PhP 10 000 = PhP 900) for 3 years, while Bank B offers PhP 927.27 interest compounded annually( PhP 10 927.27 – PhP 10 000= PhP 927.27) for 3 years. Thus, Bank B is the better option for Mr. Cruz to invest his money since it offers higher interest. Let us have another example. Example 2: Elisha is planning to open a clothing store in Naga City but needs additional funds thus decided to borrow money worth PhP 50 000. Two financial institutions offer her a good deal. ABC Bank offers a 2.5% interest rate while ECQ Bank offers the same interest rate, 2.5% but compounded annually. Both are payable for 5 years. Which is a better offer for Elisha and why? ABC Bank: Principal Interest (Is) Maturity Value Time (t) (P) Solution Answer (A) 1 (50,000)(0.025)(1) 1,250 Php 51,250 2 (50,000)(0.025)(2) 2,500 Php 52,500 3 Php 50 000 (50,000)(0.025)(3) 3,750 Php 53,750 4 (50,000)(0.025)(4) 5,000 Php 55,000 5 (50,000)(0.025)(5) 6,250 Php 56,250 ECQ Bank: Comparing the two, the interest gained after 5 years in ABC Bank is PhP 6 250 (56 250 – 50 000) while in ECQ Bank is PhP 6 570.41 (56 570.41 – 50 000). Thus, ABC Bank would be a better option for Elisha. Since it is borrowed money, the bank that offers the least interest is the best option. RO_General Mathematics_Grade 11_Q2_LP 1 3  Let’s take a closer look at another illustration where the final amount or maturity value is written as a function of time. Column A Column B 1. A(t) = 3000(1+0.03t) ; t = 2 1. A(t) = 3000(1+0.03)t ; t = 2 2. A(t) = 2000(1+0.02t) ; t = 4 2. A(t) = 2000(1+0.02)t ; t = 4 3. A(t) = 1500(1+0.06t) ; t = 3 3. A(t) = 1500(1+0.06)t ; t = 3 Now, it’s your turn to dig deeper and look for other things which make Simple Interest different from Compound Interest! ACTIVITY 1.1 Determine the given principal amount or interest, rate, and time in each problem as required. 1. In a certain bank, Angel invested PhP 88 000 in a time deposit that pays 0.5% simple interest for 3 years. 2. On the 10th birthday of her daughter, Sheina deposited Ph P150, 000 in a bank that pays 1.2% interest compounded quarterly which she plans to withdraw after 15 years. 3. An amount of PhP 30 000 was deposited by fifteen – year old students in a bank that pays 0.7% compounded daily and plans to withdraw after 10 years. 4. Ron invested PhP 25 000.00 at 16% interest for 6 years. ACTIVITY 1.2 A. Read the problem carefully and answer each of the following questions. Ms. Lovely purchased goods amounting to PhP 12 000.00. She used two credit cards to pay PhP 6 000.00 each. Credit card P charges a 3% interest rate per annum while credit card Q charges a 3% interest rate per annum compounded semi-annually. 1. What kind of interest does credit card P illustrate? 2. What kind of interest does credit card Q illustrate? 3. Which credit card will Ms. Lovely have less to pay? Why? B. Illustrate the given problem below by creating scenarios illustrating either simple or compound interest. 1. A = PhP 55 000; r = 3.9% simple interest; t = 2 ½ years 2. P = PhP 150 000; r = 2.5% interest compounded annually; t = 12 years RO_General Mathematics_Grade 11_Q2_LP 1 4  ACTIVITY 1.3 Direction. Answer each of the following. Using the same illustration in example 1 in the learning activities, sketch a graph of the difference between simple and compound interest using the Maturity Value as Y and the no. of years or time as X. Simple Interest: Principal Interest(Is) Maturity Value Time (t) (P) Solution Answer (A) 1 (10,000)(0.03)(1) 300 PhP10,300 2 (10,000)(0.03)(2) 600 PhP10,600 3 Php10 000 (10,000)(0.03)(3) 900 PhP10,900 4 (10,000)(0.03)(4) 1200 PhP11,200 5 (10,000)(0.03)(5) 1500 PhP11,500 Compound Interest: Maturity Time Principal Interest(Ic) Value (t) (P) (A) Solution Answer 1 PhP10 000.00 (10,000)(0.03)(1) 300 PhP10,300 2 PhP10,300.00 (10,000)(0.03)(1) + (10,300)(0.03)(1) 300 + 309 = 609 PhP10,609 (10,000)(0.03)(1) + (10,300)(0.03)(1) + 300 + 309 + 318.27 = 3 PhP10,609.00 PhP10,927.27 (10,609)(0.03)(1) 927.27 (10,000)(0.03)(1) + (10,300)(0.03)(1) + 300 + 309 + 318.27 + 4 PhP10,927.27 PhP11,255.09 (10,609)(0.03)(1) + (10,927.27)(0.03)(1) 327.82 = 1,255.09 (10,000)(0.03)(1) + (10,300)(0.03)(1) 300 + 309 + 318.27 + 5 PhP11,255.09 +(10,609)(0.03)(1) + (10,927.27)(0.03)(1) 327.82 + 337.65 = PhP11,592.74 +(11,255.09)(0.03)(1) 1,592.74 1. What is the graph of the function that illustrates a simple interest? (Hint: Choose either linear, exponential, or quadratic?). How about compound interest? (Hint: Choose either linear, exponential, or quadratic?). 2. Write a function for the simple interest and compound interest. Lesson 2: Simple Interest In depositing or investing my money in a bank, I will gain interest. Whereas, borrowing any amount from a bank will cost me interest. In real-life, like investing and borrowing a certain amount of money, we must know how to compute interest, maturity value, and present value in a simple interest environment, and be able to solve problems involving simple interest. This learner’s packet will help you learn how to compute the interest, maturity/future value, and present value in a simple interest environment in the simplest way. Here are some definitions or concepts that will help you compute simple interest, maturity/future value, present value, rate of interest, and time. RO_General Mathematics_Grade 11_Q2_LP 1 5  Definition Formula Terms used Simple Interest (Is) – interest that is computed on Is = Prt or P =principal the principal and then added to it. I=F–P Is = simple interest F = future value/ Principal (P) – amount of money borrowed or P = 𝐼𝑠 or invested on the origin date. 𝑟𝑡 maturity value 𝐹 r = rate P= 1+ 𝑟𝑡 t = time Rate (r) – annual rate, usually in percent, charged r= 𝐼𝑠 by the lender, or rate of increase of the 𝑃𝑡 investment. Maturity value/Future value – amount after t F = P + Is or years that the lender receives from the borrower F = P(1 + rt) on the maturity date. Time or term (t) – the amount of time in years the money is borrowed or invested; length of time t = 𝐼𝑠 between the origin and maturity dates. 𝑃𝑟 Remember: For you to easily memorize the formula, you can use the triangular figure below with the important variables( Is, P, r & t) used in simple interest. WasLearn how to solve problems involving simple interest from the succeeding examples. Sample Problems: Example 1: A Cooperative offers 3% annual simple interest rate for a particular share. How much will be earned if 150,000 pesos is invested for 2 years? Solution: Is = Prt = (150 000)(0.03)(2) = PhP 9 000.00 Example 2: A businessman invested a certain amount of money in a bank that offers 6% simple annual interest. After five years his total investment is PhP 130 000.00 which gives him an interest of PhP 30 000.00. How much did the businessman invest at the start? Given: F = PhP 130 000.00 Is = PhP 30 000.00 t = 5 years r = 6% or 0.06 RO_General Mathematics_Grade 11_Q2_LP 1 6  𝐼𝑠 𝐹 Solution 1: Using the formula P = Solution 2: Using the formula P = 𝑟𝑡 1+ 𝑟𝑡 Substitute the given to the formula Substitute the given to the formula 30 000 P = (0.06)(5) P= 130 000 1+ (0.06)(5) 30 000 = = 130 000 0.3 1.3 = 100 000 = 100 000 Therefore, the businessman had invested PhP 100 000.00 or the principal amount invested by the businessman was PhP 100 00.00 Example 3 A Cooperative offers 3% annual simple interest rate for a particular share. If you invested PhP 150 000.00 for 2 years, what is the future value of your investment? Example 4: Complete the table by finding the unknown. Principal Rate (r) Time (t) Interest (Is) (P) (year) 1 (a) 1.5% 3 1,350 2 24,000 (b) 2 1,440 3 250,000 0.5% (c) 6,250 4 500,000 3% 4 (d) RO_General Mathematics_Grade 11_Q2_LP 1 7 Solutions: 𝐼 𝐼 a. 𝑃 = 𝑟𝑡𝑠 b. 𝑟 = = 𝑃𝑡𝑠 1,350 1,440 𝑃= (0.015)(3) 𝑟= (24,000)(2) 1,350 1,440 𝑃 = 0.045 𝑟= 48,000 𝑃 = 𝑃ℎ𝑃 30 000.00 𝑟 = 0.03 𝑜𝑟 3% 𝐼𝑠 c. 𝑡 = d. 𝐼𝑠 = 𝑃𝑟𝑡 𝑃𝑟 6,250 𝑡 = (250,000)(0.005) 𝐼𝑠 = (500,000)(0.03)(4) 6,250 𝑡 = 1,250 𝐼𝑠 = ₱ 60,000 𝑡 = 5 𝑦𝑒𝑎𝑟𝑠 Let us test your understanding! ACTIVITY 2.1 ACTIVITY 1.1 following questions. Answer the 1. Jenai is investing PhP 24 000.00 in a cooperative giving 5% simple annual interest. How much interest will she earn after 2 years? For items 2-3: Use the problem below to answer the following questions. A teacher made an investment in a bank that offers 3.5% interest rate per annum. After 5 years, she earned a total interest of PhP 7000.00. 2. How much was her original investment? 3. What will be her total amount of money after 5 years? For items 4-5: Use the problem below to answer the following questions. Teacher Jonah got a loan of PhP 150 000 to buy a used car. The simple interest rate is 7.5% per annum and she paid an interest of PhP 33 750.00. 4. How many years did it take her to pay off her loan? 5. What is the annual interest? ACTIVITY 2.2 ACTIVITY 1.1 Compute the unknown in each of the following. 1. Mark borrowed ₱ 8, 500.00 with a 5% annual simple interest rate. How much interest will be earned if Mark pays after 3 years? Answer: The interest earned was __________. RO_General Mathematics_Grade 11_Q2_LP 1 8  2. When invested at an annual interest rate of 6%, an amount earned PhP 11,500.00 simple interest in two years. How much money was originally invested? Answer: The amount invested was __________. 3. If an entrepreneur applies for a loan amounting to PHP 400,000.00 in a bank, the simple interest of which is the 125,500.00 for 9 months, what interest rate is being charged? Answer: The bank charged an annual simple interest rate of __________. 4. How long will a principal amount earn interest equal to half of it at 4% simple interest? Answer: It will take __________ years for a principal to earn half of its value at a 4% simple annual interest rate. ACTIVITY 2.3 ACTIVITY Solve the 1.1 problems. Show your solutions. following 1. Sophia invested a certain amount at a 10% annual rate. After a year, the interest she received amounted to PHP 1,500.00. How much did she invest? 2. If you deposit PHP 7,500 in a bank at an annual rate of 0.5%. How much money will you have after 15 years? 3. How much money should you deposit in a bank so that it will accumulate to PhP 100,000.00 at 1% simple interest for 10 years? Lesson 3: Compound Interest Many bank savings accounts pay in COMPOUND INTEREST. In this scheme, the interest is added to the account at regular intervals, and the sum becomes the new basis for computing interest. Consequently, the interest earned at a certain time interval is automatically reinvested to bring in more interest. Here the sample problems in finding compound interest, when compound interest is computed once a year, its maturity value and present value are presented. RO_General Mathematics_Grade 11_Q2_LP 1 9 Sample Problems: Example 1: Miss Jane Alvinia invested PhP 150 000.00 in a Cooperative that offers 3% interest rate compounded annually. How much interest will she earn if she will invest the money for 2 years? Given: P = PhP 150 000.00 r = 3% or 0.03 t = 2 years a. Solve for the future value F first, since the formula in finding the compound interest requires the future value F and it was not given in the problem. F = P(1 + r)t = 150 000 (1+ 0.03)2 substitute the given = 150 000 (1.0609) simplify = 159 135.00 simplify b. Find the compound interest Ic Ic =F–P = 159 135.00 - 150 000.00 = 9 135.00 Therefore, Miss Jane Alvinia will earn an interest of PhP 9 135.00. RO_General Mathematics_Grade 11_Q2_LP 1 10  Example 2: Miss Johanna borrowed PhP 40 000.00 at 7.5% compounded annually at a bank. How much interest will be charged if she applies for a 3-year loan? Given: P = PhP 40 000.00 r = 7.5% or 0.075Bank t = 3 years Solution: a. Solve for the future value F first, since the formula in finding the compound b. Find the compound interest Ic interest requires the future value F and Ic = F – P it was not given in the problem. F = P(1 + r)t = 49 691.88 – 40 000.00 = 40 000(1 + 0.075)3 = 9 691.88 = 40 000 (1.075)3 = 49 691.88 Therefore, Miss Johanna shall pay an interest of PhP 9 691.88. Example 3: A businessman aims to have his savings grow to Php 1 000 000.00 in 5 years. How much should be his initial deposit in an account that pays 6% interest rate compounded annually? Given: F = PhP 1 000 000.00 r = 6% of 0.06 t = 5 years 𝐹 Solution 2: Using the formula P = F(1+r)-t Solution 1: Using the formula P = (1+ t 𝑟) Substitute the given to the formula Substitute the given to the formula 1 000 000 P = 1 000 000 (1 + 0.06)-5 P = (1+ 0.06) 5 = 1 000 000 (1.06)-5 1 000 000 = 5 (1.06) = 747 258.17 =747 258.17 Therefore, the businessman should have an initial deposit of PhP 747 258.17. Example 4: JV lends his friend PhP 400 000.00 at 4% compounded annually. How much did his friend owe him after 4 years? Substitute the given to the formula Given: P = PhP 400 000.00 F = P(1 + r)t r = 4% or 0.04 = 400 000 (1 + 0.04)4 t = 4 years = 400 000 (1.04)4 = 467 943.42 Therefore, JV’s friend owes him PhP467 943.42. Let us test your understanding! RO_General Mathematics_Grade 11_Q2_LP 1 11  ACTIVITY 3.1: What’s my Interest? Solve the following questions. 1. Jon Bernie is investing PhP 75,000.00 in a cooperative that gives 5% compound interest per annum. How much interest will he earn after 4 years? For items 2-3: Use the problem below to answer the following questions. A bank offers 4% compound interest per annum for an investment. After 5 years, Ednard’s total investment grows to PhP 97 332.23. 2. How much was his original investment? 3. How much interest did he earn for 5 years? For items 4-5: Use the problem below to answer the following questions. Miss Janine got a 5-year loan amounting to PhP 300 000.00. The loan is due for 5% interest rate compounded annually. 4. What is the future value of her loan? 5. How much interest will she pay after 5 years? ACTIVITY 3.2: My Present, My Interest and My Future! Answer the following questions. For items 1-3: Use Problem A below. My friend invited me to invest PhP 50 000.00 for 10 years in a food corporation. The investment will be earning 7% compound interest annually. 1. What is the present value of investment? (Find the Present Value / Principal) 2. What will be the interest of investment after 10 years? (Find the Interest) 3. How much will you get from this investment after 10 years? (Find the Future Value) For items 4-6: Use Problem B below. After 3 years, the loan that I got from a Private Lending Institution, which offers 6.5% compound interest per annum, had accumulated to PhP 301 987.41. 4. What is the present value of the loan? (Find the Present Value / Principal) 5. What will be the interest of the loan after 3 years? (Find the Interest) 6. How much will you pay after 3 years? (Find the Future Value) RO_General Mathematics_Grade 11_Q2_LP 1 12  ACTIVITY 3.3: Give My Present and My Future! ____ ___________ Answer the following questions. 1. A depositor will have PhP 125,000 in a time deposit in a bank at 2% compounded annually after 2 years. a. How much did a depositor place in this time deposit in the bank? (Give My Present Value) b. What is the future amount of this investment? (Give My Future Value) 2. A businessman had accumulated PhP 60,000 after a 5-year investment in a cooperative that offers 5% interest compounded yearly. a. How much did the businessman invest in the cooperative? (Give My Present Value) b. What is the future amount of his investment? (Give My Future Value) 3. After 6 years, Johanna shall pay PhP 100 000.00 for her loan that earns an interest of 4% compounded annually. a. What is the principal amount of Johanna’s loan? (Give My Present Value) b. What is the future amount of her loan? (Give My Future Value) - End of Practice Tasks - IV. Reflection 1. How can simple or/and compound interest be applied in your current situation? What is its implication to the less fortunate group in our country? ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ________________________________________ 2. Give your insight(s) about the importance of understanding the concept of simple interest especially if your family is applying for a loan. ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ RO_General Mathematics_Grade 11_Q2_LP 1 13  3. Interview a member of the family or a person in the community with debts or investments and reflect on it. Ask them of the advantages and disadvantages of applying for loans. Also, what are their considerations in investing their money. Write your reflection on a 1 whole sheet of short sized bond paper. (Interview may be through a phone call/text or any available online platform.) 4. What is the importance of saving money and investing it under Compound Interest scheme? Write your ideas on your journal. V. Answer Key Lesson 1: Activity 1.1: 1. Principal Amount = PhP 88 000.00 ; rate = 0.5% ; time = 3 years 2. Principal Amount = PhP 150 000.00 ; rate = 1.2% ; time = 15 years 3. Principal Amount = PhP 30 000.00 ; rate = 0.7% ; time = 10 years 4. Principal Amount = PhP 25 000.00 ; rate = 16% ; time = 6 years. Activity 1.2: A. 1. Simple Interest 2. Compound Interest 3. It is Credit card P because the interest gained in simple interest is much lower than in compound interest. B. Rubrics for Scoring: 3 – The scenario created illustrates simple or compound interest with no grammatical errors. 2 – The scenario created illustrates simple or compound interest with few grammatical errors. 1 – The scenario created does not clearly illustrate simple or compound interest with few grammatical errors. Activity 1.3: The graph can be checked through Geogebra, Desmos app, or any graphing calculator. 1. Linear, Exponential 2. Simple interest : Y = 10 000 ( 1 + 0.03x) Compound interest : Y = 10 000 ( 1 + 0.03)x RO_General Mathematics_Grade 11_Q2_LP 1 14 Lesson 2: Lesson 3: Activity 3.1 Activity 3.2: Activity 3.3: 1. Ic = PhP 16 162.97 1. P = PhP 50 000.00 1. a. P = PhP 120 146.10 2. P = PhP 80 000.00 2. Ic = PhP 48 357.00 b. F = PhP 125 000.00 3. Ic = PhP 17 332.23 3. F = PhP 98 357.00 2. a. P = PhP 47 011.57 4. F = PhP 382 884.47 4. P = PhP 250 000.00 b. F = PhP 60 000.00 5. Ic = PhP 82 884.47 5. Ic = PhP 51 987.41 3. a. P = PhP 79 031.45 6. F = PhP 301 987.41. b. F = PhP 100 000.00 VI. References Bacani, J.B. , and Soriano, J.M., (2017) General Mathematics for Grade 11, 151 Quezon City: Ephisians Publishing Inc. DepEd (2016). General Mathematics Learners Material First Edition. Department of Education. Oronce, O.A. (2016). RBS General Mathematics First Edition, Sampaloc, Manila: Rex Book Store, Inc. Reyes A., and Sulit, R. (2016), General Mathematics, Jimcyville Publications DEVELOPMENT TEAM OF THE LEARNER’S PACKET Writers : MICHEL CAPIN BISUÑA – Gainza NHS DOLORES SAPINOSO RULL – Rodriguez NHS Reviewer : JOHN EMMANUEL R. IBE – Magarao NHS Editor : JUMAR R. VELASCO – Ragay SMOHS Lay-out Artist : JHOMAR B. JARAVATA – Bula NHS Illustrators : MICHEL CAPIN BISUÑA – Gainza NHS DOLORES SAPINOSO RULL – Rodiguez NHS Language Editor : Marjorie Stanzi M. De Leon - Gainza NHS Validator : ARABELLA B. MENDEZ – Baao NHS RO_General Mathematics_Grade 11_Q2_LP 1 15