11 General Mathematics Second Quarter LEARNING ACTIVITY SHEETS PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page i COPYRIGHT PAGE Learning Activity Sheet in General Mathematics Grade 11 Copyright @ 2020 DEPARTMENT OF EDUCATION Regional Office No. 02 (Cagayan Valley) Regional Government Center, Carig Sur, Tuguegarao City, 3500 “No copy of this material shall subsist in any work of the government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. This material has been developed for the implementation of K to 12 Curriculum through the Curriculum and Learning Management Division (CLMD). It can be reproduced for educational purposes and the source must be acknowledged. Derivatives of the work including creating an edited version and enhancement of supplementary work are permitted provided all original works are acknowledged and the copyright is attributed. No work may be derived from the material for commercial purpose and profit. Consultants: Regional Director : ESTELA L. CARIÑO, EdD, CESO IV Assistant Regional Director : RHODA T. RAZON, EdD, CESO V Schools Division Superintendent : MADELYN L. MACALLING, PhD, CESO VI Assistant Schools Division Superintendents : DANTE MARCELO, PhD, CESO VI : EDNA P. ABUAN, PhD Chief Education Supervisor, CLMD : OCTAVIO V. CABASAG, PhD Chief Education Supervisor, CID : RODRIGO V. PASCUA, EdD Development Team Writers Content Editors Focal Persons : MAI RANI ZIPAGAN, PhD, GAMU RURAL HS-ISABELA : MARYJANE BUCAG, SANTO TOMAS NATIONAL HS-ISABLEA : CAYSELYN GUITERING, ALFREDA ALBANO NHS-ISABELA : CORAZON BAUTISTA, LUNA NATIONAL HS-ISABELA : CINDY LACANARIA, LUNA GENERAL COMPREHENSIVE HS-ISABELA : JAYBEL B. CALUMPIT, REGIONAL SCIENCE HS- ISABELA : JEREMAEH C. LOZANO, JONES RURAL SCHOOL-ISABELA : CHRISTIAN JULIAN, ROXAS NATIONAL HS-ISABELA : ALJON S. BUCU, PhD : MAI RANI ZIPAGAN, PhD : LEONOR BALICAO : DOMINGO PEROCHO JR., PhD : JUAN LAPPAY : INOCENCIO T. BALAG, EPS MATH : MA. CRISTINA ACOSTA, EPS LRMDS, SDO ISABELA : ISAGANI R. DURUIN, REGIONAL EPS MATH : RIZALINO G. CARONAN, REGIONAL EPS LRMDS Printed in DepEd Regional Office No. 02 Regional Government Center, Carig Sur, Tuguegarao City PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page ii Table of Contents Competency Illustrates simple and compound interests Distinguishes between simple and compound interests Computes interest, maturity value, future value, and present value in simple interest and compound interest environment Solves problems involving simple and compound interests Illustrates and distinguishes simple and general annuities Finds the future value and present value of both simple annuities and general annuities Describes the different markets for stocks and bonds Calculate the fair market of a cash flow stream that includes annuity Calculates the present value and period of deferral of a deferred annuity Analyzes the different market indices for stocks and bonds Distinguishes and solves problems involving business and consumer loans (amortization, mortgage) Illustrates and symbolizes propositions Distinguishes between simple and compound propositions Performs the different types of operations on propositions. illustrate different types of tautologies and fallacies and determine the validity of categorical syllogisms Establishes the validity and falsity of real-life arguments using logical propositions, syllogisms, and fallacies. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page Number 1 8 15 23 30 33 42 49 58 65 69 77 81 85 91 99 Page iii GENERAL MATHEMATICS Name: ______________________ Grade: ___________ Section: ____________________ Date : ___________ LEARNING ACTIVITY SHEET ILLUSTRATE SIMPLE AND COMPOUND INTEREST Background Information of Learners Interest is the cost of borrowing money where the borrower pays a fee to the lender for the money borrowed. It is usually expressed as a percentage either simple or compounded. Simple interest is an interest that is computed on principal and then added to it. On the other hand, in the compound interest the interest is computed on the principal and also on the accumulated past interests. In this learning activity sheet, we will try to illustrate how simple interest and compound interest differ. Learning Competency Illustrate Simple and Compound Interest (Quarter 2, Week 1): LC Code: M11GM-11a-1 Activity 1: BLAST from the PAST Directions: Answer the following questions. 1. 2. 3. 4. 5. What is the equivalent of 30% in decimal form? What is the equivalent of the decimal number 0.125 into percent? What percent of 24 is 12? How do we convert a percent to decimal? How do we convert a decimal to percent? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 1 Activity 2: Complete Me! Directions: Illustrate simple interest and compound interest by completing the given table. SIMPLE INTEREST (Annual) Time Principal (P) Interest (t) Rate (r) Simple Interest Solution Amount after t Answer years (maturity Value) 1 P 5,000.00 4% P 5,000.00 P 200.00 (0.04)(1) 2 P 5,000.00 4% P 5,000.00 = P 5,200.00 P 400.00 (0.04)(2) 3 P 5,000.00 4% P 5,000.00 P 5,000.00 4% P 5,000.00 (0.04)(4) P5,000.00+400.00 = P 5,400.00 P 600.00 (0.04)(3) 4 P5,000.00+200.00 P5,000.00+600.00 = P 5,600.00 P 800.00 P5,000.00+800.00 =P 5,800.00 5 6 7 8 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 2 COMPOUND INTEREST (ANNUAL) Time Amount at Rate Compound Interest Amount after t years (t) the start of (r) (maturity Value) Solution Answer P5,000.00 P 200.00 year t 1 P5,000.00 4% (0.04)(1) 2 P5,200.00 4% P5,200.00 = P 5,200.00 P 208.00 (0.04)(1) 3 P5,408.00 4% P P5,408.00 P 5,000.00 + 200.00 P 5,200+208.00 = P5,408.00 P 216.32 (0.04)(1) P5,408.00+216.32 =P 5,624.32 4 5 6 7 8 Activity 3: Dig Deeper Directions: Analyze and illustrate the given situation using the table provided and answer the questions that follow. 1. Suppose you want to invest php 10, 000.00 in the bank for 6 years. The Land bank of the Philippines (LBP) offers 12% annual simple interest rate per year. Banco De Oro offers 12 % compounded annually. Which will you choose and Why? SIMPLE INTEREST (ANNUAL) Time (t) Principal (P) Interest Simple Interest Amount after t Rate (r) Solution years Answer (maturity Value) 1 2 3 4 5 6 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 3 COMPOUND INTEREST (ANNUAL) Time Amount at Rate Compound Interest Amount after t years (t) the start of (r) (maturity Value) Solution Answer year t 1 2 3 4 5 6 Questions: 1. Which will you choose and Why? __________________________________________________________________ ____________________________________________________ 2. How do you find simple Interest? In Symbols: ________________________________________________ 3. How do you find compound Interest? In Symbols: ________________________________________________ 4. How do you find the maturity value in simple interest? In Symbols: ________________________________________________ 5. How do you find the maturity value in simple interest? In Symbols: ________________________________________________ Reflection: 1. How did you find the activity? _____________________________________________________________ ____________________________________________________________ References: DepEd General Mathematics Learner’s material DepEd Learner’s Material for Grade 9 Oronce, Orlando A., RBS General Mathematics First Edition PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 4 Answer Key Activity 1. 1. 0.3 2. 12.5% 3. 50% 4. To convert percent to decimal, drop the percent sign and move the decimal point two places to the left. 5. To convert decimal to percent, move the decimal point two places to the right. Activity 2 Time Principal SIMPLE INTEREST (Annual) Interest Simple Interest Amount after t years (t) (P) Rate (r) Solution Answer (maturity Value) P 5,000.00 4% P 5,000.00 P 1,000.00 P5,000.00+1,000.00 5 (0.04)(5) 6 P 5,000.00 4% P 5,000.00 =P 6,000.00 P 1,200.00 (0.04)(6) 7 P 5,000.00 4% P 5,000.00 =P 6,200.00 P 1,400.00 (0.04)(7) 8 P 5,000.00 4% P 5,000.00 P5,000.00+1,200.00 P5,000.00+1,400.00 =P 6,400.00 P 1,600.00 (0.04)(8) P5,000.00+1,600.00 =P 6,600.00 Time Amount at COMPOUND INTEREST (ANNUAL) Rate Compound Interest Amount after t years (t) the start of (r) Solution Answer P 5,849.29 P 233.97 (maturity Value) year t 5 P5,849.29 4% (0.04)(1) 6 P6,083.26 4% P6,083.26 = P 6,083.26 P 243.33 (0.04)(1) 7 P6,326.59 4% P 6,326.59 P6,579.65 4% P P6,579.65 (0.04)(1 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES P 6,083.26+ 243.33 = P 6,326.59 P253.06 (0.04)(1 8 P 5,849.29+ 233.97 P 6,326.59 +253.06 = P6,579.65 P263.19 P6,579.65+ 263.19 =P 6,842.84 Page 5 Activity 3: SIMPLE INTEREST (ANNUAL) Time Principal (P) (t) Interest Simple Interest Amount after t Rate (r) Solution years Answer (maturity Value) 1 P 10,000.00 12% P 10,000.00 P 1,200.00 (0.12)(1) P 10,000.00 +1,200.00= P11,200.00 2 P 10,000.00 12% P 10,000.00 P 2,400.00 (0.12)(2) 3 P 10,000.00 12% P 10,000+2,400.00 = P12,400.00 P 10,000.00 P 3,600.00 (0.12)() P 10,000+3,600.00= P13,600.00 4 P 10,000.00 12% P 10,000.00 P 4,800.00 (0.12)(4) P 10,000.00 +4,800.00= P14,800.00 5 P 10,000.00 12% P 10,000.00 P 6,000.00 (0.12)(5) 6 P 10,000.00 12% P 10,000+6,000.00 = P16,000.00 P 10,000.00 P 7,200.00 (0.12)(6) P 10,000+7,200.00= P17,200.00 COMPOUND INTEREST (ANNUAL) Time Amount at Rate Compound Interest Amount after t years (t) the start of (r) (maturity Value) Solution Answer P10,000.00 P P 10,000.00 + 1,200.00= (0.12)(1) 1,200.00 P11,200.00 P 11,200.00 P P 11,200.00+ 1,344.00 (0.12)(1) 1,344.00 =P 12,544.00 P 12,544.00 P P 12,544.00+ 1,505.28 (0.12)(1) 1,505.28 =P 14,049.28 year t 1 P 12% 10,000.00 2 P 12% 11,200.00 3 P 12,544.00 12% PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 6 4 P 12% 14,049.28 5 P 12% 15,735.19 P 14,049.28 P P 14,049.28+ 1,685.91= (0.12)(1) 1,685.91 P 15,735.19 P 15,735.19 P P 15,735.19 (0.12)(1) 1,888.22 + 1`,888.22= P 17,623.41 6 P 17,623.41 12% P 17,623.41 P P 17,623.41 + 2,114.81= (0.12)(1) 2,114.81 P 19,738.22 Questions: 1. Answers vary 2. Simple Interest is the product of the principal, rate and time I= Pr 3. A= P + I ; A= P + Prt ; A= P(1+rt) where: A =maturity value; P =principal and I= interest r =rate and t =time 4. Compound Interest: Ic= A-P 5. A=P +Pr=P(1+r) * at the end of the year A= P( 1+r)t π A= P (1+ π)kt *If compounded annually for n years * If compounding k times per year. where: A =maturity value; P =principal and I= interest r =rate and t =time; r/k= rate per compounding and kt= number of compounding PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 7 GENERAL MATHEMATICS Name: ______________________ Grade: ___________ Section: ____________________ Date : ___________ LEARNING ACTIVITY SHEET DISTINGUISHES BETWEEN SIMPLE AND COMPOUND INTEREST Background Information of Learners . In the previous illustrations of simple and compound interest, we learned that simple interest is an interest that is computed on principal and then added to it. On the other hand, compound interest the interest is computed on the principal and also on the accumulated past interests. Hence, this learning activity sheet will help you distinguish between simple interest and compound interest. Remember: Simple Interest: I= Pr Maturity Value: A= P + I ; A= P + Prt ; A= P(1+rt) Compound Interest: Ic= A-P Maturity (future) Value : A=P +Pr=P(1+r) * at the end of the year A= P( 1+r)t * Compounded annually for n years π A= P (1+ π)kt * If compounded k times per year Learning Competency Distinguishes Between Simple and Compound Interest (Quarter 2, Week 1): LC Code: M11GM-11a-2 Illustrative Examples: 1. How much interest would Maria pay to her loan amounting to php 20,000 for 3 years at 6% per year at simple interest? Given: P=20, 000.00 r= 6%= 0.06 t= 3 years Solution: I= Prt = 20,000.00 x 0.06 x 3 I= php 3, 600.00 Therefore, the interest of her loan at the end of 3 years is php 3,600.00 2. Find the future value and the amount of interest if php 20,000 was loaned for a period of 3 years with interest of 6% compounded annually. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 8 Given: A=? I=? P-php 20,000.00 r=6%=0.06 t=3 Solution: A= P(1+r)t A= 20,000 ( 1 + 0.06)3 A= 20,000 (1,06)3 = php 23,820.32 Therefore the interest at the end of 3 years is I= A-P I= 23, 820.32-20,000.00 I= 3,820.32 Activity 1: Check My Understanding! Directions: Determine whether each problem involve simple interest or compound interest. Put a check i(/) in the column. Simple interest Compound Interest 1. Mr. Edo invested php 30,000 to an account that pays 4% annually. How much is the amount interest at the end of the year? 2. What is the interest rate is being charge if Mario applies a loan amounting to php 60,000.00and pays interest of php 9,000.00 in 3 years 3. What is the interest of php 38,000.00 if invested at 4% compounded annually in 2 years? 4. What is the future value of php 18,000 at 6% in 2 years PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 9 compounded semiannually? 5. What are the rate of interest and maturity value of php 28,000.00 if its earns php 2,240.00 in a year? Activity 2: How well Do You know Me? Directions: From the given problems in activity1, answer the questions below. 1. Mr. Edo invested php 30,000 to an account that pays 4% annually. How much is the amount of interest at the end of the year? a. P=? b. r=? c. I=? d. t=? e. What kind of interest was applied, simple or compound? ______________ f. What can you infer from your answer? ___________________________________________________________ 2. What is the interest rate is being charge if Mario applies a loan amounting to php 60,000.00and pays interest of php 9,000.00 in 3 years? a. r=? b. P=? c. I=? d. t= e. What kind of interest was applied, simple or compound? _____________ f. What does your answer implies? ___________________________________________________________ 3. What is the interest of php 38,000.00 if invested at 4% compounded annually in 2 years? a. I =? b. P=? c. r=? d. t=? e. k=? f. What kind of interest was applied, simple or compound?_______ g. What did you observe from the interest in 2 years? _____________________________________________________ 4. At the end year what is the future value of php 18,000 at 6% compounded semiannually? a. A=? b. P=? c. r=? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 10 d. k=? e. What kind of interest was applied, simple or compound?____________ f. How is the interest at the end of the year? _________________________________________________ 5. What are the rate of interest and maturity value of php 28,000.00 if its earns php 2,240.00 in a year? a. r=? b. A=? c. P=? d. I=? e. t=? f. What kind of interest was applied, simple or compound? ________ g. What can you infer from your answers? _________________________________________________ Activity 3: Performance Tasks Directions: Complete the table below using the simple and compound interest. 1.Find the future value of php 15,000 at 8% compounded annually for 6 years. Time (year) Principal at the start of the year Interest Amount (At the end of the year) 1 2 3 4 5 6 Describe the data under the “Amount ( at the end of the years)” PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 11 2.Find the maturity value of php 15,000 at 8% per year at a simple interest for 6 years. Time (year) Principal Accumulated Interest Amount Due Describe the data under the “Amount Due” ___________________________________________________________________________ ___________________________________________________________ Reflection: 1.In your own understanding, how can you distinguish between simple and compound interest? Explain __________________________________________________________________ __________________________________________________________________ 2. If you were given a chance to invest your money what is your preference Simple or compound interest? Why? ___________________________________________________________________ ___________________________________________________________________ References: DepEd General Mathematics Learner’s material DepEd Learner’s Material for Grade 9 Oronce, Orlando A., RBS General Mathematics First Edition PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 12 Answer Key Activity 1. Item number 1 2 3 4 5 Activity 2 1. Simple Interest / / Compound Interest / / / a. P= 30,000.00 b. r= 4%=0.04 c. I= Prt= 30,000(0.04)(1)= 1,200.00 d. t= 1year e. simple interest 2. π° a. r = ππ = 9,000.00/60,000.00 (3)= 0.05= 5% b. P=60,000.00 c. I= 9,000.00 d. t=3 years e. simple interest 3. a. I= A-P but A=P(1+r/k)kt= ππ, πππ. ππ (π. ππ)1(2)=41,100.80 Therefore: I=3,100.80 b. P=38,000.00 c. r= 4%=0.04 d. t= 2years e. k=1 (annually) f. compound interest 4. a. A=19,096.20 b. P=18,000.00 c. r=6%=0.06 d. K= 2( semi-annually) e. t= 1 year f. compound interest PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 13 5. a. r=8%=0.08 b. A=30,240.00 c. P= 28,000.00 d. I=2,240.00 e. t= 1 year f. simple interest Activity 3: Performance Tasks 1. Time (year) Principal at the start of the year Interest 1 2 3 4 5 6 Time (year) 15 000.00 16 200.00 17 496.00 18 895.68 20 407.33 22 039.92 Principal 1 2 3 4 5 6 15 000.00 15 000.00 15 000.00 15 000.00 15 000.00 15 000.00 1 200.00 1 296.00 1 399.68 1 511.65 1 632.59 1 763.19 Accumulated Interest 1 200.00 2 400.00 3 600.00 4 800.00 6 000.00 7 200.00 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Amount (At the end of the year) 16 200.00 17 496.00 18 895.68 20 407.33 22 039.92 23 803,11 Amount Due 16 200.00 17 400.00 18 600.00 19 800.00 21 000.00 22 200.00 Page 14 GENERAL MATHEMATICS 11 Name of Learner:_____________________ Grade Level:________ Section:_____________________________ Score:_____________ LEARNING ACTIVITY SHEET SIMPLE AND COMPOUND INTEREST Background Information for Learners In this activity sheet, you examine and compare the simple and compound methods of calculating interest. This will also help you decide on what investment will pay you most over time or will give you low interest as much as possible when you borrow money. Simple Interest π°π = π·ππ where, πΌπ = simple interest π= principal π= rate π‘= term or time, in years Maturity (Future) Value π = π· + π°π or π = π·(π + ππ) where, π= maturity (future) value πΌπ = simple interest π= rate π‘= term or time, in years Example 1. Find the interest and maturity value if Sara deposits β±20,000 at a bank for 3 year at an interest rate of 4% per year. π°π = π·ππ =(20,000)(0.04)(3) π°π =2,400 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 15 π = π· + π°π or = (20,000) (2,400) π = 22,400 π = π·(π + ππ) =(20,000)[(1+(.04)(3)) π = 22,400 Thus, after 3 years, the β±20,000 deposited in the bank will earn an interest of β±2,400, therefore, the money will grow to β±22,400. Maturity (Future) Value and Compound Interest π = π·(π + π)π where, π= Principal or present value πΉ= maturity (future) value at the end of the term π= interest rate π‘= term or time, in years Compound Interest π°π = π − π· where, π= Principal or present value πΉ= maturity (future) value at the end of the term Example 2. Find the maturity value and compound interest if β±30,000 is invested at 3% compounded annually for 6 years. π = π·(π + π)π =30,000(1+0.03)6 π =35,821.57 π°π = π − π· = 35,821.57 -30,000 π°π = 5,821.57 Thus, β±30,000 invested at 3% compounded annually for 6 years will earn an interest of β±5,821.57. Therefore, the money will grow to β±35,821.57. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 16 Present Value P at Compound Interest π·= π = π(π + π)−π π (π + π) where, π= Principal or present value πΉ= maturity (future) value at the end of the term π= interest rate π‘= term or time, in years Example 3. Find the present value of β±70,000 due in 5 years at 10% compounded annually? π= = πΉ (1 + π)π‘ 70,000 (1+.10)5 =43,464.49 Therefore, an amount of β±43,464.49 was originally invested to earn β±70,000 in 5 years at 10% compounded annually. Maturity Value, Compounding m times a year π = π·(π + π)π where, πΉ = maturity (future) value π= principal π= interest rate per conversion period π‘= number of times interest is compounded Note:π = (π)π π ; π = ππ‘; (π)π =nominal interest rate (annual rate) m = frequency of conversion t = time/term in years Example 4. Find the maturity value and interest if β±15,000 is deposited in the bank at 3% compounded quarterly for 5 years. π = π·(π + π)π = 15,000(1 + .0075)20 =17,417.762 .03 j= 4 =0.0075; PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES n = mt = 4(5) = 20 Page 17 I= F-P = 17,417.762-15,000 =2,417.762 Hence, β±15,000 deposited in the bank at 3% compounded quarterly for 5 years will grow to β±17,417.762. Thus, the interest earned will be β±2,417.762. Present Value P at Compound Interest π·= π (π + π)π where, πΉ = maturity (future) value π= principal π= interest rate per conversion period π‘= number of times interest is compounded Example 5. Find the present value if β±50,000 due in 5 years if money is invested at 9% compounded semiannually. π π· = (π+π)π .09 j= 2 =0.045; n=mt=2(5)=10 50,000 =(1+0.045)10 =32,196.38 Therefore, an amount of β±32,196.38 was invested at 9% compounded semi-annually due in 5 years. Learning Competency The learner computes interest, maturity value, future value, and present value in simple interest and compound interest environment. M11GM-IIa-b-1 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 18 Exercise 1. Simple Interest Complete the table by finding the unknown. (2 points each) Principal (P) 25,000 (2.) 36,000 40,000 (6.) 65,000 250,000 Rate (r ) (1.) 5% (4.) 4.5% 10% 6% 3.5% Time (t) 5 2 6.5 (5.) 8 3 (10.) Interest (I) 10,000 9,000 7,020 9,000 120,000 (8) 131,250 Maturity Value (F) 35,000 (3.) 43,020 49,000 (7.) (9.) 381,250 Exercise 2.1. Compound Interest Complete the table by finding the unknown. (2 points each) Principal (P) 10,000 25,000 85,000 (7.) (9.) Rate (r ) 9% 3% 3.5% 5% 4.5% Time (t) 5 4 7 10 15 Interest (I) (1.) (3.) (5.) (8.) (10.) Maturity Value (F) (2.) (4.) (6.) 200,000 1,000,000 Exercise 2.2. Complete the table by finding the unknown. (2 points each) Principal (P) Nominal Rate (r) Frequency of Conversion (m) Interest rate per period Time in Years (t) Total number of conversions (n) Compound Interest (I) Compound Amount (F) 5,000 (5.) 50,000 (12.) 12% 10% 8% 5% 12 6 2 365 (1.) 2.5% (8.) (13.) 5 4 8 1 (2.) (6.) (9.) (14.) (3.) (7.) (10.) (15.) (4.) 27,130.89 (11.) 105,242.44 Exercise 3. Solve the following problems. (3 points each) 1. Ian is investing β±14,000 for 2 years. The interest rate is 5%. How much interest will Ian earn after 2 years? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 19 2. Michelle deposits β±10,000 in an account that earns 8% simple interest. How long will it take for the total amount in her account to reach β±12,000? 3. At what simple interest rate will β±5,000 amount to β±6,050 in 3 years? 4. Mark invested a certain amount at 5% simple interest per year. After 10 years, he received an interest of β±50,000. How much did he invest? 5. Matt is saving for a new car. He invests β±70,000 into an account that pays 3% interest compounded annually. How much will he have after 5 years? 6. Karla borrowed β±80,000 in a bank at 8% compounded annually, how much will she pay after 3 years? 7. How much money would you need to deposit today at 15% annual interest compounded annually to have β±1,500,000 in the account after 9 years? 8. Calculate the present value of β±100,000 payable in 25 years at 2.4% interest compounded monthly. 9. In order to have β±2,000,000 in the bank account after 20 years. How much would a person have to invest if the money will earn 8% interest compounded semi-annually? 10. Find the future value if β±50,000 is deposited in a bank at 4% interest compounded quarterly in 1 year. Exercise 4. Solve the following problems completely. (5 points each) 1. Lorraine wants to lend β±200,000 to Marie at the simple interest rate of 10% for 2 years and the same amount to James at 10% compounded annually for 2 years. Find the amount of money that Marie and James will return to Lorraine after 2 years to repay the loan. Who will pay more and by how much? 2. Calculate and compare the effective rates of interest for saving accounts paying: (a.) a nominal rate of 3.65%; compounded quarterly (b.) a nominal rate of 3.5%; compounded monthly 3. Rachel is planning to borrow β±300,00 in a bank. Bank A is offering 4% compounded semi-annually while Bank B is offering 3.5% compounded monthly and Bank C is offering 5% compounded quarterly. If she plans to borrow the money for 4 years. Which bank should she borrow the money? 4. Reflection What meaningful lessons have you learned from this topic?? ___________________________________________________________________________ ___________________________________________________________________________ Reference for Learners Verzosa, D.B, et.al (2016). General Mathematics. Quezon City, Manila PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 20 Answer Key Exercise 1. Principal (P) 25,000 (2.)90,000 36,000 40,000 (6.)150,000 65,000 250,000 Rate (r ) (1.) 8% 5% (4.) 3% 4.5% 10% 6% 3.5% Time (t) 5 2 6.5 (5.) 5 8 3 (10.) 15 Interest (I) 10,000 9,000 7,020 9,000 120,000 (8) 11,700 131,250 Rate (r ) 9% 3% 3.5% 5% 4.5% Time (t) 5 4 7 10 15 Interest (I) (1.) 5,386.24 (3.) 3,137.72 (5.)23,143.74 (8.) 77,217.35 (10.) 483,279.56 Maturity Value (F) 35,000 (3.)99,000 43,020 49,000 (7.) 270,000 (9.) 76,700 381,250 Exercise 2.1. Principal (P) 10,000 25,000 85,000 (7.) 122,782.65 (9.) 516,720.44 Maturity Value (F) (2.) 15,386.24 (4.) 28,137.72 (6.) 108,143.74 200,000 1,000,000 Exercise 2.2. Principal (P) Nominal Rate (r) Frequency of Conversion (m) Interest rate per period Time in Years (t) Total number of conversions (n) Compound Interest (I) Compound Amount (F) 5,000 15,000 50,000 100,000 12% 10% 8% 5% 12 6 2 365 1% 2.5% 4% 5 4 8 1 60 24 16 365 4,083.48 12,130.89 43,649.06 5,242.44 9,083.48 27,130.89 93,649.06 105,242.44 0.014% Exercise 3. 1. 2. 3. 4. 5. I= 1,400 2.5 years 0.07 or 7% P=30,695.66 P= 287,945.87 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES 6. P= 426,393.62 7. P=87,753.34 8. P= 54,914.058 9. P= 416,578.09 10. F=52,030.20 Page 21 Exercise 4 1. Marie (Simple Interest) P= 200,000 r= 10% t= 2 years I= Prt F= I+P = (200,000)(.10)(2) = 40,000+200,000 = 40,000 = 240,000 Therefore, Marie will repay β±240,000 to Lorraine after 2 years. James (Compound Interest) P=200,000 r=10% t= 2 years F=P(1+r)t = 200,000(1+(.10))2 = 242,000 Therefore, James will repay β±242,000. Now, β±242,000 > β±240,000, so James will pay more. He will pay 2,000 more than Marie. 2. (a. )3.65% compounded quarterly 3. Future value Bank A Bank B 4%; semi-annually 3.5%; monthly β±351,497.81 β±345,011.83 Bank C 4.5%; quarterly β±358,804.44 Since Rachel will borrow the money, she should choose Bank B with the lowest interest after 4 years. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 22 GENERAL MATHEMATICS 11 Name: _____________________ Date: ______________________ Grade Level: ____ Score: _________ Learning Activity Sheet SOLVES PROBLEMS INVOLVING SIMPLE AND COMPOUND INTERESTS Background Information for Learners Interest is a common mathematical ideas which is usually associated to business mathematics and other related areas. When we deposit money in our bank account or financial institutions, we get some extra money for our deposits. This extra money is called the interest. Similarly when we borrow money from our bank or financial institution, we have to pay them back something a little extra (other than the amount we borrowed). Again this extra amount is called interest. (Maths, 2014) Interest is charged as a percentage of the amount borrowed (or invested) for a certain fixed period before you repay (or withdraw) your borrowed (or withdrawn) amount. (Maths, 2014) Interest has two types. First is simple interest given by the formula π¨ = π·ππ, where A is the final amount, P is the principal, r is the rate and t is the time or duration. Second type π of interest is compound interest given by the formula π¨ = π·(π + π)ππ , where A is the final amount, P is the principal, r is rate, t is time or duration and n is number of times interest applied per time period. Learning Competency Solves problems involving simple and compound interests. (M11GM-IIb-2) Activity 1: SIMPLEST OF THE SIMPLE; Simple Interest Problems Directions: Answer the following problems by solving for the asked question. 1. Your mom deposited your earnings in a bank which amounted to Php4, 000. The bank offers an interest of 4.5% every year. How much your money will earn at the end of 3 years? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 23 2. You borrowed money from your classmate which amounted to Php3, 000. After 2 years, you paid your classmate an amount of Php3, 720. What is the rate of interest applies to you loan? 3. Your father bought a car which is Php200, 000 worth. He borrowed half of it to a bank with an interest rate of 13%. When your father paid the amount he borrowed, it already amounted to Php278, 000. How long is the duration of your father’s loan? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 24 Activity 2: Compound Interest Problems Directions: Answer the following problems by solving for the asked question. 1. A principal amount of Php2, 000 is placed in a savings account at 3% per annum compounded annually. How much is in the account after 7 yrs? 2. If Php3000 is placed in an account at 5% and is compounded quarterly for 5 years. How much is in the account at the end of 5 years? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 25 3. Php1200 is placed in an account at 4% compounded annually for 2 years. It is then withdrawn at the end of the two years and placed in another bank at the rate of 5% compounded annually for 4 years. What is the balance in the second account after the 4 years? Activity 3: Mixed Interest Problems Directions: Answer the following problems by solving for the asked question. 1. What would Php1000 become in a saving account at 3% per year for 3 years when the interest is not compounded (simple interest)? What would the same amount become after 3 years with the same rate but compounded annually? Which kind of interest earns more money and by how much? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 26 Guide Questions Rubrics for scoring Rubric Correctness of the Answer Detailed Procedure Description The answer presented is exact to what is agreed which means it follows the number of decimals present in the final answer to avoid confusion. The procedure was a carefully step by step procedure not ignoring the importance of each step. Weight 70% 30% Generalization Direction: Using the Venn-Euler Diagram, Illustrate the similarities and differences of the two types of interest. Simple Interest Compound Interest References for the Learners 1. Sunshine Maths (http://www.sunshinemaths.com/topics/financialmaths/introduction-to-interest/) 2. Onlinemathlearning.com (https://www.onlinemathlearning.com/simple-interestformula.html) 3. Analyzemath.com (https://www.analyzemath.com/finance/interests_problems_sol.html) PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 27 Answer Key ACTIVITY 1 1. A=? P= Php4,000 2. A=Php720 P=Php3,000 r=4.5% t=3 A=Prt A=4,000(.045)(3) A= Php540 r=? t=2 years A=Prt 720=3,000(r)2 720 = 6,000r 6,000 6,000 r=0.12 or 12% 3. A= Php78, 000 P= Php200, 000 r=13% t= ? A=Prt 78,000=200,000(.13)(t) 78,000=26,000(t) 26,000 26,000 t= 3 years ACTIVITY 2 1. A=? P= Php2,000 r=3% t=7 yrs π π¨ = π·(π + )ππ π π¨ = ππππ(π + π. ππ π ) π π¨ = π, πππ. ππ 2. A=? P= Php3,000 r=5% t=5 yrs π π¨ = π·(π + )ππ π π¨ = ππππ(π + π. ππ π ) π π¨ = π, πππ. ππ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 28 3. A=? P= Php1,200 r=4% t=2 yrs π π¨ = π·(π + )ππ π π¨ = ππππ(π + π. ππ π ) π π¨ = π, πππ. ππ A=? P= Php1,297.92 r=5% t=4 yrs π π¨ = π·(π + )ππ π π¨ = π, πππ. ππ(π + π. ππ π ) π π¨ = π, πππ. ππ ACTIVITY 3 1. SIMPLE INTEREST A=? P=1,000 r=3% t=3 yrs A=P(1+rt) A= 1000(.03)(3) A=Php1090 COMPOUND INTEREST A=? r=3% P= Php1,000 t=3 yrs π π¨ = π·(π + )ππ π π¨ = π, πππ(π + π. ππ π ) π π¨ =Php1,092.73 Conclusion: The account which is compounded annually earns more money by Php2.73 as compared to the same account with simple interest. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 29 GENERAL MATHEMATICS 11 Name of Learner: ________________________________ Section: ________________________________________ Grade Level: _________ Score: ______________ LEARNING ACTIVITY SHEET ILLUSTRATE SIMPLE AND GENERAL ANNUITIES AND DISTINGUISH BETWEEN SIMPLE AND GENERAL ANNUITIES Background Information for Learners Annuity is a sequence of payments made at equal (fixed) intervals or periods of time. Classification according to payment interval and interest period • Simple Annuity – the payment interval is the same as the interest period • General Annuity – the payment interval is not the same as the interest period Learning Competency with code The learner is able to (a) illustrate simple and general annuities; (b) distinguish between simple and general annuities (M11GM-IIc-1-2, Quarter II) Directions/Instructions: A. Fill in the blanks with the correct term. Annuity Ordinary annuity General annuity Simple annuity Annuity certain 1. An annuity in which payments begin and end at definite times is a/an ___________. 2. An annuity where the payment interval is the same as the interest period is a/an ____________. 3. An annuity where the payment interval is not the same as the interest period is a/an ______________. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 30 4. A simple annuity in which the payments are made at the end of each period is a/an ______________. 5. A sequence of payments made at equal periods is a/an _______________. B. Identify whether the given illustrates simple or general annuity. 1. A life insurance contribution paid monthly while the interest is compounded quarterly. 2. Your mom decided to join their office cooperative and agreed to contribute P1000 per month beginning in January 2020 which will earn 3% compounded monthly. 3. Your parents are planning to save for their retirement. To do this, they want to set aside a portion of their salaries and contribute monthly for their retirement funds which will earn 5% compounded quarterly. 4. Your eldest brother applied for a term life insurance. His contribution per year is P40 000 that earns 12% compounded monthly for 20 years. 5. A college educational plan earns 4% compounded quarterly and payments are made quarterly. 6. Your dad deposited all his retirement pay with bank C which will earn 4% compounded quarterly and he had an auto-credit arrangement of P20 000 per month. 7. Sir Eli deposits P10 000 on January 20, 2020 and had deposited the same amount on the same date every month. The China Bank credits 2.4% interest compounded annually to sir Eli’s account. 8. Your teacher saves P5 000 every 6 months in a bank that pays 0.25% compounded monthly. 9. Mr Manuel would like to buy a TV set payable monthly for 6 months and interest is 9% compounded semi-annually. 10. To pay his debt at 12% compounded semi-annually, Jaysenn committed for 8 quarterly payments of P24,500. Reflection What I learned in this activity ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ References: Orines, Fernando B. Next Century Mathematics (General Mathematics).Phoenix Publishing House, Inc.2016 Oronce, Orlando A. General Mathematics. Rex Book Store.2016 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 31 Answer Key A. 1. 2. 3. 4. 5. Annuity certain Simple annuity General annuity Ordinary annuity Annuity B. 1. General annuity 2. Simple annuity 3. General annuity 4. General annuity 5. Simple annuity 6. General annuity 7. General annuity 8. General annuity 9. General annuity 10. General annuity PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 32 GENERAL MATHEMATICS Name:____________________ Grade Level:____________________ Date:______________________ Score:_______________________ Learning Activity Sheet FIND THE FUTURE VALUE AND THE PRESENT VALUE OF BOTH SIMPLE ANNUITIES AND GENERAL ANNUITIES Background Information for Learners This activity serves as a learning guide for the learners. It facilitates lesson comprehension as it specifically aims for student’s mastery in solving problems about simple annuity and general annuity. Annuities may be classified according to payment interval and interest period which are simple annuity and general annuity. Simple annuity is an annuity where the payment interval is the same as the interest period while general annuity is an annuity where the payment interval is not the same as the interest period. Learning Competency: The learner find the future value and the present value of both simple annuities and general annuities. (M11GM-Ic-d-1 Activity 1 Example: Find the future value (F) and the present value (P) of this simple annuities, given the following: Periodic payment= Php 1,000,rate=5%, mode of payment=quarterly, term=2 years. Solutions: a. F = R [ (1+π)ππ‘ −1 F = 1,000 [ F = 1,000 [ π ] 5% 4(2) ) 4 5% 4 (1+ P =R[ −1 (1+0.0125)8 −1 0.0125 1−(1+π)−ππ‘ ] P = 1,000 [ ] P = 1,000 [ π 5% −4(2) ) 4 5% 4 1−(1+ 1−(1+0.0125)−8 0.0125 F = 1,000(8.3589) P = 1,000(7.5681) F = Php 8,358.9 P = Php 7,568.1 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES ] ] ] Page 33 Find the future value (F) and the present value (P) of the following simple annuities, given the table below. Periodic Payments a. b. c. d. e. Php 2, 500 Php 3,000 Php 5, 000 Php 7,500 Php 12,000 Rate 2% 3% 4% 5% 10% Mode of Payment (Compounded) Monthly quarterly quarterly Semi-annually annually Term 3 years 4 years 5 years 10 years 15 years Procedure Solve for the future value (F) and present value (P) for each item using the given in the table Guide Questions: Show a solution for each item to find the future value(F) and Present value(P) using the given in the table. Activity 2 Example: Find the future value (F) and the present value (P) of the following general annuities, given the following: Payment interval=semi-annually, regular payment= Php1,000, term=5 years, rate=6%,interest period= compounded quarterly Answer: πΉ1 πΉ1 = π π P(1+π)ππ‘ = P(1+π)ππ‘ π 6% (1+2)2(1) = (1+ 4 )4(1) π 6% (1+2) = (1+ 4 )2 π 2 = (1+0.015)2 -1 π π = 2 = 0.03 F=R[ (1+π)ππ‘ −1 F = 1,000 [ π ] (1+0.03)2(5) −1 0.03 P =R[ ] PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES 1−(1+π)−ππ‘ P = 1,000 [ π ] 1−(1+0.03)−2(5) 0.03 ] Page 34 F = 1,000 [ (1+0.03)10 −1 0.03 ] P = 1,000 [ F = 1,000(11.4639) 1−(1+0.03)−10 0.03 ] P = 1,000(8.5302) F = Php11,463.9 P = Php 8,530.2 Find the future value (F) and the present value (P) of the following general annuities, given the table below. Payment interval a. Monthly b. Quarterly c. Semiannual d. Annual Regular payment PHP 3,000 PHP 2,000 term rate Interest period (compounded) 4 years 5 years 3% 2% Quarterly Annually PHP 150, 000 10 years 5% Annually PHP 20,000 3 years 8% Semi-annually Procedure a. First convert the interest period to its equivalent interest rate for its payment interval b. Show solution for the future value (F) and Present value(P) for each item. Guide Questions: a. What is the equivalent interest rate? b. What is the Future value(F)? c. What is the Present value(P)? Activity 3 Example: Answer this problem: Your mom decided to deposit Php1,000 per month in a bank beginning January 2020 which will earn 10% compounded monthly. How much will be in your mom’s deposit at the end of December 2025? Given: Regular deposit (R) = Php1,000/month Rate (r) = 10% compounded monthly Time (t) = 5 years No.of conversions/year (m) = 12 π 10 % Interest rate per period ( j)= π = 12 = 0.0083 Solutions: F=R[ (1+π)ππ‘ −1 F=1,000[ ] π (1+0.0083)12(5) −1 0.0083 (1.0083)60 −1 F=1,000[ 0.0083 F=1,000(77.355) ] ] PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 35 F=Php 77,355 Thus, the amount in your mom’s deposit is Php 77,355 after 5 years. Directions: Answer the following problems. a. Rona started to deposit PHP 20,000 semi-annually in a fund that pays 5% compounded semi-annually. How much will be in the fund after 10 years? b. Ben is paying PHP 3,000 every 3 months for the amount he borrowed at an interest rate of 10% compounded quarterly. How much did he borrow if agreed that the loan will be paid in 3 years? c. Ronald is saving PHP 2, 000 every month by depositing it in a bank that gives an interest of 2% compounded quarterly. How much will he save in 3 years? Procedures a. Identify the given in each of the problem. b. Show solutions of each problem using the needed formula. Guide Questions a. Write all the given data in each of the problem. b. What is the formula needed? Show solutions for the answer of each problem substituting the given to the needed formula. c. Complete your answer with interpretation and with units if needed. Reflection What did I learn from this topic? ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ References: GENERAL MATHEMATICS (LM),First Edition 2016, DIWA Senior High School Series:General Mathematics, Next Century Mathematics 11 General Mathematics Copyright 2016 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 36 Answer Key (Activity 1) b. F = R [ (1+π)ππ‘ −1 π F = 2,500 [ F = 2,500 [ c. ] P =R[ 2% 12(3) ) 12 2% 12 (1+ −1 ] (1+0.001667)36 −1 0.001667 1−(1+π)−ππ‘ π P = 2,500 [ ] P = 2,500 [ 2% −12(3) ) 12 2% 12 1−(1+ F = Php 92,675 P = Php 87,282 F = 3,000 [ F = 3,000 [ π ] 3% 4(4) ) 4 3% 4 (1+ P =R[ −1 ] (1+0.0075)16 −1 ] 0.0075 1−(1+π)−ππ‘ π P = 3,000 [ P = 3,000 [ 3% −4(4) ) 4 3% 4 P = Php 45,072 π F = 5,000 [ F = 5,000 [ 4% 4(5) ) 4 4% 4 (1+ P =R[ −1 (1+0.01)20 −1 0.01 ] ] 1−(1+π)−ππ‘ π P = 5,000 [ P = 5,000 [ 4% −4(5) ) 4 4% 4 1−(1+0.01)−20 0.01 P = 5,000(18.0455) F = Php 110,095 P = Php 90,228 (1+π)ππ‘ −1 F = 7,500 [ F = 7,500 [ π ] 5% 2(10) ) 2 5% 2 (1+ P =R[ −1 (1+0.025)20 −1 0.025 1−(1+π)−ππ‘ ] P = 7,500 [ ] P = 7,500 [ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES ] 1−(1+ F = 5,000(22.019) e. F = R [ ] 0.0075 F = Php 50.796 ] ] 1−(1+0.0075)−16 P = 3,000(15.024) (1+π)ππ‘ −1 ] 1−(1+ F = 3,000(16.932) d. F = R [ ] 0.001667 P = 2,500(34.9128) (1+π)ππ‘ −1 ] 1−(1+0.001667)−36 F = 2,500(37.07) F=R[ ] π ] ] ] 5% −2(10) ) 2 5% 2 1−(1+ 1−(1+0.025)−20 0.025 ] ] Page 37 F = 7,500(25.5446576) P = 7,500(15.589) F = Php 191,585 P = Php 116,918 f. F = R [ (1+π)ππ‘ −1 π F = 12,000 [ F = 12,000 [ ] 10% 1(15) ) 1 10% 1 (1+ P =R[ −1 (1+0.01)15 −1 0.01 ] P = 12,000 [ ] 1−(1+π)−ππ‘ π 10% −1(15) ) 1 10% 1 1−(1+ P = 12,000 [ F = 12,000(16.0969) ] ] 1−(1+0.01)−15 0.01 ] P = 12,000(13.865) F = Php 193,163 P = Php 166,38 (Activity 2) a. πΉ1 πΉ1 = π π P(1+π)ππ‘ = P(1+π)ππ‘ π 3% (1+12)12(1) = (1+ 4 )4(1) π 3% (1+12) = (1+ 4 )1/3 π 12 = (1+0.0075)1/3-1 π π = 12 = 0.00249 F=R[ (1+π)ππ‘ −1 ] π F = 3,000 [ F = 3,000 [ P =R[ (1+0.00249)12(4) −1 0.00249 (1+0.00249)48 −1 0.00249 ] ] 1−(1+π)−ππ‘ P = 3,000 [ P = 3,000 [ π 1−(1+0.00249)−12(4) 1−(1+0.00249)−48 0.00249 P = 3,000(45.1895) F = Php152,760 P = Php 135,569 = π ] 0.00249 F = 3,000(50.92) b. πΉ1 ] ] πΉ1 π P(1+π)ππ‘ = P(1+π)ππ‘ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 38 π 2% (1+4)4(1) = (1+ 1 )1(1) π 2% (1+4) = (1+ 1 )1/4 π 4 = (1+0.02)1/4-1 π π = 4 = 0.00496 F=R[ (1+π)ππ‘ −1 π F = 2,000 [ F = 2,000 [ ] P =R[ (1+0.00496)4(1) −1 0.00496 (1+0.00496)4 −1 0.00496 ] π P = 2,000 [ ] P = 2,000 [ F = 2,000(4.03) ] 1−(1+0.00496)−4(1) ] 0.00249 1−(1+0.00496)−4 0.00496 ] P = 2,000(3.95) F = Php 8,060 π. πΉ1 1−(1+π)−ππ‘ P = Php 7,900 πΉ1 = π π P(1+π)ππ‘ = P(1+π)ππ‘ π 5% (1+2)2(1) = (1+ 1 )1(1) π 5% (1+2) = (1+ 1 )1/2 π 2 = (1+0.05)1/2 -1 π π = 2 = 0.024695 F=R[ (1+π)ππ‘ −1 π F = 150,000 [ F =150,000 [ ] P =R[ (1+0.024695)2(1) −1 0.024695 (1+0.024695 )2 −1 0.024695 F = 150,000(2.02469) ] 1−(1+π)−ππ‘ π P = 150,000 [ ] P = 150,000 [ ] 1−(1+0.024695)−2(1) 0.024695 1−(1+0.024695)−2 0.024695 ] ] P = 150,000(1.92828) F = Php 303,704 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES P = Php 289,242 Page 39 π. πΉ1 πΉ1 = π π P(1+π)ππ‘ = P(1+π)ππ‘ π 8% (1+1)1(1) = (1+ 2 )2(1) π 8% (1+1) = (1+ 2 )2 π 1 = (1+0.04)2 -1 π = π =.0816 F=R[ (1+π)ππ‘ −1 π F = 20,000 [ F =20,000 [ ] P =R[ (1+0.0816)1(2) −1 0.0816 (1+0.0816 )2 −1 0.0816 ] 1−(1+π)−ππ‘ P = 20,000 [ ] P = 20,000 [ π ] 1−(1+0816)−1(2) ] 0.024695 1−(1+0.0816)−2 0.0816 F = 20,000(2.0816) P = 20,000(1.77936) F = Php 41,632 P = Php 35,587 ] (Activity 3) a. Regular deposit (R) = Php20,000 semi-annually Rate (r) = 5% semi-annually Time (t) = 10 years No.of conversions/year (m) = 2 π 5% Interest rate per period ( j)= π = 2 = 0.025 Solutions: F=R[ (1+π)ππ‘ −1 ] π (1+0.025)2(10) −1 F=20,000[ 0.025 (1.025)20 −1 ] F=20,000[ 0.025 ] F=20,000(25.54) F=Php 510,800 Thus, the value in the fund is Php 510,800 after 10 years. b. Regular payment (R) = Php3,000 quarterly Rate (r) = 10% compounded quarterly Time (t) = 3 years No.of conversions/year (m) = 4 π 10% Interest rate per period ( j) = π = 4 = 0.025 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 40 Solutions: P= R [ 1−(1+π)−ππ‘ P= 3,000 [ ] π 1−(1+0.025)−4(3) 0.025 1−(1.025)−12 P= 3,000[ 0.025 P= 3,000(10.26) P= Php30,773.29 ] ] Thus, the amount borrowed by Ben is Php 30,773. c. Regular deposit (R) = Php2,000 monthly Rate (r) = 2% compounded quarterly Time (t) = 3 years No.of conversions/year (m) = 4 π 2% Interest rate per period ( j) = π= 4 = 0.005 Solutions: πΉ1 = πΉ1 π F= R [ π P(1+π)ππ‘ = P(1+π)ππ‘ π 2% (1+12)12(3) = (1+ 4 )4(3) π .02 (1+12) = (1+ 4 )1/3 π 12 = (1+0.005)1/3 -1 π π = 12 =.00166 (1+π)ππ‘ −1 F= 2,000[ F= 2,000[ ] π (1+0.00166)4(3) −1 ] 0.00166 (1.00166)12−1 0.00166 ] F= 2,000(12.11) F= Php 24,220 Thus, the savings of Ronald is Php 24,220 after 3 years. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 41 GENERAL MATHEMATICS Name of the Learner: Grade level: Section: ___ _____ Date: LEARNING ACTIVITY SHEET DIFFERENT MARKETS AND BONDS Background Information for learners There are two ways to make money in our modern world. The first way is to earn money from income by working by yourself. The other way to grow your fortune is to invest your assets so that it will increase its value over time. In this activity, you may learn the different markets for stocks and bonds to help you earn money by saving money for future use. Example of which is investing in a certain corporation, organization or company with higher possibility to gain profit. A stock market is a place where investors go to trade equity securities (i.e. shares) issued by corporations. The bond market is where investors go to buy and sell debt securities issued by corporations or governments Learning Competencies with code The learners are able to describe the different markets for stocks and bonds M11GM-11d-1, Directions/ Instructions Each activity has directions to be followed. If you have questions, you are open to contact your teacher for clarification and assistance. Activity 1. Match each term in Column A with its definition in Column B. Write your answer on the blank provided before the number. A B 1. No-par stock a. stock without par value 2. Broker b. a dealer in stocks and bonds 3. Yield c. selling price of stock PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 42 4. Broker’s commission d. a paper showing 5. Stock certificate e. rate of income on bonds 6. Shareholder f. a fee charged by a broker 7. Dividends g. anyone who owns stock 8. Market price stockholder h. profit distributed to Activity 2. Identify the components in the stock certificate below. Write your answer on the blank. 1 1 1 12 . 5 1 1 1 1 3 1 . 4 1 . 1. 1 1 1 1 1 . 1 . 2. 3. 4. 5. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 43 Answer the following questions based on the given picture above. Write your answer on the blank provided. 6. What is the name of the corporation issuing the certificate? 7.Who are the signatories of the corporation? 8. How much is the par value? 9. What is the stock certification number? 10.What is the number of shares being recorded in the certificate? Activity 3. Use the stock tables below to answer the succeeding questions. Example: 52-WKHIGH 52-WKLOW 75.5 68.9 STOCK OPEN HIGH LOW CLOSE VOLU ME 72 69.2 72 10,978 CLM BANK 72.4 1. What were the high and low prices for a share for the past 52 weeks? Answer: 75.5 and 68.9 2. What were the high and low prices for a share yesterday? Answer: 72 and 69.2 3. Give the price at which a share fared when the stock exchange closed yesterday. Answer: 72 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 44 52-WKHIGH 52-WKLOW STOCK OPEN HIGH LOW CLOSE VOLU ME 81.67 64.45 CJJ CORP. 67.54 67.54 65.35 67.25 114,900 1. What were the high and low prices for a share for the past 52 weeks? 2. What were the high and low prices for a share yesterday? 3. Give the price at which a share fared when the stock exchange closed yesterday. 52-WK- 52-WK- HIGH LOW 70 31.29 STOCK OPEN HIGH ABS-GMA CORP. 65.5 65.5 LOW CLOSE VOLU ME 64.25 58 230 66.89 1.What were the high and low prices for a share for the past 52 weeks? 2.What were the high and low prices for a share yesterday? 3.Give the price at which a share fared when the stock exchange closed yesterday. Activity 4. Find the investment by completing the table. The brokerage fee is 120 for each bond if 10 bonds are purchased, P80 for each bond if 11 to 50 are purchased, and P60 for each bond if 61 or more bonds are purchased. No. of Bonds Held Example 15 7 27 49 89 120 Market Value per Bond P3,000.00 P3,045.00 P1,980.00 P700.00 P5,050.00 P6,900.00 Brokerage Fee Investment P1,200 P46,200 1.) 2.) 3.) 4.) 5.) PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES 6.) 7.) 8.) 9.) 10.) Page 45 Reflection (The learner writes how he/she feels about the activity) References for learners Oronce, O. (2016). General Mathematics. Manila: Rex bookstore, Inc. https://www.investopedia.com/ask/answers/09/difference-between-bond-stockmarket.asp PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 46 ANSWER KEY Activity 1 1. A 2. B 3. E 4. F 5. D 6. G 7. H 8. C Activity 2 1. Number of shares 2. Corporation issuing the certificate 3. Share holder or stockholder 4. Par value 5. Certificate number 6. ABCXYZ Corporation 7. Hazel Joy P. Vergara and Joji. C. Victoriano 8. P1,000.00 9. 1496 10. 420 Activity 3 1. 81.67 and 64.45 1. 70 and 31. 29 2. 67.54 and 65.35 2. 65.5 and 64.25 3. 67.25 3. 66.89 Activity 4 Brokerage Fee PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Investment Page 47 1. P840.00 6. P22,120.00 2. P2,160.00 7. P55,620.00 3. P3,920.00 8. P38,220.00 4. P5,340.00 9. P454,790.00 5. P7,200.00 10. P835,200.00 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 48 GENERAL MATHEMATICS Name of the Learner: _______Grade level: Section: _____Date: _____ LEARNING ACTIVITY SHEET FAIR MARKET OF A CASH FLOW STREAM INCLUDES ANNUITY Background Information for learners This activity sheet will serve as a self-learning guide for the learner. This will also help you to become money wise when you plan to have your own car, lot, or even a house in the future despite of having little budget. Specifically, in this topic, you will learn to compare the offers and find out which one is cheaper or more affordable by computing or calculating the fair market value of the items. The concept for the present and future value of annuity is to solve problems about cash flows. The cash flow is a term that refers to payments received (cash inflows) or payments or deposit made (cash outflows). Cash inflows can be represented by positive numbers and cash outflows can be represented by negative numbers. The fair market value or economic value of a cash flow (payment stream) on a particular date refers to a single amount that is equivalent to the value if the payment stream at that date. This particular is called the focal date. Lets have an example Company A offers P150,000 at the end of 3 years plus P300,000 at the end of 5 years. Company B offer P25,000 at the end of each quarter for the next 5 years. Assume tat money is worth 8%compounded annually. Which offer has a better market value? Identify the given: Company A P150,000 at the end of 3 years P300,000 at the end of 5 years Company B P25,000 at the end of each quarter for the next 5 years Find: fair market value of each offer (a) Illustrate the cash flows of the two offers using time diagrams. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 49 Company A Offer: 150,000 0 1 2 25,000 25,000 1 2 3 300,000 4 5 Company B Offer: 25,000 0 25,000 3 … 20 (b). Choose a focal date and determine the values of the two offers at the focal date. Since the focal date is the start of the term, compute the present value of each offer. Company A Offer: The present value of P150,000 three years from now is P1 = F (1 + j ) − n P1 = 150,000(1 + 0.04) −6 P1 = P118,547.18 The present value of P300,000 five years from now is P2 = F (1 + j ) − n P2 = 300,000(1 + 0.04) −10 P2 = P 202,669.25 Fair Market Value (FMV) = P1 + P2 = P118,547.18 + P 202,669.25 = P321,216.43 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 50 Company B Offer: Compute for the present value of a general annuity with quarterly payments but with semiannual compounding at 8% Solve the equivalent rate, compounded quarterly, of 8% compounded semi-annually. F1 = F2 ο¦ i ( 4) Pο§ο§1 + 4 ο¨ ο¦ i ( 4) ο§ο§1 + 4 ο¨ οΆ ο·ο· οΈ οΆ ο·ο· οΈ 4 ( 5) 20 ο¦ i ( 2) = Pο§ο§1 + 2 ο¨ ο¦ 0.08 οΆ = ο§1 + ο· 2 οΈ ο¨ οΆ ο·ο· οΈ 2 ( 5) 10 1 1+ i ( 4) = (1.04) 2 4 1 i ( 4) = (1.04) 2 − 1 4 i ( 4) = 0.019803903 4 The present value of an annuity is given by P = F (1 + j ) − n P = 25,000(1 + 0.019803903 ) − 20 P = P 409,560.47 Therefore, Company B’s offer is preferable since its market value is larger. Learning Competencies with code The students are able to calculate the fair market of a cash flow stream that includes annuity. M11GM-11d-2, PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 51 ACTIVITY 1. Calculate the future value of simple ordinary annuity. 1. Reden and Redenton are twins. After the graduation and being finally able to get a good job, they plan for retirement as follows. β« Starting at the age of 20, Reden deposits P15,000.00 at the end of each year for 30 years. β« Starting at age 35, Redenton deposits P20,000 at the end of each year for 18 years. Who will have the greater amount at retirement if both annuities earn 12% per year compounded annually? 2. If you pay P100.00 at the end of each month for 25 years on account that pays interest at 9% compounded monthly, how much money do you have after 25 years? ACTIVITY 2. Calculate the present value of simple ordinary annuity. 1. Hazel Joy borrowed money to her husband to buy a car. She will repay it by making monthly payments of P10,000 per month for the next 2 years at an interest rate of 6% per year compounded monthly. How much did Hazel Joy borrow? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 52 2. Mary works very hard because she wants to have enough money in her retirement account when she reaches the age of 60. She wants to withdraw P50,000.00 every 3 months for 20 years starting 3 months after she retires. How much Mary deposited at retirement at 7.5% per year compounded quarterly for annuity? ACTIVITY 3. Calculate the fair market of a cash flow stream that includes annuity. Mr.Ribaya received two offers on a lot that he wants to sell. Mr. Ocampo has offered P50,000 and P1,000,000 lump sum payment 5 years from now. Mr. Cruz, has offered P50,000 plus P40,000 every quarter for five years. Compare the fair market values of the two offers if money can earn 5% compounded annually. Which offer has a higher market value? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 53 RUBRIC CRITERIA 1 IDENTIFYIN G GIVEN STEPS No given are correct ACCURACY None of the steps were completed /no work was shown Entire activity assignment was incorrect 2 3 At most 4 is At least 4 is correct correct given given Few steps Most steps were were completed completed thoroughly thoroughly with work with work shown shown Several step One step of of the the problem problem was was incorrect incorrect 4 TOTA L All the given are correct Every step was completed thoroughly with work solution Each step of the problem was completed and correct *Rubric is for all the activities. Reflection The learner writes how he/she feels about the activity.) References for learners Oronce, O. (2016). General Mathematics. Manila: Rex bookstore, Inc. https://www.coursehero.com/file/p7tdb9t/Calculate-the-fair-market-value-of-a-cashflow-stream-that-includes-an-annuity/ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 54 Answer key Activity 1 1. Reden Redenton Given: Given: P = 15,000 P = 20,000 j = 0.12 j = 0.12 n = 30 n = 18 Solution: FV = P Solution: (1 + j ) n − 1 j (1 + 0.12) 36 − 1 0.12 (58.1355739286 ) FV = 15,000 0.12 FV = 15,000(484.4631160717 ) FV = 15,000 FV = P 7,266,946.74 (1 + j ) n − 1 FV = P j (1 + 0.12)18 − 1 0.12 (6.689965795 ) FV = 20,000 0.12 FV = 15,000(55.7497149585 ) FV = 20,000 FV = P1,114,994 .30 2. Given: Solution: (1 + j ) n − 1 FV = P j P = 15,000 j = 0.12 n = 30 (1 + 0.0075) 54 − 1 FV = 100 0.0075 (0.4970384672 ) FV = 100 0.0075 FV = 100(66.2717956247 ) FV = P6,627.18 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 55 Activity 2. 1. Given: Solution: ο P 1 − (1 + j ) − n j PV = P = 10,000 ο ο 10,000 1 − (1 + 0.005) − 24 0.005 10,000(0.1128143311) PV = 0.005 PV = P 225,628.66 PV = j = 0.005 n = 24 ο 2. Given: Solution: ο P = 10,000 j = 0.005 n = 24 P 1 − (1 + j ) − n PV = j ο ο 50,000 1 − (1 + 0.0065) −80 PV = 0.0065 50,000(0.3925248318 ) PV = 0.0065 PV = P3,140,198.65 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES ο Page 56 Activity 3. Given: Mr. Ocampo’s offer Mr. Ocampo’s offer 50,000 down payment 50,000 down payment P1,000,000 after 5 years P40,000 every quarter for 5 years Solution 1. Choose a focal date to be the start of the term. Since the focal date is at t = 0 , compute for the present value of each offer. Mr. Ocampo’s offer Mr. Cruz offer Fair Market Value FMV = 50,000 + 783,526.17 FMV = 50,000 + 705,572.68 FMV = P833,526.17 FMV = P755,572 Solution 2. Choose the focal date to be the end of the term. Mr. Ocampo’s offer Mr. Cruz offer Fair Market Value FMV = 1000000 + 63,814.08 FMV = 900,509.40 + 63,814.08 FMV = P1,063,814.08 FMV = P964,323.48 Mr. Ocampo’s offer still has a higher value, even if the focal date is to be the end of the term. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 57 GENERAL MATHEMATICS Name of the Learner: Grade level: Section: Date: LEARNING ACTIVITY SHEET CALCULATE THE PRESENT VALUE OF AND PERIOD OF DEFERRAL OF A DEFERRED ANNUITY Background Information for learners A deferred payment annuity is an insurance product that provides future payments to the buyer rather than an immediate stream of income. An annuity is a financial contract that allows the buyer to make a lump-sum payment, or a series of payments, in exchange for receiving future periodic disbursements. A deferred payment annuity allows the investment, known as the premium, to grow both by contributions and interest before payments are initiated. A deferred payment annuity is also known as a "deferred annuity" or a "delayed annuity." The period of deferment is the time interval to the beginning of the first payment interval. *if the first payment is due at the end of a specified interval, the formula is d = m * k − 1 *if the first payment is due on the next interval the formula is d = m * k Below is the formula in solving the Present Value of a Deferred Annuity: 1 − (1 + j ) j −( d +n) PV = R −R 1 − (1 + j ) − d j PV= Present Value R=Regular payment j = rate per conversion period i (m) , where i (m ) is the annual rate and m is the number of conversion period m n = no. of paying periods n = t * m , where t is the number of years d = Deferred periods. j= PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 58 Lets have an example, Hazel availed of a cash loan that gave her an option to pay P10,000 monthly for 1 year. The first payment due after 6 months. How much is the present value of the loan if the interest rate is 12% converted monthly? Identify the given: R = P10,000 t = 1 year i ( m ) = 12%or 0.12 m = 12 Calculate the value of j , n, and d j= i ( m ) 0.12 = = 0.01 m 12 n = t *m n = 1 * 12 = 12 d = 6 − 1 = 5 periods PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 59 Substitute all the given value using the Present Value of a Deferred Annuity: 1 − (1 + j ) PV = R j −(d +n) −R 1 − (1 + j ) − d j 1 − (1 + .01) −(5+12 ) 1 − (1 + .01) −5 − 10,000 0.01 0.01 .1556225127 0.0485343124 = 10,000 − 10,000 0.01 0.01 = 10,000(15.5622512667 ) − 10,000(4.85343124 ) = 155,622.51266701 − 48,534.312393251 = P107,088.20 PV = 10,000 PV PV PV PV Learning Competencies with code The learners are able to calculate the present value of and period of deferral of a deferred annuity. M11GM-11d-3,Quarter II Week 4 Activity 1. Identify the given values of the ff. in the given situation. 1. Mr. Julian wanted to buy a new branded car. He decided to pay P20,000 monthly for 5 years starting at the end of the 2 years with an interest rate of 12% compounded monthly. R= t= i ( m) = m= j= d= n= 2. A company offers Hazel Joy Vergara a deferred payment option for the car loan with a monthly payment of P5,000 for 4 years. The payment will start at the end of 5 months at the interest of 3% compounded monthly. R= t= i ( m) = m= j= d= n= PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 60 Activity 2. Find the period of deferral in each of the following deferred annuity problem. 1. A regular payment of P500 monthly for 3 years that will start 4 months from now. 2. A payment of P100,000 every quarter for 8 years starting at the end of 2 years. 3. A semi- annual payments of P1000 for 12 years that will start 3 years from now. 4. An annual installment of 25 years, first payment after 5 years. 5. A half- year instalment of 8 years, first payment of P2,000 after 18 months. Activity 3. Calculate present value of each problem completely. 1. Mr. Julian decided to buy a house and lot for his son before the latter’s big day. A payment for every month is P24,000 for 25 years starting at the end of 4 months with an interest of 6% compounded monthly. 2. Joy gave allowance to her mother for her medicine, She withdrew P25,000 semi-annually for 10 years starting at the end of 2 years. How much is the mother‘s allowance if the interest rate is 8%converted semi-annually? PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 61 RUBRIC CRITERIA 1 IDENTIFYIN G GIVEN STEPS No given are correct ACCURACY None of the steps were completed /no work was shown Entire activity assignment was incorrect 2 3 At most 4 is At least 4 is correct correct given given Few steps Most steps were were completed completed thoroughly thoroughly with work with work shown shown Several step One step of of the the problem problem was was incorrect incorrect 4 TOTA L All the given are correct Every step was completed thoroughly with work solution Each step of the problem was completed and correct *The rubric is for Activity 3 only. Reflection (The learner writes how he/she feels about the activity.) References for learners Oronce, O. (2016). General Mathematics. Manila: Rex bookstore, Inc. http://teachtogether.chedk12.com/teaching_guides/view/31 https://www.youtube.com/watch?v=hTUQ-wqV73A https://www.investopedia.com/terms/d/deferred-payment-annuity.asp PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 62 Answer Key: Activity 1 R = P 20,000 1. t = 5 years j = 0.01 i ( m ) = 0.12 n = 60 m = 12 R = P5,000 2. t = 4 years j = 0.0025 i ( m ) = .03 n = 48 m = 12 Activity 2 1. 3 months or 3 periods 2. 7quarters or 7 periods 3. 5 semi-annuals or 5 periods 4. 4 years or 4 periods 5. 2 half-year or 2 periods Activity 3 1. Given: R = P 24,000 t = 25 years i ( m ) = 0.06 m = 12 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES j = 0.05 n = 300 k =3 Page 63 Solution: 1 − (1 + j ) PV = R j −(d +n) −R 1 − (1 + j ) − d j 1 − (1 + .005) −( 3+300 ) 1 − (1 + .005) −3 − 24,000 0.005 0.005 0.7793604883 0.0148512407 = 24,000 − 24,000 0.005 0.005 = 24,00(155.87209766 ) − 24,000(2.97024814 ) = 3,740,930.34384 − 71,285.95536 = P3,669,644.39 PV = 24,000 PV PV PV PV 2. Given: R = P 25,000 j = 0.04 t = 10 years n = 20 i ( m ) = 0.08 k =3 m=2 Solution: 1 − (1 + j ) PV = R j −( d +n) −R 1 − (1 + j ) − d j 1 − (1 + .04) −(3+ 20 ) 1 − (1 + .04) −3 − 25,000 0.04 0.04 0.594273667 0.1110036413 = 25,000 − 25,000 0.04 0.04 = 25,00(14.856841675 ) − 25,000(2.7750910325 ) = 3,740,930.34384 − 71,285.95536 = P302,043.77 PV = 25,000 PV PV PV PV PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 64 GENERAL MATHEMATICS Name of the Learner: Grade level: Section: Date: ____________ LEARNING ACTIVITY SHEET MARKET INDICES FOR STOCKS AND BONDS Background Information for learners An index is an indicator or measure of something, and in finance, it typically refers to a statistical measure of change in a securities market. In the case of financial markets, stock and bond market indices, they consist of a hypothetical portfolio of securities representing a particular market or a segment of it. Learning Competencies with code The students are able to analyze the different market indices for stocks and bonds M11GM-IIe-4 ACTIVITY 1. Directions. Find the market price of one P1,000,000.00 bond at each quoted price. Write your answer on the blank provided. Example: at 43 0.43*P1,000,000 = P430,000 at 206 1 4 2.0625*P1,000,000 = P2,062,500 a. at 38 b. at 16 c. at 89 d. at 99 3 4 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 65 e. at 106 1 2 ACTIVITY 2. Directions. Find the total cost of each stock purchased below. ( 2 points each ) No. Of Shares 86 2000 580 38 1989 Market price per Share P2.58 P0.98 P100.00 P650.50 P1,050.00 Commission Total cost P20.00 P300.00 P1,500.00 P3,000.00 P50,000.00 1. 2. 3. 4. 5. ACTIVITY 3. Answer each of the following. 1. If you bought 600 shares of CJJ Bank Corp. stock at the 52 -week low, P43.65 per share, and sold at the 52-week high , P51.20 share, a. How much money did you make on this transaction (ignoring the dividends) (2points) b. What is the broker’s commission if the broker charges 6% of the total sale price?(2points) 2. Ms. Hazel bought 378 shares of CJJ stocks at its 52-week low, P652.30 per share, and sold at the 52-week high, P1,023.00 per share. a. How much did Ms. Hazel make on this transaction, dividends not included?(2points) b. Find the broker commission if the broker charges 9.2% of the total sale price. (2points) PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 66 3. Christian bought 529 shares of CJJ stock at its 52-week low, P1,095.00 per share , and its sold at 52-week high ,P1758.80 per share. a. How much did Christian make on this transaction, (dividends not included)?(2points) b. Find the broker commission if the broker charges 8% of the total sale price.(2points) Reflection (The learner writes how he/she feels about the activity.) References for learners: Oronce, O. (2016). General Mathematics. Manila: Rex bookstore, Inc. https://www.investopedia.com/terms/i/index.asp PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 67 Answer key: Activity 1 1. 380,000 2. 160,000 3. 890,000 4. 895,000 5. 1,065,00 Activity 2 1. P241.88 2. P2260 3. P59,500 4. P27,719 5. P2,138,450 Activity 3 1. a. P4,530.00 b. P1,843.00 a. P140,124.6 b. P35,575.84 a. 351,150.2 b. 74,432.46 2. 3. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 68 GENERAL MATHEMATICS Name of Learner:_____________________________ Grade Level:__________________ Section: ____________________________________ Score: _______________________ LEARNING ACTIVITY SHEET ILLUSTRATES, DISTINGUISHES AND SOLVE PROBLEMS INVOLVING BUSINESS AND CONSUMER LOANS (AMORTIZATION AND MORTGAGE) Background Information for Learners Loans are provided to help people who are experiencing financial crisis. Loans refer to lending of things/money to individuals or organizations that is expected to be paid back on a certain time with interest. There are different types of loans that we can take depending on our needs. The various types of loans are home loans, personal loans, student loan, business loan etc. In this learning activity sheet, you will be able learn the basic concepts of loans. The business loan which is referred as the borrowed money from a bank or other lending institutions/persons that can be used to start a business or to have a business expansion and the consumer loan which is referred as the borrowed money from a bank or other lending institutions/persons that can be used for personal or family purposes. Example 1. Identify whether the following is a consumer or business loan. 1. Mr. and Mrs. Bautista borrowed money from their aunt abroad to finance the college education of their children. Solution: Consumer Loan 2. Gina plans to sell cactus online. She borrows β±10,000 from her parents to start her online selling. Solution: Business Loan 3. Mr. Marquez wants to have his own house. He went to the Pag-IBIG (Home and Development Mutual Fund) office to apply and avail for a housing loan. Solution: Consumer Loan 4. Sharon owns two (2) Green Cab Pizza carts in their barangay. She wants to put another pizza cart business in their nearby barangays so she decided to have a loan to materialize her plans. Solution: Business Loan 5. Althea has a computer shop and owns 9sets of computers. She wants to add 6 more computers so she applied and availed β±60,000 loan from a lending company. Solution: Business Loan PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 69 Example 2. Mr. Rivera borrowed β± 250,000 from a bank to purchase a residential lot. The rate of interest of his loan is 7.5% annually The loan is to be paid for 2 years. How much is to be paid after 2 years? Solution: Given: P=β± 250,000, π π = 7.5% ππ 0.075 ππ 0.075 j= π = 1 = 0.075 n= (m)(t) = (1)(2)= 2 Find: Future Amount (F) F=P (1+j)n F=250,000(1+0.075)2 F= β±288,906.25 Thus, the amount to be paid by Mr.Rivera after 2 years is β±288,906.25 Example 3. A housing loan amounting to β±870,000 requires a 20% down payment. How much is the mortgage? Solution: Given: down payment rate= 20% or 0.20 Cash price = β±870,000 Find: amount of loan or mortgage Down payment = down payment rate x cash price = 0.20 x β±870,000 = β±174,000 Amount of the Loan = Cash Price-Down Payment = β±870,000- β±174,000 = β±696,000 Thus, the amount of loan or mortgage is β±696,000 Example 4. Mrs. Lopez acquired a housing loan amounting to β±1,200,000. Her monthly amortization is β±19,942.63 for 15 years. The interest rate is 7% convertible monthly. Find the outstanding balance after the 110th payments. Solution: Given: P = β±1,200,000 R = β±19,942.63 π π = π 12 = 7% or 0.07 j n k π 12 0.07 = π = 12 = (m)(t)= (12)(15)= 180 (total number of payments) = 110 (number of payments made) Find: Outstanding balance after 110th payments Answer: Let π΅110 represents the outstanding balance after the 110th payments or present value of the remaining 70 payments (n=70) PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 70 π΅110 = π [ 1−(1+π)−π π ] = β±19,942.63 [ 0.07 −70 ) 12 0.07 12 1−(1+ ] = β±1,143,408.87 Thus, the outstanding balance after 110 payments is β±1,143,408.87 Learning Competencies Illustrates, distinguishes and solve problems involving business and consumer loans (amortization and mortgage). M11GM-IIf-1 to 3 Exercise 1: Directions: Read and analyze each situation then identify whether the given is a business loan or consumer loan. Write BL if it is a business loan or CL if it is a consumer loan. Write your answer on the space provided before the number. [1 point each] _____1. Mr. Aquino owns 2 tractors for farming. He wants to buy 2 more tractors and use these for business. He applied for a loan in a bank. _____2. Mario wants to tour his family in other places so he applied for a loan and bought a pick-up truck. _____3. Due to COVID-19 pandemic, lot of people needs to work from home through online. So, Ram decided to put up internet business because this is in demand nowadays. He then applied for a loan to purchase materials and equipment to start his business. _____4. Mr. & Mrs. Santos borrowed β±300,000 from a bank in order to build rooms for rent. _____5. Due to the hot weather, Athena's water refilling business has become popular in their barangay. So, she made a loan from a bank that can be used to expand her business in other barangay. _____6. Elsa bought a washing machine that could help her in washing their clothes through home credit. _____7. Aiyana wants to renovate their old house but she does not have savings. She went to a bank and applied for a salary loan to finance her plan. _____8. Freggie had a housing loan payable monthly for 15 years. _____9. Food is a basic necessity of a human being. So, Cory plans to have a restaurant. She borrowed money from a lending institution for the construction and operation of her business. _____10. Ariel got loan worth β±450,000 and used this amount to purchase an apartment near his workplace. Exercise 2: Directions: Analyze the given amortization procedures in the table below then fill in each blank using the guide questions below the table. Show your solution using an extra sheet of paper. [3 points each] Problem: A salary loan amounting to β±15,000 is to be paid annually for 4 years with an interest rate of 6% compounded annually. The annual amortization is β± β±4,328.87. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 71 Period 0 1 2 3 4 Totals Regular Payment (R) B β±4,328.87 β±4,328.87 β±4,328.87 G Interest Component of Payment β±900 β±694.27 D476.19 245.031 H PRINCIPAL Component of Payment β±3,428.87 C 3,852.68 4,083.84 I Outstanding Balance A β±11,571.13 7,936.53 E4,083.85 F Guide Questions: A. How much is the amount of the loan? (Outstanding balance at time 0) B. How much is the first regular annual payment? C. For the second payment, how much goes to pay the principal? D. For the third payment, how much goes to pay the interest? E. How much is the outstanding balance after the 3rd payment? F. How much should be the outstanding balance after the 4th or last payment? G. How much is the total amount of regular payment for 4 years? H. How much is the total interest paid for 4 years? I. How much is the total payment for the principal for 4 years? Exercise 3: Problem Solving Directions: Read and analyze each question. Then answer the following problems completely. Show your step-by-step solution in each item. 1. Mr. Bagain borrowed β±2,000,000 for the expansion of his farm supply business. The effective rate of interest is 8%. The loan is to be repaid in full after three years. How much is to be paid after three years? [5 points ] Solution: PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 72 2. Ramon borrowed β±1,700,000 for the purchase of a reaper harvester machine. If his monthly payment is β± 35,000 on a 5-year mortgage, find the total amount of interest. [5 points ] Solution: 3. If a 2- hectares land is to be sold for β±2,000,000 and the lender requires 30% down payment, what is the amount of the mortgage? [5 points ] Solution: 4. Beatriz got a loan amounting to β±40,000 and to be repaid in 12 months at 6% convertible monthly. How much is her monthly payment? [5 points ] Solution: 5. Mr.. Reyes is considering to pay his outstanding balance for 4 years of payment. The original amount of loan is β±450,000 payable annually in 6 years. If the interest rate is 8.5% per annum. a.) Find the regular payment annually, b) How much is the outstanding balance after 4 years of payment, c) Find the total amount of interest for 6 years. [15 points ] Solution: Reflection: What are the factors that you need to consider in applying for a loan? Why is it important to consider those factors? Explain your answer. Write your answers on the blanks provided below. ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ References of Learners Verzosa, D.B. , et.al., (2016). General Mathematics for Senior High School (First Edition). Quezon City Manila; Lexicon Press Inc. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 73 Answer Key: Exercise 1 1. BL 2. CL 3. BL 4. BL 5. BL 6. CL 7. CL 8. CL 9. BL 10. CL Exercise 2 Period 0 1 2 3 4 Totals Regular Payment (R) B. β±4,328.87 β±4,328.87 β±4,328.87 β±4,328.87 G. β±17,315.48 Interest Component of Payment β±900 β±694.27 D. β±476.19 β±245.031 H. β±2,315.49 PRINCIPAL Component of Payment β±3,428.87 C. β±3,634.6 β±3,852.68 β±4,083.84 I. β±15,000 Outstanding Balance A. β±15,000 β±11,571.13 β±7,936.53 E. β±4,083.85 F. 0 Exercise 3 1. Given: P=β±2,000,000 j=8% or 0.08 n=3 years Find: Future Value (F) F=P(1+j)n F=β±2,000,000(1+0.08)3 F=β±2,519,424 Thus, the amount to be paid after 3 years is β±2,519,424 2. The total amount paid is given by Total Amount =(35,000)(12months)(5years) =β±2,100,000 Thus, the total interest is the difference between the total amount paid and the amount of the mortgage. Total Interest = Total Amount Paid- Amount of the Mortgage = β±2,100,000- β±1,700,000 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 74 = β±400,000 The total interest of the mortgage for 5 years is β±400,000 3. Given: Cash Price=β±2,000,000 Down payment rate=30% or 0.30 Down payment= Cash Price x Down payment Rate = β±2,000,000 x 0.30 = β±600,000 Amount of the Loan = Cash price –Down payment = β±2,000,000-β±600,000 = β±1,400,000 The amount of the loan or mortgage is β±1,400,000 4. Given: P= β±40,000 i12 i12=6% or 0.06, j= π = 0.06 12 = 0.005 n=12 Find: Monthly/Regular Payment R π R = 1−(1+π)−π [ = ] π 40,000 1−(1+0.005)−12 [ ] 0.005 = β±3,442.66 The monthly payment of Beatriz is β±3,442.66 5. Given: P= β±450,000 n=6 j=8.5% or 0.085 Find: Outstanding balance after 4 years (present value of the remaining 2 payments) a. R = = π 1−(1+π)−π [ ] π 450,000 1−(1+0.085)−6 [ ] 0.085 = β±98,823.19 The regular payment annually of Mr. Reyes is β±98,823.19 b. π΅4 =R[ 1−(1+π)−π = 98,823.19 [ ] π 1−(1+0.085)−2 = β±175,027.16 0.085 ] The remaining balance of Mr. Reyes after 4th payment is β± 175,027.16 c. The total amount paid is given by PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 75 Total Amount = β±98,823.19 x 6years = β±592, 939.14 Thus, the total interest is the difference between the total amount paid and the amount of mortgage. Total Interest = Amount Paid – Amount of Mortgage = β±592, 939.14 - β±450,000 = β±142, 939.14 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 76 GENERAL MATHEMATICS Name: _____________________ Date: ______________________ Grade Level: ____________ Score: __________________ LEARNING ACTIVITY SHEET ILLUSTRATES AND SYMBOLIZES PROPOSITIONS Background Information for Learners In your everyday life, you are facing different information about people, things and events. You may wonder how the new normal can help students in learning the needed competencies. You can express your idea as “The new normal is good.” This statement is called proposition. What is a proposition? This activity sheet is a self-paced material where students can check and recheck their understanding and progress about the topic. It is an enjoyable material where ‘learning is fun’ can be experienced. This Learning Activity Sheet is intended for Senior High School students particularly Grade 11 who are taking General Mathematics subject. REASONING (Oxford illustrated Dictionary) “the intellectual faculty by which conclusions are drawn from premises. LOGIC “the study of reasoning, seeks the rules and principles of how people should reason correctly and rationally. “It is a normative science as it provides prescriptions for rational thinking.” “To discover truths is the task of all sciences; it falls to logic to discern the laws of truth… (Gottlob Frege,1956) Dear students, if you are confined to dicern from what is right or wrong, your brain starts to function and propmt you with your reasoning ability. That is Logic. When you start to express your complete idea whether it is true or false, then that is proposition. Propositions are statements in declarative form which express a single and complete idea, and bears either truth or falsity but not both. If a proposition is true, then its truth value is true, which is denoted by T; otherwise, its truth value is false, which is denoted by F. Like any other declarative sentence, it has a subject and a predicate. It is usually denoted by a small letter. “This topic is interesting.”, is a proposition. p: This topic is interesting. Learning Competency 1: The learner illustrates and symbolizes propositions .M11GMIIg-1-2 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 77 Priming Activity Directions: Determine whether each of the following statements is a proposition or not. If it is a proposition, give its truth value. 1. 2. 3. 4. 5. All birds can fly. There is life on Mars. Look out! X is an even number. What is the domain of the function? A proposition is a declarative sentence that expresses a complete idea and bears either truth or falsity. Big Idea! Key: Numbers 1 & 2 are propositions since it has a complete thought and it is a declarative sentence, although the truth value is false. Number 3 is not a proposition since the sentence is imperative. Number 4 depends on the value of x, it is neither true nor false,hence, it is not a proposition. Number 5 is a question, hence, it is not a proposition. Activity 1: “I will try It!” Tell whether the statement is a Proposition or Not. 1. 2. 3. 4. 5. All cows are black. x + 2 = 2x . x+2 = 2x, when x = -2 Look out! Wash your hands. ______________ ______________ ______________ ______________ ______________ Activity 2: “I can do It!” Tell whether the statement is a Proposition or Not. Determine the truth value. 1. 2. 3. 4. 5. COVID 19 is a Virus. 1 is an even number. x+2 = 2x, when x = -2 This statement is true. Check your answer. ______________ ______________ ______________ ______________ ______________ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES _________ _________ _________ _________ _________ Page 78 Activity 3: “I can tell why” Tell whether the statement is a Proposition or Not. If not, why? 1. Roses are red and violets are blue. _______________________________________ 2. What is a proposition? _______________________________________ 3. Wear your mask! _______________________________________ 4. Learning is fun and challenging in the new normal. _______________________________________ 5. m is a prime number. _______________________________________ Congratulations! You’ve made it this far. You may now proceed… Reflection Evaluate your understanding using declarative sentence to express your opinion regarding the topic on propositions. Answer: ___________________________________________________________________________ ___________________________________________________________________________ References DIWA Senior High School Series, p. 204 General Mathematics, p 240 Year Triangle Trigonometry, PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 79 Answer Key Activity1 1. 2. 3. 4. 5. Proposition Not Proposition Not Not Activity2 1. Proposition. True 2. Proposition. False 3. Proposition. False 4. Not a proposition since it’s neither true not false. It is a paradox. 5. Not a proposition. It is imperative. Activity 3 1. 2. 3. 4. 5. Proposition Not, because it is not a declarative statement. Not, because it is not a declarative statement. Proposition Not, because it can neither be true nor false. Module 2 (LM) BEAM Fourth Year 2 (L PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 80 GENERAL MATHEMATICS Name: _____________________ Date: ______________________ Grade Level: ____________ Score: __________________ LEARNING ACTIVITY SHEET DISTINGUISHES BETWEEN SIMPLE AND COMPOUND PROPOSITIONS Background Information for Learners Welcome back students βΊ You are now ready for the next stage…your status. Are you single or complicated? In mathematics we call it Simple proposition or Compound proposition. A simple proposition is composed of only one propositional variable. A proposition is simple if it cannot be broken down any further into other component propositions. p1: Mathematics is challenging. p2: Two is the smallest prime number. A compount proposition is composed of subpropositions and various connectives. It can be constructed using connectors, conjunctions and transitional words. Examples of propositions contain if-then, and (ο), or (ο) and not (οΎ) q1: Roses are red and violets are blue. q2: Either 2 is an even number or 2 is not the smallest prime number. SIMPLE PROPOSITION • A proposition that contains only one idea • It has only one subject and one predicate • “Math is fun.” COMPOUND PROPOSITION • A proposition that is composed of at least two simple propositions joined together by logical connectives • “Math is fun and challenging.” Learning Competency 2: The learner distinguishes between simple and compound propositions M11GM-IIg-3 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 81 Priming Activity Directions: Determine whether the following statements are simple or compound propositions. If the proposition is a compound proposition, identify logical connectors used. 1. 2. 3. 4. COVID 19 has over 2 million cases in the world. If I will study harder, then I will pass the exam. Either it is sunny in Isabela or its streets are flooded Zero is not a negative number. Key Points Numbers 1 is a simple proposition. Number 2 is compound using ‘if-then’ as connector Number 3 is compound with or as connector Number 4 uses not as a connector. Activity 1: “Check my Understanding!” Determine whether the following statements are simple or compound propositions. 1. 2. 3. 4. Either 1 is an even number or 4 is a perfect square. The immune system of my body is important in this time of pandemic. The sum of two even numbers is always even. Either logic is fun and interesting, or it is challenging. Activity 2: “Deepen your Understanding!” Determine whether the following statements are propositions. If the proposition is a compound proposition, identify the simple components and the logical connectors used. 1. 2. 3. 4. If your score is more than 2, then you will pass the subject. Ana’s average is at least 90 and she is getting an academic award. The square of an odd number is not even. A password must be at least 6 characters or it must be at least 8 characters long. Activity 3: “I Understand!” Given the following simple propositions, construct compound propositions by adding another simple proposition. Use any connector. 1. I often wash my hands. _______________________________________________ 2. I study my lessons using my phone. _______________________________________________ 3. A proposition is simple. _______________________________________________ 4. My favorite subject is Mathematics _______________________________________________ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 82 Reflection 1. What makes it simple (simple proposition)? Answer: _______________________________________________________ 2. What are the connectors that can be used to form compound proposition? Answer: _______________________________________________________ 3. Are simple propositions can be formed as compound propositions? Why Answer: _______________________________________________________ References Conceptual Math and Beyond General Mathematics, p. 203 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 83 Answer Key Activity 1 1. Compound 2. Simple 3. Simple 4. Compound Activity 2 1. p1: Your score is more than 2; p2: You will pass the subject. (If Then) 2. p1: Ana’s average is at least 90; p2: Ana is getting an academic award (and) 3. not 4. p1: A password must be at least 6 characters; p2: A password must be at least 8 characters long (or) Activity 3 (answers may vary) 1. I often wash my hands and wear mask. 2. I study my lessons using my phone or my books. 3. Either a proposition is simple or compound. 4. My favorite subject is either mathematics or English. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 84 GENERAL MATHEMATICS Name: _____________________ Date: ______________________ Grade Level: ____________ Score: __________________ LEARNING ACTIVITY SHEET PERFORMS THE DIFFERENT TYPES OF OPERATIONS ON PROPOSITIONS Background Information for Learners Dear students, finally, you are now on the last stage of this learning activity. You will learn how to construct a truth table and perform the different operations on proposition. The truth table shows all its possible truth values. Since a proposition has two possible truth values, a proposition p whould have the following truth table: p T F Truth tables can also be used to display various combinations of the truth values of two propositions p and q. . p T T F F q T F T F In addition to truth table, you will also learn NEGATION (not) which states exact opposite of a given proposition. p: I represents an imaginary number οΎp: I does not represent an imaginary number The conjunction (but, also, moreover) of two propositions is true if both component propositions are true and false if at least one of them is false. A disjunction (unless) is true if at least one of the component propositions is true and is only false if both are false. The conditional proposition (only if, implies) is false only if Q is false but P is true. Biconditional Proposition is a conjuction of two conditional propositions The truth value of Pο«Q is true whenever the two component propositions have the same truth value. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 85 Logical Operators Let p and q be arbitrary propositions Type Logical Operator Symbolic Form Read as Conjunction And pο q p and q Disjunction Or pο q p or q Conditional If…then p→ q If p, then q Biconditional If and only if pο«q p if and only if q Negation Not οΎp Not p Learning Competency 3: The learner performs the different types of operations on propositions. M11GM-IIg-4 Priming Activity 1 Directions: Express the following propositions in symbols. Assume that p is the antecedent and q is the consequent. 1. 2. 3. 4. 5. I will go to the party if and only if my parents will allow me to go. If one person reuses plastic containers, then he or she helps lessen wastes. If you are not pro-SHS, then you are not here. Either Plato is a Philosopher or Einstein is a scientist. Mathematics is not the most difficult subject. Checkpoint! Big Idea! 1. 2. 3. 4. 5. biconditional pο«q conditional p→q contrapositive οΎp→οΎq disjuction p v q Negation οΎp Activity 1: “It’s Your Turn!” Directions: Express the following propositions in symbols. Assume that p is the antecedent and q is the consequent. 1. Pia got the award and propmtly stopped using cellphones. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 86 2. 3. 4. 5. Either you will study through online or through the use of modules. Eight is not equal to negative eight. The function f is even if and only if f(x)=f(-x). Classes will start on June if and only if there is no pandemic. Activity 2: “Complete Me” Construct the truth table of the different types of logical operators. a. οΎP P T F b. c. d. _ Pο Q P Q T T T F F T F F P Q T T T F F T F F P Q T T T F F T F F PοQ P→Q PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 87 Activity 3: “Reconstruct Me” Given the following propositions below, Write the component propositions and construct the symbolic form. Example: Two lines P and Q are parallel if and only if they are coplanar and P and Q do no intersect. Component Propositions: p: Two lines P and Q are parallel. c: P and Q are coplanar. l: P and Q intersect. Symbolic form: pο«(cο(οΎl)) or pο«cοοΎl 1. If you are interested in becoming a scholar, you should fill-up the application form and submit it to our officer in-charge or to any teacher in school. 2. Upon announcement of Moderate General Community Quarantine, classes in all levels should be conducted in the new normal and children below 21 years old should stay at home. Reflection: 1. What I learned in this activity is __________________________________________. 2. I need to study more on _____________________________________________ References DIWA Senior High School Series, p. 205 General Mathematics LM, p 250 Year Triangle Trigonometry, PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 88 Answer Key Activity 1 1. Conjunction 2. Disjunction 3. Negation 4. Conditional 5. Biconditional pοq pοq οΎp p→q pο«q Activity 2 a. . P οΎP T F F T c (a) P Q PοQ T T T T F T F T T F F F (c) b P Q PοQ Module 2 dd T T T P Q P→Q T F F F T F T T T F F F T F F F T T F F T (b) (d) PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 89 Avtivity 3 1. Component Propositions: v: You are interested in becoming a scholar. f: You should fill-up the form. h: You should submit to the officer in-charge. L: You should submit it to any teacher in school. Symbolic form: v→(f ο (hο l)) 2. Component Propositions: p: There is a Moderate General Community Quarantine c: Classes in all levels are conducted in the new normal s: Children below 21 years old should stay at home Symbolic form: p→(cοs) or p→cοs Year Trigonometry, Module PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Triangle Page 90 GENERAL MATHEMATICS 11 Name of Learner: ________________________________ Section: ________________________________________ Grade Level: _________ Score: ______________ LEARNING ACTIVITY SHEET ILLUSTRATE DIFFERENT TYPES OF TAUTOLOGIES AND FALLACIES AND DETERMINE THE VALIDITY OF CATEGORICAL SYLLOGISMS Background Information for Learners A valid argument satisfies the validity condition; that is, the conclusion q is true whenever the premises p1, p2,…, pn are all true. The argument is valid if the conditional (π1 ∧ π2 ∧ … ∧ ππ ) → π is a tautology. An argument (π1 ∧ π2 ∧ … ∧ ππ ) → π, which is not valid is called a fallacy. Example 1 Prove that the argument ((p → π) Λ π) → π known as Modus Ponens is valid. Solution Show that ((p → π) Λ π ) → π is a tautology. p Q p→q (p → π) Λ π (p → π) Λ π → π T T T T T T F F F T F T T F T F F T F T Since ((p → π) Λ π) → π is a tautology, then the argument is valid. Example 2 Prove that the argument ((p → π) Λ π → π is a fallacy. This is known as the Fallacy of the Converse. Solution Show that ((p → π) Λ π → π is not a tautology using the truth table. p Q p→q (p → π) Λ π ((p → π) Λ π → π T T T T T T F F F T F T T T F F F T F T On the third row, the premises q and p → q are both true but the conclusion p is false. The given argument is a fallacy. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 91 Example 3 Determine whether the argument is valid. If triangle T1 and T2 are congruent, then they are similar. Triangles T 1 and T2 are congruent. Therefore, triangles T1 and T2 are similar. Solution The argument is valid by Modus Ponens. Furthermore, we know that from the geometry of triangles that congruent triangles are also similar (but similar are not necessary congruent). Learning Competency with code The learner is able to illustrate different types of tautologies and fallacies and determine the validity of categorical syllogisms (M11GM-IIi-1-2,Quarter II) Directions/Instructions: A. Complete the truth table for the given statement to show that the compound statement is a tautology. 1. p → (p Λ q) p T T F F q T F T F pΛ q p → (p Λ q) 2. p → (q Λ p) p T T F F Q T F T F qΛ p p → (q Λ p) PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 92 3. (p → q) ↔ (~ q → ~ p) p q p→q ~q T T T F F T F F ~q→~p (p → q) ↔ (~ q → ~ p) ~q qΛ ~q (p Λ (~ p)) Λ (q Λ (~ q)) ~p ~q ~p 4. (p Λ (~p)) Λ (q Λ (~ q)) p q ~p pΛ ~p T T T F F T F F 5. ~ (p Λ q) ↔ (~ p Λ ~ q) p Q pΛq ~ (p Λ q) T T T F F T F F ~pΛ ~q ~ (p Λ q) ↔ (~ p Λ ~ q) B. Determine whether the symbolic form of the argument is valid or a fallacy using a truth table. 1. 2. 3. 4. 5. π →π π ∴π πΛ π ∴π Λ π π Λ ~π ~π ∴~π ~π Λ π π↔π ∴π Λ π π→π π→π ∴~π → ~ π PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 93 C. Determine whether the following arguments are valid. If it is valid, identify the rule of inference which justifies the validity otherwise identify the type of fallacy exhibited by the argument. 1. All tigers are mammals. No mammals are creatures with scales. Therefore, no tigers are creatures with scales 2. All spider monkeys are elephants. No elephants are animals. Therefore, no spider monkeys are animals 3. Today isn’t a holiday. If there will be mail delivery, then today isn’t a holiday. Therefore, there will be mail delivery. 4. If today is Tuesday, then I have to finish my homework. If I have to finish my homework, then I have to go to work. Therefore, if today is Tuesday, then I have to go work. 5. If you drive a BMW, then you are a telemarketer If you are a telemarketer, then you are rolling in cash. Therefore, if you drive a BMW, then you are rolling in cash. 6. You sell used cars, or you are charming. You don’t sell used cars. Therefore, you are charming. 7. If you will bring me a cake, then today is my birthday. If today is my birthday, then you will send flowers. Therefore, if you will bring me a cake, then you will send flowers. 8. Today is Monday, or life doesn’t look bleak. Today is Monday. Therefore, life looks bleak. 9. You floss twice a day, or you don’t brush after every meal. You floss twice a day. Therefore, you brush after every meal. 10. If today is a holiday, then we’ll have a picnic. Today is holiday. Therefore, we’ll have a picnic. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 94 Reflection What I have learned in this activity ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ References: Department of Education General Mathematics (Teacher’s Guide).2016 Orines, Fernando B. Next Century Mathematics (General Mathematics).Phoenix Publishing House, Inc.2016 Oronce, Orlando A. General Mathematics. Rex Book Store.2016 http://www.math.fsu.edu PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 95 Answer Key A 1. p → (p Λ q) p T T F F q T F T F pΛ q T T T F p → (p Λ q) T T T T 2. p → (q Λ p) p T T F F q T F T F qΛ p T T T F p → (q Λ p) T T T T 3. (p → q) ↔ (~ q → ~ p) p q p→q ~q T T T F T F F T F T T F F F T T ~p F F T T ~q→~p T F T T (p → q) ↔ (~ q → ~ p) T T T T (p Λ (~ p)) Λ (q Λ (~ q)) T T T T 4. (p Λ (~ p)) Λ (q Λ (~ q)) p q ~p pΛ ~p T T F T T F F T F T T T F F T T ~q F T F T qΛ ~q T T T T 5. ~ (p Λ q) ↔ (~ p Λ ~ q) p q pΛq ~ (p Λ q) T T T F T F F T F T F T F F F T ~p F F T T ~q F T F T PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES ~pΛ ~q F T T T ~ (p Λ q) ↔ (~ p Λ ~ q) T T T T Page 96 B 1. The symbolic statement is ((p → q) Λ p) → p p q p→q p → q) Λ p ((p → q) Λ p) → p T T T T T T F F F T F T T F T F F T F T The entries in the final column of the truth table are all true, so the argument is valid. 2. The symbolic statement is (p Λ q) → (p Λ q) p q pΛ q pΛq (p Λ q) → (p Λ q) T T T T T T F T F F F T T F F F F F F T The entries in the final column of the truth table are not all true, so the argument is a fallacy. 3. The symbolic statement is ((π Λ ~π) Λ ~ π) → ~ π p q ~q pΛ ~q (π Λ ~π) Λ ~ π ~p ((π Λ ~π) Λ ~ π) → ~ π T T F T F F T T F T T T F F F T F F F T T F F T T T T T The entries in the final column of the truth table are not all true, so the argument is a fallacy. 4. The symbolic statement is (~π Λ π) Λ (π ↔ π) → (π Λ π) p q r ~p T T T T F F F F T T F F T T F F T F T F T F T F F F F F T T T T ~π Λ π π ↔ π (~π Λ π) Λ (π ↔ π) πΛπ F F F F T T F F T F T F F T F T F F F F F T F F T F T F F F F F (~π Λ π) Λ (π ↔ π) → (π Λ π) T T T T T F T T The entries in the final column of the truth table are not all true, so the argument is a fallacy. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 97 5. The symbolic statement is ((π → π) Λ (π → π)) → (~π → ~ π) p q r T T T T F F F F T T F F T T F F T F T F T F T F π → π π → π T T F F T T T T T F T T T F T T (π → π) Λ (π → π) T F F F T F T T ~q ~r F F T T F F T T F T F T F T F T ~π →~π T T F T T T F T ((π → π) Λ (π → π)) → (~π → ~ π) T T T T T T F T The entries in the final column of the truth table are not all true, so the argument is a fallacy. C 1. Valid (Modus Tollens) 2. Valid (Modus Tollens) 3. Invalid (Fallacy Of The Converse) 4. Valid (Rule Of Disjunctive Syllogism) 5. Valid (Law Of Syllogism) 6. Valid (Rule Of Disjunctive Syllogism) 7. Valid (Law Of Syllogism) 8. Invalid (Affirming The Disjunct) 9. Invalid (Affirming The Disjunct) 10. Valid (Modus Ponens) PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 98 GENERAL MATHEMATICS Name of Learner: ___________________________ Grade Level: _____________ Section:___________________________________ Date: ___________________ LEARNING ACTIVITY SHEET ESTABLISH THE VALIDITY AND FALSITY OF REAL-LIFE ARGUMENTS USING LOGICAL PROPOSITIONS, SYLLOGISMS, AND FALLACIES Background Information for Learners An argument is composed of premises and conclusion. Premises are the statements in an argument that will help you to draw or create a conclusion. You can easily identify the conclusion in an argument because of the conclusion indicators such as therefore, hence, so, thus, consequently, it is shown that and etc. π1 π2 Premises Argument ∴π Conclusion Arguments can be a valid or a fallacy. An argument is said to be valid if the truth of the premises logically supports the truth of the conclusion while, it is fallacy/invalid if the premises do not support convincing reasons for the conclusion. To test its validity, you can use the different rules of inference such as rule of simplification, rule of addition, rule of conjunction, modus ponens, modus tollens, law of syllogism, rule of disjunctive syllogism, rule of contradiction, and rule of proof cases. On the other hand, the different logical fallacies such as fallacy of the converse, fallacy of the inverse, affirming the disjunct, fallacy of the consequent, denying a conjunct, and improper transposition will help you recognize whether an argument is a fallacy or invalid. Examples: If there is a power interruption, the ISELCO will notify us. π → π The ISELCO did not notify us. ∼π So, there was no power interruption. ∴ ∼π This argument is valid through Modus Tollens If it is winter then it is cold. It is winter. Therefore, it is cold. This argument is valid through Modus Ponens π→π π ∴ ~π If it is winter then it is cold. π→π It is not winter. ~π Therefore, it is not cold. ∴ ~π This argument is invalid or fallacy using the fallacy of the Inverse PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 99 Learning Competency Establish the validity and falsity of real-life arguments using logical propositions, syllogisms, and fallacies. M11GM-IIi-3 Directions/Instructions: Read and understand the directions in each exercise. If you have any question, feel free to message your teacher for clarification and assistance. EXERCISE 1. Determine what rule of inference is used in each item. Write your answer on the blank provided. [2 points each] 1. If you are a student, then you go to school. If you go to school, then you have an allowance. Therefore, if you are a student, then you have an allowance. Answer: ___________________ 2. The statement is either true or false. The statement is not false. Therefore, the statement is true. Answer: ___________________ 3. If logic is an easy subject, then the students are happy. The students are not happy. So, logic is not an easy subject. Answer: ___________________ 4. All riders wear helmets. Allan doesn’t wear a helmet. Therefore, Allan isn’t a rider. Answer: ___________________ 5. Everybody will be happy if a scientist can discover a medicine that cures coronavirus. Everybody is happy. Thus, a scientist discovered a medicine that cures coronavirus. Answer: ___________________ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 100 Exercise 2. Determine what a logical fallacy is used in each item. Write your answer on the blank provided. [2 points each] 1. You use either Google Map or Waze to reach your destination. You used Google Map to reach your destination. Therefore, you did not use Waze to reach your destination. Answer: ___________________ 2. It is not true that classes starts in August and online class will be implemented. Classes did not start on August. Therefore, online class is implemented. Answer: ___________________ 3. If I drink coffee, it will have a sleepless night. I did not drink coffee. Thus, it won’t have a sleepless night. Answer: ___________________ 4. If there is a limited supply of foods, then I will preserve foods. I preserved foods. Hence, there was a limited supply of goods. Answer: ___________________ 5. All students wear ID. The guard wears an ID. Hence, the guard is a student. Answer: ___________________ EXERCISE 3. Identify whether the following arguments are valid or not. Put a check mark ( Μ· ) on the picture if it is valid and (x) mark if it is not valid. [2 points each] Source:https://thelogicofscience.com/2017/03/14/the-importance-oflogical-fallacies/ Source: http://fleasnobbery.blogspot.com/2011/02/syllogism.html PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 101 If you are a monkey, then you are cute. If you are cute, then you are my pet. Therefore, if you are a monkey, then you are my pet. Source: https://www.pinterest.co.uk/pin/466826317618791138/ Source:https://www.pinterest.com/pin/129267451778131847/ If you use cellphone during a storm, you will be hit by lightning. You were not hit by lightning. Therefore, you did not use cellphone during a storm. Source:https://www.quora.com/Is-it-safe-to-use-cell-phones-duringlightning You will be infected with corona virus or you will stay at home. You were not infected with corona virus. Therefore, you stayed at home. Source: https://www.istockphoto.com/vector/stay-at-home-stop-corona-virusconcept-vector-illustrator-gm1213503886-352699947 EXERCISE 4. Create a valid conclusion from each set of premises. If no valid conclusion is possible, write “no valid conclusion.” [2 points each] 1. If it rains, then the weather is cold. If the weather is cold, then I fall asleep quickly. ___________________________________________ 2. If you stay at home, then you are safe. If you are safe, then spreading of virus will be lessen. ___________________________________________ 3. All riders wear helmets. Allan wear a helmet. ___________________________________________ 4. If Ivana Alawi is a famous vlogger, then she knows how to edit videos. Ivana Alawi is not a famous vlogger. ___________________________________________ PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 102 5. The season in the Philippines is either wet or dry. The season in the Philippines is not wet. ____________________________________________ EXERCISE 5. Think and write out two valid arguments you have encountered in your daily life. [5 points each] 1. _____________________________________________________________________ _____________________________________________________________________ ______________________________________________________ 2. _____________________________________________________________________ _____________________________________________________________________ ______________________________________________________ Reflection Write your insights about the activities you have undertaken. ___________________________________________________________________________ ___________________________________________________________________________ ____________________________________________________________ References Verzosa, D. et. Al. (2016). General Mathematics. Lexicon Press Inc. https://www.stat.berkeley.edu/~stark/SticiGui/Text/reasoning.htm https://math.libretexts.org/Courses/Monroe_Community_College/MTH_220_Discrete_Math/ 2%3A_Logic/2.6_Arguments_and_Rules_of_Inference https://www.istockphoto.com/vector/stay-at-home-stop-corona-virus-concept-vector-illustratorgm1213503886-352699947 PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 103 Answer Key EXERCISE 1 1. Law of Syllogism 2. Rule of Disjunctive Syllogism 3. Modus Tollens 4. Modus Tollens 5. Modus Ponens EXERCISE 2 1. Affirming the disjunct 2. Denying a Conjunct 3. Fallacy of the inverse 4. Fallacy of the converse 5. Fallacy of converse EXERCISE 3 Source:https://thelogicofscience.com/2017/03/14/the-importance-of-logicalfallacies/ Source: http://fleasnobbery.blogspot.com/2011/02/syllogism.html If you are a monkey, then you are cute. If you are cute, then you are my pet. Therefore, if you are a monkey, then you are my pet. Source: https://www.pinterest.co.uk/pin/466826317618791138/ If you use cellphone during a storm, you will be hit by lightning. You were not hit by lightning. Therefore, you did not use cellphone during a storm. Source:https://www.quora.com/Is-it-safe-touse-cell-phones-during-lightning Source:https://www.pinterest.com/pin/129267451778131847/ You will be infected with corona virus or you will stay at home. You were not infected with corona virus. Therefore, you stayed at home. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 104 EXERCISE 4. 1. Therefore, if it rains, then I fall asleep quickly. 2. Therefore, if you stay at home, then spreading of virus will be lessen. 3. No valid conclusion 4. No valid conclusion 5. Therefore, the season in the Philippines is dry EXERCISE 5. Answers may vary. PRACTICE PERSONAL HYGIENE PROTOCOLS AT ALL TIMES Page 105
