Directions: Analyze the given lesson below. After that, solve and fill in the blanks what is missing.

GEN MATH TERMINOLOGIES QUESTION? If you have P10,000 where will you invest 1. COMPOUND INTEREST ( 1 ) your money? At simple interest or at compound This is an interest resulting from the periodic interest? Both with an annual interest rate of 10% for 4 addition of simple interest to the principal amount years 2. COMPOUND AMOUNT ( F ) This is an accumulated amount composed of SIMPLE INTEREST the principal and the compound interest. Given: P = P10.000 = 10% t = 4 years 3. COMPOUNDING/CONVERSION PERIOD (m) is the number of times in a year the interest Period Principal Rate Time Interest will be compounded. 10,000 0. 1 1000 CONVERSION NO. OF COMPOUNDING 2 10,000 0. 1 1 PERIOD COMPOUNDINGS PER YEAR PER YEAR (m) 10,000 0.1 1000 10,000 0:1 1000 Annual 1 year Semiannual IN 6 months total interest gained after 4 year's = P4000 final amount after 4 years = P14,000 Quarterly months Bimonthly 2 months in SIMPLE INTEREST for a given principal amount P, Monthly month for a time period of t years, and interest rate r the accumulated value / future value of P is given by: 4. CONVERSION PERIOD F = P(1 + rt) The time period after which the interest is F = 10000[1 + (0.10)(4)] added each time to form a new principal The TOTAL NUMBER OF CONVERSION PERIODS ( n ) F = P14000 n = t*m, COMPOUND INTEREST t = term of the loan or investment in years - is the procedure in which interest is periodically m = conversion period calculated and added to the principal. 5. ANNUAL INTEREST RATE ( r ) Given. P = P10,000 1 = 10% The interest on which the stated compound 1 = 4 years interest is based. Period Principal Rate Time Interest 6. PERIODIC INTEREST RATE ( i ) 1 10,000 0.1 1,000 The rate of interest earned in one conversion 2 11,000 0,1 1 1,100 period. 12,100 0.1 1 1,210 i = 13,310 0.1 1 1,331 m total interest gained after 4 years = P4,64) final amount after 4 years = P14,641 *If the conversion period is not indicated in the problem, assume that the conversion period is annual or m=1 *take note on the principal value every period Present worth of Future Value of FORMULAS Herlod Principal Amount, (P) Principal Amount. {F) F = P(1 +i)" P P ( 1 + 1) P = F(1 + i ) n P( 1 +1) P( 1 + 1) . (1 -1) = P(1+1): F P( 1 + 1 ) P ( 1 + 1) . ( 1 -1) = P(1-1p P = 7 P(1+1) P( 1 - 13) . ( 1 .1) = P(1+1)1 ( 1+ ( )n 1 = F -P P( 1-1) - (1 -1) = P/1 -1) 1 = P[(1 + ()" - 1] P( 1 - 1) - P( 1.1) (1-1) = P( 1-1) F = P(1 + 1)" F = 10000 ( 1 + 0.10) 4-1 Log () t = F = P14,641 m[log (1 + i)] m n = t * mGiven : What formula to use? CE = P5090 DP = P1000 If the payments follow the single sum, it is a PRESENT r = 18% = 0. IS 1 = 0.18 2 = 0.09 VALUE annuity problem. Required: Periodic Payment, R If the payments precede the single sum, it is an AMOUNT of an annuity problem. Solution: R= 1-(1+1 ) -13 Problems that involve expenses and cash equivalent are R = - 4000(0.09) 1- (1+0.09)- = P1028.37 PRESENT VALUE annuity problem. Conclusion: The semiannual payment is P1028.37 Problems that involve income or revenue are AMOUNT of an annuity problem ACTIVITY: Problem 1 1. Mr. Drei deposits Php 1 250 every end of six months Determine the amount and present value of an ordinary in an account paying 6.5% interest compounded annuity of Php 500 each quarter payable for 6 years and 9 months, if money is worth 14% compounded quarterly. semiannually. What amount is in the account at the end of 5 years? Given : R = Php500 m = 4 t = 6 yrs & 9 mos. = 6.75 2. Every three months for 4 years, a mother deposits r = 14% = 0.14 1 = 0.14/4 = 0.035 n = (6.75)(4) =27 P1ooo in a trust company for her son's education. If the Required: Amount, A? Present Value, P? money earns 15.2% compounded quarterly, how much Solution: A = R (1+0"-1 will be in the bank after the tenth deposit? 3. How much must be set aside quarterly so as to have 0.035 A = 500 (1+0.035)"-1 = Php21879.53 a fund of P2ooooo at the end of 12 years with interest P = R[1-(1+1)] at the rate of 12% converted quarterly? P = 500 1-(1+0.035)-371 0.035 = Php8642.68 THINGS YOU FOUND OUT: Conclusion: The amount of the ordinary annuity of the given 3 problem is Php21879.53 and the present value of the ordinary annuity of the given problem is Php8642.68 INTERESTING THINGS Problem 2 2 A computer set is offered for sale for Php 15 000 QUESTION YOU STILL HAVE down payment and Php 2 500 every three months for the balance for two years and 6 months. If interest is to be computed at 11% compounded quarterly, what is the cash equivalent of the computer set? Given : R = Php2500 m = 4 1 = 2 yrs & 6 mos. = 2.5 r= 11% -0.11 i= 0.11/4 = 0.0275 n = (2.5)(4) = 10 Required: Cash Equivalent, CE Solution: P = R -(1+1)-" P = 2500 1-(1+0.0275)-191 0.0275 = Php21600.19 CE = DP + P CE = 15000 + 21600.19 = Php36600.19 Conclusion: The cash equivalent of the computer set is DP + P, which is equivalent to Php36600.19 Problem 3 A water purifier costs P5ooo. it is purchased with a down payment of P1000 and 5 semiannual payments. If money is worth 18% compounded semiannually, find the semiannual payment.GEN MATH ORDINARY ANNUITY Deferred Annuity Annuities is a sequence of periodic payments made at Deferred annuity (Da) is an annuity whose first regular interval of times usually monthly, semi- payment is to start at some time later. annually, quarterly, or annually in order to pay a loan or create a fund. NOTATIONS Examples: (1) installment payments, (2) rental A = sum or amount of an ordinary annuity payments, (3) life insurance premiums, (4) weekly P = present value of an ordinary annuity wages, (5) periodic pensions, (6) mortgages, (7) car R = periodic payment payments t = term of annuity A* = sum of an annuity due Definition of Terms p* = present value of an annuity due Payment interval - the time between consecutive r = rate of an annuity payments of annuity (monthly, semi-annually, m = number of conversion period per year quarterly, annually, etc.) n = total number of conversion periods for the whole term (tm) Periodic payment (R) - amount paid during each i = periodic rate (r/m) installment period A* = sum of deferred annuity Term of annuity (t) - the time from the beginning of P . = present value of deferred annuity the first payment interval to the end of the last payment interval. The total number of conversion Present Value of an Ordinary Annuity periods in the term of annuity is given by n = tm. The present value of an annuity, denoted by P, is the Simple Annuity sum of present values of all the payments at the beginning of the term. Simple annuity is an annuity whose payment interval is the same as the conversation period m. P = R 1-(1+i)" Classification: 1, Ordinary annuity 2. Annuity due 3. Deferred annuity Amount of an Ordinary Annuity The amount of an annuity, denoted by A, is the sum of Ordinary Annuity all the values of all the periodic payments at the end of the term. Ordinary annuity (O.) is an annuity whose periodic payments are scheduled at the end of each payment A=R (1+i)" -1 interval. Loan origination Periodic Payment of an Ordinary Annuity Ist 2nd 3rd Given amount A To R = = Ai payment payment payment (1+1 )n-1 maturity ORDINARY ANNUITY Given present value P R = Pi 1- (1+1)-n Annuity Due Annuity due (Ad) is an annuity whose periodic payment Cash Equivalent is scheduled at the beginning of each payment interval. CE = DP + P Beginning of each period Where: CE = Cash Equivalent DP = Down payment P = Present value $1000 51000 51000 51000 51000 Note: Problems that involve expenses and cash equivalent are finding the present value problems and those that involve income are finding the amount of Payment paid or received at the beginning of each period annuity problems.Example #1: COMPOUND AMOUNT (F) Example #5: TERM (t) Aurora Secundo barrows Php4,800 with interest at 18% compounded quarterly. How much should How long would it take for Proboo to mature into he pay to the creditor after 4 years to pay off his P:5000 if 16% interest is computed semiannually? debt? . Given: . Given: Pre #10600 F . #15000 m a semiannually = 2 P = Php 4.800 1= 18% : 018 ma quarterly : 4 to & years time 4 4 16 * *15% -0.16 1=//m = 0.16/2 * 0.08 . Required: F: . Required: t? . Solution: F = P(1 + ()" 15000 F = 4800(1 + 0.045) 109 10680 F = Php 9707.38 . Solution: / milor 1+fil 2 log 1+0.08 . Conclusion: Aurora Secundo should pay Php 9707-38 to pay off his debt. t = 2,26 years Conclusion: It would take 2.26 years. Example #2: PRINCIPAL (P) SHORT QUIZ: How much should be deposited now at 18% compounded monthly in order to accumulate 1. Two years ago, Sostenes Cabaruan invested Php13,240 in 3 years? Phpz9,300 compounded bimonthly at 127. How much . Given: is his money now? F =Php 13,740 . 186 - 018 mamonthly = 12 2 Crispin Secundo has just been notified that the combined principal and the interest on an amount that . Required: P: he borrowed 24 months ago at 14% compounded Solution: P = F(1 + 1)"" semiannually, is now Phpoo,ooo. How much of this P = 13740(1 + 0.015)-36 amount is principal? P = Php 8039.13 3. If Php19,ooc is invested at 18% compounded . Conclusion: The amount that should be deposited now semiannually, find the compound interest in 3 years. is Php8039.13 4. At what nominal rate compounded quarterly for 8 Example #3: COMPOUND INTEREST (1) years will ? 40060 accumulate to P95296.861 5. When is P:3200 due if its present value of Proooo is Find the interest charged on a loan of Php7,000 invested a: 10* compounded quarterly? for 4 years at 112 compounded quarterly. . Given: THINGS YOU FOUND OUT Pa Php 2,000 1IK = 011 m = quarterly = 4 1/m =0 11/4 = 0.0275 m= tm . (4)(4) = 16 . Required: 1? INTERESTING THINGS . Solution: 1 = F - P 1 = P[(1 +1)" -1] 2 1 = 7000[(1 + 0.0275)16 -1] QUESTION YOU STILL HAVE 1 = Php3804-57 Conclusion: The interest charged is worth Php3804.57 Example #4: INTEREST RATE (r) If P2oooo will accumulate to P45758.55 in 14 years, what is the interest rate compounded bimonthly? . Given: P= P20000 F = P45758.55 m = bimonthly = 6 t = 14 years n = Im = (14 (6) = 84 . Required:r? . Solution: r = m 15-1 =6/4 4575853 20000 -1 r = 0.06 = 6% Conclusion: The interest rate is 6%.Il. Compound Interest: Fill in what is lacking in the table Principal Rato Php Interest Maturity Value Php 30,000 5% annually S yours (1) 8 78 8 45 (2 Php 30 . 180 19 Php 15,000 0 125% monthly 3 years (3) Php ( A)Php Php 5,000 (5) P monthly 2 yours Prip 1,000 (6) Php Php 15,000 (7) quarterly 3 months Php 150 () Php Php 12.500 (9) quarterly 2 years (10) Php Php 15 090 Php 1,500 (11 annually 5 years (12) Php Php 1,750 (13) Php 0.25% quarterly I year Php 250 (14) Php (15 ) Php 1 1 25% annually 5 yours Php 2.250 (15 ). Php ( 17 ) Php (18) monthly 1 year Php 1,000 Php 10.000 (19 ) Php 20) semi- Allenuve 25 yours Php 7,500 Php 97 .900 Ill. Future Value/Amount of an Ordinary Annuity Periodic Payment Rate Time Future Value/Amount Php 5,000 3% annually 5 years (1) Php 2,500 3 25% monthly 3 years 2) Php 1,000 6 quarterly 10 years (3) (4) 0.25% monthly 5.5 years Php 50,000 (5) 5% annually 6 years Php 25.000 1.5% quarterly 2.5 years Php 30,000 (6 IV. Present Value of an Ordinary Annuity Periodic Payment Rate Time Present Value 5% annually 10 years (1) Php 10,000 Php 500 3% monthly 30 years 2 10 years Php 1,000 3% quarterly 5 years Php 50,000 (4) 1.25% monthly 3 years Pip 25,000 (5 0.5% annually (6) 2 25% quarterly 25 years Prip 300,000