Direction: Solve for the ff. missing value. Show your Solution (5pts each for Compound Interest and 10 pts each for the rest of the number) Present Rate Time/Term Maturity/Future Compound Value Value Interest ISTI P 35, 250.15 2.5% ARAHAFTER compounded 2. semi-annually P16, 277.4 12% P135, 645 compounded 3. annually m m 4. 12 years P31 620.96 5. compounded P90 345.6 6. semi - annuallyDirection: Solve the given word problem. Show your solution. (15pts. each) 1. Mr. Diluc had P450 000 pesos on a bank which earns 8.5% interest compounded quarterly. He needed P900, 300 pesos to start an electronic business. Will 6 years be enough to accumulate such money? Explain your answer. 1 4 CHA RACTER S 2. In what sum of money will P325, 000 accumulate in 3 years and 2 months at 5% compounded semi- annually?Discussion In the field of investment Compound Interest is considered a wise strategy. Generally, compound interest is the accumulation of interest over a previous period, based on the initial capital and all the deposits that were made over time. Compound Interest is the process in which the interest is periodically calculated and added to the principal. The principal grows as the interest is added to it. This method is used in other form of investments such as savings account and bonds. In working with problems involving interest, we use the term payment period as follows: Compounding/Conversion Frequency No. of Conversions Per Year (m) Annually Semiannually 2 Quarterly Bimonthly Monthly 12 Daily 365 or 360 Formulas: ITH ERVIC Future Value Rate F = P (1+[)" or F=P(1+ -m ram -1 or r=m VP -1 Present Value The Truth (1+1) or P =F (1+1)" RSITY . Compound Interest Time logs Ic = F-P t= = m[log(1+0)] Where: 1946 1= , n=mt P - Principal r - Nominal interest rate (should always be expressed in decimal form) m - Number of conversions per year i - Rate per conversion period n -Total number of conversion periods t - Time or term of loan or investment Ic - Compound interest F - Maturity or future valueExample 1. Solve for the compound interest earned at the end of 6 years if P15,900 is invested at 9% compounded bimonthly. Given: P = P15,900 Find: F = ? r = 9% = 0.09 Ic=? t = 6 years m = 6 n = mt = 6(6) = 36 1= 1 0.09 102 = 0.015 m Solution: F = P (1+1)^n Ic= F- P = 15,900 (1 + 0.015)36 = P 27, 175.32 - P 15,900 F = P 27, 175.32 Ic = P 11, 275.32 PRESENT VALUE Example 2. In what amount must be invested right now in a savings account earnings 12% compounded semi-annually to accumulate a total of P 235, 425 after 5 = years? Given: F = P235, 425 r = 12% = 0.12 CHARA Find: P=? t = 5- years = 5.5 years m = 2 SERVICE n = mt = (5.5) (2) = 11 YEAINN 1 0.12 = 0.06 m 2 Solution: The b. P - F(1 + 1)-" 235,425 (1+0.06)11 = 235, 425(1 + 0.06)-11 235,425 1 898 298 558 P = P 124, 018.95 P = P 124, 018.95 FUTURE VALUE OR MATURITY VALUE Example 3. What will be the future value of P14,620 invested for 8 months at 18% compounded bimonthly? Given: P = P14,620 Find: F = ? r= 18% = 0.18 t = 8 months= 12 m = 6 n = mt = () (6) = 4 1 0.18 - = 0.03 Solution: F = P(1 + [)" =14,620(1 + 0.18)+ F = P 28,344.93RATE Example 4. The Future value of a ten-year, P9,550 compound interest investment certificate was P 120,426.99. What quarterly compounded nominal interest rate did the investment certificate earn? Given: F = P120,426.99 Find:r = ? P = P9,550 t = 10 years m = 4 n = tm = (10) (4) = 40 Solution: r= m =4 40 120.426.99 9.550 = 4 [ V12.610 156 021-1] 24 1.065 413 049-1) IN = 0.261 652 196 = 0.261 652 196 x 100 r = 26. 17% CHARACTER S U TIME Example 5. How long does it take if a future value of P113,200.35 and a present value of P28,300 is invested at 6% compounded monthly? Given: F = P113,200.35 Find: t = 7 VICE P = P28,300 r = 6% = 0.06 m = 12 The Truth 1= 0.06 12 0.005 VERSITY. k PH Solution: log- 0.602 061 334 1= mdog(1+0] 12 0.002 166 062] O 113.200.35 20,100 0.602 061 334 12 log(1+0.005)] 0.025 992 744 000 log 4.000 012 367 = 23.16 12[ log 1.005 ] t = 23.16 years