Lesson-26.pptx general mathematics subject

this is the ppt for lesson 26 of general mathematics grade 11 quarter 2

1. Compounding More Than Once A Year Lesson 26

2. Sometimes, interest may be compounded more than once a year. e.g. Semi-annually (compounded twice a year) Quarterly (compounded 4x a year) Monthly (compounded 12x a year)

3. Given a principal of Php 10,000.00 that earns 2% per annum. Determine its maturity value after 3 years compounded: Annually Semi-annually Quarterly

4. Compounded annually Time (t) in years Principal = Php 10,000.00 Annual interest rate = 2% compounded annually Amount at the end of the year 1 10,000 x 1.02 = 10,200 2 10200 x 1,02 = 10,404 3 10,404 x 1.02 = 10,612.08

5. Compounded Semi-annually If the interest rate per annum is 2%, then the semi-annual interest rate is 1%

6. Compounded Semi-annually Time (t) in years Principal = Php 10,000.00 Annual interest rate = 2% compounded semi-annually or 1% in every 6 months. Amount at the end of the year 0.5 10,000 x 1.01 = 10,100 1 10,100 x 1.01 = 10,201 1.5 10,201 x 1.01 = 10,303.01 2 10,303.01 x 1.01 = 10,406.04 2.5 10,406.04 x 1.01 = 10,510.10 3 10,510.10 x 1.01 = 10,615.20

7. Compounded Quarterly If the interest rate per annum is 2%, then the quarterly interest rate is 0.5% since there 4 quarters in a year.

8. Compounded Quarterly Time (t) in years Principal = Php 10,000.00 Annual interest rate = 2% compounded quarterly or 0.5% per quarter Amount at the end of the year 0.25 10,000 x 1.005 = 10,050 0.5 10,050 x 1.005 = 10,100.25 0.75 10,100.25 x 1.005 = 10,150.75 1 10,150.75 x 1.005 = 10,201.50 1.25 10,201.50 x 1.005 = 10,252.50 1.5 10,252.50 x 1.005 = 10,303.76

9. Compounded Quarterly Time (t) in years Principal = Php 10,000.00 Annual interest rate = 2% compounded quarterly or 0.5% per quarter Amount at the end of the year 1.75 10,303.76 x 1.005 = 10,355.28 2 10,355.28 x 1.005 = 10,407.06 2.25 10,407.06 x 1.005 = 10,459.10 2.5 10,459.10 x 1.005 = 10,511.40 2.75 10,511.40 x 1.005 = 10,563.96 3 10,563.96 x 1.005 = 10,616.78

10. DEFINITION OF TERMS

11. Frequency of Conversion (m) number of conversion periods in one year The time period after which the interest is added each time to form a new principal is called the conversion period. When the interest is compounded semi-annually, there are two conversion periods in a year each after 6 months. If it is compounded quarterly, there are four conversion periods.

12. Conversion or Interest Period The time between successive conversion of interest The interest period or conversion period is the time between successive conversions of interest into principal and is commonly three months, six months, or one year, where such case interest is compounded quarterly, semiannually, or annually, respectively.

13. Total Number of Conversion Periods (n) n = mt Frequency of conversion x time in years For compounded annually: Let t = 3 years m = 1 since there is only one conversion per year. n = mt = 1(3) = 3 Therefore, the total number of periods in three years is 3 if it is compounded annually.

14. Total Number of Conversion Periods (n) n = mt Frequency of conversion x time in years For compounded semi-annually: Let t = 3 years m = 2 since there are two conversions per year. n = mt = 2(3) = 6 Therefore, the total number of periods in three years is 6 if it is compounded semi-annually.

15. Total Number of Conversion Periods (n) n = mt Frequency of conversion x time in years For compounded quarterly: Let t = 3 years m = 4 since there are four conversions per year. n = mt = 4(3) = 12 Therefore, the total number of periods in three years is 12 if it is compounded quarterly.

16. Total Number of Conversion Periods (n) n = mt Frequency of conversion x time in years For compounded monthly: Let t = 3 years m = 12 since there are 12 conversions per year. n = mt = 12(3) = 36 Therefore, the total number of periods in three years is 36 if it is compounded monthly.

17. Nominal Rate ( i (m) ) - Annual rate of interest Rate of Interest for each conversion period j = annual rate of interest frequency of conversion j = i (n) /m

18. Maturity Value, Compounding m times a year

19. Sample Problem # 1 Given : 𝑃=Php50,000.00 , 𝑖 4 =0.03 , 𝑚=4 , 𝑡 =4, find F and Ic . Use the formula of maturity value 𝐹=𝑃(1+𝑗) 𝑛

20. Sample Problem #2 Find the maturity value and interest if Php 10,000.00 is deposited in a bank at 2% compounded quarterly for 5 years. Solution: j=i 4 /m = 0.02/4 = 0.005 n = mt = 4(5) = 20 conversion periods F =P(1 + j) n F = 10,000(1 + 0.005) 20 F = 11,048.96 Ic = F – P Ic = 11,048.96 – 10,000.00 Ic = 1,048.96

21. Sample Problem #3 Find the present value of Php50,000.00 due in 4 years if money is invested at 12% compounded semi-annually. Given: F = 50,000.00 ; t = 4 i 4 = 12% = 0.12 j =i 2 /m = 0.12/2 = 0.06 n = mt = 4(2) = 8 P = 50,000/(1 + 0.06) 8 P = 31,370.62