Simple Interest PDF

SH1902 Interest I. Simple Interest Simple interest is a fixed percentage of the total amount invested paid to an investor each year. Let 𝐼𝐼 be the simple interest, 𝑃𝑃 be the amount invested called the principal, 𝑟𝑟 be the fixed...

SH1902 Interest I. Simple Interest Simple interest is a fixed percentage of the total amount invested paid to an investor each year. Let 𝐼𝐼 be the simple interest, 𝑃𝑃 be the amount invested called the principal, 𝑟𝑟 be the fixed percentage or the rate, and 𝑡𝑡 be the number of years the principal is invested. Then, 𝐼𝐼 = 𝑃𝑃𝑃𝑃𝑃𝑃. The formula can be algebraically manipulated to obtain a formula when 𝑃𝑃, 𝑟𝑟, or 𝑡𝑡 is unknown. When 𝑃𝑃 is unknown When 𝑟𝑟 is unknown When 𝑡𝑡 is unknown 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝑡𝑡 = 𝑃𝑃 = 𝑟𝑟 = 𝑃𝑃𝑃𝑃 𝑟𝑟𝑟𝑟 𝑃𝑃𝑃𝑃 (in years) Table 1.1 Example: Suppose you deposited ₱10,000 in a bank that gives 10% simple interest. How much interest will you receive after three (3) years? Solution: Given: 𝑃𝑃 = ₱10,000 𝑟𝑟 = 10% = 0.1 𝑡𝑡 = 3 Solving for the interest 𝐼𝐼, we use the formula 𝐼𝐼 = 𝑃𝑃𝑃𝑃𝑃𝑃. 𝐼𝐼 = 𝑃𝑃𝑃𝑃𝑃𝑃 = ₱10,000(0.1)(3) = ₱3,000 Example: Cyndi borrowed ₱30,000 from a bank. If the interest she paid was ₱3,375 for nine (9) months, at what rate would she pay? Solution: Given: 𝐼𝐼 = ₱3,375 𝑃𝑃 = ₱30,000 Since time is in months, we first have to convert it into years. 1 year 9 3 9 months ⋅ = year = year 12 months 12 4 3 Hence, 𝑡𝑡 = 4. 06 Handout 1 *Property of STI  [email protected] Page 1 of 6  SH1902 𝐼𝐼 𝑟𝑟 = 𝑃𝑃𝑃𝑃 ₱3,375 = 3 ₱30,000 4 = 0.15 As we see, time can be expressed not only in months. But before continuing with the formulas, we must always convert time into years first. Time can also be expressed in days. In such cases, we consider two (2) ratios. Exact interest (𝐼𝐼𝑒𝑒 ) is simple interest computed based on the ratio 1 year : 365 days. Ordinary interest (𝐼𝐼𝑜𝑜 ), on the other hand, is simple interest computed based on the ratio 1 year : 360 days. Example: Miss Uy borrowed ₱80,000 from a bank. The borrowed amount earned ₱5,400 at 9% ordinary interest. How many days did it take Miss Uy to repay the loan? Solution: Given: 𝐼𝐼𝑜𝑜 = ₱5,400 𝑃𝑃 = ₱80,000 𝑟𝑟 = 9% = 0.09 𝐼𝐼 Solving for the interest 𝑡𝑡, we use the formula 𝑡𝑡 = 𝑃𝑃𝑃𝑃𝑜𝑜. 𝐼𝐼𝑜𝑜 𝑡𝑡 = 𝑃𝑃𝑃𝑃 ₱5,400 = ₱80,000(0.09) = 0.75 But note from Table 1.1 that 𝑡𝑡 is expressed in years. But the question requires time to be expressed in days. Hence, we convert 0.75 years into days using the ratio for ordinary interest. 360 days 0.75 years ⋅ = 0.75(360) days = 270 days 1 year Hence, it took 270 days before Miss Uy repaid the loan. Example: Mr. Sandoval paid the bank ₱1,200 exact interest at 8% for 146 days. How much did he borrow? Solution: Given: 𝐼𝐼𝑒𝑒 = ₱1,200 06 Handout 1 *Property of STI  [email protected] Page 2 of 6  SH1902 𝑟𝑟 = 8% = 0.08 1 2 𝑡𝑡 = 146 ⋅ = 365 5 𝐼𝐼 Solving for the principal 𝑃𝑃, we use the formula 𝑃𝑃 = 𝑟𝑟𝑟𝑟𝑒𝑒. 𝐼𝐼𝑒𝑒 𝑃𝑃 = 𝑟𝑟𝑟𝑟 ₱1,200 = 2 0.08 5 = ₱37,500 An investment grows depending on the amount of interest it accumulates over a period of time. The final amount of an investment (or debt) after interest is added is called maturity value (𝑀𝑀𝑀𝑀). By definition, 𝑀𝑀𝑀𝑀 = 𝑃𝑃 + 𝐼𝐼. Since 𝐼𝐼 = 𝑃𝑃𝑃𝑃𝑃𝑃, then 𝑀𝑀𝑀𝑀 = 𝑃𝑃 + 𝑃𝑃𝑃𝑃𝑃𝑃. Factoring out 𝑃𝑃, we arrive with an equivalent formula for the maturity value, 𝑀𝑀𝑀𝑀 = 𝑃𝑃(1 + 𝑟𝑟𝑟𝑟). 𝐼𝐼 Also, since 𝑃𝑃 = 𝑟𝑟𝑟𝑟, then 𝐼𝐼(1 + 𝑟𝑟𝑟𝑟) 𝑀𝑀𝑀𝑀 =. 𝑟𝑟𝑟𝑟 Example: Find the maturity value for each item. a. 𝑃𝑃 = ₱40,000, and 𝐼𝐼 = ₱9,600 b. 𝑃𝑃 = ₱32,000 , 𝑟𝑟 = 8%, and 𝑡𝑡 = 180 days (ordinary interest) c. 𝐼𝐼𝑒𝑒 = ₱594, 𝑟𝑟 = 9%, and 𝑡𝑡 = 73 days (ordinary interest) Solution: a. 𝑀𝑀𝑀𝑀 = 𝑃𝑃 + 𝐼𝐼 = ₱40,000 + ₱9,600 = ₱49,600 b. 𝑀𝑀𝑀𝑀 = 𝑃𝑃(1 + 𝑟𝑟𝑟𝑟) 1 = ₱32,000 1 + 0.08 180 ⋅ 360 = ₱33,280 𝐼𝐼𝑒𝑒 (1+𝑟𝑟𝑟𝑟) c. 𝑀𝑀𝑀𝑀 = 𝑟𝑟𝑟𝑟 1 ₱594 1 + 0.09 73 ⋅ = 365 1 0.09 73 ⋅ 365 = ₱33,594 06 Handout 1 *Property of STI  [email protected] Page 3 of 6  SH1902 II. Compound Interest Suppose no deposits or withdrawals were made in a savings account. After a period, the interest due is added to the previous balance and the sum then becomes the new principal for the next period. The interest that is added to the old principal to make a new principal on which interest is again calculated for the next period is called the compound interest, denoted by 𝐼𝐼𝐶𝐶. The total amount at the end of the last period is called the compound amount, denoted by 𝐴𝐴. The compound amount is computed using the formula 𝑟𝑟 𝑛𝑛 𝐴𝐴 = 𝑃𝑃 1 +. 𝑚𝑚 𝑃𝑃 is the principal or the present value 𝑟𝑟 is the annual rate 𝑚𝑚 is the number of compounding periods in a year Some common values for 𝑚𝑚 o 𝑚𝑚 = 1 if compounded annually o 𝑚𝑚 = 2 if compounded semi-annually o 𝑚𝑚 = 4 if compounded quarterly o 𝑚𝑚 = 12 if compounded monthly o 𝑚𝑚 = 365 if compounded daily 𝑛𝑛 is the total number of compounding periods The value of 𝑛𝑛 usually is the product of 𝑚𝑚 and the total number of years. The compound interest therefore is the difference between the compound amount and the principal, that is, 𝐼𝐼𝐶𝐶 = 𝐴𝐴 − 𝑃𝑃. Example: Esther deposited ₱8,000 in a savings account for four (4) years at 0.5% compounded quarterly. How much will she have in her account at the end of four (4) years? How much will be the interest of her investment? Solution: Let 𝑃𝑃 = ₱8,000, 𝑟𝑟 = 0.005, 𝑚𝑚 = 4, and 𝑛𝑛 = 4(4) = 16. Then 𝑟𝑟 𝑛𝑛 𝐴𝐴 = 𝑃𝑃 1 + 𝑚𝑚 0.005 16 = ₱8,000 1 + 4 = ₱8,161.51. Thus, Esther will have ₱8,161.51 in her account at the end of four (4) years. The interest in her investment therefore is 06 Handout 1 *Property of STI  [email protected] Page 4 of 6  SH1902 𝐼𝐼𝐶𝐶 = 𝐴𝐴 − 𝑃𝑃 = ₱8,161.51 − ₱8,000 = ₱161.51. Example: How much interest is earned after 20 years by ₱20,000 at 4% interest compounded monthly? Solution: Let 𝑃𝑃 = ₱20,000, 𝑟𝑟 = 0.04, 𝑚𝑚 = 12, 𝑛𝑛 = 12(20) = 240. Then the compound amount is 𝑟𝑟 𝑛𝑛 𝐴𝐴 = 𝑃𝑃 1 + 𝑚𝑚 0.04 240 = ₱20,000 1 + 12 = ₱44,451.64, and the compound interest is 𝐼𝐼𝐶𝐶 = 𝐴𝐴 − 𝑃𝑃 = ₱44,451.64 − ₱20,000 = ₱24,451.64. Since the maturity value of an amount invested or borrowed is the amount it is worth in the future at a given interest rate, it is also known as future value. The future value (𝐹𝐹𝐹𝐹) is the amount that matured from the principal, that is, the present value (𝑃𝑃𝑃𝑃). The relationship between the future value and the present value is similar to the relationship of 𝑟𝑟 𝑛𝑛 compound amount and principal 𝐴𝐴 = 𝑃𝑃 1 + 𝑚𝑚. We just replace 𝐴𝐴 with 𝐹𝐹𝐹𝐹 and replace 𝑃𝑃 with 𝑃𝑃𝑃𝑃 so that 𝑟𝑟 𝑛𝑛 𝐹𝐹𝐹𝐹 = 𝑃𝑃𝑃𝑃 1 +. 𝑚𝑚 Sometimes, we may be interested to know the present value in order to figure out how much money must be invested today to achieve a target amount in the future. Manipulating the equation above, we arrive with the formula for 𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹 𝑃𝑃𝑃𝑃 =. 𝑟𝑟 𝑛𝑛 1 + 𝑚𝑚 Example: How much will ₱15,000 compounded semi-annually for five (5) years grow at 6% interest? Solution: Let 𝑃𝑃𝑃𝑃 = ₱15,000, 𝑟𝑟 = 0.06, 𝑚𝑚 = 2, and 𝑛𝑛 = 2(5) = 10. Solving for the future value, 𝑟𝑟 𝑛𝑛 𝐹𝐹𝐹𝐹 = 𝑃𝑃𝑃𝑃 1 + 𝑚𝑚 0.06 10 = ₱15,000 1 + 2 = ₱20,158.75. 06 Handout 1 *Property of STI  [email protected] Page 5 of 6  SH1902 Example: If you target to accumulate ₱2,000,000 for retirement at 3% interest compounded quarterly, how much would you need to invest today to have this amount in 45 years? Solution: Let 𝐹𝐹𝐹𝐹 = ₱2,000,000, 𝑟𝑟 = 0.03, 𝑚𝑚 = 4, 𝑛𝑛 = 4(45) = 180. Then the present value should be 𝐹𝐹𝐹𝐹 𝑃𝑃𝑃𝑃 = 𝑟𝑟 𝑛𝑛 1 + 𝑚𝑚 ₱2,000,000 = 0.03 180 1 + 4 = ₱521,098.87. Thus, you need to invest ₱521,098.87 today to have ₱2,000,000 in 45 years. The formula for future value can also be algebraically manipulated to obtain a formula when interest rate (𝑟𝑟), number of compounding periods in a year (𝑚𝑚), or total number of compounding periods (𝑛𝑛) is unknown. When 𝑟𝑟 is unknown When 𝑚𝑚 is unknown When 𝑛𝑛 is unknown 1 𝑟𝑟 𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹 𝑛𝑛 𝑚𝑚 = 1 ln 𝑃𝑃𝑃𝑃 𝑟𝑟 = 𝑚𝑚 − 1 𝐹𝐹𝐹𝐹 𝑛𝑛 𝑛𝑛 = 𝑟𝑟 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 − 1 ln 1 + 𝑚𝑚 Table 1.2 Note that when time is unknown, we simply divide 𝑛𝑛 by 𝑚𝑚, that is, 𝑛𝑛 𝑡𝑡 =. 𝑚𝑚 References: Domain and range of exponential and logarithmic functions. (n.d.). In Varsity Tutors. Retrieved from https://www.varsitytutors.com/hotmath/hotmath_help/topics/domain-and-range-of- exponential-and-logarithmic-functions Chua, R., Ubarro, A., & Wu, Z. (2016). Soaring 21st century mathematics (general mathematics). Quezon City: Phoenix Publishing House. Fernando, O. (2016) Next century mathematics (general mathematics). Quezon City: Phoenix Publishing House. Lim, Y., Nocon E., Nocon, R., & Ruivivar L. (2016). Math for engaged learning (general mathematics). Quezon City: Sibs Publishing House. Melosantos, L. (2016). Math connections in the digital age (general mathematics). Quezon City: Sibs Publishing House. Zorilla, R. (2016). General mathematics for senior high school. Malabon City: Mutya Publishing House. 06 Handout 1 *Property of STI  [email protected] Page 6 of 6