![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2020\/06\/9.1-intro-piggy-bank-1024x670.jpg\" "\\"https:\/\/www.pexels.com\/photo\/money-pink-coins-pig-9660\/\\"")
\nLearning Objectives<\/p>\n\n<\/header>\n
\n\nBy the end of this section it is expected that you will be able to:\n
- \n \t
- Determine the simple interest earned on an investment or charged on a loan<\/li>\n \t
- Determine the principal amount, the interest rate, or the time for applications involving simple interest<\/li>\n \t
- Determine the maturity value of a loan that involves simple interest<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
Interest<\/strong><\/h1>\nSome people keep money at home in an easily accessible location, perhaps a piggy bank, a safe or locked box, or perhaps even a mattress. Although this provides instant access to funds it does not provide any return or earnings on this money. For that reason, most people hold their money in accounts or investments that provide some form of return or earning power.\n\nInterest<\/strong> is the price paid for the use of money.\u00a0 If you borrow money from another person or a lending institution, eventually you must pay back this amount plus the interest owing.\u00a0 When you deposit money in a bank, you are lending them money and after some time they will pay you interest on the money you lent them.\n\nThe amount of interest you will owe or receive is determined by the principal,\u00a0<\/strong> the interest rate,\u00a0<\/strong> and the time<\/strong> (the length of the loan).\u00a0 The amount of money that you lend or borrow is called the principal<\/strong>. The length<\/strong> of the loan can range between a few days to several years. The interest rate<\/strong> is stated as an annual percentage. It may be simple<\/strong> interest or compound<\/strong> interest. With simple interest<\/strong>\u00a0the interest is calculated only once<\/strong> during the entire time period of the loan or deposit.\u00a0 Simple interest is calculated solely on the principal investment or loan. With compound<\/strong> interest the interest is calculated more than once<\/strong> during the time period of the loan. It will be calculated on the principal as well as the accumulated interest This section will focus on simple interest and in the next section we will consider compound interest.\n
Simple Interest<\/strong><\/h1>\nSimple interest<\/strong> is calculated by finding the product of the principal (P<\/em>), the rate (r<\/em>), and the time (t<\/em>).\n
\n Simple Interest<\/p>\n\n<\/header>\n
\n\nThe simple interest formula<\/strong> is\u00a0 I = Prt\u00a0 <\/em>\u00a0where\n\nI = interest earned\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 r<\/em> = annual interest rate ( stated as a decimal)\n\nP<\/em> = principal\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 t = <\/em>time (in years<\/strong>)\n\n<\/div>\n<\/div>\nInterest rates are quoted for periods of one<\/strong> year and when used in a formula must be converted to a decimal fraction. The time must be expressed in the same unit of time as the interest rate so time must be stated in years or portions of a year. If you deposit money in a savings account earning 3% interest then the annual interest rate is 3% per year.\n\n EXAMPLE 1<\/p>\n\n<\/header>\n
\n\nJo borrows $2000 at an interest rate of 5% per year.\u00a0 How much interest will Jo owe after one year?\n\nSolution<\/strong>\n\nIdentify the P<\/em>, r<\/em>, and t<\/em>.\u00a0 \u00a0 \u00a0P<\/em> = $2000\u00a0 \u00a0 \u00a0r<\/em> = 5%\u00a0 = 0.05\u00a0 \u00a0 \u00a0 t<\/em> = 1 year\n\n\n
\n Here,<\/td>\n [latex]P = \\$2000[\/latex]<\/td>\n [latex]I = Prt[\/latex]<\/td>\n <\/td>\n<\/tr>\n \n <\/td>\n [latex]r = 5\\%\\, or\\, 0.05[\/latex]<\/td>\n [latex]I = 2000 \\left(5\\%\\right)\\left(1\\right)[\/latex]<\/td>\n Replace P<\/em>, r<\/em> and t<\/em> with their values<\/td>\n<\/tr>\n \n <\/td>\n [latex]t = 1\\, year[\/latex]<\/td>\n [latex]I = 2000 \\left(0.05\\right)\\left(1\\right)[\/latex]<\/td>\n Change 5% to its decimal equivalent, 0.05<\/td>\n<\/tr>\n \n <\/td>\n <\/td>\n [latex]I = 100[\/latex]<\/td>\n [latex][\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nStart with\u00a0 the formula\u00a0 \u00a0 \u00a0 I = <\/span>Prt<\/em>\n \n\nI = 2000 (5%) (1)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Replace P<\/em>, r<\/em>, and t<\/em> with their values\n\nI = 2000 (0.05) (1)\u00a0\u00a0\u00a0\u00a0\u00a0 Change 5% to its decimal equivalent, 0.05\n\nI = 100\n\nJo will pay $100<\/strong> in interest.\n\n<\/div>\n<\/div>\n<\/div>\n\n TRY IT 1<\/p>\n\n<\/header>\n
\n\nTerri borrowed $3200 at an interest rate of 4.75%. How much interest will Terri owe on the loan at the end of one year?\n\nShow answer<\/summary>[latex] I = Prt = \\$3200\\left(0.0475\\right)\\left(1\\right) = \\$55 [\/latex]\nTerri will owe $152 after one year.\n\n<\/details><\/div>\n<\/div>\nNote that the time t<\/em> is expressed in terms of\u00a0 years. When the time period is not exactly one year, the value for\u00a0 t\u00a0 <\/em>will be the fraction of the year during which interest is earned.\n\nIf the investment is made for 3 months, then t<\/em> = 3 months\/12 months = 0.25 years.\n\nIf the investment is made for 35 days then\u00a0 t = 35 days\/365 days = 7\/73 years.\n
\n EXAMPLE 2<\/p>\n\n<\/header>\n
\n\na) If an investment is made for a period of\u00a0 145 days, what portion of the year does this represent?\n\nb)\u00a0 If an investment is made for a period of\u00a0 48 weeks, what portion of the year does this represent?\n\nc)\u00a0 If an investment is made for a period of\u00a0 10 months, what portion of the year does this represent?\n\nSolution:<\/strong>\n\na)\u00a0 145 \u00f7 365= 29\/73 years\n\nb)\u00a0 48 \u00f7 52 = 12\/13 years\n\nc) 10 \u00f7 12 = 5\/6 years\n\n<\/div>\n<\/div>\n\n TRY IT 2<\/p>\n\n<\/header>\n
\n\na) If an investment is made for a period of\u00a0 220 days, what portion of the year does this represent?\n\nb)\u00a0 If an investment is made for a period of\u00a0 32 weeks, what portion of the year does this represent?\n\nc)\u00a0 If an investment is made for a period of\u00a0 2 months, what portion of the year does this represent?\n\nShow answer<\/summary>a)\u00a0 220 \u00f7 365= 44\/73 year\n\nb)\u00a0 32 \u00f7 52 = 8\/13 year\n\nc) 2 \u00f7 12 = 1\/6 year\n\n<\/details><\/div>\n<\/div>\n
\n EXAMPLE 3<\/p>\n\n<\/header>\n
\n\nDetermine the interest that will be earned on\u00a0 a deposit of $1350 at 2.8% over:\n\na) 7 months\n\nb) 25 days\n\nSolution:<\/strong>\n\na)\n\n\n
\n [latex]I = Prt[\/latex]<\/td>\n [latex]P = \\$1350[\/latex]\n\n[latex]r = 2.8\\% = 0.028[\/latex]\n\n[latex]t = 7\/12 years[\/latex]<\/td>\n<\/tr>\n \n [latex]I = \\left( 1350 \\right) \\left( 0.028 \\right) \\left( 7\/12 \\right)[\/latex]<\/td>\n <\/td>\n<\/tr>\n \n [latex]= \\$22.05[\/latex]<\/td>\n <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nInterest of $22.05 over 7 months\n\nb)\n \n\n
\n [latex]I = Prt[\/latex]<\/td>\n [latex]P = \\$1350[\/latex]\n\n[latex]r = 0.028[\/latex]\n\n[latex]t = \\frac{25}{365}[\/latex]<\/td>\n<\/tr>\n \n [latex]I = \\left( 1350 \\right) \\left( 0.028 \\right) \\left( \\frac{25}{365} \\right)[\/latex]<\/td>\n <\/td>\n<\/tr>\n \n [latex]= $2.59[\/latex]<\/td>\n <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nInterest of $2.59 over 25 days. Note that the answer is rounded to the nearest two decimal places or to the nearest cent.\n\n<\/div>\n<\/div>\n \n TRY IT 3<\/p>\n\n<\/header>\n
\n\nDetermine the interest that will be earned on\u00a0 a deposit of $2200 at 4.52% over:\n\nDetermine the interest earned after\u00a0 a) 1 month\u00a0 b) 300 days\n\nShow answer<\/summary>a)\u00a0 $8.29\u00a0 \u00a0 b)\u00a0 \u00a0$81.73\n\n<\/details><\/div>\n<\/div>\n
\n EXAMPLE 4<\/p>\n\n<\/header>\n
\n\nDetermine the interest that will be earned on\u00a0 a deposit of $4200 at 4.65% over:\n\na) [latex]1\\frac{1}{2}[\/latex] years\n\nb) 5 weeks\n\nSolution<\/strong>\n\na)\n\n\n
\n [latex]I = Prt[\/latex]<\/td>\n [latex]P = \\$4200[\/latex]\n\n[latex]r = 4.65\\% = 0.0465[\/latex]\n\n[latex]t = 1.5 years[\/latex]<\/td>\n<\/tr>\n \n [latex]I = \\left( 4200 \\right) \\left( 0.0465 \\right) \\left( 1.5 \\right)[\/latex]<\/td>\n <\/td>\n<\/tr>\n \n [latex]= \\$292.95[\/latex]<\/td>\n <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nInterest of $292.95 over [latex]1\\frac{1}{2}[\/latex]years\n\nb)\n \n\n
\n [latex]I = Prt[\/latex]<\/td>\n [latex]P = \\$4200[\/latex]\n\n[latex]r = 0.0465[\/latex]\n\n[latex]t = \\frac{5}{52}[\/latex]<\/td>\n<\/tr>\n \n [latex]I = \\left( 4200 \\right) \\left( 0.0465 \\right) \\left( \\frac{5}{52} \\right)[\/latex]<\/td>\n <\/td>\n<\/tr>\n \n [latex]= 18.78[\/latex]<\/td>\n <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nInterest of $18.78 over 5 weeks. Note that the answer is rounded to the nearest two decimal places or to the nearest cent.\n\n<\/div>\n<\/div>\n \n TRY IT 4<\/p>\n\n<\/header>\n
\n\nMax deposited $1500 in a savings account at an interest rate of\u00a0 3.28%.\n\nDetermine the interest earned after i) 3 months\u00a0 ii) 65 days\u00a0 iii) two years.\n\nShow answer<\/summary>[latex]I = Prt = \\$1500\\left(0.0328\\right)\\left(\\frac{3}{12}\\right) = \\$12.30[\/latex]\n[latex] I = Prt = \\$1500 \\left(0.0328 \\right) \\left( \\frac{65}{365}\\right) = \\$8.76[\/latex]\n[latex] I = Prt = \\$1500 \\left(0.0328 \\right) \\left(2\\right) = \\$98.40[\/latex]\n\n<\/details><\/div>\n<\/div>\n
Maturity Value<\/strong><\/h1>\nThe total amount of money due at the end of a loan period is called the maturity value<\/strong> of the loan. It is the amount to be paid on the due date of a loan or the amount to be paid to an investor at the end of the period for which an investment has been made.\n
\n Maturity Value<\/p>\n\n<\/header>\n
\n\nThe Maturity Value<\/strong> (MV) of a loan is the sum of the principal P plus the interest I.\n\nMV = P + I\n\n<\/div>\n<\/div>\nIn Example 1, Jo borrowed $2000 at an interest rate of 5%. At the end of one year Jo owed $100 in interest.\n\nThe maturity value of the loan is\u00a0 MV = P + I\u00a0 where\u00a0 P = $2000 and I = $100.\nMV = $2000 + $100 = $2100<\/p>\nThe maturity value of the loan is $2100. At the end of the year Jo will be expected to pay back $2100.\n
\n EXAMPLE 5<\/p>\n\n<\/header>\n
\n\nLinda lends Ed $500.\u00a0 Ed says he will pay her back in 60 days at 9% simple interest.\u00a0 How much interest should Linda receive?\u00a0 How much must Ed pay Linda altogether?\n\nSolution<\/strong>\n\n\n
\n [latex]P = \\$500[\/latex]<\/td>\n [latex]I = Prt[\/latex]<\/td>\n <\/td>\n<\/tr>\n \n [latex]r = 9\\% \\,or\\, 0.09[\/latex]<\/td>\n [latex]I = 500 \\left(\u00a0 0.09 \\right) \\left( \\frac{60}{365} \\right) [\/latex]<\/td>\n Replace P<\/em>, r<\/em> and t<\/em> with their values<\/td>\n<\/tr>\n \n [latex]t = 60\\, days = \\frac{60}{365} years[\/latex]<\/td>\n [latex]I = 7.39726 [\/latex]<\/td>\n Multiply<\/td>\n<\/tr>\n \n <\/td>\n [latex]I = \\$7.40 [\/latex]<\/td>\n Round to the nearest cent<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nMV = P + I = $500 + $7.40 = $507.40\n\nLinda should receive $7.40 in interest. At the end of 60 days Ed will owe Linda\u00a0 $507.40.\n\n<\/div>\n<\/div>\n \n TRY IT 5<\/p>\n\n<\/header>\n
\n\nIn order to purchase equipment, a barbershop takes out a short term loan of $3000 at a rate of 4.35%. The loan is due in 80 days.\n\nDetermine the interest that will be owed at the end of 80 days and find the maturity value of the loan.\n\nShow answer<\/summary>Interest owed is $28.60, MV is $3028.60\n\n<\/details><\/div>\n<\/div>\n
Variations On Simple Interest<\/strong><\/h1>\nThe amount of interest earned on an investment or due on a loan is calculated using I = Prt.\n\nThis formula can also be used to determine:\n
- \n \t
- the amount of\u00a0[pb_glossary id=\"6033\"]principal[\/pb_glossary] <\/strong>(P) that needs to be invested in order to earn a certain amount of\u00a0 interest over a certain period of time.<\/li>\n \t
- the interest rate<\/strong>\u00a0 (r)\u00a0 that is needed in order to earn a certain amount of interest over a given time period.<\/li>\n \t
- the amount of time<\/strong> (t) it will take in order to earn a certain amount of interest at a stated interest rate.<\/li>\n<\/ul>\nThese amounts can be determined by solving the simple interest formula <\/strong>for any of\u00a0 r, P or t.\n
\n Finding the Principal, Interest Rate, or Time<\/p>\n\n<\/header>\n
\n\n\n
\n <\/td>\n <\/td>\n where<\/td>\n<\/tr>\n \n To determine the principal<\/strong> use:<\/td>\n [latex]P = \\frac{I}{rt}[\/latex]<\/td>\n I = interest earned<\/td>\n<\/tr>\n \n To determine the interest rate<\/strong> use:<\/td>\n [latex]r = \\frac{I}{Pt}[\/latex]<\/td>\n r<\/em> = annual interest rate<\/td>\n<\/tr>\n \n To determine the time<\/strong> use:<\/td>\n [latex]t = \\frac{I}{Pr}[\/latex]<\/td>\n P<\/em> = principal<\/td>\n<\/tr>\n \n <\/td>\n <\/td>\n t = <\/em>time (in years<\/strong>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nThe following memory aid is often called the \u201cMagic Triangle\u201d, because if you cover the variable you are trying to find, the formula will magically appear!\n\n ![\"sit\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/Simple-Interest-Triangle.png\")
\n\n<\/div>\n<\/div>\nDetermining the Principal<\/strong><\/h1>\n
\n EXAMPLE\u00a0 6<\/p>\n\n<\/header>\n
\n\nA six month investment will earn 5.25%.\u00a0 How much would you need to invest if you want to earn $100 in interest?\n\nSolution<\/strong>\n\nThe principal is unknown.\u00a0 Cover P<\/em> in the Magic Triangle.\n\nP = ?\u00a0<\/em>or\u00a0![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/Simple-Interest-P-e1591835358639.png\")
appears. Use the formula: [latex] P = \\frac{I}{rt}[\/latex]\n\n[latex]I = \\$100 [\/latex]\n\n[latex]r = 5.25% = 0.0525[\/latex]\n\n\n
\n [latex]t = 6 \\, months = \\frac{6}{12}[\/latex] or [latex]0.5 \\, years[\/latex]<\/td>\n [latex]P = \\frac{100}{0.0525 (0.5)}[\/latex]<\/td>\n \u00a0Replace I<\/em>, r<\/em> and t<\/em> with their respective values<\/td>\n<\/tr>\n \n <\/td>\n [latex]P = \\frac{100}{0.02625}[\/latex]<\/td>\n \u00a0Multiply 0.0525 by 0.5<\/td>\n<\/tr>\n \n <\/td>\n [latex]P = 3809.52[\/latex]<\/td>\n \u00a0Divide 100 by 0.02625 and round answer to nearest cent<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nYou would need to invest $3809.52<\/strong>\n\n<\/div>\n<\/div>\n \n TRY IT 6<\/p>\n\n<\/header>\n
\n\nA student borrowed money from his best friend at the very low interest rate of 1.5% for a period of 9 months. At the end of 9 months the friend had earned $22.50\u00a0 in interest. Determine the original amount of the loan.\n\nShow answer<\/summary>[latex]P = \\frac{22.5}{\\left( 0.015 \\right) \\left( 0.75 \\right)} = \\$2000[\/latex]\n\n<\/details><\/div>\n<\/div>\n
Determining the Interest\u00a0<\/strong>Rate<\/b><\/span><\/h1>\n
\n EXAMPLE 7<\/p>\n\n<\/header>\n
\n\nMariko had $240 in the bank for the month of April.\u00a0 At the end of the month she had earned $0.90 in interest.\u00a0 What interest rate was the bank paying?\n\nSolution<\/strong>\n\nThe interest rate is unknown.\u00a0 Cover r<\/em> in the Magic Triangle.\n\nr<\/em> = ?<\/em> or![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/Simple-Interest-r.png\")
appears. Use the formula: [latex] r = \\frac{I}{Pt}[\/latex]\n\n\n
\n [latex]P = \\$240 [\/latex]\n\n[latex]I = \\$0.90[\/latex]\n\nt<\/em> = 1 month = [latex] \\frac{1}{12}\\,year[\/latex]<\/td>\n [latex]r = \\frac{0.90}{240\\left( \\frac{1}{12} \\right)}[\/latex]<\/td>\n<\/tr>\n \n <\/td>\n [latex]r = \\frac{0.90}{20} = 0.045[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nSince r = 0.045, the interest rate as a percentage is 4.5%<\/strong>\n\n<\/div>\n<\/div>\n \n TRY IT\u00a0 7<\/p>\n\n<\/header>\n
\n\nKris deposited $800 in an account. At the end of 6 months he had earned\u00a0 $10.40. Determine the interest rate.\n\nShow answer<\/summary>[latex] r = \\frac{10.4}{\\left( 800 \\right) \\left( 0.5 \\right)} = 0.026 = 2.6% [\/latex]\n\n<\/details><\/div>\n<\/div>\n
Determining the Time<\/strong><\/h1>\n
\n EXAMPLE 8<\/p>\n\n<\/header>\n
\n\nCarol invested $500 at an interest rate of\u00a0 6%.\u00a0 How long will it take her to earn $250 in interest?\n\nSolution<\/strong>\n\nThe time is unknown.\u00a0 Cover t<\/em> in the Magic Triangle.\n\nt<\/em> = ?<\/em> or![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/Simple-Interest-t.png\")
appears. Use [latex]t = \\frac{I}{Pr}[\/latex]\n\n\n
\n [latex]I = \\$250 [\/latex]\n\n[latex]P = \\$500[\/latex]\n\nr<\/em> = 6% = 0.06<\/td>\n [latex]t = \\frac{250}{500\\left( 0.06\\right)}[\/latex]<\/td>\n<\/tr>\n \n <\/td>\n [latex]t = \\frac{250}{30} = 8.33[\/latex] or [latex]8\\frac{1}{3}[\/latex] years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nIt will take 8.33 years<\/strong>.\n\n<\/div>\n<\/div>\n \n TRY IT 8<\/p>\n\n<\/header>\n
\n\nAn account earns 4% interest. How long will it take for a deposit of $4000 to earn $240 in interest?\n\nShow answer<\/summary>[latex]t = \\frac{240}{\\left( 4000 \\right) \\left( 0.04 \\right)} = 1.5 years[\/latex]\n\n<\/details><\/div>\n<\/div>\nWhen calculating time \"t\" using the simple interest formula, the answer will be in terms of years. Sometimes it is more reasonable to express the answer in terms of days or months.\n
\n Time Conversions<\/p>\n\n<\/header>\n
\n\nWhen converting time (in years) to months or days:\n\nTo express the time in months<\/strong> (m):\n\nMultiply the time \"t\" in years x 12 months\/year.\n\nIf time t = 0.25 years\u00a0 \u00a0then the number of months\u00a0 m\u00a0 = 0.25 years x 12 months\/year =\u00a0 3 months\n\nTo express the time in days (d):\n\nMultiply the time \"t\" in years x 365 days\/year.\n\nIf time t = 0.25 years\u00a0 \u00a0then the number of days\u00a0 d = 0.25 years x 365 days\/year =\u00a0 91.25 days\n\n<\/div>\n<\/div>\n\n EXAMPLE 9<\/p>\n\n<\/header>\n
\n\nTroy invested $4000 in an account offering 3.8%.\u00a0 How long will it take him, in days, to earn $30 in interest?\n\nSolution<\/strong>\n\n[latex]t = \\frac{I}{Pr}[\/latex] = [latex]\\frac{30}{4000\\left( 0.038 \\right)} = 0.1974 years[\/latex]\n\nTime in days = 0.1974 years x 365 days\/year = 72 days\n\nIt will take 72 days to earn $30 in interest\n\n<\/div>\n<\/div>\n\n TRY IT 9<\/p>\n\n<\/header>\n
\n\nTam invested $1875 in an account offering 4%. How long will it take her, in months, to earn $62.50 in interest?\n\nShow answer<\/summary>[latex] t = \\frac{62.50}{\\left( 1875 \\right) \\left( 0.04 \\right)} = 0.833...[\/latex] years so 0.833... years [latex]\\times[\/latex]12 = 10 months\n\n<\/details><\/div>\n<\/div>\n
Key Concepts<\/h1>\n
- \n \t
- the interest rate<\/strong>\u00a0 (r)\u00a0 that is needed in order to earn a certain amount of interest over a given time period.<\/li>\n \t
- the amount of\u00a0[pb_glossary id=\"6033\"]principal[\/pb_glossary] <\/strong>(P) that needs to be invested in order to earn a certain amount of\u00a0 interest over a certain period of time.<\/li>\n \t