{"id":1097,"date":"2017-01-11T23:51:32","date_gmt":"2017-01-11T23:51:32","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=1097"},"modified":"2022-07-02T02:38:06","modified_gmt":"2022-07-02T02:38:06","slug":"introduction-accounts-that-gain-interest","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cccs-coreq-mathforliberalarts\/chapter\/introduction-accounts-that-gain-interest\/","title":{"raw":"3.3: Annuities and Loans","rendered":"3.3: Annuities and Loans"},"content":{"raw":"<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03191150\/430367107_5671e4c5a1_b.jpg\"><img class=\"wp-image-1341 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03191150\/430367107_5671e4c5a1_b-300x161.jpg\" alt=\"\" width=\"520\" height=\"279\" \/><\/a>\r\n\r\n&nbsp;\r\n\r\nFor most of us, we aren\u2019t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. In this section, we will explore the math behind specific kinds of accounts that gain interest over time, like retirement accounts. We will also explore how mortgages and car loans, called installment loans, are calculated.\r\n\r\n&nbsp;\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nThe learning outcomes for this section include:\r\n<ul>\r\n \t<li>Calculate the balance on an annuity after a specific amount of time<\/li>\r\n \t<li>Discern between compound interest, annuity, and payout annuity given a finance scenario<\/li>\r\n \t<li>Use the loan formula to calculate loan payments, loan balance, or interest accrued on a loan<\/li>\r\n \t<li>Determine which equation to use for a given scenario<\/li>\r\n \t<li>Solve a financial application for time<\/li>\r\n<\/ul>\r\n<\/div>\r\n[embed]https:\/\/youtu.be\/b9WL_Fzi-Zs[\/embed]\r\n\r\n&nbsp;\r\n\r\n[embed]https:\/\/youtu.be\/dJ-QuHRMK7c[\/embed]\r\n<h2>Example of Solving with a Calculator:<\/h2>\r\n<b>The problem: You want to retire in 35 years with $1,000,000. Suppose you find a financial instrument with an average return of 8.5%. How much do you need to put into your financial instrument each month to reach this goal?<\/b>\r\n\r\nHere is the information we are given from the problem:\r\n\r\nN- number of periods. 35 year *12 = 420 periods\r\n\r\nI% - interest rate as a percent = 8.5%\r\n\r\nPV \u2013 present value (money you have today). No information about how much money you have today, PV = 0\r\n\r\n<strong>PMT \u2013 payment. Repeated applications of money. Don\u2019t know payment, but want to find it. <\/strong>\r\n\r\nFV \u2013 future value \u2013 amount of money you want some day down the road. 1,000,000 (no commas in calculator)\r\n\r\np\/y \u2013 payments per year = 12 (because of the key word \"monthly\")\r\n\r\nc\/y \u2013 compounding per year\u00a0(because of the key word \"monthly\")\r\n\r\n*both c\/y and p\/y will typically be the same\r\n\r\nAnswer: you need to save $385.27 each month to reach your million dollar retirement goal.\r\n\r\nThis is the calculator we can use for this problem: <a href=\"https:\/\/www.geogebra.org\/m\/jhyUqg2A\">Time Value of Money Solver<\/a>\r\n\r\nHere is a video showing how to solve this type of problem (you will need to open this video in a new window): <a href=\"https:\/\/www.loom.com\/share\/87bfd3653160422da0b5d9264b8785b8\">Example of use of Calculator<\/a>\r\n<h2>Savings Annuities<\/h2>\r\nFor most of us, we aren\u2019t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a <strong>savings annuity<\/strong>. Most retirement plans like 401k plans or IRA plans are examples of savings annuities.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\"><img class=\"aligncenter size-full wp-image-737\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\" alt=\"Glass jar labeled &quot;Retirement.&quot; Inside are crumpled $100 bills\" width=\"640\" height=\"560\" \/><\/a>\r\n\r\nAn annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship\r\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{k}\\right){{P}_{m-1}}[\/latex]<\/p>\r\nFor a savings annuity, we simply need to add a deposit, <em>d<\/em>, to the account with each compounding period:\r\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{k}\\right){{P}_{m-1}}+d[\/latex]<\/p>\r\nTaking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working in general.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. Write an explicit formula that represents this scenario.\r\n[reveal-answer q=\"747493\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"747493\"]\r\n\r\nIn this example:\r\n<ul>\r\n \t<li><em>r<\/em> = 0.06 (6%)<\/li>\r\n \t<li><em>k<\/em> = 12 (12 compounds\/deposits per year)<\/li>\r\n \t<li><em>d<\/em> = $100 (our deposit per month)<\/li>\r\n<\/ul>\r\nWriting out the recursive equation gives\r\n\r\n[latex]{{P}_{m}}=\\left(1+\\frac{0.06}{12}\\right){{P}_{m-1}}+100=\\left(1.005\\right){{P}_{m-1}}+100[\/latex]\r\n\r\nAssuming we start with an empty account, we can begin using this relationship:\r\n\r\n[latex]P_0=0[\/latex]\r\n\r\n[latex]P_1=(1.005)P_0+100=100[\/latex]\r\n\r\n[latex]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[\/latex]\r\n\r\n[latex]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[\/latex]\r\n\r\nContinuing this pattern, after <em>m<\/em> deposits, we\u2019d have saved:\r\n\r\n[latex]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[\/latex]\r\n\r\nIn other words, after <em>m<\/em> months, the first deposit will have earned compound interest for <em>m-<\/em>1 months. The second deposit will have earned interest for <em>m\u00ad<\/em>-2 months. The last month's deposit (L) would have earned only one month's worth of interest. The most recent deposit will have earned no interest yet.\r\n\r\nThis equation leaves a lot to be desired, though \u2013 it doesn\u2019t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005:\r\n\r\n[latex]1.005{{P}_{m}}=1.005\\left(100{{\\left(1.005\\right)}^{m-1}}+100{{\\left(1.005\\right)}^{m-2}}+\\cdots+100(1.005)+100\\right)[\/latex]\r\n\r\nDistributing on the right side of the equation gives\r\n\r\n[latex]1.005{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}+100{{\\left(1.005\\right)}^{m-1}}+\\cdots+100{{(1.005)}^{2}}+100(1.005)[\/latex]\r\n\r\nNow we\u2019ll line this up with like terms from our original equation, and subtract each side\r\n\r\n[latex]\\begin{align}&amp;\\begin{matrix}1.005{{P}_{m}}&amp;=&amp;100{{\\left(1.005\\right)}^{m}}+&amp;100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&amp;100(1.005)&amp;{}\\\\{{P}_{m}}&amp;=&amp;{}&amp;100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&amp;100(1.005)&amp;+100\\\\\\end{matrix}\\\\&amp;\\\\\\end{align}[\/latex]\r\n\r\nAlmost all the terms cancel on the right hand side when we subtract, leaving\r\n\r\n[latex]1.005{{P}_{m}}-{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}-100[\/latex]\r\n\r\nFactor [latex]P_m[\/latex] out of the terms on the left side.\r\n\r\n[latex]\\begin{array}{c}P_m(1.005-1)=100{{\\left(1.005\\right)}^{m}}-100\\\\(0.005)P_m=100{{\\left(1.005\\right)}^{m}}-100\\end{array}[\/latex]\r\n\r\nSolve for <em>P<sub>m<\/sub><\/em>\r\n\r\n[latex]\\begin{align}&amp;0.005{{P}_{m}}=100\\left({{\\left(1.005\\right)}^{m}}-1\\right)\\\\&amp;\\\\&amp;{{P}_{m}}=\\frac{100\\left({{\\left(1.005\\right)}^{m}}-1\\right)}{0.005}\\\\\\end{align}[\/latex]\r\n\r\nReplacing <em>m<\/em> months with 12<em>N<\/em>, where <em>N<\/em> is measured in years, gives\r\n\r\n[latex]{{P}_{N}}=\\frac{100\\left({{\\left(1.005\\right)}^{12N}}-1\\right)}{0.005}[\/latex]\r\n\r\nRecall 0.005 was <em>r\/k<\/em> and 100 was the deposit <em>d. <\/em>12 was <em>k<\/em>, the number of deposit each year.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGeneralizing this result, we get the savings annuity formula.\r\n<div class=\"textbox\">\r\n<h3>Annuity Formula<\/h3>\r\n[latex]P_{N}=\\frac{d\\left(\\left(1+\\frac{r}{k}\\right)^{Nk}-1\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\r\n \t<li><em>d<\/em> is the regular deposit (the amount you deposit each year, each month, etc.)<\/li>\r\n \t<li><em>r <\/em> is the annual interest rate in decimal form.<\/li>\r\n \t<li><em>k <\/em>is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\nIf the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.\r\n\r\n<\/div>\r\nFor example, if the compounding frequency isn\u2019t stated:\r\n<ul>\r\n \t<li>If you make your deposits every month, use monthly compounding, <em>k<\/em> = 12.<\/li>\r\n \t<li>If you make your deposits every year, use yearly compounding, <em>k<\/em> = 1.<\/li>\r\n \t<li>If you make your deposits every quarter, use quarterly compounding, <em>k<\/em> = 4.<\/li>\r\n \t<li>Etc.<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>When do you use this?<\/h3>\r\nAnnuities assume that you put money in the account <strong>on a regular schedule<\/strong> (every month, year, quarter, etc.) and let it sit there earning interest.\r\n\r\nCompound interest assumes that you put money in the account <strong>once<\/strong> and let it sit there earning interest.\r\n<ul>\r\n \t<li>Compound interest: One deposit<\/li>\r\n \t<li>Annuity: Many deposits.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nA traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA earning 6% interest, how much will you have in the account after 20 years?\r\n[reveal-answer q=\"261481\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"261481\"]\r\n\r\nIn this example,\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $100<\/td>\r\n<td>the monthly deposit<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r<\/em> = 0.06<\/td>\r\n<td>6% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 20<\/td>\r\n<td>\u00a0we want the amount after 20 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPutting this into the equation:\r\n\r\n[latex]\\begin{align}&amp;{{P}_{20}}=\\frac{100\\left({{\\left(1+\\frac{0.06}{12}\\right)}^{20(12)}}-1\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&amp;{{P}_{20}}=\\frac{100\\left({{\\left(1.005\\right)}^{240}}-1\\right)}{\\left(0.005\\right)}\\\\&amp;{{P}_{20}}=\\frac{100\\left(3.310-1\\right)}{\\left(0.005\\right)}\\\\&amp;{{P}_{20}}=\\frac{100\\left(2.310\\right)}{\\left(0.005\\right)}=\\$46200 \\\\\\end{align}[\/latex]\r\n\r\nThe account will grow to $46,200 after 20 years.\r\n\r\nNotice that you deposited into the account a total of $24,000 ($100 a month for 240 months). The difference between what you end up with and how much you put in is the interest earned. In this case it is $46,200 - $24,000 = $22,200.\r\n\r\n[\/hidden-answer]\r\n\r\nThis example is explained in detail here.\r\n\r\nhttps:\/\/youtu.be\/quLg4bRpxPA\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nA conservative investment account pays 3% interest. If you deposit $5 a day into this account, how much will you have after 10 years? How much is from interest?\r\n[reveal-answer q=\"160692\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"160692\"]\r\n<div>\r\n\r\n<em>d<\/em> = $5\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 the daily deposit\r\n\r\n<em>r<\/em> = 0.03 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3% annual rate\r\n\r\n<em>k<\/em> = 365 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re doing daily deposits, we\u2019ll compound daily\r\n\r\n<em>N<\/em> = 10 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 we want the amount after 10 years\r\n\r\n[latex]P_{10}=\\frac{5\\left(\\left(1+\\frac{0.03}{365}\\right)^{365*10}-1\\right)}{\\frac{0.03}{365}}=21,282.07[\/latex]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6691&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div>\r\n\r\n\u00a0Financial planners typically recommend that you have a certain amount of savings upon retirement. \u00a0If you know the future value of the account, you can solve for the monthly contribution amount that will give you the desired result. In the next example, we will show you how this works.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nYou want to have $200,000 in your account when you retire in 30 years. Your retirement account earns 8% interest. How much do you need to deposit each month to meet your retirement goal?\r\n[reveal-answer q=\"897790\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"897790\"]\r\n\r\nIn this example, we\u2019re looking for <em>d<\/em>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>r<\/em> = 0.08<\/td>\r\n<td>8% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re depositing monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 30<\/td>\r\n<td>30 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P30<\/em> = $200,000<\/td>\r\n<td>The amount we want to have in 30 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.\r\n\r\n[latex]\\begin{align}&amp;200,000=\\frac{d\\left({{\\left(1+\\frac{0.08}{12}\\right)}^{30(12)}}-1\\right)}{\\left(\\frac{0.08}{12}\\right)}\\\\&amp;200,000=\\frac{d\\left({{\\left(1.00667\\right)}^{360}}-1\\right)}{\\left(0.00667\\right)}\\\\&amp;200,000=d(1491.57)\\\\&amp;d=\\frac{200,000}{1491.57}=\\$134.09 \\\\\\end{align}[\/latex]\r\n\r\nSo you would need to deposit $134.09 each month to have $200,000 in 30 years if your account earns 8% interest.\r\n\r\n[\/hidden-answer]\r\n\r\nView the solving of this problem\u00a0in the following video.\r\n\r\nhttps:\/\/youtu.be\/LB6pl7o0REc\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6688&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<h3>Solving For Time<\/h3>\r\nWe can solve the annuities formula for time, like we did the compounding interest formula, by using logarithms. In the next example we will work through how this is done.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?\r\n[reveal-answer q=\"181207\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"181207\"]\r\n\r\nThis is a savings annuity problem since we are making regular deposits into the account.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $100<\/td>\r\n<td>the monthly deposit<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r<\/em> = 0.03<\/td>\r\n<td>3% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe don\u2019t know <em>N<\/em>, but we want <em>P<sub>N<\/sub><\/em> to be $10,000.\r\n\r\nPutting this into the equation:\r\n\r\n[latex]10,000=\\frac{100\\left({{\\left(1+\\frac{0.03}{12}\\right)}^{N(12)}}-1\\right)}{\\left(\\frac{0.03}{12}\\right)}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Simplifying the fractions a bit\r\n\r\n[latex]10,000=\\frac{100\\left({{\\left(1.0025\\right)}^{12N}}-1\\right)}{0.0025}[\/latex]\r\n\r\nWe want to isolate the exponential term, 1.002512<em>N<\/em>, so multiply both sides by 0.0025\r\n\r\n[latex]25=100\\left({{\\left(1.0025\\right)}^{12N}}-1\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 100\r\n\r\n[latex]0.25={{\\left(1.0025\\right)}^{12N}}-1[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Add 1 to both sides\r\n\r\n[latex]1.25={{\\left(1.0025\\right)}^{12N}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now take the log of both sides\r\n\r\n[latex]\\log\\left(1.25\\right)=\\log\\left({{\\left(1.0025\\right)}^{12N}}\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs\r\n\r\n[latex]\\log\\left(1.25\\right)=12N\\log\\left(1.0025\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide by 12log(1.0025)\r\n\r\n[latex]\\frac{\\log\\left(1.25\\right)}{12\\log\\left(1.0025\\right)}=N[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Approximating to a decimal\r\n\r\n<em>N<\/em> = 7.447 years\r\n\r\nIt will take about 7.447 years to grow the account to $10,000.\r\n\r\n[\/hidden-answer]\r\n\r\nThis example is demonstrated here:\r\n\r\nhttps:\/\/youtu.be\/F3QVyswCzRo\r\n\r\n<\/div>\r\n<h2>Payout Annuities<\/h2>\r\n<h3>Removing Money from Annuities<\/h3>\r\nIn the last section you learned about annuities. In an annuity, you start with nothing, put money into an account on a regular basis, and end up with money in your account.\r\n\r\nIn this section, we will learn about a variation called a <strong>Payout Annuity<\/strong>. With a payout annuity, you start with money in the account, and pull money out of the account on a regular basis. Any remaining money in the account earns interest. After a fixed amount of time, the account will end up empty.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02204403\/2411468004_2bae893f5e_z.jpg\"><img class=\"aligncenter size-full wp-image-744\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02204403\/2411468004_2bae893f5e_z.jpg\" alt=\"Black and white aerial shot of hands exchanging money\" width=\"640\" height=\"426\" \/><\/a>\r\n\r\nPayout annuities are typically used after retirement. Perhaps you have saved $500,000 for retirement, and want to take money out of the account each month to live on. You want the money to last you 20 years. This is a payout annuity. The formula is derived in a similar way as we did for savings annuities. The details are omitted here.\r\n<div class=\"textbox\">\r\n<h3>Payout Annuity Formula<\/h3>\r\n[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (starting amount, or principal).<\/li>\r\n \t<li><em>d<\/em> is the regular withdrawal (the amount you take out each year, each month, etc.)<\/li>\r\n \t<li><em>r<\/em> is the annual interest rate (in decimal form. Example: 5% = 0.05)<\/li>\r\n \t<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\r\n \t<li><em>N<\/em> is the number of years we plan to take withdrawals<\/li>\r\n<\/ul>\r\n<\/div>\r\nLike with annuities, the compounding frequency is not always explicitly given, but is determined by how often you take the withdrawals.\r\n<div class=\"textbox\">\r\n<h3>When do you use this?<\/h3>\r\nPayout annuities assume that you take money from the account on a regular schedule (every month, year, quarter, etc.) and let the rest sit there earning interest.\r\n<ul>\r\n \t<li>Compound interest: One deposit<\/li>\r\n \t<li>Annuity: Many deposits.<\/li>\r\n \t<li>Payout Annuity: Many withdrawals<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAfter retiring, you want to be able to take $1000 every month for a total of 20 years from your retirement account. The account earns 6% interest. How much will you need in your account when you retire?\r\n[reveal-answer q=\"261541\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"261541\"]\r\n\r\nIn this example,\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $1000<\/td>\r\n<td>the monthly withdrawal<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r <\/em>= 0.06<\/td>\r\n<td>6% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re doing monthly withdrawals, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 20<\/td>\r\n<td>\u00a0since were taking withdrawals for 20 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe\u2019re looking for <em>P<sub>0:<\/sub><\/em>\u00a0how much money needs to be in the account at the beginning.\r\n\r\nPutting this into the equation:\r\n\r\n[latex]\\begin{align}&amp;{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-20(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&amp;{{P}_{0}}=\\frac{1000\\times\\left(1-{{\\left(1.005\\right)}^{-240}}\\right)}{\\left(0.005\\right)}\\\\&amp;{{P}_{0}}=\\frac{1000\\times\\left(1-0.302\\right)}{\\left(0.005\\right)}=\\$139,600 \\\\\\end{align}[\/latex]\r\n\r\nYou will need to have $139,600 in your account when you retire.\r\n\r\nNotice that you withdrew a total of $240,000 ($1000 a month for 240 months). The difference between what you pulled out and what you started with is the interest earned. In this case it is $240,000 - $139,600 = $100,400 in interest.\r\n\r\n[\/hidden-answer]\r\n\r\nView more about this problem in this video.\r\n\r\nhttps:\/\/youtu.be\/HK2eRFH6-0U\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6687&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Evaluating negative exponents on your calculator<\/h3>\r\nWith these problems, you need to raise numbers to negative powers.\u00a0 Most calculators have a separate button for negating a number that is different than the subtraction button.\u00a0 Some calculators label this (-) , some with +\/- .\u00a0 The button is often near the = key or the decimal point.\r\n\r\nIf your calculator displays operations on it (typically a calculator with multiline display), to calculate 1.005-240 you'd type something like:\u00a0 1.005 ^ (-) 240\r\n\r\nIf your calculator only shows one value at a time, then usually you hit the (-) key after a number to negate it, so you'd hit: 1.005 yx 240 (-) \u00a0=\r\n\r\nGive it a try - you should get 1.005-240 = 0.302096\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nYou know you will have $500,000 in your account when you retire. You want to be able to take monthly withdrawals from the account for a total of 30 years. Your retirement account earns 8% interest. How much will you be able to withdraw each month?\r\n[reveal-answer q=\"494776\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"494776\"]\r\n\r\nIn this example, we\u2019re looking for <em>d<\/em>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>r<\/em> = 0.08<\/td>\r\n<td>8% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re withdrawing monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 30<\/td>\r\n<td>30 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $500,000<\/td>\r\n<td>\u00a0we are beginning with $500,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.\r\n\r\n[latex]\\begin{align}&amp;500,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.08}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.08}{12}\\right)}\\\\&amp;500,000=\\frac{d\\left(1-{{\\left(1.00667\\right)}^{-360}}\\right)}{\\left(0.00667\\right)}\\\\&amp;500,000=d(136.232)\\\\&amp;d=\\frac{500,000}{136.232}=\\$3670.21 \\\\\\end{align}[\/latex]\r\n\r\nYou would be able to withdraw $3,670.21 each month for 30 years.\r\n\r\n[\/hidden-answer]\r\n\r\nA detailed walkthrough of this example can be viewed here.\r\n\r\nhttps:\/\/youtu.be\/XK7rA6pD4cI\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6681&amp;theme=oea&amp;iframe_resize_id=mom2[\/embed]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nA donor gives $100,000 to a university, and specifies that it is to be used to give annual scholarships for the next 20 years. If the university can earn 4% interest, how much can they give in scholarships each year?\r\n[reveal-answer q=\"547109\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"547109\"]\r\n<div>\r\n\r\n<em>d<\/em> = unknown\r\n\r\n<em>r<\/em> = 0.04 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4% annual rate\r\n\r\n<em>k<\/em> = 1\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re doing annual scholarships\r\n\r\n<em>N<\/em> = 20 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 20 years\r\n\r\n<em>P0<\/em> = 100,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with $100,000\r\n\r\n[latex]100,000=\\frac{d\\left(1-\\left(1+\\frac{0.04}{1}\\right)^{-20*1}\\right)}{\\frac{0.04}{1}}[\/latex]\r\n<div>\r\n\r\nSolving for <em>d<\/em> gives $7,358.18 each year that they can give in scholarships.\r\n\r\nIt is worth noting that usually donors instead specify that only interest is to be used for scholarship, which makes the original donation last indefinitely.\u00a0\u00a0 If this donor had specified that, $100,000(0.04) = $4,000 a year would have been available.\r\n\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Loans<\/h2>\r\n<h3>Conventional Loans<\/h3>\r\nIn the last section, you learned about payout annuities. In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920.png\"><img class=\"aligncenter size-large wp-image-747\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920-1024x681.png\" alt=\"Hand holding green pen, which has just written &quot;Approved!&quot; in a circle \" width=\"1024\" height=\"681\" \/><\/a>\r\n\r\nOne great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you\u2019re acting as the bank, you pay interest. The car lender takes payments until the balance is zero.\r\n<div class=\"textbox\">\r\n<h3>Loans Formula<\/h3>\r\n[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (the principal, or amount of the loan).<\/li>\r\n \t<li><em>d <\/em> is your loan payment (your monthly payment, annual payment, etc)<\/li>\r\n \t<li><em>r<\/em> is the annual interest rate in decimal form.<\/li>\r\n \t<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\r\n \t<li><em>N<\/em> is the length of the loan, in years.<\/li>\r\n<\/ul>\r\n<\/div>\r\nLike before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.\r\n<div class=\"textbox\">\r\n<h3>When do you use this?<\/h3>\r\nThe loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.\r\n<ul>\r\n \t<li>Compound interest: One deposit<\/li>\r\n \t<li>Annuity: Many deposits<\/li>\r\n \t<li>Payout Annuity: Many withdrawals<\/li>\r\n \t<li>Loans: Many payments<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nYou can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?\r\n[reveal-answer q=\"129373\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"129373\"]\r\n\r\nIn this example,\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $200<\/td>\r\n<td>the monthly loan payment<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r <\/em>= 0.03<\/td>\r\n<td>3% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 5<\/td>\r\n<td>since we\u2019re making monthly payments for 5 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe\u2019re looking for <em>P<sub>0<\/sub><\/em>, the starting amount of the loan.\r\n\r\n[latex]\\begin{align}&amp;{{P}_{0}}=\\frac{200\\left(1-{{\\left(1+\\frac{0.03}{12}\\right)}^{-5(12)}}\\right)}{\\left(\\frac{0.03}{12}\\right)}\\\\&amp;{{P}_{0}}=\\frac{200\\left(1-{{\\left(1.0025\\right)}^{-60}}\\right)}{\\left(0.0025\\right)}\\\\&amp;{{P}_{0}}=\\frac{200\\left(1-0.861\\right)}{\\left(0.0025\\right)}=\\$11,120 \\\\\\end{align}[\/latex]\r\n\r\nYou can afford a $11,120 loan.\r\n\r\nYou will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you\u2019re paying $12,000-$11,120 = $880 interest total.\r\n\r\n[\/hidden-answer]\r\n\r\nDetails of this example are examined in this video.\r\n\r\nhttps:\/\/youtu.be\/5NiNcdYytvY\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6685&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nYou want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?\r\n[reveal-answer q=\"538293\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"538293\"]\r\n\r\nIn this example, we\u2019re looking for <em>d<\/em>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>r<\/em> = 0.06<\/td>\r\n<td>6% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re paying monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 30<\/td>\r\n<td>30 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $140,000<\/td>\r\n<td>\u00a0the starting loan amount<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.\r\n\r\n[latex]\\begin{align}&amp;140,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&amp;140,000=\\frac{d\\left(1-{{\\left(1.005\\right)}^{-360}}\\right)}{\\left(0.005\\right)}\\\\&amp;140,000=d(166.792)\\\\&amp;d=\\frac{140,000}{166.792}=\\$839.37 \\\\\\end{align}[\/latex]\r\n\r\nYou will make payments of $839.37 per month for 30 years.\r\n\r\nYou\u2019re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 - $140,000 = $162,173.20 in interest over the life of the loan.\r\n\r\n[\/hidden-answer]\r\n\r\nView more about this example here.\r\n\r\nhttps:\/\/youtu.be\/BYCECTyUc68\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6684&amp;theme=oea&amp;iframe_resize_id=mom5[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nJanine bought $3,000 of new furniture on credit. Because her credit score isn\u2019t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?\r\n[reveal-answer q=\"642704\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"642704\"]\r\n<div>\r\n\r\n<em>d<\/em> = \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 unknown\r\n\r\n<em>r<\/em> = 0.16 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16% annual rate\r\n\r\n<em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments\r\n\r\n<em>N<\/em> = 2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 years to repay\r\n\r\n<em>P0<\/em> = 3,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $3,000 loan\r\n\r\nUSING the solver, put the following values into the calculator (<a href=\"https:\/\/www.geogebra.org\/m\/jhyUqg2A\">https:\/\/www.geogebra.org\/m\/jhyUqg2A<\/a>)\r\n\r\nN = 2*12 (put in 24)\r\n\r\nI%=16\r\n\r\nPV=3000\r\n\r\nPMT = what you are solving for, leave blank\r\n\r\nFV = 0 (want to owe nothing in the end)\r\n\r\nP\/Y and C\/Y = 12 (because this is monthly).\r\n\r\nBy hand, you would do the following.\r\n\r\n[latex]\\begin{array}{c}3000=\\frac{{d}\\left(1-\\left(1+\\frac{0.16}{12}\\right)^{-2*12}\\right)}{\\frac{0.16}{12}}\\\\\\\\3000=20.42d\\end{array}[\/latex]\r\n\r\nBoth ways Solve for d (or PMT) to get monthly payments of <strong>$146.89<\/strong>\r\n\r\nTwo years to repay means $146.89(24) = $3525.36 in total payments. \u00a0This means Janine will pay $3525.36 - $3000 = $525.36 in interest.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Calculating the Balance<\/h3>\r\nWith loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\"><img class=\"aligncenter size-full wp-image-751\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\" alt=\"Pair of glasses resting on a Mortgage Loan Statement\" width=\"640\" height=\"478\" \/><\/a>\r\n\r\nTo determine the remaining loan balance after some number of years, we first need to know the loan payments, if we don\u2019t already know them. Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will <em>not<\/em> have paid off $12,000 of the loan balance.\r\n\r\nTo determine the remaining loan balance, we can think \u201chow much loan will these loan payments be able to pay off in the remaining time on the loan?\u201d\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?\r\n[reveal-answer q=\"146377\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"146377\"]\r\n\r\nTo determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we\u2019re looking for P<sub>0<\/sub> when\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $1,000<\/td>\r\n<td>the monthly loan payment<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r <\/em>= 0.06<\/td>\r\n<td>6% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 10<\/td>\r\n<td>\u00a0since we\u2019re making monthly payments for 10 more years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[latex]\\begin{align}&amp;{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-10(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&amp;{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1.005\\right)}^{-120}}\\right)}{\\left(0.005\\right)}\\\\&amp;{{P}_{0}}=\\frac{1000\\left(1-0.5496\\right)}{\\left(0.005\\right)}=\\$90,073.45 \\\\\\end{align}[\/latex]\r\n\r\nThe loan balance with 10 years remaining on the loan will be $90,073.45.\r\n\r\n[\/hidden-answer]\r\n\r\nThis example is explained in the following video:\r\n\r\nhttps:\/\/youtu.be\/fXLzeyCfAwE\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nOftentimes answering remaining balance questions requires two steps:\r\n<ol>\r\n \t<li>Calculating the monthly payments on the loan<\/li>\r\n \t<li>Calculating the remaining loan balance based on the <em>remaining time<\/em> on the loan<\/li>\r\n<\/ol>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?\r\n[reveal-answer q=\"423248\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"423248\"]\r\n\r\nFirst we will calculate their monthly payments.\r\n\r\nWe\u2019re looking for <em>d<\/em>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>r<\/em> = 0.04<\/td>\r\n<td>4% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since they\u2019re paying monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 30<\/td>\r\n<td>30 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $180,000<\/td>\r\n<td>the starting loan amount<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe set up the equation and solve for <em>d<\/em>.\r\n\r\n[latex]\\begin{align}&amp;180,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&amp;180,000=\\frac{d\\left(1-{{\\left(1.00333\\right)}^{-360}}\\right)}{\\left(0.00333\\right)}\\\\&amp;180,000=d(209.562)\\\\&amp;d=\\frac{180,000}{209.562}=\\$858.93 \\\\\\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\nNow that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $858.93<\/td>\r\n<td>the monthly loan payment we calculated above<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r <\/em>= 0.04<\/td>\r\n<td>4% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since they\u2019re doing monthly payments<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 25<\/td>\r\n<td>since they\u2019d be making monthly payments for 25 more years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[latex]\\begin{align}&amp;{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-25(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&amp;{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1.00333\\right)}^{-300}}\\right)}{\\left(0.00333\\right)}\\\\&amp;{{P}_{0}}=\\frac{858.93\\left(1-0.369\\right)}{\\left(0.00333\\right)}=\\$155,793.91 \\\\\\end{align}[\/latex]\r\n\r\nThe loan balance after 5 years, with 25 years remaining on the loan, will be $155,793.91.\r\n\r\nOver that 5 years, the couple has paid off $180,000 - $155,793.91 = $24,206.09 of the loan balance. They have paid a total of $858.93 a month for 5 years (60 months), for a total of $51,535.80, so $51,535.80 - $24,206.09 = $27,329.71 of what they have paid so far has been interest.\r\n\r\n[\/hidden-answer]\r\n\r\nMore explanation of this example is available here:\r\n\r\nhttps:\/\/youtu.be\/-J1Ak2LLyRo\r\n\r\n<\/div>\r\n<h3>Solving for Time<\/h3>\r\nRecall that we have used logarithms to solve for time, since it is an exponent in interest calculations. We can apply the same idea to finding how long it will take to pay off a loan.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nJoel is considering putting a $1,000 laptop purchase on his credit card, which has an interest rate of 12% compounded monthly. How long will it take him to pay off the purchase if he makes payments of $30 a month?\r\n[reveal-answer q=\"427851\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"427851\"]\r\n<div>\r\n\r\n<em>d<\/em> = $30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The monthly payments\r\n\r\n<em>r<\/em> = 0.12 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12% annual rate\r\n\r\n<em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments\r\n\r\n<em>P0<\/em> = 1,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $1,000 loan\r\n\r\nWe are solving for <em>N<\/em>, the time to pay off the loan\r\n\r\n[latex]1000=\\frac{30\\left(1-\\left(1+\\frac{0.12}{12}\\right)^{-N*12}\\right)}{\\frac{0.12}{12}}[\/latex]\r\n<div>\r\n\r\nSolving for <em>N<\/em> gives 3.396. It will take about 3.4 years to pay off the purchase.\r\n\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>FYI<\/h3>\r\nHome loans are typically paid off through an amortization process, <strong>amortization<\/strong> refers to paying off a debt (often from a loan or mortgage) over time through regular payments.\u00a0An <b>amortization schedule<\/b> is a <b>table<\/b> detailing each periodic payment on an\u00a0<b>amortizing<\/b> loan as generated by an <b>amortization calculator<\/b>.\r\n\r\nIf you want to know more, click on the link below to view the website\u00a0\u201cHow is an Amortization Schedule Calculated?\u201d by MyAmortizationChart.com. This website provides a brief overlook of Amortization Schedules.\r\n<ul>\r\n \t<li><a href=\"http:\/\/www.myamortizationchart.com\/articles\/how-is-an-amortization-schedule-calculated\/\" target=\"_blank\" rel=\"noopener\">How is an Amortization Schedule Calculated?<\/a><\/li>\r\n<\/ul>\r\n<h2>Which Formula to Use?<\/h2>\r\n&nbsp;\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03001020\/chimpthink.png\"><img class=\"wp-image-1325 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03001020\/chimpthink-300x205.png\" alt=\"\" width=\"373\" height=\"254\" \/><\/a>\r\n\r\nNow that we have surveyed the basic kinds of finance calculations that are used, it may not always be obvious which one to use when you are given a problem to solve.\u00a0Here are some hints on deciding which equation to use, based on the wording of the problem.\r\n<h3>Loans<\/h3>\r\nThe easiest types of problems to identify are loans.\u00a0 Loan problems almost always include words like <strong>loan<\/strong>, <strong>amortize<\/strong> (the fancy word for loans), <strong>finance<\/strong> (i.e. a car), or <strong>mortgage<\/strong> (a home loan). Look for words like monthly or annual payment.\r\n\r\nThe loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.\r\n<div class=\"textbox\">\r\n<h3>Loans Formula<\/h3>\r\n[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (the principal, or amount of the loan).<\/li>\r\n \t<li><em>d <\/em>is your loan payment (your monthly payment, annual payment, etc)<\/li>\r\n \t<li><em>r<\/em> is the annual interest rate in decimal form.<\/li>\r\n \t<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\r\n \t<li><em>N<\/em> is the length of the loan, in years.<\/li>\r\n<\/ul>\r\n<\/div>\r\n&nbsp;\r\n<h3>Interest-Bearing Accounts<\/h3>\r\nAccounts that gain interest fall into two main categories. \u00a0The first is on where you put money in an account once and let it sit, the other is where you make regular payments or withdrawals from the account as in a retirement account.\r\n\r\n<span style=\"text-decoration: underline;\"><strong>Interest<\/strong><\/span>\r\n<ul>\r\n \t<li>If you're letting the money sit in the account with nothing but interest changing the balance, then you're looking at a <strong>compound interest<\/strong> problem. Look for words like compounded, or APY.\u00a0Compound interest assumes that you put money in the account <strong>once<\/strong> and let it sit there earning interest.<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>COMPOUND INTEREST<\/h3>\r\n[latex]P_{N}=P_{0}\\left(1+\\frac{r}{k}\\right)^{Nk}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\r\n \t<li><em>P<sub>0 <\/sub><\/em>is the starting balance of the account (also called initial deposit, or principal)<\/li>\r\n \t<li><em>r<\/em> is the annual interest rate in decimal form<\/li>\r\n \t<li><em>k<\/em> is the number of compounding periods in one year\r\n<ul>\r\n \t<li>If the compounding is done annually (once a year), <em>k<\/em> = 1.<\/li>\r\n \t<li>If the compounding is done quarterly, <em>k<\/em> = 4.<\/li>\r\n \t<li>If the compounding is done monthly, <em>k<\/em> = 12.<\/li>\r\n \t<li>If the compounding is done daily, <em>k<\/em> = 365.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<ul>\r\n \t<li>The exception would be bonds and other investments where the interest is not reinvested; in those cases you\u2019re looking at <strong>simple interest<\/strong>.<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>SIMPLE INTEREST OVER TIME<\/h3>\r\n[latex]\\begin{align}&amp;I={{P}_{0}}rt\\\\&amp;A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}rt={{P}_{0}}(1+rt)\\\\\\end{align}[\/latex]\r\n<ul>\r\n \t<li><em>I<\/em> is the interest<\/li>\r\n \t<li><em>A<\/em> is the end amount: principal plus interest<\/li>\r\n \t<li>[latex]\\begin{align}{{P}_{0}}\\\\\\end{align}[\/latex] is the principal (starting amount)<\/li>\r\n \t<li><em>r<\/em> is the interest rate in decimal form<\/li>\r\n \t<li><em>t<\/em> is time<\/li>\r\n<\/ul>\r\nThe units of measurement (years, months, etc.) for the time should match the time period for the interest rate.\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<strong><span style=\"text-decoration: underline;\">Annuities<\/span><\/strong>\r\n<ul>\r\n \t<li>If you're putting money <strong><em>into<\/em><\/strong> the account on a regular basis (monthly\/annually\/quarterly) then you're looking at a <strong>basic annuity<\/strong> problem.\u00a0 Basic annuities are when you are saving money.\u00a0 Usually in an annuity problem, your account starts empty, and has money in the future. Annuities assume that you put money in the account <strong>on a regular schedule<\/strong> (every month, year, quarter, etc.) and let it sit there earning interest.<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>ANNUITY FORMULA<\/h3>\r\n[latex]P_{N}=\\frac{d\\left(\\left(1+\\frac{r}{k}\\right)^{Nk}-1\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\r\n \t<li><em>d<\/em> is the regular deposit (the amount you deposit each year, each month, etc.)<\/li>\r\n \t<li><em>r <\/em>is the annual interest rate in decimal form.<\/li>\r\n \t<li><em>k <\/em>is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\nIf the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.\r\n\r\n<\/div>\r\n&nbsp;\r\n<ul>\r\n \t<li>If you're pulling money <em><strong>out<\/strong><\/em> of the account on a regular basis, then you're looking at a <strong>payout annuity<\/strong> problem.\u00a0 Payout annuities are used for things like retirement income, where you start with money in your account, pull money out on a regular basis, and your account ends up empty\u00a0in the future.\u00a0Payout annuities assume that you take money from the account on a regular schedule (every month, year, quarter, etc.) and let the rest sit there earning interest.<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>PAYOUT ANNUITY FORMULA<\/h3>\r\n[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (starting amount, or principal).<\/li>\r\n \t<li><em>d<\/em> is the regular withdrawal (the amount you take out each year, each month, etc.)<\/li>\r\n \t<li><em>r<\/em> is the annual interest rate (in decimal form. Example: 5% = 0.05)<\/li>\r\n \t<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\r\n \t<li><em>N<\/em> is the number of years we plan to take withdrawals<\/li>\r\n<\/ul>\r\n<\/div>\r\nRemember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and then determine what approach will best allow you to solve the problem.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nFor each of the following scenarios, determine if it is a compound interest problem, a savings annuity problem, a payout annuity problem, or a loans problem. Then solve each problem.\r\n<ol>\r\n \t<li>Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it for college, withdrawing money for tuition and expenses each quarter. How much can she take out each quarter if she has 3 years of school left?[reveal-answer q=\"160930\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"160930\"]This is a payout annuity problem. She can pull out $1833.60 a quarter.[\/hidden-answer]<\/li>\r\n \t<li>Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an account earning 3% interest compounded monthly. How much will he have saved up after 3 years?[reveal-answer q=\"160931\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"160931\"]This is a savings annuity problem. He will have saved up $7,524.11[\/hidden-answer]<\/li>\r\n \t<li>Keisha is managing investments for a non-profit company. \u00a0 \u00a0 \u00a0 They want to invest some money in an account earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much should Keisha deposit into the account?\u00a0[reveal-answer q=\"160932\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"160932\"]This is compound interest problem. She would need to deposit $22,386.46.[\/hidden-answer]<\/li>\r\n \t<li>Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly payments of $100, how much new equipment can she buy?\u00a0[reveal-answer q=\"160933\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"160933\"]This is a loans problem. She can buy $4,609.33 of new equipment[\/hidden-answer]<\/li>\r\n \t<li>How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in two years?[reveal-answer q=\"160934\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"160934\"]This is a savings annuity problem. You would need to save $200.46 each month[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn the following video, we present more examples of how to use the language in the question to determine which type of equation to use to solve a finance problem.\r\n\r\nhttps:\/\/youtu.be\/V5oG7lLTECs\r\n\r\nIn the next video example, we show how to solve a finance problem that has two stages, the first stage is a savings problem, and the second stage is a withdrawal problem.\r\n\r\nhttps:\/\/youtu.be\/CNkvwMuLuis\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=67287&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=67306&amp;theme=oea&amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=67133&amp;theme=oea&amp;iframe_resize_id=mom5[\/embed]\r\n\r\n<\/div>","rendered":"<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03191150\/430367107_5671e4c5a1_b.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1341 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03191150\/430367107_5671e4c5a1_b-300x161.jpg\" alt=\"\" width=\"520\" height=\"279\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p>For most of us, we aren\u2019t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. In this section, we will explore the math behind specific kinds of accounts that gain interest over time, like retirement accounts. We will also explore how mortgages and car loans, called installment loans, are calculated.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>The learning outcomes for this section include:<\/p>\n<ul>\n<li>Calculate the balance on an annuity after a specific amount of time<\/li>\n<li>Discern between compound interest, annuity, and payout annuity given a finance scenario<\/li>\n<li>Use the loan formula to calculate loan payments, loan balance, or interest accrued on a loan<\/li>\n<li>Determine which equation to use for a given scenario<\/li>\n<li>Solve a financial application for time<\/li>\n<\/ul>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Finding annuities using the TVM Solver\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/b9WL_Fzi-Zs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Amortized loans: Calculating payments, interest, and amortization schedule using technology\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/dJ-QuHRMK7c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Example of Solving with a Calculator:<\/h2>\n<p><b>The problem: You want to retire in 35 years with $1,000,000. Suppose you find a financial instrument with an average return of 8.5%. How much do you need to put into your financial instrument each month to reach this goal?<\/b><\/p>\n<p>Here is the information we are given from the problem:<\/p>\n<p>N- number of periods. 35 year *12 = 420 periods<\/p>\n<p>I% &#8211; interest rate as a percent = 8.5%<\/p>\n<p>PV \u2013 present value (money you have today). No information about how much money you have today, PV = 0<\/p>\n<p><strong>PMT \u2013 payment. Repeated applications of money. Don\u2019t know payment, but want to find it. <\/strong><\/p>\n<p>FV \u2013 future value \u2013 amount of money you want some day down the road. 1,000,000 (no commas in calculator)<\/p>\n<p>p\/y \u2013 payments per year = 12 (because of the key word &#8220;monthly&#8221;)<\/p>\n<p>c\/y \u2013 compounding per year\u00a0(because of the key word &#8220;monthly&#8221;)<\/p>\n<p>*both c\/y and p\/y will typically be the same<\/p>\n<p>Answer: you need to save $385.27 each month to reach your million dollar retirement goal.<\/p>\n<p>This is the calculator we can use for this problem: <a href=\"https:\/\/www.geogebra.org\/m\/jhyUqg2A\">Time Value of Money Solver<\/a><\/p>\n<p>Here is a video showing how to solve this type of problem (you will need to open this video in a new window): <a href=\"https:\/\/www.loom.com\/share\/87bfd3653160422da0b5d9264b8785b8\">Example of use of Calculator<\/a><\/p>\n<h2>Savings Annuities<\/h2>\n<p>For most of us, we aren\u2019t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a <strong>savings annuity<\/strong>. Most retirement plans like 401k plans or IRA plans are examples of savings annuities.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-737\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\" alt=\"Glass jar labeled &quot;Retirement.&quot; Inside are crumpled $100 bills\" width=\"640\" height=\"560\" \/><\/a><\/p>\n<p>An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{k}\\right){{P}_{m-1}}[\/latex]<\/p>\n<p>For a savings annuity, we simply need to add a deposit, <em>d<\/em>, to the account with each compounding period:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{k}\\right){{P}_{m-1}}+d[\/latex]<\/p>\n<p>Taking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working in general.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. Write an explicit formula that represents this scenario.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q747493\">Show Answer<\/span><\/p>\n<div id=\"q747493\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example:<\/p>\n<ul>\n<li><em>r<\/em> = 0.06 (6%)<\/li>\n<li><em>k<\/em> = 12 (12 compounds\/deposits per year)<\/li>\n<li><em>d<\/em> = $100 (our deposit per month)<\/li>\n<\/ul>\n<p>Writing out the recursive equation gives<\/p>\n<p>[latex]{{P}_{m}}=\\left(1+\\frac{0.06}{12}\\right){{P}_{m-1}}+100=\\left(1.005\\right){{P}_{m-1}}+100[\/latex]<\/p>\n<p>Assuming we start with an empty account, we can begin using this relationship:<\/p>\n<p>[latex]P_0=0[\/latex]<\/p>\n<p>[latex]P_1=(1.005)P_0+100=100[\/latex]<\/p>\n<p>[latex]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[\/latex]<\/p>\n<p>[latex]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[\/latex]<\/p>\n<p>Continuing this pattern, after <em>m<\/em> deposits, we\u2019d have saved:<\/p>\n<p>[latex]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[\/latex]<\/p>\n<p>In other words, after <em>m<\/em> months, the first deposit will have earned compound interest for <em>m-<\/em>1 months. The second deposit will have earned interest for <em>m\u00ad<\/em>-2 months. The last month&#8217;s deposit (L) would have earned only one month&#8217;s worth of interest. The most recent deposit will have earned no interest yet.<\/p>\n<p>This equation leaves a lot to be desired, though \u2013 it doesn\u2019t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005:<\/p>\n<p>[latex]1.005{{P}_{m}}=1.005\\left(100{{\\left(1.005\\right)}^{m-1}}+100{{\\left(1.005\\right)}^{m-2}}+\\cdots+100(1.005)+100\\right)[\/latex]<\/p>\n<p>Distributing on the right side of the equation gives<\/p>\n<p>[latex]1.005{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}+100{{\\left(1.005\\right)}^{m-1}}+\\cdots+100{{(1.005)}^{2}}+100(1.005)[\/latex]<\/p>\n<p>Now we\u2019ll line this up with like terms from our original equation, and subtract each side<\/p>\n<p>[latex]\\begin{align}&\\begin{matrix}1.005{{P}_{m}}&=&100{{\\left(1.005\\right)}^{m}}+&100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&100(1.005)&{}\\\\{{P}_{m}}&=&{}&100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&100(1.005)&+100\\\\\\end{matrix}\\\\&\\\\\\end{align}[\/latex]<\/p>\n<p>Almost all the terms cancel on the right hand side when we subtract, leaving<\/p>\n<p>[latex]1.005{{P}_{m}}-{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}-100[\/latex]<\/p>\n<p>Factor [latex]P_m[\/latex] out of the terms on the left side.<\/p>\n<p>[latex]\\begin{array}{c}P_m(1.005-1)=100{{\\left(1.005\\right)}^{m}}-100\\\\(0.005)P_m=100{{\\left(1.005\\right)}^{m}}-100\\end{array}[\/latex]<\/p>\n<p>Solve for <em>P<sub>m<\/sub><\/em><\/p>\n<p>[latex]\\begin{align}&0.005{{P}_{m}}=100\\left({{\\left(1.005\\right)}^{m}}-1\\right)\\\\&\\\\&{{P}_{m}}=\\frac{100\\left({{\\left(1.005\\right)}^{m}}-1\\right)}{0.005}\\\\\\end{align}[\/latex]<\/p>\n<p>Replacing <em>m<\/em> months with 12<em>N<\/em>, where <em>N<\/em> is measured in years, gives<\/p>\n<p>[latex]{{P}_{N}}=\\frac{100\\left({{\\left(1.005\\right)}^{12N}}-1\\right)}{0.005}[\/latex]<\/p>\n<p>Recall 0.005 was <em>r\/k<\/em> and 100 was the deposit <em>d. <\/em>12 was <em>k<\/em>, the number of deposit each year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Generalizing this result, we get the savings annuity formula.<\/p>\n<div class=\"textbox\">\n<h3>Annuity Formula<\/h3>\n<p>[latex]P_{N}=\\frac{d\\left(\\left(1+\\frac{r}{k}\\right)^{Nk}-1\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\n<li><em>d<\/em> is the regular deposit (the amount you deposit each year, each month, etc.)<\/li>\n<li><em>r <\/em> is the annual interest rate in decimal form.<\/li>\n<li><em>k <\/em>is the number of compounding periods in one year.<\/li>\n<\/ul>\n<p>If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.<\/p>\n<\/div>\n<p>For example, if the compounding frequency isn\u2019t stated:<\/p>\n<ul>\n<li>If you make your deposits every month, use monthly compounding, <em>k<\/em> = 12.<\/li>\n<li>If you make your deposits every year, use yearly compounding, <em>k<\/em> = 1.<\/li>\n<li>If you make your deposits every quarter, use quarterly compounding, <em>k<\/em> = 4.<\/li>\n<li>Etc.<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>When do you use this?<\/h3>\n<p>Annuities assume that you put money in the account <strong>on a regular schedule<\/strong> (every month, year, quarter, etc.) and let it sit there earning interest.<\/p>\n<p>Compound interest assumes that you put money in the account <strong>once<\/strong> and let it sit there earning interest.<\/p>\n<ul>\n<li>Compound interest: One deposit<\/li>\n<li>Annuity: Many deposits.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA earning 6% interest, how much will you have in the account after 20 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q261481\">Show Answer<\/span><\/p>\n<div id=\"q261481\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example,<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $100<\/td>\n<td>the monthly deposit<\/td>\n<\/tr>\n<tr>\n<td><em>r<\/em> = 0.06<\/td>\n<td>6% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 20<\/td>\n<td>\u00a0we want the amount after 20 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Putting this into the equation:<\/p>\n<p>[latex]\\begin{align}&{{P}_{20}}=\\frac{100\\left({{\\left(1+\\frac{0.06}{12}\\right)}^{20(12)}}-1\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&{{P}_{20}}=\\frac{100\\left({{\\left(1.005\\right)}^{240}}-1\\right)}{\\left(0.005\\right)}\\\\&{{P}_{20}}=\\frac{100\\left(3.310-1\\right)}{\\left(0.005\\right)}\\\\&{{P}_{20}}=\\frac{100\\left(2.310\\right)}{\\left(0.005\\right)}=\\$46200 \\\\\\end{align}[\/latex]<\/p>\n<p>The account will grow to $46,200 after 20 years.<\/p>\n<p>Notice that you deposited into the account a total of $24,000 ($100 a month for 240 months). The difference between what you end up with and how much you put in is the interest earned. In this case it is $46,200 &#8211; $24,000 = $22,200.<\/p>\n<\/div>\n<\/div>\n<p>This example is explained in detail here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Saving Annuities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/quLg4bRpxPA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>A conservative investment account pays 3% interest. If you deposit $5 a day into this account, how much will you have after 10 years? How much is from interest?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q160692\">Show Answer<\/span><\/p>\n<div id=\"q160692\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p><em>d<\/em> = $5\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 the daily deposit<\/p>\n<p><em>r<\/em> = 0.03 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3% annual rate<\/p>\n<p><em>k<\/em> = 365 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re doing daily deposits, we\u2019ll compound daily<\/p>\n<p><em>N<\/em> = 10 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 we want the amount after 10 years<\/p>\n<p>[latex]P_{10}=\\frac{5\\left(\\left(1+\\frac{0.03}{365}\\right)^{365*10}-1\\right)}{\\frac{0.03}{365}}=21,282.07[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6691&#38;theme=oea&#38;iframe_resize_id=mom1\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6691&amp;theme=oea&amp;iframe_resize_id=mom1<\/a><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div>\n<p>\u00a0Financial planners typically recommend that you have a certain amount of savings upon retirement. \u00a0If you know the future value of the account, you can solve for the monthly contribution amount that will give you the desired result. In the next example, we will show you how this works.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>You want to have $200,000 in your account when you retire in 30 years. Your retirement account earns 8% interest. How much do you need to deposit each month to meet your retirement goal?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q897790\">Show Answer<\/span><\/p>\n<div id=\"q897790\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example, we\u2019re looking for <em>d<\/em>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>r<\/em> = 0.08<\/td>\n<td>8% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re depositing monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 30<\/td>\n<td>30 years<\/td>\n<\/tr>\n<tr>\n<td><em>P30<\/em> = $200,000<\/td>\n<td>The amount we want to have in 30 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.<\/p>\n<p>[latex]\\begin{align}&200,000=\\frac{d\\left({{\\left(1+\\frac{0.08}{12}\\right)}^{30(12)}}-1\\right)}{\\left(\\frac{0.08}{12}\\right)}\\\\&200,000=\\frac{d\\left({{\\left(1.00667\\right)}^{360}}-1\\right)}{\\left(0.00667\\right)}\\\\&200,000=d(1491.57)\\\\&d=\\frac{200,000}{1491.57}=\\$134.09 \\\\\\end{align}[\/latex]<\/p>\n<p>So you would need to deposit $134.09 each month to have $200,000 in 30 years if your account earns 8% interest.<\/p>\n<\/div>\n<\/div>\n<p>View the solving of this problem\u00a0in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Savings annuities - solving for the deposit\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/LB6pl7o0REc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6688&#38;theme=oea&#38;iframe_resize_id=mom1\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6688&amp;theme=oea&amp;iframe_resize_id=mom1<\/a><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<h3>Solving For Time<\/h3>\n<p>We can solve the annuities formula for time, like we did the compounding interest formula, by using logarithms. In the next example we will work through how this is done.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q181207\">Show Answer<\/span><\/p>\n<div id=\"q181207\" class=\"hidden-answer\" style=\"display: none\">\n<p>This is a savings annuity problem since we are making regular deposits into the account.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $100<\/td>\n<td>the monthly deposit<\/td>\n<\/tr>\n<tr>\n<td><em>r<\/em> = 0.03<\/td>\n<td>3% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We don\u2019t know <em>N<\/em>, but we want <em>P<sub>N<\/sub><\/em> to be $10,000.<\/p>\n<p>Putting this into the equation:<\/p>\n<p>[latex]10,000=\\frac{100\\left({{\\left(1+\\frac{0.03}{12}\\right)}^{N(12)}}-1\\right)}{\\left(\\frac{0.03}{12}\\right)}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Simplifying the fractions a bit<\/p>\n<p>[latex]10,000=\\frac{100\\left({{\\left(1.0025\\right)}^{12N}}-1\\right)}{0.0025}[\/latex]<\/p>\n<p>We want to isolate the exponential term, 1.002512<em>N<\/em>, so multiply both sides by 0.0025<\/p>\n<p>[latex]25=100\\left({{\\left(1.0025\\right)}^{12N}}-1\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 100<\/p>\n<p>[latex]0.25={{\\left(1.0025\\right)}^{12N}}-1[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Add 1 to both sides<\/p>\n<p>[latex]1.25={{\\left(1.0025\\right)}^{12N}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now take the log of both sides<\/p>\n<p>[latex]\\log\\left(1.25\\right)=\\log\\left({{\\left(1.0025\\right)}^{12N}}\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs<\/p>\n<p>[latex]\\log\\left(1.25\\right)=12N\\log\\left(1.0025\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide by 12log(1.0025)<\/p>\n<p>[latex]\\frac{\\log\\left(1.25\\right)}{12\\log\\left(1.0025\\right)}=N[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Approximating to a decimal<\/p>\n<p><em>N<\/em> = 7.447 years<\/p>\n<p>It will take about 7.447 years to grow the account to $10,000.<\/p>\n<\/div>\n<\/div>\n<p>This example is demonstrated here:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Use logs to find the time it takes an annuity to grow\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/F3QVyswCzRo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<h2>Payout Annuities<\/h2>\n<h3>Removing Money from Annuities<\/h3>\n<p>In the last section you learned about annuities. In an annuity, you start with nothing, put money into an account on a regular basis, and end up with money in your account.<\/p>\n<p>In this section, we will learn about a variation called a <strong>Payout Annuity<\/strong>. With a payout annuity, you start with money in the account, and pull money out of the account on a regular basis. Any remaining money in the account earns interest. After a fixed amount of time, the account will end up empty.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02204403\/2411468004_2bae893f5e_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-744\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02204403\/2411468004_2bae893f5e_z.jpg\" alt=\"Black and white aerial shot of hands exchanging money\" width=\"640\" height=\"426\" \/><\/a><\/p>\n<p>Payout annuities are typically used after retirement. Perhaps you have saved $500,000 for retirement, and want to take money out of the account each month to live on. You want the money to last you 20 years. This is a payout annuity. The formula is derived in a similar way as we did for savings annuities. The details are omitted here.<\/p>\n<div class=\"textbox\">\n<h3>Payout Annuity Formula<\/h3>\n<p>[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (starting amount, or principal).<\/li>\n<li><em>d<\/em> is the regular withdrawal (the amount you take out each year, each month, etc.)<\/li>\n<li><em>r<\/em> is the annual interest rate (in decimal form. Example: 5% = 0.05)<\/li>\n<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\n<li><em>N<\/em> is the number of years we plan to take withdrawals<\/li>\n<\/ul>\n<\/div>\n<p>Like with annuities, the compounding frequency is not always explicitly given, but is determined by how often you take the withdrawals.<\/p>\n<div class=\"textbox\">\n<h3>When do you use this?<\/h3>\n<p>Payout annuities assume that you take money from the account on a regular schedule (every month, year, quarter, etc.) and let the rest sit there earning interest.<\/p>\n<ul>\n<li>Compound interest: One deposit<\/li>\n<li>Annuity: Many deposits.<\/li>\n<li>Payout Annuity: Many withdrawals<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>After retiring, you want to be able to take $1000 every month for a total of 20 years from your retirement account. The account earns 6% interest. How much will you need in your account when you retire?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q261541\">Show Answer<\/span><\/p>\n<div id=\"q261541\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example,<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $1000<\/td>\n<td>the monthly withdrawal<\/td>\n<\/tr>\n<tr>\n<td><em>r <\/em>= 0.06<\/td>\n<td>6% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re doing monthly withdrawals, we\u2019ll compound monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 20<\/td>\n<td>\u00a0since were taking withdrawals for 20 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We\u2019re looking for <em>P<sub>0:<\/sub><\/em>\u00a0how much money needs to be in the account at the beginning.<\/p>\n<p>Putting this into the equation:<\/p>\n<p>[latex]\\begin{align}&{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-20(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&{{P}_{0}}=\\frac{1000\\times\\left(1-{{\\left(1.005\\right)}^{-240}}\\right)}{\\left(0.005\\right)}\\\\&{{P}_{0}}=\\frac{1000\\times\\left(1-0.302\\right)}{\\left(0.005\\right)}=\\$139,600 \\\\\\end{align}[\/latex]<\/p>\n<p>You will need to have $139,600 in your account when you retire.<\/p>\n<p>Notice that you withdrew a total of $240,000 ($1000 a month for 240 months). The difference between what you pulled out and what you started with is the interest earned. In this case it is $240,000 &#8211; $139,600 = $100,400 in interest.<\/p>\n<\/div>\n<\/div>\n<p>View more about this problem in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Payout Annuities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/HK2eRFH6-0U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6687&#38;theme=oea&#38;iframe_resize_id=mom1\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6687&amp;theme=oea&amp;iframe_resize_id=mom1<\/a><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Evaluating negative exponents on your calculator<\/h3>\n<p>With these problems, you need to raise numbers to negative powers.\u00a0 Most calculators have a separate button for negating a number that is different than the subtraction button.\u00a0 Some calculators label this (-) , some with +\/- .\u00a0 The button is often near the = key or the decimal point.<\/p>\n<p>If your calculator displays operations on it (typically a calculator with multiline display), to calculate 1.005-240 you&#8217;d type something like:\u00a0 1.005 ^ (-) 240<\/p>\n<p>If your calculator only shows one value at a time, then usually you hit the (-) key after a number to negate it, so you&#8217;d hit: 1.005 yx 240 (-) \u00a0=<\/p>\n<p>Give it a try &#8211; you should get 1.005-240 = 0.302096<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>You know you will have $500,000 in your account when you retire. You want to be able to take monthly withdrawals from the account for a total of 30 years. Your retirement account earns 8% interest. How much will you be able to withdraw each month?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q494776\">Show Answer<\/span><\/p>\n<div id=\"q494776\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example, we\u2019re looking for <em>d<\/em>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>r<\/em> = 0.08<\/td>\n<td>8% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re withdrawing monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 30<\/td>\n<td>30 years<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>0<\/sub><\/em> = $500,000<\/td>\n<td>\u00a0we are beginning with $500,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.<\/p>\n<p>[latex]\\begin{align}&500,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.08}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.08}{12}\\right)}\\\\&500,000=\\frac{d\\left(1-{{\\left(1.00667\\right)}^{-360}}\\right)}{\\left(0.00667\\right)}\\\\&500,000=d(136.232)\\\\&d=\\frac{500,000}{136.232}=\\$3670.21 \\\\\\end{align}[\/latex]<\/p>\n<p>You would be able to withdraw $3,670.21 each month for 30 years.<\/p>\n<\/div>\n<\/div>\n<p>A detailed walkthrough of this example can be viewed here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Payout annuity - solve for withdrawal\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/XK7rA6pD4cI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6681&#38;theme=oea&#38;iframe_resize_id=mom2\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6681&amp;theme=oea&amp;iframe_resize_id=mom2<\/a><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>A donor gives $100,000 to a university, and specifies that it is to be used to give annual scholarships for the next 20 years. If the university can earn 4% interest, how much can they give in scholarships each year?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q547109\">Show Answer<\/span><\/p>\n<div id=\"q547109\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p><em>d<\/em> = unknown<\/p>\n<p><em>r<\/em> = 0.04 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4% annual rate<\/p>\n<p><em>k<\/em> = 1\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re doing annual scholarships<\/p>\n<p><em>N<\/em> = 20 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 20 years<\/p>\n<p><em>P0<\/em> = 100,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with $100,000<\/p>\n<p>[latex]100,000=\\frac{d\\left(1-\\left(1+\\frac{0.04}{1}\\right)^{-20*1}\\right)}{\\frac{0.04}{1}}[\/latex]<\/p>\n<div>\n<p>Solving for <em>d<\/em> gives $7,358.18 each year that they can give in scholarships.<\/p>\n<p>It is worth noting that usually donors instead specify that only interest is to be used for scholarship, which makes the original donation last indefinitely.\u00a0\u00a0 If this donor had specified that, $100,000(0.04) = $4,000 a year would have been available.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Loans<\/h2>\n<h3>Conventional Loans<\/h3>\n<p>In the last section, you learned about payout annuities. In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-747\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920-1024x681.png\" alt=\"Hand holding green pen, which has just written &quot;Approved!&quot; in a circle\" width=\"1024\" height=\"681\" \/><\/a><\/p>\n<p>One great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you\u2019re acting as the bank, you pay interest. The car lender takes payments until the balance is zero.<\/p>\n<div class=\"textbox\">\n<h3>Loans Formula<\/h3>\n<p>[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (the principal, or amount of the loan).<\/li>\n<li><em>d <\/em> is your loan payment (your monthly payment, annual payment, etc)<\/li>\n<li><em>r<\/em> is the annual interest rate in decimal form.<\/li>\n<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\n<li><em>N<\/em> is the length of the loan, in years.<\/li>\n<\/ul>\n<\/div>\n<p>Like before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.<\/p>\n<div class=\"textbox\">\n<h3>When do you use this?<\/h3>\n<p>The loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.<\/p>\n<ul>\n<li>Compound interest: One deposit<\/li>\n<li>Annuity: Many deposits<\/li>\n<li>Payout Annuity: Many withdrawals<\/li>\n<li>Loans: Many payments<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q129373\">Show Answer<\/span><\/p>\n<div id=\"q129373\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example,<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $200<\/td>\n<td>the monthly loan payment<\/td>\n<\/tr>\n<tr>\n<td><em>r <\/em>= 0.03<\/td>\n<td>3% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 5<\/td>\n<td>since we\u2019re making monthly payments for 5 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We\u2019re looking for <em>P<sub>0<\/sub><\/em>, the starting amount of the loan.<\/p>\n<p>[latex]\\begin{align}&{{P}_{0}}=\\frac{200\\left(1-{{\\left(1+\\frac{0.03}{12}\\right)}^{-5(12)}}\\right)}{\\left(\\frac{0.03}{12}\\right)}\\\\&{{P}_{0}}=\\frac{200\\left(1-{{\\left(1.0025\\right)}^{-60}}\\right)}{\\left(0.0025\\right)}\\\\&{{P}_{0}}=\\frac{200\\left(1-0.861\\right)}{\\left(0.0025\\right)}=\\$11,120 \\\\\\end{align}[\/latex]<\/p>\n<p>You can afford a $11,120 loan.<\/p>\n<p>You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you\u2019re paying $12,000-$11,120 = $880 interest total.<\/p>\n<\/div>\n<\/div>\n<p>Details of this example are examined in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Car loan\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/5NiNcdYytvY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6685&#38;theme=oea&#38;iframe_resize_id=mom1\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6685&amp;theme=oea&amp;iframe_resize_id=mom1<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>You want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q538293\">Show Answer<\/span><\/p>\n<div id=\"q538293\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example, we\u2019re looking for <em>d<\/em>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>r<\/em> = 0.06<\/td>\n<td>6% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re paying monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 30<\/td>\n<td>30 years<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>0<\/sub><\/em> = $140,000<\/td>\n<td>\u00a0the starting loan amount<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.<\/p>\n<p>[latex]\\begin{align}&140,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&140,000=\\frac{d\\left(1-{{\\left(1.005\\right)}^{-360}}\\right)}{\\left(0.005\\right)}\\\\&140,000=d(166.792)\\\\&d=\\frac{140,000}{166.792}=\\$839.37 \\\\\\end{align}[\/latex]<\/p>\n<p>You will make payments of $839.37 per month for 30 years.<\/p>\n<p>You\u2019re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 &#8211; $140,000 = $162,173.20 in interest over the life of the loan.<\/p>\n<\/div>\n<\/div>\n<p>View more about this example here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Calculating payment on a home loan\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BYCECTyUc68?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6684&#38;theme=oea&#38;iframe_resize_id=mom5\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6684&amp;theme=oea&amp;iframe_resize_id=mom5<\/a><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>Janine bought $3,000 of new furniture on credit. Because her credit score isn\u2019t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q642704\">Show Answer<\/span><\/p>\n<div id=\"q642704\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p><em>d<\/em> = \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 unknown<\/p>\n<p><em>r<\/em> = 0.16 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16% annual rate<\/p>\n<p><em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments<\/p>\n<p><em>N<\/em> = 2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 years to repay<\/p>\n<p><em>P0<\/em> = 3,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $3,000 loan<\/p>\n<p>USING the solver, put the following values into the calculator (<a href=\"https:\/\/www.geogebra.org\/m\/jhyUqg2A\">https:\/\/www.geogebra.org\/m\/jhyUqg2A<\/a>)<\/p>\n<p>N = 2*12 (put in 24)<\/p>\n<p>I%=16<\/p>\n<p>PV=3000<\/p>\n<p>PMT = what you are solving for, leave blank<\/p>\n<p>FV = 0 (want to owe nothing in the end)<\/p>\n<p>P\/Y and C\/Y = 12 (because this is monthly).<\/p>\n<p>By hand, you would do the following.<\/p>\n<p>[latex]\\begin{array}{c}3000=\\frac{{d}\\left(1-\\left(1+\\frac{0.16}{12}\\right)^{-2*12}\\right)}{\\frac{0.16}{12}}\\\\\\\\3000=20.42d\\end{array}[\/latex]<\/p>\n<p>Both ways Solve for d (or PMT) to get monthly payments of <strong>$146.89<\/strong><\/p>\n<p>Two years to repay means $146.89(24) = $3525.36 in total payments. \u00a0This means Janine will pay $3525.36 &#8211; $3000 = $525.36 in interest.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>Calculating the Balance<\/h3>\n<p>With loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-751\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\" alt=\"Pair of glasses resting on a Mortgage Loan Statement\" width=\"640\" height=\"478\" \/><\/a><\/p>\n<p>To determine the remaining loan balance after some number of years, we first need to know the loan payments, if we don\u2019t already know them. Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will <em>not<\/em> have paid off $12,000 of the loan balance.<\/p>\n<p>To determine the remaining loan balance, we can think \u201chow much loan will these loan payments be able to pay off in the remaining time on the loan?\u201d<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q146377\">Show Answer<\/span><\/p>\n<div id=\"q146377\" class=\"hidden-answer\" style=\"display: none\">\n<p>To determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we\u2019re looking for P<sub>0<\/sub> when<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $1,000<\/td>\n<td>the monthly loan payment<\/td>\n<\/tr>\n<tr>\n<td><em>r <\/em>= 0.06<\/td>\n<td>6% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 10<\/td>\n<td>\u00a0since we\u2019re making monthly payments for 10 more years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>[latex]\\begin{align}&{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-10(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1.005\\right)}^{-120}}\\right)}{\\left(0.005\\right)}\\\\&{{P}_{0}}=\\frac{1000\\left(1-0.5496\\right)}{\\left(0.005\\right)}=\\$90,073.45 \\\\\\end{align}[\/latex]<\/p>\n<p>The loan balance with 10 years remaining on the loan will be $90,073.45.<\/p>\n<\/div>\n<\/div>\n<p>This example is explained in the following video:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Calculate remaining balance on loan from payment\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fXLzeyCfAwE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Oftentimes answering remaining balance questions requires two steps:<\/p>\n<ol>\n<li>Calculating the monthly payments on the loan<\/li>\n<li>Calculating the remaining loan balance based on the <em>remaining time<\/em> on the loan<\/li>\n<\/ol>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q423248\">Show Answer<\/span><\/p>\n<div id=\"q423248\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we will calculate their monthly payments.<\/p>\n<p>We\u2019re looking for <em>d<\/em>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>r<\/em> = 0.04<\/td>\n<td>4% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since they\u2019re paying monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 30<\/td>\n<td>30 years<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>0<\/sub><\/em> = $180,000<\/td>\n<td>the starting loan amount<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We set up the equation and solve for <em>d<\/em>.<\/p>\n<p>[latex]\\begin{align}&180,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&180,000=\\frac{d\\left(1-{{\\left(1.00333\\right)}^{-360}}\\right)}{\\left(0.00333\\right)}\\\\&180,000=d(209.562)\\\\&d=\\frac{180,000}{209.562}=\\$858.93 \\\\\\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Now that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $858.93<\/td>\n<td>the monthly loan payment we calculated above<\/td>\n<\/tr>\n<tr>\n<td><em>r <\/em>= 0.04<\/td>\n<td>4% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since they\u2019re doing monthly payments<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 25<\/td>\n<td>since they\u2019d be making monthly payments for 25 more years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>[latex]\\begin{align}&{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-25(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1.00333\\right)}^{-300}}\\right)}{\\left(0.00333\\right)}\\\\&{{P}_{0}}=\\frac{858.93\\left(1-0.369\\right)}{\\left(0.00333\\right)}=\\$155,793.91 \\\\\\end{align}[\/latex]<\/p>\n<p>The loan balance after 5 years, with 25 years remaining on the loan, will be $155,793.91.<\/p>\n<p>Over that 5 years, the couple has paid off $180,000 &#8211; $155,793.91 = $24,206.09 of the loan balance. They have paid a total of $858.93 a month for 5 years (60 months), for a total of $51,535.80, so $51,535.80 &#8211; $24,206.09 = $27,329.71 of what they have paid so far has been interest.<\/p>\n<\/div>\n<\/div>\n<p>More explanation of this example is available here:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-11\" title=\"Calculate remaining balance on a mortgage\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-J1Ak2LLyRo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<h3>Solving for Time<\/h3>\n<p>Recall that we have used logarithms to solve for time, since it is an exponent in interest calculations. We can apply the same idea to finding how long it will take to pay off a loan.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>Joel is considering putting a $1,000 laptop purchase on his credit card, which has an interest rate of 12% compounded monthly. How long will it take him to pay off the purchase if he makes payments of $30 a month?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q427851\">Show Answer<\/span><\/p>\n<div id=\"q427851\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p><em>d<\/em> = $30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The monthly payments<\/p>\n<p><em>r<\/em> = 0.12 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12% annual rate<\/p>\n<p><em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments<\/p>\n<p><em>P0<\/em> = 1,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $1,000 loan<\/p>\n<p>We are solving for <em>N<\/em>, the time to pay off the loan<\/p>\n<p>[latex]1000=\\frac{30\\left(1-\\left(1+\\frac{0.12}{12}\\right)^{-N*12}\\right)}{\\frac{0.12}{12}}[\/latex]<\/p>\n<div>\n<p>Solving for <em>N<\/em> gives 3.396. It will take about 3.4 years to pay off the purchase.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>FYI<\/h3>\n<p>Home loans are typically paid off through an amortization process, <strong>amortization<\/strong> refers to paying off a debt (often from a loan or mortgage) over time through regular payments.\u00a0An <b>amortization schedule<\/b> is a <b>table<\/b> detailing each periodic payment on an\u00a0<b>amortizing<\/b> loan as generated by an <b>amortization calculator<\/b>.<\/p>\n<p>If you want to know more, click on the link below to view the website\u00a0\u201cHow is an Amortization Schedule Calculated?\u201d by MyAmortizationChart.com. This website provides a brief overlook of Amortization Schedules.<\/p>\n<ul>\n<li><a href=\"http:\/\/www.myamortizationchart.com\/articles\/how-is-an-amortization-schedule-calculated\/\" target=\"_blank\" rel=\"noopener\">How is an Amortization Schedule Calculated?<\/a><\/li>\n<\/ul>\n<h2>Which Formula to Use?<\/h2>\n<p>&nbsp;<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03001020\/chimpthink.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1325 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/03001020\/chimpthink-300x205.png\" alt=\"\" width=\"373\" height=\"254\" \/><\/a><\/p>\n<p>Now that we have surveyed the basic kinds of finance calculations that are used, it may not always be obvious which one to use when you are given a problem to solve.\u00a0Here are some hints on deciding which equation to use, based on the wording of the problem.<\/p>\n<h3>Loans<\/h3>\n<p>The easiest types of problems to identify are loans.\u00a0 Loan problems almost always include words like <strong>loan<\/strong>, <strong>amortize<\/strong> (the fancy word for loans), <strong>finance<\/strong> (i.e. a car), or <strong>mortgage<\/strong> (a home loan). Look for words like monthly or annual payment.<\/p>\n<p>The loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.<\/p>\n<div class=\"textbox\">\n<h3>Loans Formula<\/h3>\n<p>[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (the principal, or amount of the loan).<\/li>\n<li><em>d <\/em>is your loan payment (your monthly payment, annual payment, etc)<\/li>\n<li><em>r<\/em> is the annual interest rate in decimal form.<\/li>\n<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\n<li><em>N<\/em> is the length of the loan, in years.<\/li>\n<\/ul>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>Interest-Bearing Accounts<\/h3>\n<p>Accounts that gain interest fall into two main categories. \u00a0The first is on where you put money in an account once and let it sit, the other is where you make regular payments or withdrawals from the account as in a retirement account.<\/p>\n<p><span style=\"text-decoration: underline;\"><strong>Interest<\/strong><\/span><\/p>\n<ul>\n<li>If you&#8217;re letting the money sit in the account with nothing but interest changing the balance, then you&#8217;re looking at a <strong>compound interest<\/strong> problem. Look for words like compounded, or APY.\u00a0Compound interest assumes that you put money in the account <strong>once<\/strong> and let it sit there earning interest.<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>COMPOUND INTEREST<\/h3>\n<p>[latex]P_{N}=P_{0}\\left(1+\\frac{r}{k}\\right)^{Nk}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\n<li><em>P<sub>0 <\/sub><\/em>is the starting balance of the account (also called initial deposit, or principal)<\/li>\n<li><em>r<\/em> is the annual interest rate in decimal form<\/li>\n<li><em>k<\/em> is the number of compounding periods in one year\n<ul>\n<li>If the compounding is done annually (once a year), <em>k<\/em> = 1.<\/li>\n<li>If the compounding is done quarterly, <em>k<\/em> = 4.<\/li>\n<li>If the compounding is done monthly, <em>k<\/em> = 12.<\/li>\n<li>If the compounding is done daily, <em>k<\/em> = 365.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<ul>\n<li>The exception would be bonds and other investments where the interest is not reinvested; in those cases you\u2019re looking at <strong>simple interest<\/strong>.<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>SIMPLE INTEREST OVER TIME<\/h3>\n<p>[latex]\\begin{align}&I={{P}_{0}}rt\\\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}rt={{P}_{0}}(1+rt)\\\\\\end{align}[\/latex]<\/p>\n<ul>\n<li><em>I<\/em> is the interest<\/li>\n<li><em>A<\/em> is the end amount: principal plus interest<\/li>\n<li>[latex]\\begin{align}{{P}_{0}}\\\\\\end{align}[\/latex] is the principal (starting amount)<\/li>\n<li><em>r<\/em> is the interest rate in decimal form<\/li>\n<li><em>t<\/em> is time<\/li>\n<\/ul>\n<p>The units of measurement (years, months, etc.) for the time should match the time period for the interest rate.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><strong><span style=\"text-decoration: underline;\">Annuities<\/span><\/strong><\/p>\n<ul>\n<li>If you&#8217;re putting money <strong><em>into<\/em><\/strong> the account on a regular basis (monthly\/annually\/quarterly) then you&#8217;re looking at a <strong>basic annuity<\/strong> problem.\u00a0 Basic annuities are when you are saving money.\u00a0 Usually in an annuity problem, your account starts empty, and has money in the future. Annuities assume that you put money in the account <strong>on a regular schedule<\/strong> (every month, year, quarter, etc.) and let it sit there earning interest.<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>ANNUITY FORMULA<\/h3>\n<p>[latex]P_{N}=\\frac{d\\left(\\left(1+\\frac{r}{k}\\right)^{Nk}-1\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\n<li><em>d<\/em> is the regular deposit (the amount you deposit each year, each month, etc.)<\/li>\n<li><em>r <\/em>is the annual interest rate in decimal form.<\/li>\n<li><em>k <\/em>is the number of compounding periods in one year.<\/li>\n<\/ul>\n<p>If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<ul>\n<li>If you&#8217;re pulling money <em><strong>out<\/strong><\/em> of the account on a regular basis, then you&#8217;re looking at a <strong>payout annuity<\/strong> problem.\u00a0 Payout annuities are used for things like retirement income, where you start with money in your account, pull money out on a regular basis, and your account ends up empty\u00a0in the future.\u00a0Payout annuities assume that you take money from the account on a regular schedule (every month, year, quarter, etc.) and let the rest sit there earning interest.<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>PAYOUT ANNUITY FORMULA<\/h3>\n<p>[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (starting amount, or principal).<\/li>\n<li><em>d<\/em> is the regular withdrawal (the amount you take out each year, each month, etc.)<\/li>\n<li><em>r<\/em> is the annual interest rate (in decimal form. Example: 5% = 0.05)<\/li>\n<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\n<li><em>N<\/em> is the number of years we plan to take withdrawals<\/li>\n<\/ul>\n<\/div>\n<p>Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and then determine what approach will best allow you to solve the problem.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>For each of the following scenarios, determine if it is a compound interest problem, a savings annuity problem, a payout annuity problem, or a loans problem. Then solve each problem.<\/p>\n<ol>\n<li>Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it for college, withdrawing money for tuition and expenses each quarter. How much can she take out each quarter if she has 3 years of school left?\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q160930\">Show Answer<\/span><\/p>\n<div id=\"q160930\" class=\"hidden-answer\" style=\"display: none\">This is a payout annuity problem. She can pull out $1833.60 a quarter.<\/div>\n<\/div>\n<\/li>\n<li>Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an account earning 3% interest compounded monthly. How much will he have saved up after 3 years?\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q160931\">Show Answer<\/span><\/p>\n<div id=\"q160931\" class=\"hidden-answer\" style=\"display: none\">This is a savings annuity problem. He will have saved up $7,524.11<\/div>\n<\/div>\n<\/li>\n<li>Keisha is managing investments for a non-profit company. \u00a0 \u00a0 \u00a0 They want to invest some money in an account earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much should Keisha deposit into the account?\u00a0\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q160932\">Show Answer<\/span><\/p>\n<div id=\"q160932\" class=\"hidden-answer\" style=\"display: none\">This is compound interest problem. She would need to deposit $22,386.46.<\/div>\n<\/div>\n<\/li>\n<li>Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly payments of $100, how much new equipment can she buy?\u00a0\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q160933\">Show Answer<\/span><\/p>\n<div id=\"q160933\" class=\"hidden-answer\" style=\"display: none\">This is a loans problem. She can buy $4,609.33 of new equipment<\/div>\n<\/div>\n<\/li>\n<li>How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in two years?\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q160934\">Show Answer<\/span><\/p>\n<div id=\"q160934\" class=\"hidden-answer\" style=\"display: none\">This is a savings annuity problem. You would need to save $200.46 each month<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p>In the following video, we present more examples of how to use the language in the question to determine which type of equation to use to solve a finance problem.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-12\" title=\"Identifying type of finance problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/V5oG7lLTECs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next video example, we show how to solve a finance problem that has two stages, the first stage is a savings problem, and the second stage is a withdrawal problem.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-13\" title=\"Multistage finance problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/CNkvwMuLuis?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=67287&#38;theme=oea&#38;iframe_resize_id=mom1\">https:\/\/www.myopenmath.com\/multiembedq.php?id=67287&amp;theme=oea&amp;iframe_resize_id=mom1<\/a><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=67306&#38;theme=oea&#38;iframe_resize_id=mom2\">https:\/\/www.myopenmath.com\/multiembedq.php?id=67306&amp;theme=oea&amp;iframe_resize_id=mom2<\/a><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=67133&#38;theme=oea&#38;iframe_resize_id=mom5\">https:\/\/www.myopenmath.com\/multiembedq.php?id=67133&amp;theme=oea&amp;iframe_resize_id=mom5<\/a><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1097\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Outcomes and Introduction. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Bad Credit? We can help!. <strong>Authored by<\/strong>: BookMama. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.flickr.com\/photos\/myloonyland\/430367107\">https:\/\/www.flickr.com\/photos\/myloonyland\/430367107<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: 2.0<\/li><li>Math in Society. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Retirement. <strong>Authored by<\/strong>: Tax Credits. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/flic.kr\/p\/bH1jrv\">https:\/\/flic.kr\/p\/bH1jrv<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Savings Annuities. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/quLg4bRpxPA\">https:\/\/youtu.be\/quLg4bRpxPA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Savings annuities - solving for the deposit. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/LB6pl7o0REc\">https:\/\/youtu.be\/LB6pl7o0REc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 6691, 6688. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li> Determining The Value of an Annuity. <strong>Authored by<\/strong>: Sousa, James (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/DWFezRwYp0I\">https:\/\/youtu.be\/DWFezRwYp0I<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Determining The Value of an Annuity on the TI84. <strong>Authored by<\/strong>: Sousa, James (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/OKCZRZ-hWH8\">https:\/\/youtu.be\/OKCZRZ-hWH8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Payment. <strong>Authored by<\/strong>: Sergio Marchi. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/flic.kr\/p\/4F6pqy\">https:\/\/flic.kr\/p\/4F6pqy<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-nd\/4.0\/\">CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives <\/a><\/em><\/li><li>Payout Annuities. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/HK2eRFH6-0U\">https:\/\/youtu.be\/HK2eRFH6-0U<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Payout annuity - solve for withdrawal. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/XK7rA6pD4cI\">https:\/\/youtu.be\/XK7rA6pD4cI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 6681, 6687. <strong>Authored by<\/strong>: David Lippman. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Payout Annuity Formula - Part 1, Part 2. <strong>Authored by<\/strong>: Sousa, James (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/zxRFWEj5v5w,%20https:\/\/youtu.be\/h3CWS9Rs7-s\">https:\/\/youtu.be\/zxRFWEj5v5w,%20https:\/\/youtu.be\/h3CWS9Rs7-s<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>approved-finance-business-loan-1049259. <strong>Authored by<\/strong>: InspiredImages. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/approved-finance-business-loan-1049259\/\">https:\/\/pixabay.com\/en\/approved-finance-business-loan-1049259\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Car loan. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/5NiNcdYytvY\">https:\/\/youtu.be\/5NiNcdYytvY<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculating payment on a home loan. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BYCECTyUc68\">https:\/\/youtu.be\/BYCECTyUc68<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 6684, 6685. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Identifying type of finance problem. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/V5oG7lLTECs\">https:\/\/youtu.be\/V5oG7lLTECs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Multistage finance problem. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/CNkvwMuLuis\">https:\/\/youtu.be\/CNkvwMuLuis<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 67287, 67306, 67133. <strong>Authored by<\/strong>: Abert, Rex. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Outcomes and Introduction\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Bad Credit? 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