{"id":71096,"date":"2021-04-05T10:03:53","date_gmt":"2021-04-05T15:03:53","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=71096"},"modified":"2026-03-25T11:53:26","modified_gmt":"2026-03-25T16:53:26","slug":"compound-interest","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/compound-interest\/","title":{"rendered":"Compound Interest Formula"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_ByFdeDiyClo\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_ByFdeDiyClo\" data-source-videoID=\"ByFdeDiyClo\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Compound Interest Formula Video\" height=\"720\" width=\"1280\" class=\"size-full\" data-matomo-title = \"Compound Interest Formula\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_ByFdeDiyClo:hover {cursor:pointer;} img#videoThumbnailImage_ByFdeDiyClo {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/08\/Compound-interest-thumb.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_ByFdeDiyClo\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_ByFdeDiyClo\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Compound Interest Formula\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_ByFdeDiyClo\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_ByFdeDiyClo\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_ByFdeDiyClo\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction VSR_Function() {\n  var x = document.getElementById(\"VSR\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"VSR_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"VSR\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Calculating_Simple_Interest\" class=\"smooth-scroll\">Calculating Simple Interest<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Calculating_Compound_Interest_(Example_1)\" class=\"smooth-scroll\">Calculating Compound Interest (Example 1)<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Calculating_Compound_Interest_(Example_2)\" class=\"smooth-scroll\">Calculating Compound Interest (Example 2)<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Calculating_Compound_Interest_(Example_3)\" class=\"smooth-scroll\">Calculating Compound Interest (Example 3)<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Practice_Questions\" class=\"smooth-scroll\">Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"overview\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"overview\">Overview<\/label><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"overview-spoiler\">\n<p>Compound interest is interest that is calculated on both the money deposited and the interest earned from that deposit. The formula for compound interest is \\(A=P(1+\\frac{r}{n})^{nt}\\), where \\(A\\) represents the final balance after the interest has been calculated for the time, \\(t\\), in years, on a principal amount, \\(P\\), at an annual interest rate, \\(r\\). The number of times in the year that the interest is compounded is \\(n\\).<\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example 1:<\/h3>\n<p>Jasmine deposits $520 into a savings account that has a 3.5% interest rate compounded monthly. What will be the balance of Jasmine\u2019s savings account after two years? To find the balance after two years, \\(A\\), we need to use the formula, \\(A=P(1+\\frac{r}{n})^{nt}\\). The principal, \\(P\\), in this situation is the amount Jasmine used to start her account, $520. The rate, \\(r\\), as stated in the problem, is 3.5% (or 0.035 as a decimal) and compounded monthly, so \\(n=12\\). Since we are looking for the balance of the account after two years, 2, is the time, \\(t\\). \\(A=520(1+\\frac{0.035}{12})^{12(2)}\\) \\(A=557.65\\) The balance of Jasmine\u2019s account after 2 years is $557.65.<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example 2:<\/h3>\n<p>Lex has $1,780.80 in his savings account that he opened 6 years ago. His account has an annual interest rate of 6.8% compounded annually. How much money did Lex use to open his savings account? To find the principal, \\(P\\) we can use the same formula, \\(A=P(1+\\frac{r}{n})^{nt}\\). We have the balance of the account, \\(A\\), after 6 years, which is $1,780.80. The interest rate, \\(r\\), is 6.8% (or 0.068 as a decimal) and is compounded annually, so \\(n=1\\). The time, \\(t\\), is 6, since we know he opened his account 6 years ago. Plug in the known values into the formula and solve for the missing variable, \\(P\\).<\/p>\n<div id=\"pqs\">\u00a0<\/div>\n<p>\\(1{,}780.80=P(1+\\frac{0.068}{1})^{1(6)}\\) \\(1{,}780.80=1.484P\\) \\(1{,}200=P\\) The principal amount Lex used to open his account 6 years ago is $1,200.<\/p>\n<p><a class=\"class_names\" style=\"color: black;\" href=\"https:\/\/www.mometrix.com\/university\/mathcr\/?utm_source=academy&amp;utm_medium=inline&amp;utm_campaign=academy-mu-ads&amp;utm_content=mathcr-test\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-57671 size-full\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/imcr20-New.png\" alt=\"Click here for 20% off of Mometrix Math College Readiness Online Course. Use code: IMCR20\" width=\"728\" height=\"90\" \/><\/a><\/p>\n<\/div>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Dominic and William each started the year by depositing $1,000 into savings accounts at their banks that each offer 4% interest rate. Even though neither of them made additional deposits, at the end of the year, Dominic had $1,040 in his account, while William had $1,040.60. How is this possible?<\/p>\n<p>While Dominic and William both received a 4% interest rate, the fact that they use two different banks informs us that the interest on one of their savings accounts must compound more frequently than the other.<\/p>\n<p>In fact, it turns out that Dominic\u2019s account only collects simple interest, while William\u2019s account has interest that compounds quarterly.<\/p>\n<h2><span id=\"Calculating_Simple_Interest\" class=\"m-toc-anchor\"><\/span>Calculating Simple Interest<\/h2>\n<p>\nSimple interest can be calculated using the formula \\(i=Prt\\), which states that the interest is equal to the principal times the annual interest rate times the number of years. Using this formula, we can find that Dominic collects $40 after one year, bringing his balance to $1,040.<\/p>\n<p style=\"text-align: center\">\\(i = Prt\\)<\/p>\n<p style=\"text-align: center\">\\($40=$1,000\\times 0.04\\times 1\\)<\/p>\n<p><h2><span id=\"Calculating_Compound_Interest_(Example_1)\" class=\"m-toc-anchor\"><\/span>Calculating Compound Interest (Example 1)<\/h2>\n<p>\nBut how did William end up with 60 extra cents? Well, we know that he still gets a 4% interest rate, but his interest is calculated and paid quarterly, every three months. And whatever interest he gets one quarter, he is able to gain interest on <em>that<\/em> interest in the following quarters.<\/p>\n<p>This is what makes compound interest so interesting! Even if William never deposits anything else into his account, his interest will continue to grow larger and larger with each quarter. While slow at first, this forms exponential growth that becomes more and more powerful over time.<\/p>\n<h3><span id=\"Balance_After_Quarter_1\" class=\"m-toc-anchor\"><\/span>Balance After Quarter 1<\/h3>\n<p>\nLet\u2019s work through the arithmetic of it. We know that William started the year with a principal of $1,000 and receives a 4% interest rate.<\/p>\n<p style=\"text-align: center\">\\(P = 1,000\\)<\/p>\n<p style=\"text-align: center\">\\(r=0.04\\)<\/p>\n<p>Since his interest is compounded quarterly, he will receive a piece of that 4% interest every three months. At the end of the first quarter, William receives one-fourth of the 4%, which is 1%, applied to his current balance of $1,000. <\/p>\n<p>Then, this first-quarter interest is added to his starting balance. So, we can represent his balance at the end of the first quarter as \\(Q_1\\) being equal to \\(P+P(\\frac{r}{4})\\), or equivalently, \\(P(1+\\frac{r}{4})\\).<\/p>\n<p style=\"text-align: center\">\\(Q_1=P(1+\\dfrac{r}{4})=1,000(1+\\dfrac{0.04}{4})\\) \\(=1,000(1+0.01)=1,010\\)<\/p>\n<p>So after the first quarter, William\u2019s balance is $1,010.<\/p>\n<h3><span id=\"Balance_After_Quarter_2\" class=\"m-toc-anchor\"><\/span>Balance After Quarter 2<\/h3>\n<p>\nWhat will his balance be at the end of the second quarter?<\/p>\n<p>To determine that amount, we want to apply the one-fourth of the 4% interest to William\u2019s newest balance, \\(Q_1\\). So this time, we are going to have \\(Q_2\\), the balance at the end of the second quarter, equal to \\(Q_1\\) plus interest.<\/p>\n<p>In other words, we can basically reuse the formula for \\(Q_1\\), but we are going to write \\(Q_1\\) in place of \\(P\\) now, because \\(Q_1\\) is the new \u201cprevious balance.\u201d<\/p>\n<p style=\"text-align: center\">\\(Q_2=Q_1(1+\\dfrac{r}{4})=1,000(1+0.01)\\)<\/p>\n<p style=\"text-align: center\">\\(Q_2=1,010(1.01)\\)<\/p>\n<p style=\"text-align: center\">\\(Q_2=1,020.10\\)<\/p>\n<p>This would give us a new balance of $1,020.10.<\/p>\n<p>Before we move on to the third quarter balance, I want to point something out. When we wrote that \\(Q_2=Q_1(1+\\frac{r}{4})\\), we were equivalently saying that \\(Q_2=P(1+\\frac{r}{4})(1+\\frac{r}{4})\\) because we already defined \\(Q_1\\) as being equal to \\(P(1+\\frac{r}{4})\\).<\/p>\n<p>This expression can be slightly condensed to \\(Q_2=P(1+\\frac{r}{4})^2\\).<\/p>\n<h3><span id=\"Balance_After_Quarter_3\" class=\"m-toc-anchor\"><\/span>Balance After Quarter 3<\/h3>\n<p>\nAll right, let\u2019s look at the end of the third quarter. William\u2019s balance at the end of the third quarter, \\(Q_3,\\) should then be equal to \\(Q_2\\) plus interest.<\/p>\n<p>Since we just said that \\(Q_2\\) is equal to \\(P(1+\\frac{r}{4})^2\\), we can use that in the calculation of \\(Q_3\\).<\/p>\n<p style=\"text-align: center\">\\(Q_3=Q_2(1+\\dfrac{r}{4})\\) \\(=P(1+\\dfrac{r}{4})^2(1+\\dfrac{r}{4})\\) \\(=P(1+\\dfrac{r}{4})^3\\)<\/p>\n<p>We now have that \\(Q_3\\) is equal to the initial balance, \\(P\\), times the quarterly interest cubed.<\/p>\n<p>Plugging in William\u2019s starting balance of $1,000 and the interest rate, \\(r=0.04\\), we get that \\(Q_3=$1,030.30\\).<\/p>\n<p>In this quarter, the interest was greater than in the previous by an increment of ten cents.<\/p>\n<h3><span id=\"Balance_After_Quarter_4\" class=\"m-toc-anchor\"><\/span>Balance After Quarter 4<\/h3>\n<p>\nThe fourth-quarter balance, \\(Q_4\\), can be calculated by taking \\(Q_3\\) and adding on the rest of the interest for the year.<\/p>\n<p style=\"text-align: center\">\\(Q_4=Q_3(1+\\dfrac{r}{4})\\) \\(=P(1+\\dfrac{r}{4})^3(1+\\dfrac{r}{4})\\) \\(=P(1+\\dfrac{r}{4})^4\\)<\/p>\n<p>Plugging in the numerical values for \\(P\\) and \\(r\\), we get that William\u2019s ending balance is indeed $1,040.60.<\/p>\n<p>In the final quarter, his interest grew again, and in the end he had 60 cents more than Dominic. While 60 cents is relatively inconsequential, that difference would grow exponentially from year to year! In fact, after ten years, the difference would be over $80, and after twenty years the difference would be over $400!<\/p>\n<p>Through working that calculation, we saw that the balance at the end of each interest payment was equal to the starting principal times \\(1+\\frac{r}{4}\\) raised to whatever power corresponded to that payment. We can generalize this and say that the ending balance, which we\u2019ll call \\(A\\), after \\(t\\) years for an account with quarterly compounding interest is equal to \\(P(1+\\frac{r}{4})^{4t}\\), since there are four payments per year.<\/p>\n<p>But since some banks may compound interest only twice a year, or perhaps offer it twelve times a year, we can generalize this formula even further. If we let \\(n\\) equal the number of interest payments per year, then the ending balance of a compound interest savings account will be as follows:<\/p>\n<p style=\"text-align: center\">\\(A=P(1+\\frac{r}{n})^{nt}\\)<\/p>\n<h2><span id=\"Calculating_Compound_Interest_(Example_2)\" class=\"m-toc-anchor\"><\/span>Calculating Compound Interest (Example 2)<\/h2>\n<p>\nLet\u2019s use the generalized formula in another example.<\/p>\n<p>Marlise\u2019s savings account has $500 at the beginning of the year. Her bank pays 5% interest annually, which is compounded daily. What is her balance after five years if she does not deposit any more money?<\/p>\n<p>This is a bit of an extreme example, because it\u2019s very unlikely to find a bank that pays 5% annual interest and compounds daily. Let\u2019s work it out though and see what Marlise\u2019s ending balance would be.<\/p>\n<p>In this case, \\(P=500\\), \\(r=0.05\\), \\(n=365\\) (because the interest compounds daily), and \\(t=5 \\text{ years}\\).<\/p>\n<p>Plugging all of these into the compound interest formula, we get:<\/p>\n<p style=\"text-align: center\">\\(A=P(1+\\frac{r}{n})^{nt}\\) \\(=500(1+\\dfrac{0.05}{365})^{365\\times 5}\\)<\/p>\n<p>With the help of a calculator, we can determine that this comes out to $642.00. That\u2019s $142 in growth!<\/p>\n<h2><span id=\"Calculating_Compound_Interest_(Example_3)\" class=\"m-toc-anchor\"><\/span>Calculating Compound Interest (Example 3)<\/h2>\n<p>\nLet\u2019s work through one more example.<\/p>\n<p>For the past 50 years, Mrs. Baker has been collecting interest on an initial deposit. Her savings account pays 3% interest annually, and compounds monthly. If her balance right now is $8,946.62, how much was her initial deposit?<\/p>\n<p>This time, we know the ending balance, \\(A\\), but do not know the amount of the initial deposit, \\(P\\). We can solve for \\(P\\), though, by rearranging our compound interest formula.<\/p>\n<p style=\"text-align: center\">\\(A=P(1+\\frac{r}{n})^{nt}\\)<\/p>\n<p>Dividing both sides by \\((1+\\frac{r}{n})^{nt}\\), we can get \\(P\\) all by itself.<\/p>\n<p style=\"text-align: center\">\\(P=\\dfrac{A}{(1+\\frac{r}{n})^{nt}}\\)<\/p>\n<p>Now, let\u2019s plug in what we know. We know that Mrs. Baker has $8,946.62 as her ending amount, \\(A\\).<\/p>\n<p style=\"text-align: center\">\\(P=\\dfrac{8,946.62}{(1+\\frac{r}{n})^{nt}}\\)<\/p>\n<p>We also know that her annual interest rate r is equal to 3%, or 0.03, and that her interest compounds monthly, so \\(n=12\\).<\/p>\n<p style=\"text-align: center\">\\(P=\\dfrac{8,946.62}{(1+\\frac{0.03}{12})^{12(t)}}\\)<\/p>\n<p>Finally, we know that she\u2019s been collecting this interest for 50 years, so \\(t=50\\).<\/p>\n<p style=\"text-align: center\">\\(P=\\dfrac{8,946.62}{(1+\\frac{0.03}{12})^{12(50)}}\\)<\/p>\n<p>With the help of a calculator, we can determine that all this is equal to 2,000, so Mrs. Baker started her account with $2,000 fifty years ago, and it has now grown to nearly five times that much.<\/p>\n<p style=\"text-align: center\">\\(P=2,000.00\\)<\/p>\n<p>Now that we\u2019ve used the compound interest formula to find both the ending and starting balances of accounts, you may be unsurprised that it can also be used to solve for the annual interest rate, r, or t, the number of years passed!<\/p>\n<p>While the compound interest formula is certainly more complex than the simple interest formula, you can commit it to memory with just a few minutes of dedicated focus, and the calculations are quick if you have access to a calculator.<\/p>\n<p>Einstein is credited with saying that \u201ccompound interest is the most powerful force in the universe.\u201d With a little practice, you\u2019ll soon have this powerful force mastered!<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\">Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA teacher wants to invest $30,000 into an account that compounds annually. The interest rate at this bank is 1.8%. How much money will be in the account after 6 years?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">$43,389.35<\/div><div class=\"PQ\"  id=\"PQ-1-2\">$35,389.35<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">$33,389.35<\/div><div class=\"PQ\"  id=\"PQ-1-4\">$37,389.35<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Use the compound interest formula to solve this problem.<\/p>\n<p style=\"text-align:center;\">\\(A=P(1+\\frac{r}{n})^{nt}\\)<\/p>\n<p>From here, simply plug in each value and simplify in order to isolate the variable \\(A\\).<\/p>\n<p style=\"text-align:center; line-height: 45px;\">\n\\(A=30{,}000(1+\\frac{0.018}{1})^{1\u00d76}\\)<br \/>\n\\(A=30{,}000(1.018)^6\\)<br \/>\n\\(A=$33{,}389.35\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nAn investment earns 3% each year and is compounded monthly. Calculate the total value after six years from an initial investment of $5,000.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">$5,114.74<\/div><div class=\"PQ\"  id=\"PQ-2-2\">$4,984.74<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-3\">$5,984.74<\/div><div class=\"PQ\"  id=\"PQ-2-4\">$2,984.74<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Once again, use the compound interest formula to solve this problem.<\/p>\n<p style=\"text-align:center;\">\n\\(A=P(1+\\frac{r}{n})^{nt}\\)<\/p>\n<p>From here, simply plug in each value and simplify in order to isolate the variable \\(A\\).<\/p>\n<p style=\"text-align:center; line-height: 45px;\">\n\\(A=5,000(1+\\frac{0.03}{12})^{12\u00d76}\\)<br \/>\n\\(A=5,000(1.36)^{72}\\)<br \/>\n\\(A=$5{,}984.74\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nKristen wants to have $2,000,000 for retirement in 45 years. She invests in a mutual fund and pays 8.5% each year, compounded quarterly. How much should she deposit into the mutual fund initially?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">$47,421.08<\/div><div class=\"PQ\"  id=\"PQ-3-2\">$35,421.08<\/div><div class=\"PQ\"  id=\"PQ-3-3\">$43,421.08<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">$45,421.08<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Once again, use the compound interest formula to solve this problem.<\/p>\n<p style=\"text-align:center;\">\n\\(A=P(1+\\frac{r}{n})^{nt}\\)<\/p>\n<p>From here, simply plug in each value and simplify in order to isolate the variable \\(P\\).<\/p>\n<p style=\"text-align:center; line-height: 45px;\">\n\\(2,000,000=P(1+\\frac{0.085}{4})^{4\u00d745}\\)<br \/>\n\\(2,000,000=P(1.02125)^{180}\\)<br \/>\n\\(P=$45{,}421.08\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSean invests $50,000 into an index annuity that averages 6.5% per year, compounded semi-annually. After 9 years how much will be in his account?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-4-1\">$88,918.29<\/div><div class=\"PQ\"  id=\"PQ-4-2\">$89,918.29<\/div><div class=\"PQ\"  id=\"PQ-4-3\">$68,918.29<\/div><div class=\"PQ\"  id=\"PQ-4-4\">$81,918.29<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Once again, use the compound interest formula to solve this problem.<\/p>\n<p style=\"text-align:center;\">\n\\(A=P(1+\\frac{r}{n})^{nt}\\)<\/p>\n<p>From here, simply plug in each value and simplify in order to isolate the variable \\(A\\).<\/p>\n<p style=\"text-align:center; line-height: 45px;\">\n\\(A=50{,}000(1+\\frac{0.065}{2})^{2\u00d79}\\)<br \/>\n\\(A=50{,}000(1.0325)^{18}\\)<br \/>\n\\(A=$88{,}918.29\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nCalculate the interest rate for an account that started with $5,000 and now has $13,000 and has been compounded annually for the past 12 years.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">9.288%<\/div><div class=\"PQ\"  id=\"PQ-5-2\">11.288%<\/div><div class=\"PQ\"  id=\"PQ-5-3\">3.288%<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-4\">8.288%<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Once again, use the compound interest formula to solve this problem.<\/p>\n<p style=\"text-align:center;\">\n\\(A=P(1+\\frac{r}{n})^{nt}\\)<\/p>\n<p>From here, simply plug in each value and simplify in order to isolate the variable \\(r\\).<\/p>\n<p style=\"text-align:center; line-height: 45px;\">\n\\(13{,}000=5{,}000(1+\\frac{r}{1})^{1\u00d712}\\)<br \/>\n\\(13{,}000=5{,}000(1+\\frac{r}{1})^{12}\\)<br \/>\n\\(2.6=(1+\\frac{r}{1})^{12}\\)<\/p>\n<p>To get rid of the exponent 12, find the 12th root of both sides. This means raising both sides of the equation to the power of 112.<\/p>\n<p style=\"text-align:center; line-height: 45px;\">\n\\(2.6^{\\frac{1}{12}}=((1+\\frac{r}{1})^{12})^{\\frac{1}{12}}\\)<br \/>\n\\(1.08288=1+r\\)<\/p>\n<p>Subtract 1 from both sides.<\/p>\n<p style=\"text-align:center; line-height: 35px;\">\n\\(r=0.08288\\) (as a decimal)<br \/>\n\\(r=8.288\\%\\) (as a percent)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-i\/\">Return to Algebra I Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra I Videos<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-71096","1":"page","2":"type-page","3":"status-publish","5":"page_category-math-advertising-group","6":"page_category-math-videos","7":"page_type-video","8":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/71096","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=71096"}],"version-history":[{"count":9,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/71096\/revisions"}],"predecessor-version":[{"id":279301,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/71096\/revisions\/279301"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=71096"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}