Skip to main content
Mathematics LibreTexts

8.1: Principal, Rate, Time

  • Page ID
    22110
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Your investments may be at risk if stock and bond markets slump, as a story in the Globe and Mail predicts. You wonder if you should shift your money into relatively secure short-term investments until the market booms again. You consider your high-interest savings account, but realize that only the first $60,000 of your savings account is insured. Perhaps you should put some of that money into treasury bills instead.

    Looking ahead, what income will you live on once you are no longer working? As your career develops, you need to save money to fund your lifestyle in retirement. Some day you will have $100,000 or more that you must invest and reinvest to reach your financial retirement goals.

    To make such decisions, you must first understand how to calculate simple interest. Second, you need to understand the characteristics of the various financial options that use simple interest. Armed with this knowledge, you can make smart financial decisions!

    The world of finance calculates interest in two different ways:

    1. Simple Interest. A simple interest system primarily applies to short-term financial transactions, with a time frame of less than one year. In this system, which is explored in this chapter, interest accrues but does not compound.
    2. Compound Interest. A compound interest system primarily applies to long-term financial transactions, with a time frame of one year or more. In this system, which the next chapter explores, interest accrues and compounds upon previously earned interest.

    Simple Interest

    In a simple interest environment, you calculate interest solely on the amount of money at the beginning of the transaction. When the term of the transaction ends, you add the amount of the simple interest to the initial amount. Therefore, throughout the entire transaction the amount of money placed into the account remains unchanged until the term expires. It is only on this date that the amount of money increases. Thus, an investor has more money or a borrower owes more money at the end.

    clipboard_e15d095435d30a0fe6d724eccbe584119.png

    The figure illustrates the concept of simple interest. In this example, assume $1,000 is placed into an account with 12% simple interest for a period of 12 months. For the entire term of this transaction, the amount of money in the account always equals $1,000. During this period, interest accrues at a rate of 12%, but the interest is never placed into the account. When the transaction ends after 12 months, the $120 of interest and the initial $1,000 are then combined to total $1,120.

    A loan or investment always involves two parties—one giving and one receiving. No matter which party you are in the transaction, the amount of interest remains unchanged. The only difference lies in whether you are earning or paying the interest.

    • If you take out a personal loan from a bank, the bank gives you the money and you receive the money. In this situation, the bank earns the simple interest and you are being charged simple interest on your loan. In the figure, this means you must pay back not only the $1,000 you borrowed initially but an additional $120 in interest.
    • If you place your money into an investment account at the bank, you have given the money and the bank has received the money. In this situation, you earn the simple interest on your money and the bank pays simple interest to your investment account. In the figure, this means the bank must give you back your initial $1,000 at the end plus an additional $120 of interest earned.

    The Formula

    The best way to understand how simple interest is calculated is to think of the following relationship:

    Amount of simple interest = How much at What simple interest rate for How long

    Notice that the key variables are the amount, the simple interest rate, and time. Formula 8.1 combines these elements into a formula for simple interest.

    clipboard_ea95119aba7fd1e1e112d708d9c18af2e.png

    Formula 8.1

    How It Works

    Follow these steps when you calculate the amount of simple interest:

    Step 1: Formula 8.1 has four variables, and you need to identify three for any calculation involving simple interest. If necessary, draw a timeline to illustrate how the money is being moved over time.

    Step 2: Ensure that the simple interest rate and the time period are expressed with a common unit. If they are not already, you need to convert one of the two variables to the same units as the other.

    Step 3: Apply Formula 8.1 and solve for the unknown variable. Use algebra to manipulate the formula if necessary.

    Assume you have $500 earning 3% simple interest for a period of nine months. How much interest do you earn?

    Step 1: Note that your principal is $500, or \(P=\$ 500\). The interest rate is assumed to be annual, so \(r=3 \%\) per year. The time period is nine months.

    Step 2: Convert the time period from months to years: \(t=\dfrac{9}{12}\).

    Step 3: According to Formula 8.1, \(I=\$ 500 \times 3 \% \times=\$ 11.25\). Therefore, the amount of interest you earn on the $500 investment over the course of nine months is $11.25.

    Important Notes

    Recall that algebraic equations require all terms to be expressed with a common unit. This principle remains true for Formula 8.1, particularly with regard to the interest rate and the time period. For example, if you have a 3% annual interest rate for nine months, then either

    • The time needs to be expressed annually as \(\dfrac{9}{12}\) of a year to match the yearly interest rate, or
    • The interest rate needs to be expressed monthly as \(\dfrac{3\%}{12} = 0.25\%\) per month to match the number of months.

    It does not matter which you do so long as you express both interest rate and time in the same unit. If one of these two variables is your algebraic unknown, the unit of the known variable determines the unit of the unknown variable. For example, assume that you are solving Formula 8.1 for the time period. If the interest rate used in the formula is annual, then the time period is expressed in number of years.

    clipboard_e60f1adb238a6a2ccc54349f78f18d084.png

    Paths To Success

    Four variables are involved in the simple interest formula, which means that any three can be known, requiring you to solve for the fourth missing variable. To reduce formula clutter, the triangle technique illustrated here helps you remember how to rearrange the formula as needed.

    Exercise \(\PageIndex{1}\): Give It Some Thought
    1. If you have a debt for which you haven’t made any payments, yet it is being charged simple interest on the principal, would you owe more or less money in the future?
    Answer
    1. More, because the interest is earned and therefore is added to your savings account.
    2. More, because you owe the principal and you owe the interest, which increases your total amount owing
    Example \(\PageIndex{1}\): How Much Interest Is Owed?

    Julio borrowed $1,100 from Maria five months ago. When he first borrowed the money, they agreed that he would pay Maria 5% simple interest. If Julio pays her back today, how much interest does he owe her?

    Solution

    Calculate the amount of interest that Julio owes Maria (\(I\)).

    What You Already Know

    Step 1:

    The terms of their agreement are as follows: \(P = \$1,100, r = 5\%, t = 5\) months

    How You Will Get There

    Step 2:

    The rate is annual, and the time is in months. Convert the time to an annual number.

    clipboard_ebfe5286cc936e016a353aa5f87b0932f.png

    Step 3:

    Apply Formula 8.1.

    Perform

    Step 2:

    Five months out of 12 months in a year is \(\dfrac{5}{12}\) of a year, or \(t=\dfrac{5}{12}\).

    Step 3:

    \[I=\$ 1,100 \times 5 \% \times \dfrac{5}{12}=\$ 1,100 \times 0.05 \times 0.41 \overline{6}=\$ 22.92 \nonumber \]

    For Julio to pay back Maria, he must reimburse her for the $1,100 principal borrowed plus an additional $22.92 of simple interest as per their agreement.

    Example \(\PageIndex{2}\): Do You Know Your Interest Rate?

    A $3,500 investment earned $70 of interest over the course of six months. What annual rate of simple interest did the investment earn?

    Solution

    Calculate the annual interest rate (\(r\)).

    What You Already Know

    Step 1:

    The principal, interest amount, and time are known: \(P = \$3,500, I = \$70, t = 6\) months

    How You Will Get There

    Step 2:

    The computed interest rate needs to be annual, so you must express the time period annually as well.

    clipboard_ebe9c2d62de5e7cf3c32175d7d2fd3936.png

    Step 3:

    Apply Formula 8.1, rearranging for \(r\).

    Perform

    Step 2:

    Six months out of 12 months in a year is \(\dfrac{6}{12}\) of a year, or \(t=\dfrac{6}{12}\).

    Step 3:

    \[\$ 70=\$ 3,500 \times r \times \dfrac{6}{12} \nonumber \]

    \[r=\dfrac{\$ 70}{\$ 3,500 \times \dfrac{6}{12}}=\dfrac{\$ 70}{\$ 1,750}=0.04 \text { or } 4 \% \nonumber \]

    For $3,500 to earn $70 simple interest over the course of six months, the annual simple interest rate must be 4%.

    Example \(\PageIndex{3}\): What Did You Start with?

    What amount of money invested at 6% annual simple interest for 11 months earns $2,035 of interest?

    Solution

    Calculate the amount of money originally invested, which is known as the present value or principal, symbolized by \(P\).

    What You Already Know

    Step 1:

    The interest rate, time, and interest earned are known: \(r = 6\%, t = 11\) months, \(I = \$2,035\)

    How You Will Get There

    Step 2:

    Convert the time from months to an annual basis to match the interest rate.

    clipboard_e39c7f886d1cfa026fd9c70fe4c62c5d8.png

    Step 3:

    Apply Formula 8.1, rearranging for \(P\).

    Perform

    Step 2:

    Eleven months out of 12 months in a year is \(\dfrac{11}{12}\) of a year, or \(t=\dfrac{11}{12}\).

    Step 3:

    \[\$ 2,035=P \times 6 \% \times \dfrac{11}{12} \nonumber \]

    \[P=\dfrac{\$ 2,035}{6 \% \times \dfrac{11}{12}}=\dfrac{\$ 2,035}{0.06 \times 0.91 \overline{6}}=\$ 37,000 \nonumber \]

    To generate $2,035 of simple interest at 6% over a time frame of 11 months, $37,000 must be invested.

    Example \(\PageIndex{4}\): how long?

    For how many months must $95,000 be invested to earn $1,187.50 of simple interest at an interest rate of 5%?

    Solution

    Calculate the length of time in months (\(t\)) that it takes the money to acquire the interest.

    What You Already Know

    Step 1:

    The amount of money invested, interest earned, and interest rate are known: \(P = \$95,000.00, I = \$1,187.50, r = 5\%\)

    How You Will Get There

    Step 2:

    Express the time in months. Convert the interest rate to a "per month" format.

    clipboard_e456746c1c2c00b4c76fffd9d68b1abc2.png

    Step 3: Apply Formula 8.1, rearranging for \(t\).

    Perform

    Step 2:

    5% per year converted into a monthly rate is \(r=\dfrac{0.05}{12}\)

    Step 3:

    \[\$ 1,187.50=\$ 95,000 \times \dfrac{0.05}{12} \times t \nonumber \]

    \[t=\dfrac{\$ 1,187.50}{\$ 95,000 \times \dfrac{0.05}{12}}=\dfrac{\$ 1,187.50}{\$ 95,000 \times 0.0041 \overline{6}}=3 \text { months } \nonumber \]

    For $95,000 to earn $1,187.50 at 5% simple interest, it must be invested for a three-month period.

    Time and Dates

    In the examples of simple interest so far, the time period equaled an exact number of months. While this is convenient in many situations, financial institutions and organizations calculate interest based on the exact number of days in the transaction, which changes the interest amount.

    To illustrate this, assume you had money saved for the entire months of July and August, where \(t=\dfrac{2}{12}\) or \(t=0.1\) of a year. However, if you use the exact number of days, the 31 days in July and 31 days in August total 62 days. In a 365-day year that is \(t=\frac{62}{365} \text { or } t=0.169863\) of a year. Notice a difference of 0.003196 occurs. Therefore, to be precise in performing simple interest calculations, you must calculate the exact number of days involved in the transaction.

    Figuring Out the Days

    In practice, when counting the number of days in a transaction you include the first day but not the last day. This is because interest is calculated on the daily closing balance, not on interim balances throughout a day. For example, if you borrowed $500 on June 3 and paid it back with interest on June 5, the closing balance on June 3 and June 4 is $500. However, on June 5 a zero balance is restored. Therefore, you count June 3 and 4 as days of interest but not June 5, so you will owe two days of interest, not three.

    There are three common ways to calculate the number of days involved in a transaction.

    Method 1: Use Your Knuckles

    You can use your knuckles as an aid to remembering which months of the year have more or fewer days.

    clipboard_ee3fba98a0056daa26db6cab7dccf3be8.png

    Put your knuckles side by side and start on your outermost knuckle. Every knuckle represents a month with 31 days. Every valley between the knuckles represents a month that does not have 31 days (30 in all except February, which has 28 or 29). Now start counting months. Your first knuckle (pinkie finger) is January, which has 31 days. Next is your first valley, which is February and it does not have 31 days. Your next knuckle (ring finger) is March, which has 31 days … and so on. When you get to your last knuckle (your index finger), move to the other hand’s first knuckle (an index finger again). Notice that July and August are the two months back to back with 31 days.

    If you are trying to calculate the number of days between March 20 and May 4, use your knuckles to recall that March has 31 days and April has 30 days. The calculations are illustrated in the table here.

    Start Date End Date Days between Dates
    March 20 March 31 31 − 20 = 11 days
    March 31 (which precedes April 1, so it is day "0" for April) April 30 30 − 0 = 30 days
    April 30 (which precedes May 1, so it is day "0" for May) May 4 4 − 0 = 4 days
    TOTAL 11 + 30 + 4 = 45 days

    Why does the end date of one line become the start date of the next line? Recall that you count the first day, but not the last day. Therefore, on the first line March 31 has not been counted yet and must be counted on the second line. Ultimately a transaction extending from March 20 to May 4 involves 45 days as the time period.

    Method 2: Use A Table of Serial Numbers

    In the table on the next page, the days of the year are assigned a serial number. The number of days between any two dates in the same calendar year is simply the difference between the serial numbers for the dates.

    Using the example in Method 1, when the dates are within the same calendar year, find the serial number for the later date, May 4 (Day 124) and the serial number for the earlier date, March 20 (Day 79). Then, find the difference between the two serial numbers (\(124-79=45\)days).

    clipboard_ef58e1c67ba54a04e3baac4671ff4aa9c.png

    *For Leap years, February 29 becomes Day 60 and the serial number for each subsequent day is increased by 1.

    Method 3: Use The BAII Plus Date Function

    Your BAII Plus calculator is programmed with the ability to calculate the number of days involved in a transaction according to the principle of counting the first day but not the last day. The function "Date" is located on the second shelf above the number one. To use this function, open the window by pressing 2nd 1.

    clipboard_e3685eea0efad7992134d47c79835a307.png

    • Scroll amongst the four display lines by using your \(\uparrow\) and \(\downarrow\) arrows. There are four variables:
      1. DT1 is the starting date of the transaction.
      2. DT2 is the ending date of the transaction.
      3. DBD is the days between dates, which is your \(t\) in Formula 8.1.
      4. ACT or 360 is the method of calculating the days between dates. In ACT (for “actual”), the calculator calculates the DBD based on the correct number of days in the month. In 360, the DBD is based on treating every month as having 30 days (\(30 \times 12=360\) days). For all calculations, set the calculator to the ACT mode. Toggle this setting by pressing 2nd SET.
    • To enter a date into the calculator, the calculator uses the format of MM.DDYY where:
      • MM is the numerical value of the month; you do not need to enter the leading zero on months with only one digit.
      • DD is the day of the month; you must always enter this value with two digits using a leading zero for days 1 through 9 in a month.
      • YY is the last two digits of the year; you must always enter this with two digits. If your question does not involve a year, select an appropriate year of your choice.
    • As long as you know any two of the DT1, DT2, or DBD variables, you can calculate the third. Enter any two of the three variables, pressing ENTER after each and ensuring the window shows the output variable you are seeking, and then press CPT.

    Using the example in Method 1, to calculate the time from March 20 to May 4 you first need to choose a year arbitrarily, perhaps 2011. Therefore, perform the following sequence:

    \[\text{2nd DATE}\nonumber \]

    \[\text{2nd CLR Work (to erase previous calculations)}\nonumber \]

    \[3.2011 \text{ENTER }\downarrow \text{(this is March 20, 2011)}\nonumber \]

    \[5.0411 \text{ENTER }\downarrow \text{(this is May 4, 2011)}\nonumber \]

    \[\text{CPT}\nonumber \]

    \[\text{Answer: }45\nonumber \]

    Note that the answer is the same as the one you arrived at manually using your knuckles.

    Important Notes

    When solving for \(t\), decimals may appear in your solution. For example, if calculating \(t\) in days, the answer may show up as 45.9978 or 46.0023 days; however, interest is calculated only on complete days. This occurs because the interest amount (\(I\)) used in the calculation has been rounded off to two decimals. Since the interest amount is imprecise, the calculation of \(t\) is imprecise. When this occurs, round \(t\) off to the nearest integer.

    Things To Watch Out For

    Miscalculating the number of days in February has to be one of the most common errors. To help you remember how many days are in February, recall your knowledge of leap years.

    • February has 28 days except for leap years, in which it has 29 days.
    • Leap years are those years that are evenly divisible by four or 400. However, those years that are evenly divisible by 100 (excepting years that are evenly divisible by 400) are not leap years. Here are some examples to illustrate this rule:
      1. The year 1900 was not a leap year, since it is evenly divisible by 100 and not by 400.
      2. The year 1996 was a leap year, since it is evenly divisible by four and not by 100.
      3. The year 2000 was one of those exception years, and it was a leap year since it is evenly divisible by 400.
      4. The year 2004 was once again a leap year since it is evenly divisible by four and not by 100. Thus, in this century 2004, 2008, 2012, 2016, 2020, 2024, and so on by fours are all leap years with 29 days.
      5. The year 2100 will not be a leap year, since it is evenly divisible by 100.

    Note that when it comes to the total number of days in a year, simple interest calculations ignore the 366th day in a leap year. Therefore, assume a year has 365 days in all calculations.

    Paths To Success

    When entering two dates either on the calculator or in Excel, the order in which you key in the dates is not important. If you happen to put the last day of the transaction into the first field (DT1 or Start Date) and the start day of the transaction into the last field (DT2 or End Date), then the number of days will compute as a negative number since the dates are reversed. Ignore the negative sign in these instances. Using the example, on your calculator if you had keyed May 4 into DT1 and March 20 into DT2, the DBD computes as −45, in which you ignore the negative sign to determine the answer is 45 days.

    Example \(\PageIndex{5}\): Time Using Dates

    Azul borrowed $4,200 from Hannah on November 3, 2011. Their agreement required Azul to pay back the loan on April 14, 2012, with 8% simple interest. In addition to the principal, how much interest does Azul owe Hannah on April 14?

    Solution

    Calculate the amount of simple interest (\(I\)) that Azul owes Hannah on his loan.

    What You Already Know

    Step 1:

    The principal, simple interest rate, and dates are known: \(P = \$4,200, r = 8\%\); Start Date = November 3, 2011; End Date = April 14, 2012

    How You Will Get There

    Step 1 (continued):

    Calculate the number of days in the transaction. Note that this transaction involves February 2012, which is a leap year.

    Step 2:

    The rate is annual and the time is in days. Convert the time to an annual number.

    clipboard_e3242294930554a518c001400a2b3b41e.png

    Step 3:

    Apply Formula 8.1.

    Perform

    Step 1 (continued):

    Start Date End Date Days between Dates
    November 3, 2011 November 30, 2011 30 − 3 = 27
    November 30, 2011 December 31, 2011 31 − 0 = 31
    December 31, 2011 January 31, 2012 31 − 0 = 31
    January 31, 2012 February 29, 2012 29 − 0 = 29
    February 29, 2012 March 31, 2012 31 − 0 = 31
    March 31, 2012 April 14, 2012 14 − 0 = 14
    TOTAL 27 + 31 + 31 + 29 + 31 + 14 = 163 days

    Step 2:

    163 days out of 365 days in a year is \(t=\dfrac{163}{365}\)

    Step 3:

    \[I=\$ 4,200 \times 8 \% \times \dfrac{163}{365}=\$ 4,200 \times 0.08 \times 0.446575=\$ 150.05 \nonumber \]

    To get the number of days, use the DATE function:

    DT1 DT2 DBD Mode
    11.0311 4.1412 Answer: 163 ACT

    Azul borrowed the money for a period of 163 days and owes Hannah $150.05 of simple interest on his loan, in addition to repaying the principal of $4,200.

    Example \(\PageIndex{6}\): Figuring Out Dates Using Time

    On September 13, 2011, Aladdin decided to pay back the Genie on his loan of $15,000 at 9% simple interest. If he paid the Genie the principal plus $1,283.42 of interest, on what day did he borrow the money from the Genie?

    Solution

    To determine the starting date, the time involved in this transaction, or \(t\), must be calculated.

    What You Already Know

    Step 1:

    The principal, interest amount, simple interest rate, and end date are known: \(P = \$15,000.00, I = \$1,283.42, r = 9\%\); End Date = September 13, 2011

    How You Will Get There

    Step 2:

    The time is in days, but the rate is annual. Convert the rate to a daily rate.

    clipboard_e1b985d3e2ea4f3f6cb9dfed5b298a5df.png

    Step 3:

    Apply Formula 8.1, rearranging for \(t\).

    Step 3 (continued):

    Using the End Date, use the t to determine the Start Date.

    Perform

    Step 2:

    Convert the interest rate to daily: \(r=\dfrac{9 \%}{365}\).

    Step 3:

    \[t=\dfrac{\$ 1,283.42}{\$ 15,000 \times \dfrac{0.09}{365}}=346.998741 \text { days } \rightarrow 347 \text { days } \nonumber \]

    Step 3 (continued):

    The number of days is close to a year. Go back 365 days and start at September 13, 2010. Notice that 347 days is 365 − 347 = 18 days short of a full year. Adding 18 days onto September 13, 2010, equals October 1, 2010. This is the start date.

    Calculator Instructions

    To calculate the start date, use the DATE function:

    DT1 DT2 DBD Mode
    Answer: 10-01-2010 9.1311 347 ACT

    If Aladdin owed the Genie $1,283.42 of simple interest at 9% on a principal of $15,000, he must have borrowed the money 347 days earlier, which is October 1, 2010.

    Variable Interest Rates

    Not all interest rates remain constant. There are two types of interest rates:

    1. Fixed. A fixed interest rate is an interest rate that is unchanged for the duration of the transaction.
    2. Variable. A variable interest rate is an interest rate that is open to fluctuations over the duration of a transaction. This variable interest rate is usually tied to the prime rate, which is an interest rate set by the Bank of Canada that usually forms the lowest lending rate for the most secure loans. Most people and companies are usually charged the prime rate plus a percentage (typically 0.5% to 5%) to arrive at the variable interest rate. The Bank of Canada periodically adjusts this rate because of economic and financial considerations in Canada.

    In all of the previous examples in this section, the interest rate (\(r\)) was fixed for the duration of the transaction. When \(r\) fluctuates, you must break the question into time periods or time fragments for each value of \(r\). In each of these time fragments, you calculate the amount of simple interest earned or charged and then sum the interest amounts to form the total interest earned or charged.

    clipboard_e94ce12b15f08f10e364fcf0f48a35d10.png

    This figure illustrates this process with a variable rate that changes twice in the course of the transaction. To calculate the total interest amount, you need to execute Formula 8.1 for each of the time fragments. Then total the interest amounts from each of the three time fragments to calculate the interest amount (\(I\)) for the entire transaction.

    Example \(\PageIndex{7}\): A Floating Rate

    Julio borrowed $10,000 on May 17 at a variable interest rate of prime + 3% when the prime rate was 2¼%. On June 22 the prime rate increased by ¼%, and on September 2 it increased again by ½%. How much interest will Julio pay when he repays his loan on October 4?

    Solution

    Calculate the amount of interest that Julio owes for the entire time frame of his loan, or \(I\), from May 17 to October 4.

    What You Already Know

    Step 1:

    There is an initial interest rate and then it changes twice over the course of the transaction. That means there are three time periods and three different interest rates. The principal is also known: \(P = \$10,000\).

    Date Interest Rate
    May 17 to June 22 2¼% + 3% = 5¼%
    June 22 to September 2 5¼% + ¼% = 5½%
    September 2 to October 4 5½% + ½% = 6%

    How You Will Get There

    Step 1 (continued):

    Calculate the number of days for each of the three time periods.

    Step 2:

    The rates are annual while each of the times are in days. Convert each of the times to an annual number.

    clipboard_edd5a2a89a550aa9d6a3e684c8631a662.png

    Step 3:

    For each time period, apply Formula 8.1.

    Step 3 (continued):

    Sum the three interest amounts to arrive at the total interest paid.

    Dates Calculations Total # of Days Convert \(t\) to annual Calculate interest
    Step 1 (continued) Step 2 Step 3
    May 17 to June 22 May 17 to May 31 = 14 days May 31 to June 22 = 22 days 14 + 22 = 36 days \(\dfrac{36}{365}\) \(\begin{aligned}
    I_{1}&=\$ 15,000(0.0525)\left(\dfrac{36}{365}\right) \\
    &=\$ 77.671232
    \end{aligned}\)
    June 22 to September 2 June 22 to June 30 = 8 days June 30 to July 31 = 31 days July 31 to August 31 = 31 days August 31 to Sept 2 = 2 days
    8 + 31 + 31 + 2 = 72 days
    \(\dfrac{72}{365}\) \(\begin{aligned}
    I_{2}&=\$ 15,000(0.055)\left(\dfrac{72}{365}\right) \\
    &=\$ 162.739726
    \end{aligned}\)
    September 2 to October 4 Sept 2 to Sept 30 = 28 days Sept 30 to October 4 = 4 days 28 + 4 = 32 \(\dfrac{32}{365}\) \(\begin{aligned}
    I_{3}&=\$ 15,000(0.06)\left(\dfrac{32}{365}\right) \\
    &=\$ 78.904109
    \end{aligned}\)
    Step 3 (continued): TOTAL $319.32

    Calculator Instructions

    Arbitrarily pick the year 2011 and use the DATE function:

    Time Fragment DT1 DT2 DBD Mode
    1 5.1711 6.2211 Answer: 36 ACT
    2 6.2211 9.0211 Answer: 72 ACT
    3 9.0211 10.0411 Answer: 32 ACT

    Julio owes $319.32 of simple interest on his October 4 repayment date.

    Contributors and Attributions


    This page titled 8.1: Principal, Rate, Time is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jean-Paul Olivier via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.