How long will it take a sum of money to increase its value by 25% if it is invested at 5% per annum by simple interest?

Principal: The money which we deposit in or the lower from the bank or the money learned called the principal.

Rate of interest: The interest paid on $ 100 for one year is called the rate per cent per year or rate per cent per annum.

Time: The period of time for which the money is lent or invested.

Interest: Additional money paid by the borrowed to the lender for using the money is called interest.

Simple Interest: If the interest is calculated uniformly on the original principal throughout the lone period, it is called simple interest.

Amount: The total money paid back to the lender is called the amount.

Calculate Simple Interest

Formula to calculate Simple Interest?

If P denotes the principal ($), R denotes the rate (percentage p.a.) and T denotes time (years), then:-

S.I = (P × R × T)/100

R = (S.I × 100)/(P × T)

P = (S.I × 100)/(R × T)

T = (S.I × 100)/(P × R)

If the denotes the amount, then A = P + S.I

Note:

● When we calculated the time period between two dates, we do not could the day on which money is deposited but we count the day on which money is retuned.

● Time is always taken according to the per cent rat.

● For converting time in days into years, divide th number of days by 365 (for ordering or lap year.)

● For converting time in month into years, divide th number of month by 12 (for ordering or lap year.)

Examples to find or calculate simple interest when principal, rate and time are known

Calculate Simple Interest

Find the simple interest on:

(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.

Solution:

R = 5% p.a.

T = 3 years 4 months = 40/12 years = 10/3 years Therefore, S.I = (P × R × T)/100 = (900 × 5 × 10)/(100 × 3) = $ 150

Amount = P + S.I = $ 900 + $ 150 = $ 1050

(b) $ 1000 for 6 months at 4% per annum. Find the amount also.

Solution:

R = 4% p.a.

T = 6 months = 6/12 years S.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20 Therefore, A = P + I = $( 1000 + 20) = $ 1020

(c) $ 5000 for 146 days at 15¹/₂% per annum.

Solution:

(d) $ 1200 from 9ᵗʰ April to 21ˢᵗ June at 10% per annum.

Solution:

= 73 days [April = 21, May = 31, Jun = 21, 73 days]

= 73/365 years

S.I = (1200 × 10 × 73)/(100 × 365) = $ 24

Examples to find or calculate Time when Principal, S.I and Rate are known

Calculate Simple Interest

1. In how much time dose $ 500 invested at the rate of 8% p.a. simple interest amounts to $ 580.

Solution:

Therefore S.I = A - P = $ (580 - 500) = $ 80

Therefore T = (100 × S.I)/(P × R) = (100 × 80)/(500 × 3) = 2 years

2. In how many years will a sum of $ 400 yield an interest of $ 132 at 11% per annum?

Solution:

Calculate Simple Interest

3. In how many years will a sum double itself at 8 % per annum?

Solution:

So , S.I. = A - P = 2P – P = P

T = (100 × S.I)/(P × R) = ( 100 × P)/(P × 8) = 25/2 = 121/2 years

Calculate Simple Interest

4. In how many years will simple interest on certain sum of money at 6 1/4% Per annum be 5/8 of itself?

Solution:

T = 10 years

Examples to find or calculate Rate per cent when Principal, S.I. and Time are known

1. Find at what rate of interest per annum will $ 600 amount to $ 708 in 3 years.

Solution:

Calculate Simple Interest

2. Simple interest on a certain sum is 36/25 of the sum. Find the Rate per cent and time if they are both numerically equal.

Solution:

Calculate Simple Interest

3. At what rate per cent per annum will $ 6000 produce $ 300 as S.I. in 1 years?

Solution:

4. At what rate per cent per annum will a sum triple itself in 12 years ?

Solution:

Examples to find or calculate Principal when Rate, Time and S.I. are known

Calculate Simple Interest

1. What sum will yield $ 144 as S.I. in 21/2 years at 16% per annum?

Solution:

2. A some amount to $ 2040 in 21/2 years at , P = ?

Solution:

51x = 2040 × 40 ---- x = (2040 × 40)/51 = $ 1600

Calculate Simple Interest

3. A certain sum amounts to $ 6500in 2 years and to $ 8750 in 5 years at S.I. Find the sum and rate per cent per annum.

Solution:

4. Divide $ 6500 in to two parts , such that if one part is lent out at 9% per annum and other at 10% per annum, the total yearly income is $ 605

Solution:

But given that total S.I.= $ 605

So, (6500 - x)/100 =605 -----65000 - x = 60500

----- 65000 – 60500 = x ---- x = $ 4500

Now, second part = 6500 – x = 6500 – 4500 = $ 2000

Hence, first part = $ 2000 and second part = $ 4500

5. When the rate of interest in a bank is increased from 9% to 10% per annum; A person deposits $ 500 more into his account. If the annual interest now Received by him is $ 150more then before, find his original deposit.

Solution:

Then, S.I. on $ x for 1 year at (10 - 9 )% = 1% per annum + S.I. on $ 500

For I year at 10% per annum = $ 15

----- ( x × 1 × 1)/100 + ( 500 × 10 × 1)/100 = 150

----- x/(100 ) + 50 = 150 ---- x/(100 ) + 150 – 50 ----- x/(100 ) + 100

----- x = 100 × 100 = $ 10,000

Therefore, the original deposit is $ 10,000.

Calculate Simple Interest

● Simple Interest

What is Simple Interest?

Calculate Simple Interest

Practice Test on Simple Interest

● Simple Interest - Worksheets

Simple Interest Worksheet

7th Grade Math Problems

8th Grade Math Practice 

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Learning Outcomes

• Calculate one-time simple interest, and simple interest over time
• Determine APY given an interest scenario
• Calculate compound interest

We have to work with money every day. While balancing your checkbook or calculating your monthly expenditures on espresso requires only arithmetic, when we start saving, planning for retirement, or need a loan, we need more mathematics.

Simple Interest

Discussing interest starts with the principal, or amount your account starts with. This could be a starting investment, or the starting amount of a loan. Interest, in its most simple form, is calculated as a percent of the principal. For example, if you borrowed $100 from a friend and agree to repay it with 5% interest, then the amount of interest you would pay would just be 5% of 100: $100(0.05) = $5. The total amount you would repay would be $105, the original principal plus the interest.

[latex]\begin{align}&I={{P}_{0}}r\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}r={{P}_{0}}(1+r)\\\end{align}[/latex]

• I is the interest
• A is the end amount: principal plus interest
• [latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] is the principal (starting amount)
• r is the interest rate (in decimal form. Example: 5% = 0.05)

A friend asks to borrow $300 and agrees to repay it in 30 days with 3% interest. How much interest will you earn?

The following video works through this example in detail.

One-time simple interest is only common for extremely short-term loans. For longer term loans, it is common for interest to be paid on a daily, monthly, quarterly, or annual basis. In that case, interest would be earned regularly.

For example, bonds are essentially a loan made to the bond issuer (a company or government) by you, the bond holder. In return for the loan, the issuer agrees to pay interest, often annually. Bonds have a maturity date, at which time the issuer pays back the original bond value.

Suppose your city is building a new park, and issues bonds to raise the money to build it. You obtain a $1,000 bond that pays 5% interest annually that matures in 5 years. How much interest will you earn?

Further explanation about solving this example can be seen here.

We can generalize this idea of simple interest over time.

[latex]\begin{align}&I={{P}_{0}}rt\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}rt={{P}_{0}}(1+rt)\\\end{align}[/latex]

• I is the interest
• A is the end amount: principal plus interest
• [latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] is the principal (starting amount)
• r is the interest rate in decimal form
• t is time

The units of measurement (years, months, etc.) for the time should match the time period for the interest rate.

Interest rates are usually given as an annual percentage rate (APR) – the total interest that will be paid in the year. If the interest is paid in smaller time increments, the APR will be divided up.

For example, a 6% APR paid monthly would be divided into twelve 0.5% payments.
[latex]6\div{12}=0.5[/latex]

A 4% annual rate paid quarterly would be divided into four 1% payments.
[latex]4\div{4}=1[/latex]

Treasury Notes (T-notes) are bonds issued by the federal government to cover its expenses. Suppose you obtain a $1,000 T-note with a 4% annual rate, paid semi-annually, with a maturity in 4 years. How much interest will you earn?

This video explains the solution.

A loan company charges $30 interest for a one month loan of $500. Find the annual interest rate they are charging.

Compound Interest

With simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called compounding.

Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?

The 3% interest is an annual percentage rate (APR) – the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn [latex]\frac{3%}{12}[/latex]= 0.25% per month.

In the first month,

• P0 = $1000
• r = 0.0025 (0.25%)
• I = $1000 (0.0025) = $2.50
• A = $1000 + $2.50 = $1002.50

In the first month, we will earn $2.50 in interest, raising our account balance to $1002.50.

In the second month,

• P0 = $1002.50
• I = $1002.50 (0.0025) = $2.51 (rounded)
• A = $1002.50 + $2.51 = $1005.01

Notice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that compounding interest gives us.

Calculating out a few more months gives the following:

We want to simplify the process for calculating compounding, because creating a table like the one above is time consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if Pm represents the amount of money after m months, then we could write the recursive equation:

P0 = $1000

Pm = (1+0.0025)Pm-1

You probably recognize this as the recursive form of exponential growth. If not, we go through the steps to build an explicit equation for the growth in the next example.

Build an explicit equation for the growth of $1000 deposited in a bank account offering 3% interest, compounded monthly.

View this video for a walkthrough of the concept of compound interest.

While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If N is the number of years, then m = N k. Making this change gives us the standard formula for compound interest.

[latex]P_{N}=P_{0}\left(1+\frac{r}{k}\right)^{Nk}[/latex]

• PN is the balance in the account after N years.
• P0 is the starting balance of the account (also called initial deposit, or principal)
• r is the annual interest rate in decimal form
• k is the number of compounding periods in one year
  • If the compounding is done annually (once a year), k = 1.
  • If the compounding is done quarterly, k = 4.
  • If the compounding is done monthly, k = 12.
  • If the compounding is done daily, k = 365.

The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest. 

In the next example, we show how to use the compound interest formula to find the balance on a certificate of deposit after 20 years.

A certificate of deposit (CD) is a savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit $3000 in a CD paying 6% interest, compounded monthly. How much will you have in the account after 20 years?

A video walkthrough of this example problem is available below.

Let us compare the amount of money earned from compounding against the amount you would earn from simple interest

As you can see, over a long period of time, compounding makes a large difference in the account balance. You may recognize this as the difference between linear growth and exponential growth.

When we need to calculate something like [latex]5^3[/latex] it is easy enough to just multiply [latex]5\cdot{5}\cdot{5}=125[/latex].  But when we need to calculate something like [latex]1.005^{240}[/latex], it would be very tedious to calculate this by multiplying [latex]1.005[/latex] by itself [latex]240[/latex] times!  So to make things easier, we can harness the power of our scientific calculators.

Most scientific calculators have a button for exponents.  It is typically either labeled like:

^ ,   [latex]y^x[/latex] ,   or [latex]x^y[/latex] .

To evaluate [latex]1.005^{240}[/latex] we’d type [latex]1.005[/latex] ^ [latex]240[/latex], or [latex]1.005 \space{y^{x}}\space 240[/latex].  Try it out – you should get something around 3.3102044758.

You know that you will need $40,000 for your child’s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?

It is important to be very careful about rounding when calculating things with exponents. In general, you want to keep as many decimals during calculations as you can. Be sure to keep at least 3 significant digits (numbers after any leading zeros). Rounding 0.00012345 to 0.000123 will usually give you a “close enough” answer, but keeping more digits is always better.

To see why not over-rounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.

If we first compute r/k, we find 0.05/12 = 0.00416666666667

Here is the effect of rounding this to different values:

If you’re working in a bank, of course you wouldn’t round at all. For our purposes, the answer we got by rounding to 0.00417, three significant digits, is close enough – $5 off of $4500 isn’t too bad. Certainly keeping that fourth decimal place wouldn’t have hurt.

View the following for a demonstration of this example.

In many cases, you can avoid rounding completely by how you enter things in your calculator. For example, in the example above, we needed to calculate [latex]{{P}_{30}}=1000{{\left(1+\frac{0.05}{12}\right)}^{12\times30}}[/latex]

We can quickly calculate 12×30 = 360, giving [latex]{{P}_{30}}=1000{{\left(1+\frac{0.05}{12}\right)}^{360}}[/latex].

Now we can use the calculator.

Using your calculator continued

The previous steps were assuming you have a “one operation at a time” calculator; a more advanced calculator will often allow you to type in the entire expression to be evaluated. If you have a calculator like this, you will probably just need to enter:

1000 ×  ( 1 + 0.05 ÷ 12 ) yx 360 =

Solving For Time

Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.

Often we are interested in how long it will take to accumulate money or how long we’d need to extend a loan to bring payments down to a reasonable level.

If you invest $2000 at 6% compounded monthly, how long will it take the account to double in value?

Get additional guidance for this example in the following: