Math 7 · 7.RP.A.3

Simple interest

Money left with a bank or lent to someone comes back with something extra. How much extra, and how do you work it out?

You need

Two percent changes one after the otherMath 7

Where you are 3 in Percent in money situations

Fractions, ratio and percent3

Percent8

Percent in money situations3

Markup and margin Discounts, sales tax, tips and commission

Simple interest

When you reach for it

Money can be rented. Put it in a savings account and the bank uses it, so the bank pays you for the use. Borrow it to buy a laptop or a bike and you pay the lender for the use. The payment for the use of money is called interest, and it is set as a percent of the money, for each year the money is out.

Reach for the simple interest formula whenever a sum of money is lent or deposited at a stated rate, the interest is worked out on the original sum only, and you want to know one of four things: how much interest, how much in total, what rate, or how long. It is the most direct use of a percent of a quantity there is, repeated once for every year.

The rates in this lesson are examples chosen to make the arithmetic clear. Real rates vary from lender to lender and from year to year.

The lesson runs over two days. Part 1 names the five quantities, builds the formula I = Prt and finds the total. Part 2 runs the formula backwards to find a missing rate, time or starting sum, handles rates given per month, and sets simple interest beside compound interest.

The procedure

Part 1: five words

Every interest problem has the same cast.

The principal, P, is the sum of money deposited or borrowed at the start, in dollars.
The rate, r, is the percent of the principal paid as interest for each year. A rate of 3\% per year means 3\% of the principal for every year. In the formula it is written as a decimal or fraction: 3\% = 0.03.
The time, t, is how long the money is out, in years. Eight months is \displaystyle \frac{8}{12} = \frac{2}{3} of a year.
The interest, I, is the extra money, in dollars. A saver earns it; a borrower owes it.
The total amount, A, is the principal and the interest together: what the saver has at the end, or what the borrower must pay back.

Simple interest is interest worked out on the principal only. The interest is never added to the principal to earn interest of its own; that other arrangement is compound interest, in Part 2.

Part 1: why the formula is I = Prt

Take 600 dollars in a savings account paying 3\% per year simple interest. One year’s interest is 3\% of 600 dollars:

600 \times 0.03 = 18

so 18 dollars. In the second year the interest is worked out on the principal again, which is still 600 dollars, so the second year also brings 18 dollars. So does the third, and every year after.

Figure 1Simple interest on 600 dollars at 3\% per year. Every year adds the same 18 dollars, because every year’s interest is 3\% of the same 600 dollars.

Each year adds one year’s interest, P \times r, and every year adds the same amount. Over t years that amount is added t times, and adding the same thing t times is multiplying it by t:

I = P \times r \times t = Prt

The order of the three factors does not matter, since multiplication is commutative and associative; Prt is the usual way to write it.

The same reasoning covers part of a year. Interest builds up evenly through the year, so half a year earns half a year’s interest and \displaystyle \frac{2}{3} of a year earns \displaystyle \frac{2}{3} of it. With P and r fixed, the interest is proportional to the time, and I = Prt still holds with t a fraction.

Two things must match. The rate is a rate per year, so the time must be in years. And the rate goes in as a decimal, because “3\% of” means “0.03 times”.

Part 1: the total amount

The total at the end is what was there at the start plus the interest:

A = P + I = P + Prt

Both terms on the right have the factor P, so the distributive property lets you take it out:

A = P(1 + rt)

The second form does the job in one multiplication. For the 600 dollars at 3\% over 4 years, rt = 0.03 \times 4 = 0.12, so A = 600 \times 1.12 = 672. The factor 1 + rt says “the whole principal, plus rt of it again”, which is a percent increase multiplier for an increase of 12\%.

Part 2: finding a missing rate, time or principal

I = Prt connects four quantities. Know any three and the fourth is fixed. To find it, substitute the three you know, multiply the known numbers together, and divide both sides of the equation by that product, which keeps the equation true. This is solving for one variable, done with numbers in place of the other letters. Solved once with letters, the three versions are

r = \frac{I}{Pt} \qquad t = \frac{I}{Pr} \qquad P = \frac{I}{rt}

but there is no need to memorize them: substituting and dividing gets there every time.

A problem may give the total paid back instead of the interest. Subtracting the principal, I = A - P, turns it back into an interest problem. If the principal is the unknown, use A = P(1 + rt) instead and divide by 1 + rt.

A rate found this way comes out as a decimal per year. Multiply by 100 to write it as a percent.

Part 2: rates per month and rates per year

Some lenders quote a rate per month. The formula is the same with every quantity in months: interest is principal times rate per month times number of months. The rule is the one from Part 1: the time is counted in the same unit the rate is quoted for.

Comparing a monthly rate with a yearly one needs them in the same unit. Under simple interest a year is twelve months of the same interest, so

\text{rate per year} = 12 \times \text{rate per month}

1\% per month is 12\% per year, and 18\% per year is 1.5\% per month.

Part 2: simple against compound

Most savings accounts work differently. With compound interest each year’s interest is added to the balance, and the next year’s interest is worked out on that larger balance: interest is earned on interest. Each year the balance is multiplied by 1 + r, so after t years it is P(1 + r)^t. That is the same change repeated, exactly as in two percent changes one after the other.

Here is 2{,}000 dollars at 6\% per year, both ways, with compound totals rounded to the nearest cent.

Year	Simple interest total (dollars)	Compound interest total (dollars)
1	2{,}120	2{,}120.00
2	2{,}240	2{,}247.20
3	2{,}360	2{,}382.03
5	2{,}600	2{,}676.45
10	3{,}200	3{,}581.70
20	4{,}400	6{,}414.27

After one year the two are equal, since both have earned 6\% of 2{,}000 dollars and nothing else. In the second year the compound account earns 6\% of 2{,}120 dollars, which is 127.20 dollars, against 120 dollars for simple interest. The extra 7.20 dollars is 6\% of the first year’s 120 dollars of interest. From then on the gap widens every year.

Figure 22{,}000 dollars at 6\% per year. Simple interest grows by the same amount every year, a straight line. Compound interest grows by the same multiplier every year, and bends away upward.

Simple interest adds the same amount each year, so its graph is a straight line. Compound interest multiplies by the same factor each year, so its graph curves upward. Working out compound interest in full, including interest added monthly or daily, is its own lesson.

Worked examples

Examples 1 and 2 go with Part 1. Examples 3 to 7 go with Part 2.

Example 1

Interest on a savings account

Adaeze deposits 600 dollars in a savings account that pays 3\% per year simple interest. She leaves it there for 4 years and adds nothing. Name the principal, rate and time, and find the interest and the total in the account at the end.

Solution

The principal is the sum deposited, P = 600. The rate per year as a decimal is r = 0.03. The time in years is t = 4. Substitute into I = Prt:

I = 600 \times 0.03 \times 4 = 18 \times 4 = 72

The interest is 72 dollars. The total is the principal plus the interest, 600 + 72 = 672, so 672 dollars.

Check. One year’s interest is 3\% of 600 dollars, which is 18 dollars, and four equal years make 72 dollars. These are the balances in the time line above.

Example 2

A time in months

Cressida puts 1{,}500 dollars into a 9-month certificate of deposit, an account that pays a fixed rate in return for leaving the money untouched until the end. It pays 4.2\% per year simple interest. How much interest does she earn, and what does she collect at the end?

Solution

The rate is per year, so the time must be in years:

t = \frac{9}{12} = 0.75

Then P = 1500, r = 0.042, and

I = 1500 \times 0.042 \times 0.75 = 63 \times 0.75 = 47.25

She earns 47.25 dollars and collects 1500 + 47.25, which is 1{,}547.25 dollars.

Check. A full year would earn 1500 \times 0.042 = 63 dollars. Nine months is three quarters of a year, and three quarters of 63 dollars is 47.25 dollars.

Example 3

Finding the rate on a family loan

Bartholomew borrows 1{,}200 dollars from his aunt. After 3 years he pays her back the 1{,}200 dollars and 126 dollars of interest, as they agreed. What yearly simple interest rate did they agree on?

Solution

Here I = 126, P = 1200 and t = 3, and r is unknown. Substitute and multiply the known numbers:

\begin{aligned} 126 &= 1200 \times r \times 3 \\ 126 &= 3600r \end{aligned}

Divide both sides by 3600:

r = \frac{126}{3600} = 0.035

As a percent that is 3.5\% per year.

Check. 3.5\% of 1{,}200 dollars is 42 dollars a year, and three years of 42 dollars is 126 dollars.

Example 4

Finding the time

Yevgenia has 2{,}500 dollars in a savings account paying 2.4\% per year simple interest.

(a) How long until the interest reaches 210 dollars?

(b) How long until the account holds 2{,}800 dollars?

Solution

(a) I = 210, P = 2500, r = 0.024, and t is unknown:

\begin{aligned} 210 &= 2500 \times 0.024 \times t \\ 210 &= 60t \\ t &= 3.5 \end{aligned}

It takes 3.5 years, three and a half. The 60 in the middle line has a meaning of its own: it is one year’s interest in dollars, so the question was really “how many years of 60 dollars make 210 dollars?”

(b) A total of 2{,}800 dollars means interest of 2800 - 2500 = 300 dollars. One year brings 60 dollars, so 60t = 300 and t = 5. It takes 5 years.

Example 5

Finding the principal from the total

Cressida’s grandfather bought a savings bond that paid 3\% per year simple interest. After 6 years it was worth 944 dollars, interest included. What did he pay for it?

Solution

The 944 dollars is a total, and the principal is the unknown, so use A = P(1 + rt). First the factor:

1 + rt = 1 + 0.03 \times 6 = 1.18

so 944 = 1.18P. Divide both sides by 1.18:

P = \frac{944}{1.18} = 800

He paid 800 dollars.

Why divide, and not take a percent of 944? The 18\% of interest is 18\% of the principal, so the 944 dollars is 118\% of the principal: 944 = 1.18P. The total was made by multiplying the principal by 1.18, and dividing by 1.18 is what undoes that multiplication.

Check. 800 \times 0.18 = 144, and 800 + 144 = 944.

Example 6

A monthly rate against a yearly rate

Dagfinn needs 480 dollars for a used guitar and will repay it in one payment after 10 months. One lender offers 1.25\% per month simple interest. Another offers 14\% per year simple interest. Which costs him less, and by how much?

Solution

The first rate is per month, so count the time in months:

I = 480 \times 0.0125 \times 10 = 6 \times 10 = 60

The second rate is per year, so count the time in years, \displaystyle \frac{10}{12} of a year:

I = 480 \times 0.14 \times \frac{10}{12} = 67.2 \times \frac{10}{12} = 56

The yearly offer costs 56 dollars and the monthly one 60 dollars, so the yearly offer is cheaper by 4 dollars.

The rates explain why. 1.25\% per month is 12 \times 1.25\% = 15\% per year, which is more than 14\%. Once both are in the same unit, the comparison can be read straight off the rates, for any amount and any time.

Example 7

Simple against compound, year by year

Zubair has 3{,}000 dollars to save for 3 years. One account pays 5\% per year simple interest. Another pays 5\% per year compound interest, added at the end of each year. Find the total in each after 3 years.

Solution

Simple interest is 5\% of 3{,}000 dollars, which is 150 dollars, every year:

A = 3000(1 + 0.05 \times 3) = 3000 \times 1.15 = 3450

Compound interest is worked out on the balance at the start of each year:

Year	Balance at the start (dollars)	Interest, 5\% of it (dollars)	Balance at the end (dollars)
1	3{,}000.00	150.00	3{,}150.00
2	3{,}150.00	157.50	3{,}307.50
3	3{,}307.50	165.375	3{,}472.875

Rounded to the nearest cent, the compound account ends at 3{,}472.88 dollars. The same number comes from one multiplication by the repeated multiplier, 3000 \times 1.05^3 = 3000 \times 1.157625.

The compound account ends 22.88 dollars ahead. Year 1 is identical. In year 2 it earns 7.50 dollars more, which is 5\% of year 1’s 150 dollars of interest. In year 3 it earns 15.375 dollars more, interest on two years of earlier interest.

Boundaries

The rate and the time must use the same unit. A rate per year needs the time in years; a rate per month needs it in months. Mixing them multiplies the interest by 12 or divides it by 12.

The rate goes in as a decimal or a fraction, not as the percent number. 4\% is 0.04. Putting in 4 makes the interest a hundred times too large.

Simple interest means the principal never changes. The formula covers one fixed sum left for the whole time. A deposit, a withdrawal, a part repayment, or interest being added to the balance all change the principal, and then each stretch of time with its own principal needs its own calculation.

Most real savings accounts and many loans compound. For them I = Prt is exact over the first period only, and after that it gives too little. Simple interest is still used for some short loans, some bonds and some loans between family or friends.

Months as twelfths of a year is one convention. Some lenders count the time in days and divide by 365, or by 360. The formula stays the same; only the fraction of a year changes, and so the interest can come out slightly different.

Finding a rate, time or principal means dividing, so the other quantities cannot be zero. If no time passes, or there is no principal, there is no interest, and nothing can be learned about the rate.

Money is rounded to the cent at the end. Rounding a rate or a part-year time early can move the answer by more than a cent.

Where this goes next

I = Prt is a formula with four letters, and running it backwards in Part 2 is a first case of solving a formula for one of its variables, a move used across algebra and science.

With the principal and rate fixed, the total A = P + (Pr)t grows by the same amount for every year, so plotted against time it is a straight line: a starting value of P and a slope of Pr dollars per year. That is a linear function in context.

Compound interest is the other kind, and the one most accounts use. The full method, including interest added monthly or daily, is compound interest and compounding periods, and the curve in the second figure is a first look at exponential growth: growth by a fixed multiplier rather than a fixed amount.

Exercises

Part 1: interest and the total

For each, find the interest and the total amount at the end.

750 dollars is deposited in a savings account at 4\% per year simple interest for 3 years.
2{,}400 dollars is deposited at 3.5\% per year simple interest for 2 years.
5{,}000 dollars is put into an 18-month certificate of deposit at 4.8\% per year simple interest.
960 dollars is saved for 7 months at 5\% per year simple interest.
Adaeze borrows 380 dollars from her grandfather to buy a sewing machine, at 6\% per year simple interest, and repays it all after 2 years.
A savings bond bought for 200 dollars pays 2.75\% per year simple interest and is cashed after 4 years.
A loan of 3{,}200 dollars at 7.5\% per year simple interest is repaid after 4 years. Use A = P(1 + rt).
Cressida lends Dagfinn 150 dollars at 4\% per year simple interest. He repays her after 6 months. How much does she get back?

Part 2: the rate, the time or the principal

A deposit of 800 dollars earns 54 dollars of simple interest over 3 years. What is the yearly rate?
Bartholomew borrows 2{,}000 dollars from his mother and, after 4 years, pays her back 2{,}360 dollars in all. What yearly simple interest rate did they agree on?
How long does 1{,}600 dollars at 2.5\% per year simple interest take to earn 100 dollars?
How long does 3{,}000 dollars at 4\% per year simple interest take to grow to 3{,}600 dollars?
A deposit earns 121 dollars of simple interest in 2 years at 5.5\% per year. How much was deposited?
A savings bond paying 3\% per year simple interest is worth 763 dollars, interest included, after 3 years. What was paid for it?

Part 2: per month and per year

(a) Write 0.9\% per month as a simple interest rate per year. (b) Write 21\% per year as a simple interest rate per month.
Yevgenia borrows 620 dollars from her brother for 9 months. He offers either 1\% per month or 10\% per year, simple interest either way. Find the interest under each offer. Which should she take, and how much does it save her?
What simple interest rate per month makes 400 dollars earn 30 dollars in 5 months?

Part 2: simple against compound

500 dollars is saved for 3 years at 10\% per year. Find the total under simple interest and under compound interest added once a year, and the difference.
4{,}000 dollars is saved for 2 years at 5\% per year. Find the total under simple interest and under compound interest added once a year, and say where the difference comes from.

Reasoning

At 8\% per year simple interest, how many years does it take any principal to double? Explain why the answer does not depend on the principal.
Show that 1{,}000 dollars for 6 years and 2{,}000 dollars for 3 years, both at 4\% per year simple interest, earn the same interest. Explain from the formula why doubling the principal and halving the time always does this.
A lender charges 2\% per month simple interest on a loan repaid after 3 years. What percent of the principal is the interest? Explain why the time in I = Prt must be 36 here, not 3.
Explain why simple and compound interest at the same yearly rate give the same total after one year, and why the compound total is larger after every later year.

Answers

Interest 90 dollars, total 840 dollars. 750 \times 0.04 = 30 a year, and 30 \times 3 = 90.
Interest 168 dollars, total 2{,}568 dollars. 2400 \times 0.035 = 84 a year, and 84 \times 2 = 168.
Interest 360 dollars, total 5{,}360 dollars. 18 months is t = 1.5 years; 5000 \times 0.048 = 240 a year, and 240 \times 1.5 = 360.
Interest 28 dollars, total 988 dollars. One year earns 960 \times 0.05 = 48, and 7 months is \displaystyle \frac{7}{12} of a year: \displaystyle 48 \times \frac{7}{12} = 28.
Interest 45.60 dollars, total 425.60 dollars. 380 \times 0.06 \times 2 = 380 \times 0.12 = 45.60.
Interest 22 dollars, total 222 dollars. 200 \times 0.0275 = 5.50 a year, and 5.50 \times 4 = 22.
Total 4{,}160 dollars, of which 960 dollars is interest. rt = 0.075 \times 4 = 0.3, so A = 3200 \times 1.3 = 4160.
She gets back 153 dollars. 6 months is t = 0.5, so I = 150 \times 0.04 \times 0.5 = 3.
2.25\% per year. 54 = 800 \times r \times 3 = 2400r, so \displaystyle r = \frac{54}{2400} = 0.0225.
4.5\% per year. The interest is 2360 - 2000 = 360, and 360 = 2000 \times r \times 4 = 8000r, so r = 0.045.
2.5 years. One year earns 1600 \times 0.025 = 40 dollars, and 40t = 100 gives t = 2.5.
5 years. The interest is 3600 - 3000 = 600; one year earns 3000 \times 0.04 = 120, and 120t = 600 gives t = 5.
1{,}100 dollars. 121 = P \times 0.055 \times 2 = 0.11P, so \displaystyle P = \frac{121}{0.11} = 1100.
700 dollars. 1 + rt = 1 + 0.03 \times 3 = 1.09, so 763 = 1.09P and \displaystyle P = \frac{763}{1.09} = 700.
(a) 10.8\% per year, since 12 \times 0.9\% = 10.8\%. (b) 1.75\% per month, since 21\% \div 12 = 1.75\%.
55.80 dollars at 1\% per month and 46.50 dollars at 10\% per year; the yearly offer saves her 9.30 dollars. Monthly: 620 \times 0.01 \times 9 = 55.80. Yearly: 9 months is 0.75 of a year, and 620 \times 0.10 \times 0.75 = 46.50. Also 1\% per month is 12\% per year, more than 10\%.
1.5\% per month. 30 = 400 \times r \times 5 = 2000r, so \displaystyle r = \frac{30}{2000} = 0.015.
Simple 650 dollars, compound 665.50 dollars, a difference of 15.50 dollars. Simple: 500 \times (1 + 0.10 \times 3) = 500 \times 1.3 = 650. Compound: 500 \times 1.1^3 = 500 \times 1.331 = 665.50.
Simple 4{,}400 dollars, compound 4{,}410 dollars. Simple: 4000 \times 1.10 = 4400. Compound: 4000 \times 1.05^2 = 4000 \times 1.1025 = 4410. The extra 10 dollars is 5\% of the first year’s 200 dollars of interest, earned in the second year because that interest was added to the balance.
12.5 years. Doubling means the interest equals the principal: P \times 0.08 \times t = P. Dividing both sides by P (not zero) leaves 0.08t = 1, so t = 12.5. The P cancels, so every principal takes the same time: each year adds 8\% of the principal, and it takes \displaystyle \frac{100}{8} = 12.5 years of 8\% to make 100\%.
Both earn 240 dollars: 1000 \times 0.04 \times 6 = 240 and 2000 \times 0.04 \times 3 = 240. The interest is the product P \times r \times t. Doubling P multiplies the product by 2, and halving t multiplies it by \displaystyle \frac{1}{2}, so together they multiply it by \displaystyle 2 \times \frac{1}{2} = 1 and leave it unchanged.
72\% of the principal. The rate is 2\% for each month, so t counts months, and 3 years is 36 months: I = P \times 0.02 \times 36 = 0.72P. With t = 3 the formula would count only three months of interest. The same answer comes from turning the rate into 24\% per year and keeping t = 3 years: P \times 0.24 \times 3 = 0.72P.
In the first year both accounts earn the rate times the same principal, since no interest has been added yet, so the totals agree. Under compound interest that first year’s interest joins the balance, so from the second year on the interest is the rate times a balance larger than the principal, while simple interest is still the rate times the principal. Each later year the compound account earns more than the simple one, so its total stays ahead and the gap grows.