Simple Interest

When you borrow or save money, interest is the extra amount paid for using that money. With simple interest, the interest is calculated only on the original amount, so you earn the same amount every year. It’s the simplest model of how money grows, and a good starting point before compound interest.

Key ideas

The simple interest formula

$$I = Prt$$

• $I$ is the interest earned (or owed), in dollars.
• $P$ is the principal, the original amount invested or borrowed.
• $r$ is the annual interest rate, as a decimal: $4\% = 0.04$.
• $t$ is the time in years.

The total amount

The amount $A$ is the principal plus the interest:

$$A = P + I = P(1 + rt)$$

Time that isn’t in whole years

The rate is per year, so $t$ must be in years:

• months: divide by $12$ (for example, $9$ months is $t = \tfrac{9}{12} = 0.75$)
• days: divide by $365$ (for example, $73$ days is $t = \tfrac{73}{365} = 0.2$)

Simple interest is linear

Each year adds the same interest, $Pr$. So the amounts at the end of each year form an arithmetic sequence with common difference $d = Pr$, and the graph of $A$ against $t$ is a straight line.

Worked examples

Example 1: Interest and amount

Find the interest and the total amount when $2500 is invested at $4\%$ per year simple interest for $3$ years.

Solution.

$$I = Prt = 2500(0.04)(3) = 300$$ $$A = P + I = 2500 + 300 = 2800$$

The interest is $300, and the amount is $2800.

Example 2: Time in months

Find the amount when $800 is invested at $3.5\%$ per year simple interest for $9$ months.

Solution. $t = \tfrac{9}{12} = 0.75$ years.

$$I = 800(0.035)(0.75) = 21$$

The amount is $800 + 21 = 821$, or $821.

Example 3: Finding the rate

$1200 grows to $1380 in $5$ years with simple interest. What is the annual rate?

Solution. The interest is $1380 - 1200 = 180$.

$$180 = 1200(r)(5) = 6000r \quad\Rightarrow\quad r = \frac{180}{6000} = 0.03$$

The rate is $3\%$ per year.

Example 4: An arithmetic sequence

$5000 is invested at $6\%$ per year simple interest. Write the amount after $n$ years as an arithmetic sequence, and find when it reaches $8000.

Solution. Each year earns $5000(0.06) = 300$. The amounts are $5300, $5600, $5900, \dots, an arithmetic sequence with $d = 300$:

$$A_n = 5000 + 300n$$ $$5000 + 300n = 8000 \quad\Rightarrow\quad 300n = 3000 \quad\Rightarrow\quad n = 10$$

It reaches $8000 after $10$ years.

Common mistakes

Using the percentage instead of the decimal. $4\%$ is $0.04$. Using $4$ makes the interest $100$ times too big.

Using months or days as $t$. The rate is per year, so $9$ months is $t = 0.75$, not $t = 9$.

Mixing up $I$ and $A$. $I$ is just the interest. $A$ is what you have in total. Read which one the question asks for.

Forgetting to subtract to find the interest. In Example 3, the interest is $1380 - 1200 = 180$, not $1380$.

Practice

1. (Warm-up) Find the simple interest on $600 at $5\%$ per year for $4$ years.

Solution

$I = 600(0.05)(4) = 120$, so $120.

2. (Warm-up) Find the amount when $1500 is invested at $2.5\%$ per year simple interest for $2$ years.

Solution

$I = 1500(0.025)(2) = 75$, so $A = 1575$, or $1575.

3. (Warm-up) Write each time in years.

• (a) $18$ months
• (b) $73$ days

Solution

(a) $\tfrac{18}{12} = 1.5$ years.

(b) $\tfrac{73}{365} = 0.2$ years.

4. (Core) Find the amount when $950 is invested at $4.2\%$ per year simple interest for $146$ days.

Solution

$t = \tfrac{146}{365} = 0.4$ years.

$$I = 950(0.042)(0.4) = 15.96$$

$A = 950 + 15.96 = 965.96$, so $965.96.

5. (Core) How long does it take $2000 at $5\%$ per year simple interest to earn $350?

Solution

$$350 = 2000(0.05)t = 100t \quad\Rightarrow\quad t = 3.5$$

It takes $3.5$ years.

6. (Core) What principal earns $84 in interest in $2$ years at $3.5\%$ per year simple interest?

Solution

$$84 = P(0.035)(2) = 0.07P \quad\Rightarrow\quad P = 1200$$

The principal is $1200.

7. (Core) $4000 is invested at $3\%$ per year simple interest. List the amounts after $1$, $2$, and $3$ years, and write a general term for the amount after $n$ years.

Solution

Each year earns $4000(0.03) = 120$. The amounts are $4120, $4240, $4360: an arithmetic sequence with $d = 120$.

$$A_n = 4000 + 120n$$

8. (Challenge) At $8\%$ per year simple interest, how long does it take an investment to double?

Solution

Doubling means the interest equals the principal: $I = P$.

$$P = P(0.08)t \quad\Rightarrow\quad 1 = 0.08t \quad\Rightarrow\quad t = 12.5$$

It takes $12.5$ years, whatever the starting amount.

9. (Challenge) $10 000 is split between two accounts paying simple interest: one at $3\%$ per year and one at $5\%$ per year. The total interest after one year is $420. How much is in each account?

Solution

Let $x$ dollars be in the $3\%$ account, so $10\,000 - x$ is in the $5\%$ account.

$$\begin{aligned} 0.03x + 0.05(10\,000 - x) &= 420 \\ 0.03x + 500 - 0.05x &= 420 \\ -0.02x &= -80 \\ x &= 4000 \end{aligned}$$

$4000 is in the $3\%$ account and $6000 in the $5\%$ account. Check: $120 + 300 = 420$. ✓