Simple Interest

Percent

In this lesson part, we will look at solving simple interest problems and see how simple interest relates to arithmetic sequences. Before we do so, here is a quick reminder of working with percentages.

The word percent (symbolized as \(\%\)) means "out of one hundred."
We can see that from the word itself:

• per means "out of" (think of a phrase like kilometres per hour), and
• cent means "hundred" (think of a word like century, meaning one hundred years).

Since percent is a measure out of \(100\), something like \(27\) percent can be written as

• a percent: \(27\%\)
• a fraction: \(\dfrac{27}{100}\)
• a decimal: \(0.27\)

We can calculate the percentage of a number by multiplying. For example, we can calculate \(25\% \) of \(32\):

\(\begin{align*} 25\% \text{ of } 32 &= 25\% \times 32\\ &=0.25 \times 32\\ &= 8 \end{align*}\)

We will make use of calculations involving percentages as we study simple interest.

Example 2

Solution

The word annual is commonly used in financial settings; it means yearly.

1. To determine the value of the account after \(5\) years, we first determine the interest earned in \(5\) years.

   We are given the following:

   • The principal invested, in dollars, is \(P=1500\).
   • The annual interest rate is \(r=4\%\) or \(r=0.04\).
   • The time in years is \(t=5\).

   This is enough information to determine \(I\) and then \(A\):

   \(\begin{align*} I&=Prt\\ &= 1500(0.04)(5)\\ &= 300 \end{align*}\)

   The accumulated amount after \(5\) years is:

   \(\begin{align*} A&= P+I\\ &= 1500+300\\ &=1800 \end{align*}\)

   Therefore, the accumulated amount in the account after \(5 \) years is \($1800\).

2. To show that the annual balances form an arithmetic sequence, we first calculate the interest earned in one year:

   \(\begin{align*} 4\% \text{ of } 1500 &= 0.04 \times 1500\\ &= 60 \end{align*}\)

   Since the account pays simple interest, the amount of interest earned in each year is the same.

   Thus, the sequence of balances is defined recursively as:

   \(\begin{align*} &t_1=1500\\ &t_n=t_{n-1}+60 \end{align*}\)

   for \(n \ge 2\).

   This leads to the sequence of balances:

   \(1500,~1560,~1620,~1680, \ldots\)

   This sequence is arithmetic with common difference \(60\).

3. To show that the annual balances grow linearly, consider this table of annual balances:

   Year	Balance at End of Year (in dollars)
   \(0\)	\(1500\)
   \(1\)	\(1560\)
   \(2\)	\(1620\)
   \(3\)	\(1680\)
   \(4\)	\(1740\)
   \(5\)	\(1800\)

   Since the balances increase by \($60\) each year, the first differences are all \(60\).

   Therefore, the balances grow linearly with time.

4. If \(f(x)\) is the balance in the account at time \(x\), in years, then \(f(x)\) is the accumulated amount after \(x\) years:

   \(\begin{align*} f(x) &= P+I\\ &= P+Prt\\ &= 1500+1500(0.04)(x)\\ &= 1500+60x \end{align*}\)

   Notice that we again have shown that this is a linear relationship.

Simple Interest, Arithmetic Sequences, and Linear Relations

Since simple interest always grows at a constant rate, we can make the following statements connecting simple interest, arithmetic sequences, and linear relationships.

Example 3

Solution

The accumulated amount, \(A\) or \(f(t)\), is given by:

\(\begin{align*} A &= P+I\\ &= P+Prt\\ &= P(1+rt) \end{align*}\)

Comparing \(f(t)=1300(1+0.03t)\) and \(A=P(1+rt)\), we see that:

• \(P=1300\), and
• \(r=0.03\) or \(3\%\).

Therefore, \(f(t)=1300(1+0.03t)\) represents the accumulated amount after \(t\) years when \($1300\) is invested in a simple interest account paying \(3\%\) per year.

Example 4

Solution — Part A

We are given the following:

• The interest owing, in dollars, is \(I=150\).
• The principal borrowed, in dollars, is \(P=1250\).
• The annual interest rate is \(r=0.04\).

This is enough information to determine the time of the loan, \(t\):

\(\begin{align*} I&=Prt\\ t &= \dfrac{I}{Pr}\\ &= \dfrac{150}{1250(0.04)}\\ &= 3 \end{align*}\)

Therefore, the money is borrowed for \(3\) years.

Solution — Part B

We are given the following:

• The interest earned, in dollars, is \(I=153\).
• The monthly interest rate is \(r=0.015\).
• The investment lasts for \(1\) year. Since \(r\) is a monthly interest rate, we require time in months, so \(t=12\).

This is enough information to determine the principal, \(P\):

\(\begin{align*} I&=Prt\\ P &= \dfrac{I}{rt}\\ &= \dfrac{153}{0.015(12)}\\ &= 850 \end{align*}\)

Therefore, the principal is \($850\).

Example 5

Solution

Vijay already has a principal of \($1200\); to reach his goal, he requires an accumulated amount of \($1500\). Thus, he requires the interest earned to be \(I=300\).

Therefore, we are given the following:

• The interest earned, in dollars, is \(I=300\).
• The principal, in dollars, is \(P=1200\).
• Since he needs this money in two and a half years, the time of the investment in years is \(t=2.5\).

This is enough information to determine the rate, \(r\):

\(\begin{align*} I&=Prt\\ r &= \dfrac{I}{Pt}\\ &= \dfrac{300}{1200(2.5)}\\ &= 0.1 \end{align*}\)

Expressing \(r=0.1\) as a percentage, we see that Vijay requires an annual simple interest rate of \(10\%\). Vijay's financial goal seems unreasonable since it is unlikely that he will be able to invest money in a short-term investment that will yield such a high interest rate.