2.1 Simple interest (Page 2/2)

Siyavula textbooks: grade 10 Page 2 / 2

If you look carefully at the similarities between the two answers above, we can generalise the result. If you invest your money ( $P$ ) in an account which pays a rate of interest ( $i$ ) for a period of time ( $n$ years), then, using the symbol $A$ for the Closing Balance:

A = P (1 + i \cdot n)

As we have seen, this works when $n$ is a fraction of a year and also when $n$ covers several years.

Interest Calculation

Annual Rates means Yearly rates. and p.a.(per annum) = per year

If I deposit R1 000 into a special bank account which pays a Simple Interest of 7% for 3 years, how much will I get back at the end of this term?

Determine what is given and what is required
- opening balance, $P = R 1 000$
- interest rate, $i = 7 %$
- period of time, $n = 3 years$
We are required to find the closing balance (A).
Determine how to approach the problem
We know from [link] that:

$A = P (1 + i \cdot n)$
Solve the problem
$\begin{matrix} A & = & P (1 + i \cdot n) \\ = & R 1 000 (1 + 3 \times 7 %) \\ = & R 1 210 \end{matrix}$
Write the final answer
The closing balance after 3 years of saving R1 000 at an interest rate of 7% is R1 210.

If I deposit R30 000 into a special bank account which pays a Simple Interest of 7.5%, for how many years must I invest this amount to generate R45 000?

Determine what is given and what is required
- opening balance, $P = R 30 000$
- interest rate, $i = 7, 5 %$
- closing balance, $A = R 45 000$
We are required to find the number of years.
Determine how to approach the problem
We know from [link] that:

$A = P (1 + i \cdot n)$
Solve the problem
$\begin{matrix} A & = & P (1 + i \cdot n) \\ R 45 000 & = & R 30 000 (1 + n \times 7, 5 %) \\ (1 + 0, 075 \times n) & = & \frac{45000}{30000} \\ 0, 075 \times n & = & 1, 5 - 1 \\ n & = & \frac{0, 5}{0, 075} \\ n & = & 6, 6666667 \\ n & = & 6 years 8 months \end{matrix}$
Write the final answer
The period is 6 years and 8 months for R30 000 to generate R45 000 at a simple interest rate of 7,5%. If we were asked for the nearest whole number of years, we would have to invest the money for 7 years.

Other applications of the simple interest formula

Troy is keen to buy an additional hard drive for his laptop advertised for R 2 500 on the internet. There is an option of paying a 10% deposit then making 24 monthly payments using a hire-purchase agreement where interest is calculated at 7,5% p.a. simple interest. Calculate what Troy's monthly payments will be.

Determine what is given and what is required
A new opening balance is required, as the 10% deposit is paid in cash.
- 10% of R 2 500 = R250
- new opening balance, $P = R 2 500 - R 250 = R 2 250$
- interest rate, $i = 7, 5 %$
- period of time, $n = 2 years$
We are required to find the closing balance (A) and then the monthly payments.
Determine how to approach the problem
We know from [link] that:

$A = P (1 + i \cdot n)$
Solve the problem
$\begin{matrix} A & = & P (1 + i \cdot n) \\ = & R 2 250 (1 + (2 \times 7, 5 %)) \\ = & R 2 587, 50 \\ Monthly payment & = & 2587, 50 \div 24 \\ = & R 107, 81 \end{matrix}$
Write the final answer
Troy's monthly payments = R 107,81

Many items become less valuable as they are used and age. For example, you pay less for a second hand car than a new car of the same model. The older a car is the less you pay for it. The reduction in value with time can be due purely to wear and tear from usage but also to the development of new technology that makes the item obsolete, for example, new computers that are released force down the value of older models. The term we use to descrive the decrease in value of items with time is depreciation .

Depreciation, like interest can be calculated on an annual basis and is often done with a rate or percentage change per year. It is like ”negative” interest. The simplest way to do depreciation is to assume a constant rate per year, which we will call simple depreciation. There are more complicated models for depreciation but we won't deal with them here.

Seven years ago, Tjad's drum kit cost him R12 500. It has now been valued at R2 300. What rate of simple depreciation does this represent ?

Determine what is given and what is required
- opening balance, $P = R 12 500$
- period of time, $n = 7 years$
- closing balance, $A = R 2 300$
We are required to find the interest rate( $i$ ).
Determine how to approach the problem
We know from [link] that:

$A = P (1 + i \cdot n)$

Therefore, for depreciation the formula will change to:

$A = P (1 - i \cdot n)$
Solve the problem
$\begin{matrix} A & = & P (1 - i \cdot n) \\ R 2 300 & = & R 12 500 (1 - 7 \times i) \\ i & = & 0, 11657 . . . \end{matrix}$
Write the final answer
Therefore the rate of depreciation is $11, 66 %$

Simple interest

An amount of R3 500 is invested in a savings account which pays simple interest at a rate of 7,5% per annum. Calculate the balance accumulated by the end of 2 years.
Calculate the simple interest for the following problems.
1. A loan of R300 at a rate of 8% for l year.
2. An investment of R225 at a rate of 12,5% for 6 years.
I made a deposit of R5 000 in the bank for my 5 year old son's 21st birthday. I have given him the amount of R 18 000 on his birthday. At what rate was the money invested, if simple interest was calculated ?
Bongani buys a dining room table costing R 8 500 on Hire Purchase. He is charged simple interest at 17,5% per annum over 3 years.
1. How much will Bongani pay in total ?
2. How much interest does he pay ?
3. What is his monthly installment ?

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Source: OpenStax, Siyavula textbooks: grade 10 maths [ncs]. OpenStax CNX. Aug 05, 2011 Download for free at http://cnx.org/content/col11239/1.2

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