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Consumer Arithmetic

Number and Algebra : Module 22Year : 9

June 2011

PDF Version of module

Assumed Knowledge

• Facility with the arithmetic of fractions, decimals and percentages.
• Facility with converting fractions and decimals to percentages and vice versa.
• Facility with simple percentage problems.
• Familiarity with the basic operations on a calculator, including the power button xy.
• Familiarity with rounding numbers to a given number of decimal places, the nearest whole number, to the nearest ten, and so on.
• Facility with rounding money to the nearest dollar and to the nearest cent.
• Familiarity with the use of spreadsheets.

Motivation

For most people, one of the primary applications of arithmetic is in dealing with money and in coping with questions involving financial computations. For example, when we are offered a 25% discount in a shop, we ought to be able to work out, at least roughly, what this will be and make sure we receive it. Similar calculations also apply to bank interest and commissions. If we are offered a 12% commission on sales of $3000, how much should we receive? Such simple problems should be able to be easily solved.

When buying a car, using a credit card, or deciding how good a particular ‘bargain’ is,
we all need to be competent in our understanding of commercial or consumer arithmetic. Otherwise we run the serious risk of wasting our money or, even worse, being at the mercy of unscrupulous individuals. This is particularly important when we come to the ‘big ticket items’ of life such as purchasing a car, a home or planning superannuation. It is also important when dealing with our tax returns.

This module will apply the arithmetical skills related to percentages, fractions, addition, subtraction, multiplication and division. Thus it gives practical applications of the ideas
and methods of arithmetic that students have been studying since primary school.

Content

In most of the modules from F–8 we present the material so that the ideas can be understood and the exercises solved without using a calculator. For the work in this module, we assume that the reader has a calculator at hand, and that the answers to computations are rounded in a way that is appropriate in the given context. When presenting problems to school students it is wise for the teacher to specify carefully how they wish their students to round up or down, whether to the nearest dollar, or nearest cent and so on. Note that rounding at multiple stages in a calculation may alter an answer by a small amount. For this reason it is better to only round at the last step.

In the following sections, we will, for the most part, apply only two basic ideas. Finding a percentage of a number, and increasing or decreasing a number by a given percentage. Finding simple interest or commissions all involve multiplying a given number by a factor which will depend on the given percentage. Other problems often simply involve reversing the process. This can be done using division. For example, to increase a number by 15%, we multiply the number by 1.15, to reverse this process we simply divide by 1.15. We will see below that these ideas simplify a number of problems that students have traditionally found difficult.

Percentages

A percentage is a way of writing a fraction with a denominator of 100.
Thus 10% means which can be simplified to or as a decimal, 0.1.

Since 100% =1, to convert a fraction or decimal to a percentage, we multiply by 100% .

For example,

= × 100% = 66% and 3.2 = 3.2 × 100% = 320%.

It is useful to be familiar with a number of commonly used fractions and their
percentage equivalents.

Fraction	percentage	Fraction	percentage
	50%		60%
	>25%		80%
	75%		12.5%
	33%		37.5%
	66%		62.5%
	20%		87.5%
	40%		100%

Percentage of a quantity

When told, for example, that 84% of a group of 250 people surveyed watched the cricket match on Saturday, it is natural to ask what exact number this was. To do this we find
84% of 250, which is × 250 = 210 people.

Also, if we know, for example, that 37 of the 50 people in the swimming club attended training, then it is useful to express this number as a percentage of the whole. This is
done by expressing 37 as a fraction of 50 and converting it to a percentage.
Thus, × 100 = 74%, so we can say that 74% of the club attended training.

Reversing the Process

To find 15% of a given quantity, we multiply the quantity by 15%. To reverse the process, we divide by 15%. Thus, for example, if 15% of Paul’s take home pay is $90, then his take home pay was

90 ÷ 15% = 90 ÷ 0.15 = $600.

This is a simple but important algorithm that is very useful in consumer arithmetic.

Commission

A commission is a fee charged for services rendered. For example, a real estate agent will charge a commission on the selling price of a house (in addition to such things as the fee for advertising the house), and a financial advisor will charge a commission for managing a portfolio of investments. Commissions are generally stated in the form of percentages.

For example, a gallery charges a commission of 6.5% for selling a painting. If they sold a painting for $180 000, how much did the owner receive?

	Commission	= 180 000 × 6.5%
		= 180 000 × 0.065
		= 11 700.

Hence the owner receives 180 000 − 11 700 = $168 300.

This can also be calculated as 180 000 × 93.5%

On the other hand, suppose that the gallery received $3640 commission for the sale of a painting. What was the selling price of the painting?

This is a reverse process, so

	Selling price	= commission ÷ 6.5%
		= 3640 ÷ 0.065 = $56 000.

Profit and Loss

A simple business model is to consider sales and costs. Sales refer to the income of the business and costs may include wages and rent.

Thus we can define

Income − Expenditure = Profit/Loss OR Sales − Costs = Profit/Loss.

Clearly, if the difference above is positive then we have a profit and if it is negative, a loss.

Income Tax

Income tax rates are often progressive. This means that the more you earn, the higher the rate of tax on each extra dollar you earn. Taxable income (which is the income after deductions) is divided into tax brackets. Since these can change from year to year, we will assume the following income tax brackets as an example.

• There is no tax on the first $12 000 that a person earns in the given year.
• From $12 001 to $30 000 the tax rate is 15c for each dollar over $12 000.
• From $30 001 to $75 000, the tax rate is 25c for each dollar over $30 000.
• Over $75 000, the tax rate is 35c for each dollar over $75 000.

Based on this schedule, the tax payable for a person earning, for example, $62 300 is calculated as follows:

Tax on first $12 000 = $0

Tax on next $18 000 = 18 000 × 0.15 = $2700

Tax on the remaining $32 300 = 32 300 × 0.25 = $8075

Hence the total tax is $10 775.

Simple Interest

When money is borrowed, the borrower normally repays the amount they have borrowed plus another amount that is called the interest. The amount of interest usually depends on:

• the amount borrowed, called the principal,
• the rate at which the interest is charged,
• the time for which the money is borrowed.

The rate is almost always expressed as a percentage for each year. For example, the rate might be 5% per annum, or 5% p.a. The phrase per annum is the Latin for for a year.

In simple interest transactions, interest is paid on the original amount borrowed. It is calculated by finding the amount of interest on the principal in one year and multiplying it by the number of years. Thus, if P is the principal, r is the interest rate, expressed as a percentage per annum, and T is the number of years, then we have the well-known simple interest formula

I = PrT.

If the interest rate is expressed in terms of some other period of time, for example per month, then we will need to adjust T accordingly.

Thus, for example, the simple interest on $12 000 for 6 years at 8.25% is given by

	I	= PrT
		= 12 000 × 0.0825 × 6
		= $5940.

Since the formula contains four pronumerals, specifying any three of them allows the fourth to be found.

Percentage Increase and Decrease

When a quantity is increased or decreased, the change is often expressed as a percentage of the original amount. The key to finding percentage increases and decreases is to combine two steps into one. Thus, for example, to increase 30 by 5% we can find 5% of 30, which is 1.5, and add it on to 30 to produce 31.5.

It is better, however, to combine the two steps and find 105% (= 100% + 5%) of 30,
which is

1.05 × 30 = 31.5.

Hence, to increase a quantity by a% we multiply by (100 + a)%.

This key idea will be used repeatedly in the remaining work in this module.

Percentage decrease can also be calculated in one step. For example, if we wish to decrease a quantity by 30%, then we obtain 100 − 30 = 70% of the amount we started with.

Hence, to decrease a quantity by a% we multiply by (100 − a)%.

Again, this key idea will be used repeatedly in the remaining work in this module.

To find the rate of increase or decrease

Reversing the Process

Using the idea above, we can now solve problems which at first sight might appear rather hard. For example, after increasing my salary by 7%, I now receive $890 per week. What was my salary before the increase?

To increase by 7% we multiply by 107%. Hence to reverse the process, we divide by 107%. Thus, my original salary was

890 ÷ 107% = 890 ÷ 1.07 ≈ $831.78.

The GST

In 1999 the Australian Government introduced a Goods and Services Tax, or GST for short. This tax applies to nearly all goods and services in Australia.

The current rate is 10% on the pre-tax price of the goods or service.

• When GST is to be added to the pre-tax price, we multiply by 110% = 1.1.
• If the quoted price includes the GST, the pre-tax price is obtained by dividing by 1.1.

Discounts

It is very common for a shop to discount the price of an item. Discounts are normally expressed as a percentage of the original price. Hence finding the price after a discount is the same as decreasing a quantity by a given percentage. As we have seen before, the process can also be reversed.

Repeated Increase and decrease

The method of increasing or decreasing a quantity by a given percentage is very useful when several increases or decreases occur.

For example, if the population of a town is 30 500 and increases by 6% then 7% and finally 4.5% over a three year period, then the new population is 30 500 × 1.06 × 1.07 × 1.045 ≈ 36 150. Notice that to find the overall percentage increase on the original, we calculate 1.06 × 1.07 × 1.045 ≈ 1.185 and so the percentage increase is about 18.5%. This is somewhat larger than the sum of the three increases, which is 17.5%, and illustrates the fact that percentages cannot always be added.

The same method can be applied to repeated decrease.

The reverse method can again be applied.

Compound Interest

When an amount of money, called the principal, is invested in a bank, the bank generally pays interest at the end of each month and adds that interest to the account balance. Thus the balance, on which the interest is calculated, increases each month. This process is referred to as compound interest. In some situations the interest is paid yearly.

Adding on interest to a given amount is another example of increasing by a certain percentage and so the previous method again applies.

While there is a formula for calculating the final balance for a compound interest problem, when introducing compound interest it is best to avoid using the formula in the first instance.

The bank has paid $1566.96 in interest and this corresponds to a simple interest rate of

(1566.96 ÷ 2000) × 100 ÷ 8 = 9.7935%, per annum.

Note then that when compound interest is being used, a 12% interest rate per year is not the same as a 1% interest rate per month, calculated monthly. The two rates are, however, close.

Reversing the Process

As always, the process can be reversed.

Depreciation

Depreciation occurs when the value of an asset reduces as time passes. Assets such as cars, office furniture and computers typically lose value over time. Accountants generally assume that such assets decrease in value every year by the same amount, called the depreciation rate. For taxation purposes, the tax laws fix the depreciation rate for many standard assets.

Once again there is a formula for calculating the final value after depreciation, but it is best, when introducing the idea, to avoid using a formula in the first instance.

The process can also be reversed as the following exercise will show.

Compound interest formula

We saw above how to compute compound interest via an iterative process. There is a straightforward formula for finding the value of an investment of $P, at an interest rate r per unit of time, over n units of time.

After 1 unit of time, the balance is P(1 + r).

After 2 units of time, the balance is P(1 + r)(1 + r) = P(1 + r)2.

Continuing the pattern, we can see that the formula for the amount A after n units of time is given by

A = P(1 + r)n.

Depreciation Formula

Earlier we used an iterative process to find the value to which an asset depreciates over a period of time. As with compound interest, there is a formula for calculating this.

Suppose we have an asset, originally worth $P, which depreciates at a percentage rate r per annum for n years.

After 1 year, the value of the asset is P(1 − r).

After 2 years, the value of the asset is P(1 − r)(1 − r) = P(1 − r)2.

Continuing the pattern, we can see that the formula for the value A of the asset after n years is given by

A = P(1 − r)n.

Spreadsheet applications

Spreadsheets provide an excellent way to model financial calculations based on compound interest, because the successive steps in the compounding can be displayed in the successive lines of the spreadsheet.

When a formula is copied in a spreadsheet, the user can control whether each cell reference is an absolute reference of a relative reference. Relative references allow the recursive processes of financial calculations to be modelled easily and quickly.

Algebra is essential for spreadsheets, and nowadays spreadsheets are probably the most common application for algebra. For the simple applications of this module, cell references are used as pronumerals. There is no benefit in using a ‘name manager’ to define pronumerals.

This is a first treatment, and accordingly, interest is only calculated annually and installments are only paid annually. Spreadsheets, however, have an extremely large maximum size, and can easily cope with monthly or daily calculations.

Compound interest

Superannuation

Housing loans

Links Forward

To deal with more complicated financial problems such as superannuation and time payments, we need to understand what a geometric series is and how to find the formula for the sum of a geometric series.

A geometric sequence (or geometric progression) is a list of numbers in which each is obtained by multiplying the previous one by a constant multiplier called the common ratio. For example, the sequence 2, 6, 18, 54, 162, … is an example of a geometric sequence in which the common ratio is 3. When we add the numbers in a geometric sequence up to some point, we obtain a geometric series.

Hence the series 2 + 6 + 18 + 54 = 80 is the sum of the first four terms of the geometric sequence given above.

Geometric series

As above, we define a geometric series to be the sum of numbers in which each term is a constant multiple of the previous one. We will right the first term as a and the constant ratio as s. Hence the series can be written as

Sn = a + as + as2 +… + asn − 1,

and contains n terms. Series such as this arise naturally in financial (and many other) problems. We would like to obtain a simple formula for this sum. To do this we multiply both sides by s, thus

sSn = as + as2 + … +asn.

If we now subtract these equations, most of the terms cancel and we obtain

sSn − Sn = asn − a.

Factoring, (s − 1)Sn = a(sn − 1), and so Sn = .

This is the formula for the sum to n terms of the geometric series with first term a and constant ratio s.

Superannuation

Superannuation is a way of saving for retirement. Money is regularly invested over a long period of time and (compound) interest is paid.

Suppose that I am to invest in a superannuation scheme for 30 years which pays 6% p.a. I put $3000 each year into the scheme, and (for the sake of simplicity) we will suppose that the interest is calculated yearly. Thus, the first $3000 will be invested for the full 30 years at 6% p.a. Hence it will grow to accrue to $3000 × 1.0630.

At the beginning of the next year, the next $3000 will be invested for 29 years and so will accrue to $3000 × 1.0629 and so on. Writing the amounts from smallest to largest, the total value of my investment will this be:

A = $3000 × 1.06 + $3000 × 1.062 + … + $3000 × 1.0630.

This is a geometric series with a = $3000 × 1.06, and s = 1.06 and n = 30.

Substituting into the formula, the value of the sum is

A = ≈ $237 174.56.

This example illustrates both the value of regular saving and the power of compound interest − $90 000 invested becomes roughly $237 000 over 30 years.

Time payments

Dealing with time payments is somewhat more difficult and the reader may wish to skip over this section. On the other hand, those who are paying off loans may find it useful.
It is, of course, too advanced for junior high school. Spreadsheets were used in a previous section.

When a loan is taken out, the amount borrowed (the principal) is generally repaid, along with the interest accrued, in equal installments. The interest depends on the balance owing and so will decrease over time as more payments are made.

Finding the amount of each repayment to be made, given that we wish to pay out the loan in a given time, appears at first sight to be a difficult problem, but it can be solved using geometric series.

Suppose, for example, that a car loan for $7000 is taken out at an interest rate of 6.2% p.a. (calculated monthly) and that the loan is to be repaid in 5 years with monthly repayments. How much should each monthly payment be, in order to pay out the loan in this time?

Since we are working in months, the period of the loan is 5 × 12 = 60 months, and the interest rate per month is approximately 6.2% per month. (Recall that for compound interest, a 12% p.a. rate is not exactly equal to a 1% per month rate.)

Hence r ≈ 0.0051667. To simplify the notation we will write

R = 1 + r = 1.0051667.

Let P be the amount of each repayment, and Ak be the amount owing after k payments. After each month we owe R times the amount owing in the previous month (the balance plus the interest) minus P (the monthly repayment).

Hence

		A1 = 7000R − P
and so		A2 = R(7000R − P) − P = 7000R2 − P(1 + R).

Continuing this process,

A3 = R[7000R2 − P(1 + R)] − P = 7000R3 − P(1 + R + R2) (check this step carefully.)

From here we can see where the pattern is leading, so after n months,

An = 7000Rn− P(1 + R + R2 + … + Rn − 1).

The sum inside the parentheses is a geometric series consisting of n terms with a = 1 and s = R. We can thus apply the formula for the sum of a geometric series to obtain

An= 7000Rn− P.

Now after 60 months, the loan has been repaid and so An=0, when

substituting the values of R = 1.0051667, n = 60, An=0, we obtain
0 = 7000 × 1.00516670 − P, and we can solve this for P to obtain

P ≈ $135.98.

Hence, we need to pay $135.98 per month in order to pay out the loan.

Note then that we have to repay a total of 60 × 135.98 = $8158.80 and so the total interest paid is $1158.80.

History

It appears that the concept of 'money’ arose in very ancient times via the need for an intermediary stage in the barter process. Thus, if I wish to exchange some barley for someone’s fruit, but the barley harvest has not yet come, I can hand over some other good such as metal, cattle, salt or wine, which I can then redeem when the barley has been harvested. This is referred to as commodity money. Thus metal, which is very durable, came, in time, to represent the basis for our modern concept of money.

In the first half of the seventh century BC, Pheidon of Argos, on the Greek mainland, is believed to have set up a system of standardized weights and measures, and to have begun the use of stamped coinage. Although this was the first use of coinage in Europe, the earliest known coins stem from Ionia (part of modern day Turkey) and by 650 BC stamped coinage was in relatively common usage in Lydia in Asia Minor from where the concept spread to most parts of Greece and beyond. Coins were typically minted by some official government process with the amount of precious metal being carefully controlled to equal the 'value’ of the coin. In time, the 'authority’ of the government stamp enabled the coin to be devalued in terms of the precious metal it contained, while nevertheless retaining official value. This led to the notion of representative money.

The concept of lending money at interest is an old one and is mentioned in the ancient Code of Hammurabi (about 1760 BC). The Greek word for interest, tokos, is related to the word for 'giving birth’. Thus the ancients thought of money 'giving birth’ to money in the form of interest.

Banking in Europe arose after the time of the Crusades, since paying for soldiers, weapons, transport etc across large distances required the transfer of large amounts of money. This was achieved in the form of bills of exchange. By the time of the Industrial Revolution, the church rules forbidding usury, the lending of money at interest, appear to have been either forgotten or reinterpreted and the lending of money at interest became very widespread.

It is worth commenting that the father of the famous Fibonacci (Leonardo of Pisa, c. 1202 AD) was a business man and that the young Leonardo assisted his father in his bookkeeping at an extremely young age. Many of the problems in Fibonacci’s Liber Abaci involve calculations with interest and percentages and are often couched in a financial setting.

There is some debate as to the origin of the % sign, however the word percentage is derived from the Italian per cento meaning out of a hundred.

It is interesting to note that the number e (≈ 2.71828 to 5 decimal places) which occurs as the base of the natural logarithm in Calculus, was known at an earlier date, since it arises from a simple problem in compound interest. Suppose we invest say $1 at 100% p.a.,
(a rather generous rate indeed!), compounded over n equal units of time during a one year period. The formula , where P = 1 is the amount invested, r = 1 is the interest rate per annum, n is the number of units of time, and A is the amount we have, becomes . If we calculate this quantity for larger and larger values of, we get closer and closer to the number 2.71828…. Using limit techniques we can show that this limit is equal to the number e. Thus, the most interest we can make from this scenario is about $1.72 (!)

While there were earlier works written on the arithmetic of money, percentages and simple interest, it was in 1613 that Richard Witt produced a book entirely devoted to compound interest.

In 1847 the French Central Bank attempted to control interest rates through the manipulation of the money supply. This practice continues today via central banks such as the Reserve Bank of Australia.

Today, modern banking is a very complicated activity which employs highly trained mathematicians as financial and risk analysts.

In 1973 the Black-Scholes model for financial markets and derivative investments was put forward. This involved some quite sophisticated Mathematics that makes use of partial differential equations and is very important in the modern financial world. Modern actuarial and financial studies at tertiary level involve material from a number of branches of advanced Mathematics.

ANSWERS TO EXERCISES

EXERCISE 1

62.5%

EXERCISE 2

$1425.60

EXERCISE 3

a $29,700

b 24.75%

EXERCISE 4

$32

EXERCISE 5

a $480.70

b 12.6%

EXERCISE 6

$25.00

EXERCISE 7

12.7% rounded to one decimal place

EXERCISE 8

a $96,000

b $128,000

EXERCISE 9

59% gain, rounded to one per cent.

EXERCISE 10

a $17,958.56

b $31,384.28

EXERCISE 11

a $389,927.27

b $378,015.00

EXERCISE 12

a The loan will be paid off on Dec 31 of the 13th year with a credit of $16,110.35.

b The loan will be paid off on Dec 31 of the 15th year with a credit of $471.75.

 

The Improving Mathematics Education in Schools (TIMES) Project 2009-2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations.

The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations.

© The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE-EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
https://creativecommons.org/licenses/by-nc-nd/3.0/