9A 9B 9C 9D 9E Depreciation Financial arithmetic · 2016. 7. 19. · 9E Depreciation 9 areaS oF StUdy • Applications of simple interest and compound interest formulas • Appreciation

388

9A Simple interest 9B Compound interest, infl ation and

appreciation 9C Savings and credit card accounts 9D Purchase options 9E Depreciation

9

areaS oF StUdy

Applications of simple interest and compound • interest formulasAppreciation and depreciation of assets, • including investment of money, capital gains of physical assets, and depreciation of assets by infl ation

Cash fl ow in common savings and credit • accounts including interest calculationsComparison of purchase options including cash, • credit card, bank loan, time payments (hire purchase) and store cardsDepreciation of assets•

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Digital doc10 Quick Questions

Financial arithmetic

Simple interestFinancial arithmeticAt various stages of our lives, we will be faced with making important and informed decisions regarding our fi nances. We may fi nd ourselves asking:

If I wish to purchase a car, should • I borrow from the bank or accept a payment plan offered by the car dealer?If I wish to invest $5000, how much • interest will I earn annually?If I wish to borrow a sum of $100 000 • from a bank or fi nance company, how much money will I need to repay in total?

To answer these questions, we need to have a basic understanding of ‘money matters’. This chapter will look at some of these fi nancial issues, including the fundamental concepts of interest, appreciation and depreciation, the cash fl ow in savings accounts and a comparison of purchase options available (including the hidden fl aws of some of these options).

Simple interestWhen you lend money for a certain period of time (a term deposit) to a bank, building society, or other fi nancial institution, you expect to be rewarded by eventually getting your money back, plus an extra amount, commonly known as interest (I ).

9a

maths Quest 11 Standard General mathematics for the ti-nspire

389Chapter 9 Financial arithmetic

Similarly, if you borrow money from any institution by taking out a loan or mortgage, you must pay back the original sum plus interest.

The following examples deal with simple interest, that is, interest that is paid only on the original sum of money invested or borrowed.

The formula used to calculate simple interest is given by:

IPRT=100

whereI = interest ($)P = principal ($) — the sum of money borrowed or investedR = rate of interest (% p.a.) — per annum (per year)T = term of interest (years) — the period of time for which the sum of money is to

be borrowed or invested

The sum of the principal, P, and the interest, I, is called the total amount and is denoted by the symbol A.

The formula used to calculate the total amount is given by:

A = P + Iwhere:

A = total amount at the end of the term ($)P = principal ($)I = simple interest ($)

Worked example 1

Calculate the amount of simple interest, I, earned and the total amount, A, at the end of the term if:a $12 000 is invested for 5 years at 9.5% p.a.b $2500 is invested for 3 months at 4.5% p.a.

think Write

a 1 Write the formula for simple interest. a IPRT=100

2 Write the known values of the variables. P = $12 000 R = 9.5%T = 5 years

3 Substitute the values into the given formula. I = × ×12 000 9 5 5100

.

4 Evaluate. = 570 000100

= 5700

5 Answer the question and include the appropriate unit.

The amount of interest earned is $5700.

6 Write the formula for the total amount. A = P + I

7 Substitute the values for P and I. = 12 000 + 5700

8 Evaluate. = 17 700


The total amount at the end of the term is $17 700.

b 1 Write the formula for simple interest. b IPRT=100

390

2 Write the known values of the variables.Note: T must be expressed in years, so divide 3 months by 12 months.

P = $2500 R = 4.5%T = 3 months

= 312

or 0.25 years


. .

4 Evaluate and round the answer to 2 decimal places.

= 2812 5100

.

= 28.13


The amount of interest earned is $28.13.

6 Write the formula for the total amount. A = P + I

7 Substitute the values for P and I. = 2500 + 28.13

8 Evaluate. = 2528.13


The total amount at the end of the term is $2528.13.

Worked example 2

Lilay is saving for a new computer. How long, to the nearest month, will it take her to earn $650 simple interest on $8375 that is invested at 6.25% p.a.?

think Write

1 Write the formula for simple interest. IPRT=100

2 Write the known values of the variables. I = $650P = $8375R = 6.25%

3 Substitute the values into the given formula. 6508375 6 25

100= × ×. T



4

5 Write the solution. Solving 6508375 6 25

100= × ×. t

for t gives

T = 1.24 179 104

6 Round the answer to 4 decimal places. T = 1.2418 years

7 Convert 0.2418 years to months by multiplying the decimal by 12.

T = 1 year and (0.2418 × 12) months= 1 year and 2.9016 months≈ 1 year and 3 months

8 Answer the question. It will take approximately 1 year and 3 months to earn $650 in simple interest.

Other examples of investments involving simple interest include investment bonds and debentures. Investment bonds are offered by both the state and federal governments and by government-owned companies.

Interest earned on investment bonds can be paid at various intervals, for example monthly, quarterly, every six months (semi-annually) or yearly. Bonds are traded on fi nancial markets. That is, they can be sold prior to the term expiry date (also known as bond maturity).

Debentures are similar to investment bonds but are issued by private companies to investors to raise capital. At the end of the term, the principal (or face value) is returned to the investor, while the interest earned is paid at various intervals.

Worked example 3

After comparing investment options from a variety of institutions, Lynda and Jason decided to invest their $18 000 in State Government bonds at 7.75% p.a. The investment is for 5 years and the interest is paid semi-annually (every six months). Calculate how much interest:a they receive in every paymentb will be received in total.

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Worked example 3

On a Calculator page, press:• MENU b• 3: Algebra 3• 1: Solve 1Complete the entry line as:

solve 6508375 6 25

100= × ×

.,

tt

Then press ENTER ·.

392

think Write

a 1 Write the formula for simple interest. a IPRT=100

2 Write the known values of the variables.Note: T must be expressed in years; divide 6 months by 12 months.

P = $18 000 R = 7.75%T = 6 months

= 612

or 0.5 years


. .

4 Evaluate. = 697 50100

= 697.5


Lynda and Jason receive $697.50 in interest every 6 months.


2 Write the known values of the variables. P = $18 000 R = 7.75%T = 5 years


.

4 Evaluate. = 697 500100

= 6975


Lynda and Jason will receive a total of $6975 in interest.

b Alternative method b

1 Multiply the interest received in each 6 month period by the number of 6 month periods in 5 years, that is, multiply $697.50 by 10.

Interest obtained each 6 months = $697.50Number of payments to be received = 10Total interest received = $697.50 × 10

= $6975

2 Answer the question. Lynda and Jason will receive a total of $6975 in interest.

Simple interest is given by 1. IPRT=100

.

The total amount is given by 2. A = P + I.When calculating simple interest, the interest earned is the same for each time period.3.

rememBer

Simple interest 1 We 1 For each of the following calculate:

i the amount of simple interest, I, earned ii the total amount, A, at the end of the term.a $1200 for 1 year at 10.5% p.a.b $8320 for 3 years at 6.45% p.a.

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c $960 for 2 12 years at 9.20% p.a.

d $3730 for 4 34 years at 7.39% p.a.

e $18 400 for 100 days at 6.89% p.a. (Hint: Write 100 days as a fraction of a year.)

f $2460 for 4 years and 3 months at 5.75% p.a.

g $126 000 for 6 months at 8.35% p.a.h $1998 for 40 days at 4.89% p.a.i $44 600 for 500 days at 11.15% p.a.j $4835 for 2 years and 5 months at

4.95% p.a.k $9862 for 2 years and 7 months at

11.25% p.a.l $115 for 1 year and 2 months at 3.125% p.a.m $14 995 for 5 months at 6.67% p.a.n $5000 for 3 months at 5.25% p.a.o $1750 for 90 days at 7.45% p.a.p $7920 for 120 days at 8.26% p.a.

2 Calculate the simple interest paid per year on the following investments.a $3690 at 11% p.a. b $22 400 at 6.85% p.a.

c $620 at 14 23% p.a. d $52 290 at 3.19% p.a.

e $7810 at 5.25% p.a. f $790 at 6 13% p.a.

g $1575 at 8 14

% p.a. h $49 500 at 7.62% p.a.

i $63 720 at 3.35% p.a. j $112 000 at 10.5% p.a.k $11 400 at 2.75% p.a. l $16 000 at 4.8% p.a.m $561 at 7.2% p.a. n $8471 at 5.86% p.a.o $27 512 at 9.21% p.a. p $13 635 at 2.9% p.a.

3 Calculate the simple interest that has to be paid if $4650 is invested on term deposit for 180 days at 5.75% p.a.

4 We2 How long, to the nearest month, will it take to earn $1950 simple interest if $16 325 is invested at 9.75% p.a.?

5 Norman borrowed $3500 for 8 months at 11 34

% p.a. simple interest. Calculate the total amount Norman must repay at the end of the term of the loan.

6 mC a The total value of an investment of $3500 after 2 years and 6 months if simple interest is paid at the rate of 5% per annum is:

A $437.50 B $826.25 C $1870.50D $3937.50 E $3975.50

b What sum, to the nearest dollar, must be invested for one year at 6% per annum simple interest in order to earn $1200 interest?A $21 200 B $2000 C $20 000D $12 000 E $12 200

c Rodney borrowed $3500 from the student credit union for 2 years at 8% per annum simple interest with repayments made by equal instalments at the end of each calendar month. The amount Rodney has to pay at the end of each month, to the nearest dollar, is:A $169 B $221 C $280D $23 E $61

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Digital docSkillSHEET 9.1

Time conversions — simple interest

exam tip Although Australian currency has a fi ve cent coin as its lowest denomination, fi nancial calculations are expected to be correct to a stated accuracy. Unless stated otherwise, answers correct to the nearest cent are expected, and these should not be rounded to the nearest fi ve cents. Domestic bills are often paid by electronic means and are usually calculated and debited correct to the nearest cent. Sums of money correct to fractions of a cent are common in fi nancial transactions. The price of petrol is given correct to the nearest tenth of a cent. The exchange rate for the Australian dollar is often quoted correct to the nearest hundredth of a cent. Students must carry out calculations to the nearest cent or an appropriate accuracy as specifi ed in the question.

[Assessment report 2007]

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Digital docSpreadsheet 118

Simple interest

394

7 Find how many months it takes to produce the simple interest in each of the following investments:

Investment ($) Simple interest ($) Interest rate (%)

a 850 65 8.25

b 1260 120 9.45

c 16480 2800 11.5

d 3450 600 6.65

e 2900 330 4.95

8 Determine how much money, to the nearest dollar, must be invested to produce:a $2120 simple interest over 4 years at 7.65% p.a.b $360 simple interest over 1 year at 5.55% p.a.c $8200 simple interest over 6 months at 11.25% p.a.d $935 simple interest over 1 year and 2 months at 6.25% p.a.

9 Find the rate of simple interest per annum charged for the following loans:a $3500 for 1.5 years with interest of $375b $8000 for 2 years with interest of $1800c $6200 for 9 months with interest of $406.88d $2450 for 1 year and 2 months with interest of $227.24.

10 We3 Sue and Harry invested $14 500 in State Government bonds at 8.65% p.a. The investment is for 10 years and the interest is paid semi-annually (that is, every six months). Calculate how much interest:a they receive every paymentb will be received in total.

11 Anna invested $85 000 in Ski International debentures. She earns 7.25% p.a., which is paid quarterly for one year.a Calculate how much interest:

i she receives quarterly ii will be received in total.

b Would Anna receive the same amount of interest over a 3-year period if it were paid annually rather than quarterly?

12 Mrs Williams invested $60 000 in government bonds at 7.49% p.a. with interest paid semi-annually (that is, every 6 months).a How much interest is she paid each 6 months?b How much interest is she paid over 3 1

2 years?

c For how long would she need to invest the money to earn a total of $33 705 in interest?

Compound interest, inf lation and appreciationCompound interestThe effect of compounding (which oil billionaire J. P. Getty called the ‘eighth wonder in the world’ and theoretical physicist Albert Einstein described as ‘the driving force of the Universe’) is a secret of fi nancial wealth creation.

As mentioned previously, when we are dealing with simple interest, the interest is the same for each time period.

9B

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Compound interest



In calculating compound interest, though, the principal on which interest is calculated is increased by adding (or reinvesting) the interest at the end of each interest period during the term. Therefore, interest is calculated on the principal as well as the interest over each time period. This is the process by which our savings accounts earn interest. The difference compounding makes can be illustrated in the following table. Let us consider an amount of $1000 to be invested for a period of 5 years at an interest rate of 10% p.a. We will compare the interest earned using (i) simple interest and (ii) compound interest.

Simple interest Compound interest

Initial principal, P = $1000 Initial principal, P = $1000

Rate of interest, R = 10% Rate of interest, R = 10%

Interest for Year 110% of $1000 I1 = $100


Principal at the beginning of Year 2P2 = $1000

Principal at the beginning of Year 2P2 = $1000 + $100 = $1100










Interest for Year 410% of $1331 I4 = $133.10


Principal at the beginning of Year 5P5 = $1331 + $133.10 = $1464.10


Interest for Year 510% of $1464.10 I5 = $146.41

The simple interest earned over a 5 year period is $500.

The compound interest earned over a 5 year period is $610.51.

The previous table has illustrated how, as the principal for compound interest increases periodically, that is, P = $1000, $1100, $1210, $1331, $1464.10, . . . , so does the interest, that is, I = $100, $110, $121, $133.10, $146.41, . . . , while both the principal for simple interest and the interest earned remain constant, that is, P = $1000 and I = $100. The difference of $110.51 between the compound interest and simple interest earned in a 5-year period represents the interest earned on added interest.

If we were to place the set of data obtained in two separate tables and represent each set graphically — as the total amount of the investment, A, versus the year of the investment, n — we would find that:

396

1. The simple interest investment is represented by a straight line as shown below.

n A A

n

1500

1400

1300

1200

1100

1000

10 2 3 4 5

0 1000

1 1100

2 1200

3 1300

4 1400

5 1500

2. The compound interest investment is represented by an exponential graph as shown below.

n A A

n

1610.51

1464.101331121011001000

10 2 3 4 5

0 1000

1 1100

2 1210

3 1331

4 1464.10

5 1610.51

The above data may be entered into a CAS calculator and used to plot each of the graphs. To graph the data from the tables above using the CAS calculator, follow these steps.


1. Open a Lists & Spreadsheet page.2. For the simple interest investment,

enter the data into the spreadsheet as follows.

Enter the values of n into column A and values of A into column B.

Label column A: n and column B: amount.

3. To plot the data, open a Data & Statistics page.

Tab to each axis to select ‘Click to add variable’. Place ‘n’ on the horizontal axis and ‘amount’ on the vertical axis. The graph should appear as shown.


Formula for compound interestWe will now derive a mathematical relationship for compound interest. As seen from the example illustrated previously, the initial amount:

A0 = P = $1000.

At the end of Year 1: At the end of Year 2: At the end of Year 3:

A1 = P + I1 A2 = A1 + I2 A3 = A2 + I3

= P + 10%P = A1 + 0.1A1 = A2 + 0.1A2

= +P P10100

= A1(1 + 0.1) = A2(1 + 0.1)

= P(1 + 0.1)(1 + 0.1) = P(1 + 0.1)2 (1 + 0.1)

= P + 0.1P = P(1 + 0.1)2 = P(1 + 0.1)3

= P(1+ 0.1)1

Following the pattern emerging:A4 = P(1 + 0.1)4

A5 = P(1 + 0.1)5

Generally:

A Pr

n

= +1100

4. Repeat steps 1 to 3 for the compound interest investment.

398

whereA = the amount at the end of n compounding periods ($)P = principal ($)r = rate of interest per period (%)n = number of compounding periods

Note: In the formula for compound interest, r is the rate of interest per period, not per annum and n is the number of compounding periods, not years. It refl ects the fact that compounding occurs not only on an annual basis but can be more frequent: that is, semi-annually (half-yearly), quarterly (every three months), monthly, weekly or daily.

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Worked example 4

Tamara has $15 000 to invest for 3 years. She considers the following options: i a term deposit at 5.25% p.a. compounded annually ii shares paying a dividend rate of 5.08% p.a. compounded quarterly iii a building society paying a return of 5.4% p.a. compounded monthly iv a business venture with guaranteed return of 7.3% p.a. compounded daily.All the investments are equally secure. Advise Tamara which option to take.

think Write

1 Write the formula for compound interest. A Pr

n

= +

1100

2 The value of P will be the same for all calculations. Substitute it into the formula.

P = $15 000

Ar

n

= +

15000 1100

3 Write the known values of the variables. r and n for each investment.

For investment i: i r = 5.25%n = 3

For investment ii: the interest rate is given as per annum, but we need to know the rate per quarter; hence, we divide 5.08 by 4. The interest is compounded quarterly, that is, 4 times in one year. Therefore in 3 years, interest will be compounded 12 times.

ii r = 5 084.

= 1.27% per quartern = 4 × 3

= 12

For investment iii: the interest rate is given as per annum, but we need to know the rate per month; hence, we divide 5.4 by 12. The interest is compounded monthly, that is, 12 times in one year. Therefore in 3 years, interest will be compounded 36 times.

iii r = 5 412.

= 0.45% per monthn = 12 × 3

= 36

For investment iv: the interest rate is given as per annum, but we need to know the rate per day; hence, we divide 7.3 by 365. The interest is compounded daily, that is, 365 times in one year. Therefore in 3 years, interest will be compounded 1095 times.

iv r = 7 3365

.

= 0.02% per dayn = 365 × 3

= 1095



4

5

6 Write the value of each investment correct to the nearest cent.

Investment i: Amount = $17 488.70Investment ii: Amount = $17 452.63Investment iii: Amount = $17 631.49Investment iv: Amount = $18 672.06

7 Compare amounts from options i to iv and answer the question.

The best option for Tamara is iv, as she receives the greatest amount of interest in the same period of time.

inflationOne of the measures of how an economy is performing is the rate of inflation. Inflation is the rise in prices within an economy and is generally measured as a percentage. In Australia this percentage is called the Consumer Price Index (CPI). By looking at the inflation rate we can estimate what the cost of various goods and services will be at some time in the future.

To estimate the future price of an item one year ahead, we increase the price of an item by the rate of inflation.

Worked example 5

The cost of a new car is $68 000. If the inflation rate is 6%, estimate the price of the car after one year.

think Write

1 Write the relationship between the future price and current price of the car.

Future price = current price + 6% current price

For investment i, substitute the values of the pronumerals into the formula and evaluate.On a Calculator page, complete the entry line as:

15 000 1100

× +

rn

| r = 5.25 and n = 3


To find the value of investments ii, iii and iv, re-enter the expression for investment i, change the values of r and n; then press ENTER ·.

400

2 Substitute the values into the given formula. Future price = 68 000 + 6100

× 68 000

3 Evaluate. = 68 000 + 4080= 72 080

4 Answer the question and include the appropriate unit. The price of the car after one year is $72 080.

When calculating the future cost of an item several years ahead, the method of calculation is the same as for compound interest. This is because we are adding a percentage of the cost to the cost each year.

Worked example 6

The cost of a wide-screen high-definition plasma television is $4999. If the average inflation rate is 4%, estimate the cost of the television after five years (to the nearest dollar).

think Write

1 Write the formula for compound interest and any known values.Note: P is the original price, r is the inflation rate and n is the number of years.

A Pr

n

= +

1100

P = 4999r = 4%n = 5

2 Substitute the values into the given formula. A = +

4999 14

100

5

3 Evaluate. = 4999(1 + 0.04)5

= 4999(1.04)5

= 6082.047 859


The cost of the television after five years is $6082.

Inflation causes prices to rise and therefore money loses its value; that is, the amount we are able to purchase (buying power) decreases.

appreciationA similar calculation can be made to anticipate the future value of collectable items, such as stamp collections, memorabilia from special occasions, land, paintings and antiques. This type of item increases in value over time if it becomes rare, and rises at a rate much greater than inflation. The amount by which an item grows in value over time is known as appreciation.

Worked example 7

Gianluca purchases a rare stamp for $360. It is anticipated that the value of the stamp will rise 9% per year. Calculate the value of the stamp after 15 years, correct to the nearest $10.

think Write

1 Write the formula for compound interest and any known values.Note: The compound interest formula is used as the price of the stamp increases by 9% every year.

A Pr

n

= +

1100

P = 360r = 9%n = 15



2 Substitute the values into the given formula. A = +

360 19

100

15

3 Evaluate. = 360(1 + 0.09)15

= 360(1.09)15

= 1311.293 6854 Answer the question and include the appropriate

unit.The value of the stamp after 15 years (to the nearest $10) is $1310.

If the difference between the amounts received for an asset is greater than the amount originally paid for the asset, a capital gain is obtained. When a capital gain or profit occurs as a result of selling certain assets such as shares and investment properties, then a capital gains tax must be paid to the Australian Taxation Office. The calculations required to determine the actual capital gain are beyond the scope of this course and will be covered in Further Mathematics.

Worked example 8

Linda and James bought a house for $112 000 in an area where house prices rise (appreciate) on average 4% per year. They decided to hold on to their house until its value is at least $200 000. How many years should Linda and James wait until they sell their current house?

think Write

1 Write the formula for compound interest and any known values.Note: The compound interest formula is used as the price of the house increases by 4% every year.

A Pr

n

= +

1100

P = $112 000r = 4%n = $200 000

2 Substitute the values into the given formula. 200 000 112 000 14

100= +

n

3

4 Write the solution. Solving 200 000 112 000 14

100= +

n

for n gives

n = 14.7 834 766

5 Answer the question. Linda and James should wait for approximately 15 years before selling their house for the desired price.


solve 200 000 112000 14

100= +

n

n,


402

Note: If solving equations where the power is unknown manully, logarithms are used. Consider the calculations below:

200 000 112000 14

100= +

n

200 000112000

112000 1 04112000

= ( . )n

2514

1 04= ( . )n

log log ( . )10 102514

1 04

= n

= n × log10 (1.04)

log

log ( . )log ( . )

log

10

10

10

2514

1 041 04

= ×n

110 1 04( . )

0 2518119730 01703339..

= n

n = 14.78

Compound interest formula is given by 1. A Pr

n

= +

1100

.

When dealing with compound interest, the interest is calculated on the principal as 2. well as the interest over each time period.Inflation is the measure of the rate at which prices increase.3. The inflation rate is given as a percentage and is called the Consumer Price Index.4. To estimate the cost of an item after a particular year or number of years, we use the 5. compound interest formula, using P as the original price, r as the inflation rate and n as the number of years.Rare items such as collectables and memorabilia increase in value as time goes on at a 6. rate that is usually greater than inflation.The amount by which an item grows in value over time is known as appreciation.7. A capital gain occurs when the amount received for an asset is greater than the amount 8. originally paid for the asset.If solving equations where the unknown is a power, take the logarithm to base 10 of 9. both sides of the equation; that is,

an = blog10 (a)n = log10 (b)

n × log10 (a) = log10 (b)

nba

= log ( )log ( )

10

10

rememBer

Compound interest, inf lation and appreciation 1 We4 Rosemary has $25 000 to invest for 5 years. She considers the following options:

a a term deposit at 6.75% p.a. compounded annuallyb shares paying a dividend rate of 5.15% p.a. compounded quarterly

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c a building society paying a return of 5.3% p.a. compounded monthlyd a business venture with guaranteed return of 6.4% p.a. compounded daily. (Assume there

is only one leap year in the given 5-year period.)All the investments are equally secure. Advise Rosemary which option to take.

2 Warren wishes to invest $10 000. He has four investment alternatives:a simple interest at 9% p.a.b compound interest at 8% p.a.c compound interest at 7 1

2% p.a. adjusted quarterly

d compound interest at 7% p.a. adjusted daily.Calculate the amount of his investment at the end of 7 years for each, and hence advise Warren on which investment alternative will produce the greatest return.

3 For each of the following, calculate the interest on the investment compounded: i annually ii semi-anually iii monthly iv daily.

(Hint: The compound interest can be found as a difference between the compounded amount and the principal.)a $2500 for 2 years at 4.95% p.a. b $960 for 1 year at 4.50% p.a.

c $7850 for 3 12 years at 5.85% p.a. d $162 750 for 2 1

2 years at 5.45% p.a.

e $14 000 for 4 years at 9.65% p.a. f $125 800 for 20 years at 11.95% p.a. 4 mC $10 000 is invested at a rate of 10% per annum compounding half yearly. The value, in

dollars, of this investment after five years, is given by:A 10 000 × 0.10 × 5 B 10 000 × 0.05 × 10 C 10 000 × 0.0510

D 10 000 × 1.0510 E 10 000 × 1.105 [©VCAA 2007]

5 Bettina and Brendan invested $1000 for their son Matthew on his first birthday. They placed the money in a term deposit with an interest rate of 5.2% p.a. compounded monthly. Calculate the amount of Matthew’s investment on his 21st birthday.

6 Over the last 3 years a comprehensive hospital cover from ‘TakeCare’ private medical insurance rose at an average of 9.5% and currently costs $1980 per year. With this rate of increase continued, what would be the insurance premium after another 3 years?

(Hint: The increase in premium compounds each year.)

7 We5 The cost of a motorcycle is $20 000. If the inflation rate is 4%, estimate the cost of the motorcycle after one year.

8 We6 An electrical guitar is priced at $850 at the beginning of 2005.a If the infl ation rate is 3.3% p.a., estimate the cost of the guitar at the beginning of 2006.b The government predicts infl ation to fall to 2.7% in 2006. Estimate the cost of the guitar

at the beginning of 2007.

9 The cost of a litre of milk is $1.70. If the inflation rate is an average 6%, estimate the cost of a litre of milk after 10 years.

10 A daily newspaper costs $1.00. With an average inflation rate of 3.4%, estimate the cost of a newspaper after 5 years (to the nearest 5c).

11 mC At the start of each year Joe’s salary increases to take inflation into account. Inflation averaged 2% per annum last year and 3% per annum the year before that. Joe’s salary this year is $42 000. Joe’s salary two years ago, correct to the nearest dollar, would have been:A $39 900 B $39 925 C $39 926D $39 976 E $39 977 [©VCAA 2007]

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Digital docSkillSHEET 9.2

Conversions — compound

interest

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Compound interest

exam tip An incorrect solution strategy, which was used by the 42% of students, was to fi rst depreciate his current salary of $42 000 by 3% and then by 2%. The error here is that, for example, a 3% depreciation of this year’s salary is not equal to a 3% increase in the previous year’s salary. This is because the percentages are applied to different bases.


404

12 We7 A ring bought for $240 appreciates (increases in value) by 14% p.a. If this rate of appreciation continued, what would be the value of the ring after 25 years?

13 Merv bought a shirt signed by the Australian cricket team after it won the 2007 World Cup for $250. If the value of the shirt continues to increase by 25% p.a., calculate the value of the shirt (to the nearest $10) in 2017.

14 Peter purchased a rare bottle of wine for $420. If the value of the wine is predicted to increase at 8% p.a., estimate the value of the wine in 15 years (to the nearest $10).

15 The 1968 Australian 2 cent coin is very rare. If a coin collector purchased one in 2006 for $500 and the value of the coin increases by 20% per year, calculate the value in 2020 (to the nearest $10).

16 We8 Gaetano bought a house for $450 000 in an area where house prices appreciate an average 3% per year. He decided to hold on to his house until its value is at least $650 000. How many years should he wait until he sells his current house?

17 A painting is currently valued at $4500 and is known to appreciate an average 8% per year. After how many years would the painting have a value of at least $25 000?

18 According to the last census, Whitehorse Marsh has a population of 23 600. The population increases at a constant rate of 3.5% p.a. How long, to the nearest month, will it take for the population of Whitehorse Marsh to reach the 4 0 000 mark?

19 mC a A 1.25 L bottle of soft drink costs $1.50. With the annual infl ation rate over the next 4 years expected to be 4%, the price of the bottle then will be approximately:

A $1.54 B $6.00 C $1.75 D $2.11 E $1.84

b What is the amount, rounded to the nearest $100, to be invested for 6 years and compounded semi-annually at 8% p.a.? The value of the investment at the end of the term is $15 000.A $24 500 B $9400 C $11 000 D $900 E $8500

20 Find the required annual rate for the following investments with compound interest.

Value of the investment at the end

of the period ($) Principal ($)Compounding

periods (n)

a 8 800 8 000 2

b 62 000 50 000 6

c 24 000 16 500 5

d 15 700 15 000 3

e 626 479 4

f 1 700 1 655 2

g 60 000 53 570 5

h 35 000 23 900 7

i 125 000 96 890 4

j 269 000 210 000 8

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Digital docWorkSHEET 9.1



Savings and credit card accountsSavings accountsWhen considering opening a savings account it is important to understand the types of accounts available and how each of them operate. Money going into an account (credit or deposit) or coming out of (debit or withdrawal) of an account is called a transaction. Transactions in savings accounts can be recorded in a passbook or on statements that are posted out monthly.

Worked example 9

For the given statement below calculate:a the total creditsb the total debitsc the balance after each transactiond the final balance for the period of the statement.

Date Transaction detail Debit Credit Balance

30 Nov 2009 Opening balance $6846.74 CR

2 Dec Credit interest 8.04

ABWDL Northland 200.00

State Govt Tax GDT 5.10

3 Dec ABWDL Jam Factory 350.00

9 Dec ABWDL Jam Factory 80.00

10 Dec Salary 1628.92

19 Dec ABWDL Camberwell 420.00

24 Dec Salary 1628.92

26 Dec ABWDL Chadstone 628.54

31 Dec 2009 Closing balance

think Write

a 1 Look at the credit column of the statement and add each of the entries.

a Total credits = 8.04 + 1628.92 + 1628.92= 3265.88


The total credits for the given month are $3265.88.

b 1 Look at the debit column of the statement and add each of the entries.

b Total debits = 200 + 5.10 + 350 + 80 + 420 + 628.54 = 1683.64


The total debits for the given month are $1683.64.

9C

406

c 1 Write the opening balance and take note of the first transaction. If the transaction appears in the credit column, it is added on. If the transaction appears in the debit column, it is subtracted.

c Transaction 1 ⇒ 6846.74 + 8.04 = 6854.78Transaction 2 ⇒ 6854.78 − 200 = 6654.78Transaction 3 ⇒ 6654.78 − 5.10 = 6649.68Transaction 4 ⇒ 6649.68 − 350 = 6299.68Transaction 5 ⇒ 6299.68 − 80 = 6219.68Transaction 6 ⇒ 6219.68 + 1628.92 = 7848.60Transaction 7 ⇒ 7848.60 − 420 = 7428.60Transaction 8 ⇒ 7428.60 + 1628.92 = 9057.52Transaction 9 ⇒ 9057.52 − 628.54 = 8428.98

2 Perform each calculation and enter answers in the ‘Balance’ column of the table.


30 Nov 2009 Opening balance $6846.74 CR

2 Dec Credit interest 8.04 $6854.78 CR

ABWDL Northland 200.00 $6654.78 CR

State Govt Tax GDT

5.10 $6649.68 CR

3 Dec ABWDL Jam Factory

350.00 $6299.68 CR

9 Dec ABWDL Jam Factory

80.00 $6219.68 CR

10 Dec Salary 1628.92 $7848.60 CR

19 Dec ABWDL Camberwell

420.00 $7428.60 CR

24 Dec Salary 1628.92 $9057.52 CR

26 Dec ABWDL Chadstone 628.54 $8428.98 CR

31 Dec 2009 Closing balance $8428.98 CR

d 1 Look at the bottom of the balance column of the table from part c.

d


The final balance for the period of the statement is $8428.98.

Two methods used by banks for calculating interest on savings accounts are: (i) minimum monthly balance and(ii) minimum daily balance.Once calculated, the interest is added to the account at specific times; for example, if interest is paid quarterly, it will be deposited into the account on the first day of March, June, September and December. Let’s investigate how the two methods of calculating interest work and which is the better option.

minimum monthly balanceAs the name suggests, the bank looks at the balances of the savings account for each month. It then calculates the interest on the smallest balance that appears each month.



Worked example 10

The passbook page below shows all the transactions for the month of November. Find the interest that will be earned in November if the bank pays 6.8% p.a. simple interest on the minimum monthly balance.

Date Deposit Withdrawal Balance

4 November 582.75 $789.63

12 November 300.00 $489.63

18 November 582.75 $1072.38

22 November 200.00 $872.38

28 November 160.00 $1032.38

think Write

1 Look at the passbook page carefully and state the smallest balance for November.Note: The balances at the beginning and end of the month must be checked even though they may not appear on the passbook page.

The balance on the 1–3 November, if shown, would have been $789.63 – $582.75 = $206.88. Hence, the minimum monthly balance is $206.88.

2 Write the formula for simple interest. IPRT=100

3 List the values of P, R and T. P = 206.88R = 6.8% p.a.T = 1 month

= 112

or 0 083. years

4 Substitute the values into the given formula. IPRT=100

5 Evaluate. = × ×206 88 6 8 0 083100

. . .

= 1.172 32


The interest earned for November was $1.17.

minimum daily balanceIn this case, the bank looks at the various balances of the savings account for each month. The number of days the balance is maintained is used to calculate the interest.

Worked example 11

Use the passbook page from worked example 10 and the minimum daily balance method to find the interest that will be earned in November if the bank pays 6.8% p.a. simple interest on the minimum daily balance.

408

think Write

1 Set up a table showing each new balance and the number of days the balance applies. Look at all running balances including those for 1 and 30 November.

Date Balance

Number of days the balance applies

Simple interest calculations

IPRT==100

Interest earned

1 November $206.88 3206 88 6 8

100

3365

. .× ×$0.1156

4 November $789.63 8 789 63 6 8

100

8365

. .× × $1.1769

12 November $489.63 6489 63 6 8

100

6365

. .× ×$0.5473

18 November $1072.38 41072 38 6 8

100

4365

. .× ×$0.7991

22 November $872.38 6872 38 6 8

100

6365

. .× ×$0.9752

28 November $1032.38 31032 38 6 8

100

3365

. .× ×$0.5770

Total $4.1911

2 Calculate the simple interest for each balance. As the interest rate is in % per annum, express the number of days as a fraction of a year; for example, 3 days = 3

365 of

a year.

3 Sum the interest. Note: The calculations are to hundredths of a cent for accuracy.

4 Round to the nearest cent. $4.1911 ≈ $4.195 Answer the question

and include the appropriate unit.

The interest earned for November was $4.19.

In comparison, the minimum daily balance method offers more interest than the minimum monthly balance method because it credits the customer for all monies in the account.

Credit cards and accountsCredit cards allow you to purchase goods and services without paying for them on the spot. They can also be used for obtaining cash advances, paying bills and making purchases over the phone or on the Internet.

Common credit cards used in Australia include MasterCard, Visa, Diners Club and American Express.

When applying for a MasterCard or Visa, a customer is given a choice of having either an interest-free period (usually up to 55 days) for a small annual fee (around $30), or no fee payable and no interest-free period (with the interest rate usually being lower for the second option). Each cardholder is offered a certain limit of credit.

A monthly statement showing all transactions for the previous month is issued for every cardholder. Upon receiving a monthly statement, a customer may decide to pay the bank in full by the due date indicated on the statement and hence not have to pay any interest with an interest-free period card. Alternatively, the customer may choose to make the minimum payment only. In this case interest will be charged on the unpaid balance. The minimum payment is usually a certain percentage of the unpaid balance or a certain fixed amount — whichever is larger. Variations in interest rates occur from time to time and cardholders are notified of these changes in advance.



annual ratesThe annual percentage rates for some standard credit cards are:• 16.90% for MasterCard/Visa (up to 55 interest-free days with an annual fee)• 15.25% for MasterCard/Visa (no interest-free days with no annual fee).

minimum paymentsFor the examples and exercises in this chapter, minimum credit card account payments are calculated as follows.

Balance less than $25If the closing balance of the statement is less than $25, then the minimum payment is the same as the closing balance.

Balance more than $25Where the closing balance is greater than $25, the payment to be made is the greater of:1. $25, or2. 1.5% of the closing balance (rounded down to the nearest $1 if the balance exceeds $1700).Note: If the closing balance is greater than the credit limit of the card, then the minimum payment must also include the excess of the balance over the credit limit.

Worked example 12

Find the minimum payment due for each of the following balances using the information supplied previously.a $23.40 b $1836.25 c $280.10 d $1960 with the credit limit being $1900

think Write

a Since the closing balance is under $25, it should be paid in full.

a As $23.40 < $25,the amount due = $23.40.

b 1 Since the closing balance is over $1700, calculate 1.5% of it.

b Amount due = 1.5% of $1836.25

= 1 5100

. × 1836.25

= $27.54

2 Round down to the nearest dollar. Rounded down to the nearest dollar, the amount due is $27.

c Since the closing balance is above $25but below $1700, the minimum paymentis $25.

c $25 < $280.10 < $1700Therefore payment due = $25

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Worked example 12

410

d 1 Since the closing balance is above $1700, calculate 1.5% of it and round down to the nearest dollar.

d 1.5% of $1960 = 1 5100

. × 1960

= 29.40= 29

2 Calculate the excess of the closing balance above the credit limit.

The excess of the closing balance above the credit limit = $1960 − $1900

= $60

3 Add the two amounts. 29 + 60 = 89. The amount due is $89.

For the ‘no interest-free period’ option the interest charged on the outstanding amount of each purchase and cash advance is charged from the date of the purchase (or cash advance) and until the purchase (or cash advance) is repaid in full. The same is true for cash advances, obtained with ‘55 days interest-free period’ cards. The procedure is explained below.

No interest-free days cardsInterest is charged on the outstanding amount of each purchase and cash advance from the date the purchase or cash advance is debited to your card account until you repay the purchase or cash advance.

Interest is calculated for a statement period in three steps:• first, the outstanding balances are averaged over the statement period;• then the average is multiplied by the daily percentage rate applying to your card

account; and• finally, the result from the prior step is multiplied by the number of days in the

statement period.The result from the last step is the amount of interest charged to your card account in the statement period.

When is interest debited?Your card account is debited on the last day of each statement period with the interest calculated during that statement period up to and including that last day.

Worked example 13

For a ‘no interest-free period’ credit card, calculate the interest charged on the average outstanding daily balance of $220 with the interest rate of 15.25% p.a. if the statement covers a 30-day period.

think Write

1 Calculate the daily percentage rate. Daily interest rate = annual interest rate

365

= 15 25365

.

= 0.041 78%

2 Calculate the daily interest charged on the outstanding balance.

Interest = 220 × 0 04178

100.

= 0.091 92



3 Find the interest charged over the 30-day period and round to the nearest cent.

Total interest for 30 days= 0.091 92 × 30= 2.757 48= 2.76 (to the nearest cent)

For ‘up to 55 days interest free’ credit cards, no interest is charged if the amount is paid in full by the due date, which is usually 25 days from the date of the statement. If the closing balance is not repaid in full by the due date, the cardholder then temporarily loses the interest-free option. The interest is usually charged on the outstanding balance from the day of the first purchase (that is, it is backdated!) until the outstanding balance is paid in full. Any purchases made before the balance is fully repaid are also added to the total. Thus, if the balance is not paid in full by the due date, the card is effectively a ‘no interest-free period’ card, but with the higher interest rate being applied.

Worked example 14

For a ‘55 days interest free’ credit card, calculate the amount of interest charged on an outstanding balance of $450 which was repaid 10 days after the due date, given that the first purchase was made on the first day of the 30-day statement period and the annual percentage interest rate was 16.90%. (Assume that no other purchases were made after the end of the statement.)

think Write

1 Calculate the length of time for which the interest is charged, keeping in mind that it is charged from the date of the first purchase and until the balance was repaid.

The number of days from the first purchase to the last day of statement = 30 (as the purchase was made on the first day and the period covers 30 days).The number of days from the date of the statement to the due date = 25.The number of days from the due date to the date of actual payment = 10.Total days = 30 + 25 + 10 = 65

2 Calculate the daily interest rate. Daily interest rate = 16 90365. %

= 0.046 301%

3 Find the interest charged on $450 over the period of 65 days and round to the nearest cent.

Interest = $450 × 0 046 301

100.

× 65

= $13.54

Worked example 15

Kerry has a credit card with an interest-free period, and interest is then charged on the outstanding balance at a rate of 18% p.a. Kerry pays a $1200 bill for her council rates on her credit card.a Kerry pays $600 by the due date. What is the outstanding balance on the card?b Calculate the interest Kerry must then pay for the second month.c An alternative credit card charges 12% p.a. interest with no interest-free period. Calculate the

interest that Kerry would have been charged on the first month.d Calculate the balance owing after Kerry pays $600 and then calculate the interest for the second

month.e Which credit card would be the cheapest to use for this bill?

412

think Write

a Subtract the repayment from the balance. a Balance owing = $1200 − $600= $600

b Use the simple interest formula to calculate one month’s interest.

b Interest = 600 18

100

112

× ×

= $9.00

c Use the simple interest formula to calculate the first month’s interest.

c Interest = 1200 12

100

112

× ×

= $12.00

d 1 Add the interest to the amount of the bill and subtract the repayment.

d Balance owing = $1200 + $12 − $600= $612

2 Use the simple interest formula to calculate the second month’s interest.

Interest = 612 12

100

112

× ×

= $6.12

e Add the two months of interest together for the second card and compare with the interest for the first card.

e The interest on the second card is $18.12; therefore, the card with the interest-free period is cheaper in this case.

Two methods used by banks for calculating interest on savings accounts are:1. (a) minimum monthly balance(b) minimum daily balance.The minimum daily balance offers the best rate of interest as it credits the customer for 2. all monies in the account.A credit card is a source of an instant loan to the cardholder.3. The two options for Visa or MasterCard are:4. (a) no annual fee and no interest-free period(b) an annual fee and a specific interest-free period.The bank requires a minimum monthly payment that is usually the greater of a certain 5. fixed amount or a specific percentage of the closing balance.For all transactions made with a ‘no interest-free period’ card and for cash advances 6. obtained with an ‘up to 55 days interest-free period’ credit card, the interest is calculated from the date of the first purchase.For interest-free period credit cards, if the closing balance is paid in full by the due 7. date indicated on the statement, no interest is incurred. Otherwise, interest is charged until the balance is repaid.

rememBer

Savings and credit card accounts 1 We9 For the given statement below calculate:

a the total creditsb the total debitsc the balance after each transactiond the final balance for the period of the statement.

exerCiSe

9C




30 Sep 2009 Opening balance $12 526.07 CR

1 Oct ABWDL Chadstone 65.89

ABWDL Northland 400.00

State Govt Tax GDT 12.85

5 Oct Salary 1895.89

8 Oct ABWDL Jam Factory 750.93

16 Oct ABWDL Camberwell 590.86

19 Oct Salary 1895.89

27 Oct ABWDL Northland 100.00

29 Oct ABWDL Chadstone 1628.54

31 Oct 2009 Closing balance

2 We 10 A bank savings passbook showed that the opening balance for the month was $4862.95. In that given month Joseph paid the following bills out of the account: electricity $42.89, telephone $260.78 and rent $1086.14. Joseph’s wage of $2656.24 for the month was also deposited into the account. The bank pays 4.8% p.a. simple interest on the minimum monthly balance.a What was Joseph’s minimum monthly balance?b If interest is paid monthly into the account, how much interest did Joseph earn that month?

3 The passbook page below shows all the transactions for the month of August. Find the interest that will be earned in August if the bank pays 5.6% p.a. simple interest on the minimum monthly balance.

Date Deposit Withdrawal Balance

5 August 639.21 948.75

11 August 786.50

19 August 639.21

21 August 424.18

25 August 80.75

31 August 71.60

4 We 11 Use the passbook page from question 3 and the minimum daily balance method to find the interest that will be earned in August if the bank pays 5.6% p.a. simple interest on the minimum daily balance.

5 For the month of September, Luciana received $5.80 interest on her savings account. Luciana’s minimum balance in September was $625.60. What was the per annum (p.a.) simple interest rate offered by the bank?

6 Marcella receives the following statement from her bank. Due to a computer error the interest and balance were not calculated. Marcella contacted the bank and was informed that she received interest at a rate of 5 1

4% p.a. paid monthly on her minimum monthly balance.

414

Copy out Marcella’s statement and fill in the balance and interest payments.

Date Transaction Debit Credit Balance

1 August 2009 Balance B/F 4172.90 CR

4 August 2009 Deposit 178.15

8 August 2009 Cheque 3192 628.50

17 August 2009 Wages 1191.28

23 August 2009 Cheque 3193 486.75

31 August 2009 Interest

3 September 2009 Cheque 3195 511.29

8 September 2009 Deposit 42.60

13 September 2009 Wages 1191.28

20 September 2009 ATM 500.00

23 September 2009 Cheque 3194 840.00

30 September 2009 Interest

2 October 2009 Deposit 17.80

10 October 2009 ATM 120.00

16 October 2009 Wages 1191.28

24 October 2009 Cheque 3196 789.70

31 October 2009 Interest

7 Using the bank statement from question 6, another bank offers to show Marcella that minimum daily balance credited each quarter is more rewarding. The interest is still 5 1

4% p.a. but is

credited only at the end of each quarter; that is, on 31 October. Calculate:a the interest for the quarter ending Octoberb the increase in interest earned using the minimum daily balance method.

8 Roberto has the following income (credits) and expenses (debits) for June and July.Income: $1254.68 salary each fortnight beginning 8 June $528.37 income tax refund on 29 JulyExpenses: $828.40 rent on 30 June and 30 July $532.65 car registration on 15 June $71.57 electricity on 7 July $599 gym membership on 10 July $138.57 Visa account on 24 JulyDraw up a statement (as for question 6) for Roberto, remembering that he receives 3.5% interest paid on the last day of each month on the minimum monthly balance in the account and the opening balance for June is $1262.80.

9 mC A bank statement for the month of October is shown below.

Date Description of transaction Debit Credit Balance

01 Oct Opening balance 853.92

01 Oct Withdrawal — Internet banking 380.00 473.92

16 Oct Deposit — Cheque 518.15 992.07

18 Oct Credit card payment 125.56 866.51

23 Oct Withdrawal — Internet banking 250.00 616.51

31 Oct Closing balance 616.51

Interest on this account is calculated at a rate of 0.15% per month on the minimum monthly balance.



The interest payment for the month of October will be:A $0.19 B $0.57 C $0.71D $0.92 E $1.28 [©VCAA 2006]

10 We 12 The XYZ Bank requires the minimum payment of credit card balances to be:(a) the closing balance if it is under $25, or (b) the greatest of:

(i) the excess of the closing balance over the credit limit, or (ii) $25, or(iii) if the closing balance is greater than the credit limit of the card, then the minimum

payment must include this excess.Calculate the minimum payments on each of the following balances.a $17.50 b $26.49c $147.42 d $785.00e $1326.12 f $2312.58g $3489.60 h $1954.00 with a limit of $1900i $2320.48 with a limit of $2300 j $3080.00 with a limit of $3000

11 We 13 For a ‘no interest-free period’ credit card, calculate the interest charged on an average outstanding daily balance of $430 with a percentage interest rate of 14.01% p.a. if the statement covers a 30-day period.

12 An ‘up to 55 days interest-free’ credit cardholder used his card on 15 March to obtain a cash advance of $365, which he repaid on 20 March. What was the amount of interest charged on the cash advance at the rate of 15.01% p.a.?

13 Here is some information extracted from a monthly credit card statement:Statement begins: 1 April; Statement ends: 30 April; Payment due date: 25 May.

Date Transaction Details Amount

3 Apr HBA $180.00

8 Apr Myer Chadstone $89.00

16 Apr Optus $252.25

22 Apr Coles Glen Huntleigh $112.90

30 Apr Sportsgirl City $69.95

a Calculate the interest-free period for each of the above transactions.b Complete the following sentence: ‘To make full use of the “up to 55 days interest-free”

option, the purchases should be made at the _______________ of the statement period.’

14 We 14 For a ‘55 days interest-free’ credit card, calculate the amount of interest charged on an outstanding balance of $625 that was repaid a fortnight after the due date, given that the first purchase was made on the first day of the 30-day period and the annual percentage rate was 14.98%. (Assume that no other purchases were made after the end of the statement in question.)

15 Study the statement for the ‘55 days interest-free period’ credit card on the following page and answer these questions.a What is the length of the period of time covered by this statement?b What was the closing balance of the previous statement?c Did the payment of the previous statement balance incur any interest charges? Explain

your answer.d Explain how the minimum amount due was calculated.e Explain how the amount of available credit was calculated.

16 The closing balance for the statement in question 14 was repaid in full on 20 June. Find the amount of interest charged, if:a no further purchases were made until that dateb a further $300 was spent on 31 May.

416

Date

29 Apr

30 Apr

3 May

8 Apr

4 May

4 May

5 May

Reference Number

74900052MENTAJ

89101123XYZ

FIZ3456ROGERDUTY

72345670J4U00ABCD

12345678GOODILUV

789108ABCD1234

7654321XYZWRST

MR JOHN CITIZEN

Transaction Details

Payment received - thank you

Interest charges

Goverment duties - last month

Travel Wide Melbau

Books & Musical World Carlton AU

SCUD Shoes Noble Park AU

Groovy Music Nth Mlbourne AU

Amount (A) $

22.10-

2.50

0.32

296.18

47.00

128.00

176.00

2345 6789 1234 9299

PAGE 1 OF 1

1234 4321 5618 3765

$0.00 $22.10 + $650.00 - $22.10 + $650.00Overdue/Over limit Opening Balance

Available credit $350Credit limit $1000

Daily percentage rate .04630Annual percentage rate 16.90

1234

New charges Payments/refunds Closing Balance

SPECIMEN STATEMENT ONLY – USED FOR PURPOSE OF ILLUSTRATIONValid as At 05/2009.

8 APR 2009

5 MAY 2009

30 MAY 2009

13 2121

30 MAY 2009

Mr John Citizen123 Sample stSampleville NSW 1111

Credit Card Statement

Bank of Australia

17 We 15 Kai has two credit cards. One has an interest-free period and interest is then charged on the outstanding balance at a rate of 18% p.a. The other has no interest-free period with interest added from the date of purchase at a rate of 14% p.a. Kai has $1500 worth of bills to pay in the coming month and intends to use one of the cards to pay them. Kai intends to then pay the balance off in monthly instalments of $500.a If Kai uses the card with the interest-free period and pays $500 by the due date, what is

the outstanding balance on the card?b Calculate the interest Kai must then pay for the second month.



c Calculate the balance owing at the end of the second month and the balance owing at the end of the third month, at which time Kai pays off the entire balance.

d Calculate the interest payable in the first month if Kai uses the card without the interest-free period.

e Calculate the balance owing after Kai pays $500, and then calculate the interest for the second month.

f Calculate the balance owing at the end of the second month and the balance owing at the end of the third month, at which time Kai pays off the entire balance.

g Which card should Kai use for these bills?

purchase optionsWhen purchasing goods and services, there are various methods of payment. In this exercise we will compare purchase options and discuss some of the advantages and disadvantages associated with these options.

Cash purchasesWhen purchasing goods with cash, the customer owns the goods outright and no further payments are required. Some retailers may also offer customers a discount if they purchase goods with cash.

Worked example 16

An electrical store offers its customers a 10% discount if goods are fully paid for in cash on the day. How much will Gianluca pay for a DVD recorder with a marked price of $699 if he pays for it in cash?

think Write

1 Formulate an equation to calculate the new price of the DVD recorder.

New price = marked price − 10% of marked price

2 Substitute the known values into the rule. = 699 − 10% × 699

3 Evaluate. = 699 − 10100

× 699

= 699 − 69.9= 629.1


Gianluca will pay $629.10 for the DVD recorder.

Credit cards and store cardsAs discussed earlier, a credit card works as a pre-approved loan up to an amount agreed upon by the customer and the lending institution. The card can then be used until the amount of the debt reaches the limit. Credit cards may be issued by banks, major department stores (also known as store cards), various corporations and credit unions.

Worked example 17

Peter has a $2000 outstanding balance on his Myer store card. The interest rate charged is 26.7% p.a. on the balance unpaid by the due date.a If Peter pays $300 by the due date, calculate the balance owing.b Calculate the interest owing for the next month.c What will the balance owing be on Peter’s next credit card statement?d What will the total amount owing be on the credit card after another month’s interest is added?

9d

418

think Write

a 1 Calculate the balance owing, that is, the difference between the outstanding balance and the payment made.

a Balance owing = outstanding balance − payment made = 2000 − 300 = 1700


The balance owing is $1700.


2 List the values of P, R and T. P = 1700R = 26.7% p.a.T = 1 month

= 112

or 0 083. years

3 Substitute the values into the given formula.

IPRT=100

4 Evaluate. = × ×1700 26 7 0 083100

. .

= 37.825


The interest owing for the next month is $37.83.

c 1 Calculate the new balance owing, that is, the sum of the original balance and the interest.

c Balance owing = original balance + interest owed= 1700 + 37.83= 1737.83


The new balance owing is $1737.83.

d 1 Repeat steps 1 to 5 of part b. d IPRT=100

P = 1737.83R = 26.7% p.a.T = 1 month

= 112

or 0 083. years

IPRT=100

= × ×1737 83 26 7 0 083100

. . .

= 38.666 717 5The interest owing for the next month is $38.67.

2 Calculate the total amount owing. Total amount owing = unpaid balance + interest owed= 1737.83 + 38.67= 1776.50


The total amount owing is $1776.50.



hire-purchase, f lat rate and real rate of interesthire-purchaseDo you want to buy the latest model of a computer but cannot afford it? Not a problem! Just pay a small initial amount (deposit) and then (at a later date, say in 6 or 12 months’ time) start paying regular, very affordable amounts per week or per month (instalments). This ‘brilliant’ idea, promoted as ‘hire-purchase’ by many large retail outlets and car yards, means you need not go to a bank to arrange a loan because the place of purchase will take care of it for you. All you need to do is enjoy your new possession and make the required instalments regularly. ‘For how long?’ you may ask. You might not find the answer to this question in large print in the glossy advertisements! You would be sure, however, to ask about it before you signed the contract. Before you consider a hire-purchase plan, think about the following facts.

Fact 1A computer (car, TV, furniture, and so on) bought under a hire-purchase contract is not owned by the purchaser until the last cent has been paid. The purchaser of the goods ‘hires’ them for the duration of the contract and if the conditions of the contract are not met, the goods may be repossessed with no refund of the money already paid.

Fact 2Under a hire-purchase contract the loan taken must be repaid with a certain number of instalments. The instalments contain principal and interest, and very often the interest is more than the actual principal! The goods bought under hire-purchase are, generally, depreciable goods with real value. By the time the contract has expired, they are worth significantly less than the total amount paid.

Worked example 18

A computer with a cash price of $4800 can be bought under a hire-purchase scheme for 25% deposit and thirty $148 monthly instalments. Calculate:a the total amount paid under the schemeb the total amount of interest paidc the flat rate of interest (that is, simple interest) per year charged.

think Write

a 1 Write the formula for the total amount to be paid.

a Amount = deposit + monthly instalments

2 Calculate the deposit by determining 25% of $4800.

Deposit = 25% of the purchase priceDeposit = 0.25 × 4800

= $1200

3 Calculate the amount paid as instalments by multiplying the monthly amount by the number of instalments.

Instalments = 148 × 30= $4440

4 Calculate the total amount, A, by adding the values obtained in steps 2 and 3.

A = 1200 + 4440= 5640

5 Answer the question. The total amount to be paid under the scheme is $5640.

420

b 1 Write the formula for the interest paid.Note: To obtain the interest paid, subtract the cash price from the total amount.

b I = A − cash price

2 Write the known values of the variables.

A = $5640Cash price = $4800


I = 5640 − 4800

4 Evaluate. = 840

5 Answer the question. The total amount of interest paid is $840.

c 1 Write the formula for flat rate (simple) interest.

c IPRT=100

2 Write the known values of the variables.Note: The principal amount is obtained by subtracting the initial deposit from the purchase price.Months are to be converted into years.

I = 840P = 4800 − 1200

= 3600T = 30 months

= 3012

or 2.5 years

3 Substitute the values into the given equation.

840 = 3600 2 5

100× ×R .

4

5 Write the solution. Solving 840 = 3600 2 5100× ×r . for r gives

r = 9.333 333 33

6 Answer the question. The flat rate of interest is 9 13%.

Flat rate and real rate of interestAs seen from worked example 18, a flat rate of interest is charged on the total amount owed over the whole period of repayment even though the principal owing is reduced by each instalment.



solve 8403600 2 5

100= × ×

rr

.,



Let us investigate further and calculate the rate of interest per payment after the 10th and 20th payments were made from worked example 18. The monthly instalment of $148 consists of two parts: the repayment of the principal and the payment of the interest. We know the principal amount owing is $3600, as calculated in part c step 2.

Therefore the principal amount repaid in each instalment is 360030

= $120. This leaves the

amount of $148 − $120 = $28 to be paid for interest per instalment.

After 10 paymentsThe principal owed: P10 = 3600 − 120 × 10 = $2400The interest paid: I10 = $28

The rate of interest per payment: R10 = 282400

100×

= 116% per payment

The corresponding rate of interest per annum: R10 = 116 × 12

= 14% p.a.

After 20 paymentsThe principal owed: P20 = 3600 − 120 × 20 = $1200The interest paid: I20 = $28

The rate of interest per payment: R20 = 28

1200100×

= 213% per payment

The corresponding rate of interest per annum: R20 = 213 × 12

= 28% p.a.

The calculations above suggest that the actual interest rate charged is much higher than the flat rate. This actual interest rate is calculated as an average rate of interest per year and is called the real rate of interest or the effective rate of interest. The real rate of interest is considered to be the rate of interest that is paid on the average amount owing. There are two computational formulas that may be used to determine the real or effective rate of interest.

When the flat rate of interest is not known, the following formula may be used to determine the real or effective rate of interest.

Ref = 24

1I

P m( )++ × 100

==++

24001I

P m( )where

Ref = real or effective rate of interest (% p.a.)I = total interest paid on the loan ($)P = total principal owed ($)m = number of monthly instalments

$148

$120 $28Principal Interest

422

Ref can also be expressed in terms of, R, the flat (simple) rate of interest:

Ref =2

1Rn

n ++where

Ref = real or effective rate of interest (% p.a.)R = flat rate of interest (% p.a.)n = the total number of instalments to be made.

We will determine the real (or effective) rate of interest using both formulas and the figures obtained from worked example 18.

Ref = 24001I

P m( )+Ref = 2

1Rn

n +

= 2400 8403600 30 1

×+( )

= 2 9 30

30 1

13

× ×+

= 2 016 000111600

= 56031

= 18.06% = 18.06%

Note: In this case the answers obtained for the value of Ref, using both methods, were equal. This was due to the instalments being paid monthly. If, however, the instalments are not paid monthly, the Ref values will differ slightly.

Comparing the flat rate of interest (normally advertised under hire-purchase) obtained in worked example 18, R = 9 1

3%, and the effective interest rate Ref ≈ 18.06%, we see that:

Ref ≈ 2R.

Therefore, the interest claimed to be charged under a hire-purchase agreement (the flat rate of interest) is much less (almost half) the real (effective) interest rate actually paid. This is a good matter for consideration before entering into a hire-purchase contract!

Worked example 19

A hire-purchase plan to buy a new car priced at $46 890 offers a $19 500 deposit to be paid along with 260 weekly payments, each of $159. Find the effective rate of interest.

think Write

1 Write the formula for the effective rate of interest.Note: This formula is used since R, the flat rate, is not known.

Ref = 2400

1I

P m( )+

2 Determine the values of each of the variables.

(a) Calculate I. The total interest paid on the loan is the difference between the total amount repaid and the cash price of the car.

I = A − cash price= (deposit + instalments) − cash price= (19 500 + 260 × 159) − 46 890= (19 500 + 41 340) − 46 890= 60 840 − 46 890= $13 950

(b) Calculate P. The principal amount to be paid is obtained by subtracting the deposit from the cash price.

P = Cash price − depositP = 46 890 − 19 500

= $27 930



(c) Calculate m. The number of monthly instalments is obtained by converting weeks to years (that is, dividing 260 by 52) and then converting years to months (that is,

multiplying 26052

by 12).

m = 26052

× 12

= 5 × 12= 60 months

3 Substitute the values I, P and m into the given formula.

Substitute I = $13 950, P = $27 390, m = 60

Ref = 2400 1395027390 60 1

×+( )

4 Evaluate. = 33480 0001670 790

= 20.04

5 Answer the question. The effective rate of interest is 20.04%.

personal loansAs seen with hire-purchase plans, the final amount paid for goods compared to their original price is quite staggering. Other deferred payment plans that allow you to take products home without paying an initial deposit or offer ‘affordable weekly/monthly repayment’ plans also have higher, hidden interest costs and penalties, none of which are advertised in glossy brochures. It is imperative, therefore, that the consumer analyses and compares each of the available options and perhaps also considers taking out a personal loan for the amount required.

Personal loans are usually obtained when purchasing items such as a car, home entertainment system or to pay for a holiday. They are also better options than time payments such as hire-purchase and deferred-payment plans. Personal loans require regular payments to be made at set intervals. Interest and fees associated with setting up the loan are factored into the repayments.

The most common method of calculating interest on personal loans is a reducing interest method, in which interest is calculated on the amount still owing rather than the entire amount borrowed.

Using the CaS calculator finance solverProblems involving reducing balance loans can be easily solve using the CAS calculator’s Finance Solver.

Here N represents total number of repayments. For example, if repayments are made monthly over 2 years, N = 12 × 2 = 24.

I (%) represents interest rate per year.

424

PV stands for ‘present value’. This is the value of the loan. (If you borrowed the money from the bank, enter the value of your loan as a positive number.)

Pmt represents the amount of each payment. (If you are repaying the loan to the bank, enter the amount of payment as negative number.)

FV stands for ‘future value’, that is, the value of you loan at some stage in the future. If you wish to repay your loan in full, FV = 0.

PpY means number of payments per year. If for example, you pay monthly, PpY = 12.

The finance solver can be used to find any one of the above values, provided that all other values are known. To find the unknown value, fill in all known values, place the cursor where the entry for the unknown value should be and press ENTER ·.

Worked example 20

A person has borrowed $900 from a bank at 15% p.a., calculated on a reducing monthly balance. If he agrees to repay $150 per month, calculate:a the number of monthly repayments needed to repay the loanb the total amount repaidc the amount of interest paid.

think Write

a 1 a

2 The number of monthly repayments (N) is over 6 (6.275 823 381 952). Therefore, 7 repayments are needed.

Seven monthly repayments are needed to repay the loan of $900 in full.

b 1 b


On a Calculator page, press:• MENU b• 3: Algebra 3• C: Finance Solver CComplete the entry lines as:I (%): 15PV: 900Pmt: −150FV: 0PpY: 12Then return to the top line labelled N and press ENTER ·.

Since the required number of repayments is not a whole number, the last repayment will be less that $150. Find the amount still owed after the 6th repayment.Complete the entry lines as:N: 6I (%): 15PV: 900Pmt: −150PpY: 12Then return to the line labelled FV and press ENTER ·.


2 Write the amount owed after the 6th repayment.

After the 6th repayment has been made, the amount still owed is $41.05.

3 Calculate the percentage rate of interest per month. To find the amount of the last payment: find 101.25% of the amount still owed at the beginning of the month.

Monthly interest = 1512

%

= 1.25%Amount to be repaid, including interest

= 101.25% of $41.05

= ×101 25100

41 05.

.

= $41.56As $41.56 < $150, a repayment of only $41.56 and no further repayments are needed.

4 The total amount repaid consists of 6 payments of $150 each and a 7th payment of $41.56. Find the total amount repaid.

Total amount repaid = 6 × $150 + $41.56= $941.56

c Find the interest paid by subtracting the amount borrowed from the total amount paid.

c Interest = amount borrowed − amount paid = $941.56 − $900 = $41.56

The calculations for the preceding example can be done on a spreadsheet and summarised in a table as shown below.

Month

Amount still owing at

beginning of month

($)

Interest charged on

amount owing (1.25%)

($)

Amount owing

together with interest

($)

Monthly repayment

($)

Amount owing at end of month

($)

1 900.00 11.25 911.25 150 761.25

2 761.25 9.52 770.77 150 620.77

3 620.77 7.76 628.53 150 478.53

4 478.53 5.98 484.51 150 334.51

5 334.51 4.18 338.69 150 188.69

6 188.69 2.36 191.05 150 41.05

7 41.05 0.51 41.56 41.56 nil

Total 41.56 941.56

The above table shows all the information required to answer the three questions in worked example 20: the number of repayments and the totals of both interest and repayments.

426

When purchasing goods with cash, the customer owns the goods outright and no 1. further payments are required.Credit cards and store cards are sources of an instant loan to the cardholder.2. Credit cards and store cards are repaid monthly with a minimum payment.3. There are many different types of credit cards: some have an interest-free period; 4. others charge interest from the date of purchase. This leads to varying interest rates.Hire-purchase allows customers to take goods home after paying an initial deposit. 5. They are then required to make regular payments. The goods belong to the store until all payments have been made.Flat rate is the simple rate of interest charged on the original sum borrowed.6. Real or effective rate is the rate of interest being paid on the average principal 7. outstanding.

Ref = +2400

1I

P m( )

=+

21

Rnn

where Ref = real or effective rate is the rate of interest (% p.a.)

I = total interest paid on the loan ($)P = total principal owed ($)m = number of monthly instalmentsR = flat rate of interest (% p.a.)n = total number of instalments to be made

The relationship between the flat rate of interest, 8. R, and the effective rate, Ref, is Ref ≈ 2R.The most common method of calculating interest on personal loans is the reducing-9. interest method.A reducing-interest loan has interest calculated on the amount still owing.10. Every time a payment is made, the amount of interest to be paid is reduced.11.

rememBer

purchase options 1 We 16 An electrical store offers its customers an 8% discount if goods are fully paid for in

cash on the day. How much will Robyn pay for a digital camera with a marked price of $729 if she pays for it in cash?

2 A department store offers its customers a 5% discount if goods are fully paid for in cash on the day. How much will Laurien pay for a home theatre projector with a marked price of $2499 if she pays for it in cash?

3 A department store offers its customers a 7% discount if goods are fully paid for in cash on the day. How much will Jan pay for an iPod with a marked price of $499 if she pays for it in cash?

4 mC Mervyn bought a new lawn mower at a sale. First, there was a 20% discount from the original price. Then, an $80 trade-in for his old mower was subtracted from this reduced price. This left Mervyn with $368 to pay for the new lawn mower. The original price of the new lawn mower was:A $468.00 B $537.50 C $540.00 D $560.00 E $580.00

[©VCAA 2006]

exerCiSe

9d



5 Helen has a credit card with an outstanding balance of $1850. If the interest rate is 18% p.a., calculate the amount of interest that Helen will be charged for one month if the balance is not paid by the due date.

6 Martin buys office equipment for $1450 with his store card. The card has no interest-free period and interest is charged at a rate of 15% p.a. Calculate one month’s interest on this purchase.

7 We17 Fiona has a $2000 outstanding balance on her Myer store card. The interest rate charged is 21% p.a. on the balance unpaid by the due date.a If Fiona pays $200 by the due date, calculate the balance owing.b Calculate the interest owing for the next month.c What will the new balance owing be on Fiona’s next credit card statement?d What will be the total amount owing on the credit card after another month’s interest is

added?

8 Noel has a credit card that charges interest at a rate of 12% p.a. but has no interest-free period. He makes a purchase of $1750 on the credit card.a After one month, Noel’s credit card statement arrives. What will be the outstanding

balance on the statement?b The minimum repayment will be 5% of the outstanding balance. Calculate the amount

that Noel will owe if he makes the minimum payment.c In the next month Noel makes purchases totalling $347.30. Calculate the interest charged

and the balance owing for the next month’s statement.

9 We18 A second-hand car with a cash price of $25 200 can be bought under a hire-purchase scheme for 20% deposit and one hundred and twenty $350 monthly instalments. Calculate:a the total amount paid under the schemeb the total amount of interest paidc the flat rate of interest (that is, simple interest) per year charged.

10 A used caravan with a cash price of $8990 is purchased under a hire-purchase agreement. A 15% deposit is made and the balance plus interest is to be paid in 104 equal instalments of $130 each.a Calculate the initial deposit.b Calculate the total amount paid under the agreement.c Calculate the total amount of interest paid.d Copy and complete the following sentence. If the payments are to be made over a

six-year period, the instalments should be paid every ___ .e Calculate the flat rate of interest charged per year.

11 A new dishwashing machine is purchased for $985 on a hire-purchase contract with no deposit paid, 6 months interest free, and 10.95% per annum flat interest over 18 equal monthly instalments. Calculate:a the amount of interest chargedb the total amount to be repaidc the amount of each monthly instalment.

12 We19 A hire-purchase plan to buy a new TV and cabinet at $1390 offers a $250 deposit to be paid along with 104 equal weekly payments of $13.50 each. Find the effective rate of interest.

13 a Calculate the real (effective) rate of interest for question 9.b Comment on the results.

14 mC A $2000 lounge suite was sold under a hire-purchase agreement. A deposit of $200 was paid. The balance was to be paid in 36 equal monthly instalments of $68. The annual flat rate of interest applied to this agreement is:A 10.0% B 11.4% C 12.0% D 22.4% E 36.0%

[©VCAA 2006]

428

15 mC a A set of bedroom furniture can be bought for $2600 cash or on hire-purchase by paying a deposit of 10% of the purchase price and then $65 per month for 4 years. The amount that the customer would save by paying cash instead of hire-purchase is:

A $3380 B $780C $260 D $3120E $520

b Ken bought a second-hand car from a car yard. The car, with a cash price of $18 300, was bought for $5000 deposit and a weekly payment of $125 for 3 years. The total amount of interest, in dollars, that Ken paid was:A $4200 B $2400C $2600 D $6200E $5200

16 We20 Peri borrows $1200 from a bank at 13.5% p.a., calculated on a reducing monthly balance. If he is to pay $120 eachmonth, calculate:a the number of monthly repayments needed to repay the loanb the total amount repaidc the amount of interest paid.

17 Mila takes out a personal loan of $4500 to buy new furniture. The loan is at 11.3% p.a., calculated on the reducing monthly balance. Mila agrees to pay $150 each month.a Calculate the amount still owed at the end of each month for the fi rst

6 months.b Calculate the amount of the loan that is repaid during that period of

time.c Express the amount in b as a percentage of the total repayments made

over the 6 month period.

depreciationAs mentioned earlier, items that represent scarce resources such as land, collectables, paintings and antiques normally appreciate in value over time. Most of the goods that we consume, such as household items, cars, electrical and electronic appliances, plant machinery and equipment, fi xtures and furnishings lose (depreciate in) their value over time.

Depreciation means that items reduce in value from the time they are acquired. The price for which an item was originally bought is called the purchase price of the item. The portion of the purchase price an item loses in value due to depreciation is called its accumulated depreciation. The value of an item at any given time is called its book value. This information may be summarised by the following equation:

Book value = purchase price − accumulated depreciation

Items used for income producing purposes are called assets. Scrap value is the value that an asset is expected to have at the end of its useful life. The useful life of some assets is expected to be longer than others. For example, we expect buildings to last 40 years and longer, whereas computers have a life expectancy of about 3 years. This is refl ected in the depreciation rate, which varies for different assets. The depreciation rate normally is expressed as a percentage of the purchase price.

The recorded form of changing asset values due to depreciation is called a depreciation schedule. Depreciation schedules are available from the taxation department.

eBookpluseBookplus


Reducing-interest loans

eBookpluseBookplus

Digital docWorkSHEET 9.2

9e



There are a number of ways in which depreciation can be calculated, as shown in the fl owchart below.

Depreciation

According to time According to use

Straight-linedepreciation

Reducing-balancedepreciation

Unit costdepreciation

Straight-line depreciationThe straight-line depreciation method, also known as the fl at rate depreciation method or the constant depreciation method, allocates an equal amount of depreciation to each time period (normally one year) in the asset’s useful life — that is, the asset’s value loses the same amount each year. The formulas associated with the straight-line depreciation method are:

1. Book value = purchase price − accumulated depreciation B.V. = P − D where:

B.V. = book value of asset ($)D = accumulated depreciation ($)P = purchase price ($)R = depreciation rate (% p.a.)T = period of depreciation (years)

= −PPRT100

= −

PRT

1100

2. RP SnP

= − ×100where

R = depreciation rate (% p.a.)P = purchase price ($)S = scrap value ($)n = number of time periods

Note: The formula for calculating the accumulated depreciation, D, is similar to the formula of simple interest.


Tutorialint-

Worked example 21

A company car purchased for $39 600 depreciates at 12% per annum by straight-line depreciation.a Calculate the book value of the car

after 3 years.b Calculate the total depreciation

over the fi rst 6 years.c How long will it take until the car

reaches its scrap value of $4500?d Prepare a depreciation schedule.e Draw a graph which depicts the

depreciation of the car over its useful life and comment on the shape of the graph.

0901

430

think Write/draW

a 1 Write the formula for book value. a B V PRT

. . = −

1100


P = $39 600R = 12%T = 3 years


B.V. = 39 600 112 3100

− ×

= 39 600 136

100−

= 39 60064

100

4 Evaluate. = 25 3445 Answer the question and include the

appropriate unit.The book value of the car after 3 years will be $25 344.

b 1 Write the formula for depreciation. b DPRT=100


P = $39 600R = 12%T = 6 years


D = × ×39600 12 6100

4 Evaluate. = 2851200100

= 28 5125 Answer the question and include the

appropriate unit.The total depreciation of the car over the first 6 years is $28 512.

c 1 Write the formula for the depreciation rate.

c RP SnP

= − ×100


P = $39 600R = 12%S = $4500


1239600 4500

39600100= − ×

n

4



solve 1239600 4500

39600100= − ×

n

n,



5 Write the solution. Solving 1239600 4500

39600100= − ×

n for n gives

n = 7.3864 years

6 Convert the answer from years to years and months by multiplying the decimal portion by 12.

n = 7 years and (0.3864 × 12) months= 7 years and 4.6364 months≈ 7 years and 5 months

d 1 Rule up a table with 4 columns titled Period n (years), Depreciation ($), Accumulated depreciation ($) and Book value ($). List the period of time from 1 year to 7 years and 5 months (see table below in step 5).

d

2 Calculate the depreciation.

Write the formula for simple interest. DPRT=100

Substitute the values into the given formula.

D139600 12 1

100= × ×

Evaluate.

Note: The car will depreciate by the same amount each full year. In 5

12

of a year the car will depreciation

by 512

× $4752.

= 475200100

= $4752

3 Calculate the accumulated depreciation.Multiply the depreciation value obtained in step 2 by the specific year.

Year 1: 1 × 4752 = $4752Year 2: 2 × 4752 = $9504Year 3: 3 × 4752 = $14 256 and so on

4 Calculate the book value.Subtract the accumulated depreciation from the purchase price.

Year 1: 39 600 − 4752 = $38 848Year 2: 39 600 − 9504 = $30 096Year 3: 39 600 − 14 256 = $25 344 and so on

5 Enter all of the information into the table.

This schedule could have been prepared easily using a spreadsheet as each of the calculations is repetitive and uses the same formulas in each respective column.

Period n (years)

Dep. ($)

Accum. dep. ($)

Book value ($)

1 4752 4 752 34 848

2 4752 9 504 30 096

3 4752 14 256 25 344

4 4752 19 008 20 592

5 4752 23 760 15 840

6 4752 28 512 11 088

7 4752 33 264 6 336

7 512

1980 35 100 4 500

432

e 1 Rule up a set of axes.Label the horizontal axis n and scale appropriately.Label the vertical axis Book value and scale appropriately.

e

n

Boo

k va

lue

($ ’

000)

10

20

30

40

1 2 3 4 5 6 7 75—12

39 600

4500

2 Plot the points obtained from the table above.Note: The first point on the graph should be (0, 39 600). This indicates that initially, that is, at n = 0, the value of the car was $39 600.

3 Draw a straight line connecting the points.Note: Alternatively the data may have been entered into a CAS calculator and then plotted as demonstrated on page 396.

4 Comment on the shape of the graph. The graph is a straight line with a negative gradient. The line intersects the vertical axis at $39 600. As the car depreciates at a constant rate of $4752 each year, the value of the gradient must be −4752. The equation of the straight line is B.V. = 39 600 − 4752n.

reducing-balance depreciationWe have just observed the similarity in calculating straight-line depreciation and simple interest. Both are fixed percentages of a fixed amount. The difference, however, is that interest is added to the principal, whereas depreciation reduces the cost of an asset.

The reducing-balance method, also known as the diminishing value depreciation method or the reduced value method, can be compared with compound interest. Both are characterised by applying a fixed percentage to an amount that changes at the beginning of each time period. As the book value of an asset declines from period to period, so does depreciation.

Compared with straight-line depreciation, in which the item’s value decreases by the same amount each time period, reducing-balance depreciation decreases by a constant percentage of the current value for each time period.

The formula for the book value of an asset that depreciates under the reducing-balance method is:

B V PR

n

. . = −

1100

whereB.V. = book value of depreciated item ($)

P = purchase price ($)R = allowed annual rate of depreciation (% p.a.)n = number of depreciation periods (years)

Note: The formula for calculating the book value is similar to the formula of compound interest.



Worked example 22

A construction company purchased a new scrapper for $86 000. It depreciates at 12% per annum reducing-balance.a Calculate the book value of the scrapper after 3 years.b Calculate the total depreciation over the first 5 years of service.c After how many years does the scrapper have a value below $20 000?d After how many years does the scrapper reach its scrap value of $8000?e Prepare a depreciation schedule for the first 6 years of the scrapper’s use.f Draw a graph that depicts the depreciation of the scrapper over its useful life, and comment on the

shape of the graph.

think Write/draW

a 1 Write the formula for book value. a B V PR

n

. . = −

1100


P = $86 000 R = 12% n = 3 years


B V. . = −

86000 112

100

3

4 Evaluate. =

8600088

100

3

= 86 000(0.88)3

= 58 606.59


The value of the scrapper after 3 years is $58 606.59.

b 1 Write the formula for total depreciation.

b Total depreciation = P − B.V.

= − −

P PR

n

1100


P = $86 000 R = 12% n = 5 years


Total dep. = − −

86000 86000 112

100

5

4 Evaluate. = −

86000 8600088

100

5

= 86 000 − 86 000(0.88)5

= 86 000 − 45 384.94= 40 615.06


The total depreciation over the first 5 years of service is $40 615.06.

c 1 Write the formula for book value. c B V PR

n

. . = −

1100


B.V. = $20 000 P = $86 000 R = 12%


20 000 86000 112

100= −

n

434

4

5 Write the solution. Solving 20 000 86000 112

100= −

n

for n gives

n = 11.4103 years

6 Convert 11.4103 years into years and months by multiplying the decimal portion by 12.


7 Answer the question. The scrapper will have a value below $20 000 in approximately 11 years and 5 months.

d 1 Repeat steps 1 to 5 performed in part c.

d B V PR

n

. . = −

1100

B.V. = $8000 P = $86 000 R = 12%

8000 86000 112

100= −

n

Write the solution. Solving 8000 86000 112

100= −

n

for n gives

n = 18.5781 years

2 Convert 18.5781 years into years and months by multiplying the decimal portion by 12.



On a Calculator page, press:MENU • b3: Algebra • 31: Solve • 1

Complete the entry line as:

solve 20 000 86000 112

100= −

n

n,


On a Calculator page, press:MENU • b3: Algebra • 31: Solve • 1

Complete the entry line as:

solve 8000 86 000 112

100= −

n

n,



3 Answer the question. The scrapper will reach its scrap value in approximately 18 years and 7 months.

e 1 Rule up a table with 4 columns titled Period n (years), Depreciation ($), Accumulated depreciation ($) and Book value ($). List the period of time from 1 year to 6 years (see table at step 8).

e

2 Calculate the depreciation for the first year by multiplying the purchase price by 12%.

Year 1 Depreciation = 12% of $86 000

= ×12100

86000

= $10 320

Place this value in the Depreciation and Accumulated depreciation columns (see table at step 8).

Year 1 Acc. dep. = $10 320

3 Calculate the book value for the first year by subtracting the depreciation from the purchase price. Place the value in the book value column (see table below right).Or use the formula.

Year 1 B.V. = P − D= 86 000 − 10 320= $75 680

4 Calculate the depreciation for the second year by multiplying the previous year’s book value by 12%. Place this value in the relevant column (see table below right).

Year 2 Depreciation = 12% of $75 680

= ×12100

75 680

= $9081.60

5 Add the depreciation for the current year to the accumulated depreciation of the previous year.

Year 2 Acc. dep. = 10 320 + 9081.60= $19 401.60

6 Calculate the book value for the second year by subtracting the accumulated depreciation (for that year) from the purchase price. Place the value in the book value column (see table at right).Or use the formula.

Year 2 B.V. = P − D= 86 000 − 19 401.60= $66 598.40

Period n

(years)Dep. ($)

Accum. dep. ($)

Book Value ($)

1 10 320.00 10 320.00 75 680.00

2 9 081.60 19 401.60 66 598.40

3 7 991.81 27 393.41 58 606.59

4 7 032.79 34 426.20 51 573.80

5 6 188.86 40 615.06 45 384.94

6 5 446.19 46 061.25 39 938.75

7 Repeat steps 4 to 6 for Years 3, 4, 5 and 6.

8 Check that the information has been clearly entered into the table.Again, this schedule could have been prepared easily using a spreadsheet as each of the calculations is repetitive and uses the same formulas in each respective column.

436

f 1 Rule up a set of axes.Label the horizontal axis n and scale appropriately.Label the vertical axis Book value and scale appropriately.

f

2 Plot the points obtained from the table above.Note: The fi rst point on the graph should be (0, 86 000). This indicates that initially, that is, at n = 0, the value of the scrapper was $86 000.

n

4050

Boo

k va

lue

($ ’

000)

607080

6543210

3 Draw a curve connecting the points.Note: Alternatively the data may have been entered into a CAS calculator and then plotted as demonstrated on page 396.

4 Comment on the shape of the graph. The graph is an exponential curve. As time increases, the book value decreases by different amounts.

Unit cost depreciationUnit cost depreciation, which is also called unit value depreciation, applies to machinery that performs some repetitive job. The method relates depreciation to use rather than to time, as previously considered methods do. Under unit cost depreciation, the cost of the machinery (minus its scrap value) is divided by the estimated number of production units expected from the machinery during its useful life. Production units might be expressed as kilometres, operating hours and so on.Unit cost depreciation U.C.D. =

total depreciation

expected life of asset

= purchase price – scrap value


= −P SE


Tutorialint-0902

Worked example 23

Calculate the book value of an $8900 photocopier with a scrap value of $150, if it has an expected life of 500 000 photocopies and has been used 64 000 times.

think Write

1 Write the formula for unit cost depreciation. U C DP S

E. . . = −

2 Write the known values of the variables. P = $8900S = $150E = 500 000 copies

3 Substitute the values into the given formula. U C D. . . = −8900 150500 000



4 Evaluate. = 8750500 000

= $0.0175/copy

5 Interpret the answer. The unit cost depreciation is $0.0175 per copy; that is, the depreciation rate of the photocopier is $0.0175 (or 1.75c) per copy.

6 Calculate the accumulated depreciation after 64 000 copies have been made.

Accum. dep. = 0.0175 × 64 000= 1120

7 Interpret the answer. The accumulated depreciation for 64 000 copies is $1120.

8 Calculate the book value of the photocopier by subtracting the accumulated depreciation from the purchase price.

B.V. = P − Accum. dep.= 8900 − 1120= 7780

9 Answer the question. The book value of the photocopier after being used 64 000 times is $7780.

Worked example 24

An industrial machine, valued at $95 200 when new, depreciates at a rate of 0.035% of the initial cost per 14-hour working shift.a Calculate the book value of the machine after 7 years of operation

(the machine is in use 6 days per week, 48 weeks per year).b Estimate how many years it will take for the machine to be completely

written off (that is, to have no scrap value).

think Write

a 1 Write the formula for book value. a B.V. = P − D


P = $95 200


B.V. = 95 200 − D

4 Calculate the accumulated depreciation.Multiply the percentage depreciation rate by the purchase price of the item in order to determine the depreciation cost per one 14-hour working shift.Calculate the accumulated depreciation for the machine working 6 days a week, 48 weeks a year for 7 years.

Depreciation cost =0 035100.

× 95 200

= 33.32The depreciation cost per one working shift of 14 hours is $33.32.Accum. dep. = 33.32 × 6 × 48 × 7

= 67 173.12The accumulated depreciation in 7 years is $67 173.12.

5 Substitute the accumulated depreciation into the book value formula.

B.V. = 95 200 − 67 173.12= 28 026.88

6 Answer the question. The book value of the machine after 7 years of operation is $28 026.88.

b 1 Write the formula for book value. b B.V. = P − D

eBookpluseBookplus

Tutorialint-0903

Worked example 24

438


B.V. = 0P = $95 200

Let x be the required number of years.D = 33.32 × 6 × 48 × x

= 9596.16x


0 = 95 200 − 9596.16x

4 Transpose the equation to make x the subject. Add 9596.16x to both sides of the equation.Divide both sides of the equation by 9596.16.

0 + 9596.16x = 95 200 − 9596.16x + 9596.16x9596.16x = 95 200

9596 169596 16

952009596 16

.. .

x =

x = 9.9206 years

5 Convert 9.9206 years to years and months by multiplying the decimal portion by 12.

x = 9 years and (0.9206 × 12) months= 9 years and 11.047 months≈ 10 years

6 Answer the question. It will take approximately 10 years for the machine to have no scrap value.

The book value of an item at a specific time is defined as the difference between the 1. purchase price and the accumulated depreciation at that time.

Straight-line depreciation

The book value is given by 2. B V PRT

. . = −

1100

.

The depreciation rate is given by 3. RP SnP

= − ×100.

Straight-line depreciation may be illustrated graphically as a straight line with a 4. negative gradient.

Reducing-balance depreciation

The book value is given by 5. B V PRT

n

. . .= −

1100

Reducing-balance depreciation may be illustrated graphically by an exponential curve.6.

Unit cost depreciationUnit cost depreciation 7. =

purchase price – scrap value


U.C.D. = −P SE

rememBer

depreciation 1 We21 A machine costing $286 000 depreciates at 16.5% per annum, by straight-line

depreciation.a Calculate the book value of the machine after the first 2 years of service.b Calculate the total depreciation over the first 5 years.c How long will it take until the machine reaches its scrap value of $10 000?

exerCiSe

9e



d Prepare a depreciation schedule. (Using a spreadsheet is recommended.)e Draw a graph that depicts the depreciation of the machine over its useful life and

comment on the shape of the graph.

2 A building crane was purchased by a construction company for $450 000 and is expected to have a useful life for 19 years. Calculate the crane’s value at the end of 16 years of service if its recommended straight-line depreciation rate is 5.0% per annum.

3 A factory writes off 4% per annum for a building that has a current book value of $310 000. If the adopted method to calculate depreciation is straight-line, what is the book value of the factory building in 11 years’ time?

4 mC The yearly depreciation expense of a computer worth $5200, at a flat rate of 30% per annum, is:A $3640 B $1560 C $1810 D $364 E $1650

5 mC Given that D = accumulated depreciation, P = purchase price of an asset and T = period of depreciation, then the constant rate depreciation, R, can be calculated using the formula:

A RDPT

=100

B RPT

D=

100C R

DPT

= 100

D RP

DT= 100 E R

TPD

= 100

6 Find the accumulated depreciation of the following assets.a Purchase price $2600, straight-line depreciation at 20% p.a. for 2 yearsb Purchase price $18 200, flat rate (straight) depreciation at 18% p.a. for 4 yearsc Purchase price $136 000, constant depreciation at 5.25% p.a. for 13 yearsd Purchase price $15 612, straight-line depreciation at 12% p.a. for 5 years

7 What is the book value of the following items which are to be depreciated?

Purchase price P($)

Period n (years)

Depreciation R (%)

Book valueB.V. ($)

a 12 500 5 10

b 6 321 3 5

c 989 2 3.75

d 172 4 1.75

e 15 675 7 4

f 10 221 4 17

g 75 000 5 11.25

h 455 000 7 13.5

i 878 750 10 8.9

j 98 000 5 12

8 We22 A clothing factory purchased a new machine for $21 600 with reducing-balance depreciation at 16% per annum.a Calculate the book value of the machinery after 4 years.b Calculate the total depreciation over the first 6 years of use.c After how many years does the machinery reach a value below $10 000?d After how many years does the machinery reach its scrap value of $3000?e Prepare a depreciation schedule for the first 6 years of the machinery’s use.f Draw a graph which depicts the depreciation of the machinery over its useful life and

comment on the shape of the graph.

440

9 An industrial steam-cleaner costs $3200 with diminishing value at 15% p.a. Calculate:a the amount of depreciation during the first year of serviceb the amount of depreciation during the eighth year, and express this amount as a

percentage of the cost.

10 A motorbike is purchased by a courier company for $11 600 with reducing-balance depreciation at 18% p.a. The trade-in value of the motorbike is expected to be $1500.a Find the book value of the motorbike after 3 years.b Calculate the accumulated depreciation over this period.c After how many years will the motorbike be traded in for a new one?

11 A company writes off 22% p.a. for reduced value depreciation of its machinery with a current book value of $606 820. What was the purchase cost of the machinery 3 years ago?

12 mC A new computer station cost $5490. With 26% p.a. reducing value depreciation, the station’s book value at the end of the third year will be, to the nearest dollar, close to:A $3265 B $2225 C $1684 D $3082 E $2811

13 mC The value of a new photocopier is $8894. Its value depreciates by 26% in the first year, 21% in the second year and 16% reducing balance in the remaining 7 years. The scrap value of the photocopier, to the nearest dollar, is:A $827 B $369 C $1134 D $1534 E $1814

14 Camera equipment purchased for $150 000 will have a scrap value of $9000 in 5 years.a Find the annual depreciation under: i a straight-line depreciation ii a reducing-balance depreciation.

b Graph the results for each type of depreciation on the same set of axes.

15 We23 Calculate the book value of an $11 500 printer with a scrap value of $200 if it has an expected life of 800 000 print-outs and has been used 72 000 times.

16 Calculate the value of a printer after it has printed 1.8 × 109 pages. The printer was purchased for $56 000, and had an estimated scrap value of $3500 with an expected operational value of 3 × 109 pages printed over its useful life.

17 We24 An industrial food processor, valued at $67 800 when new, depreciates at a rate of 0.028% of the initial cost per 12-hour working shift.



a Calculate the book value of the food processor after 3 years of operation (the machine is in use 6 days per week, 49 weeks per year).

b Estimate how many years it will take for the food processor to be completely written off (that is, to have no scrap value).

18 mC The unit cost depreciation in cents per hour for a machine bought for $16 800 with scrap value $1200 designed for 72 000 hours of operation is close to:A 22 c/h B 34 c/h C 45 c/h D 56 c/h E 2.17 c/h

19 mC A press bought for $186 000 depreciates at 16c per newspaper printed. The number of newspapers that can be printed before the press reaches its scrap value of $2000 is:A 150 000 B 1 500 000 C 1 150 000 D 115 000 E 11 500 000

20 mC A car is valued at $30 000 when new. Its value is depreciated by 25 cents for each kilometre it travels. The number of kilometres the car travels before its value depreciates to $8 000 is:A 32 000 B 55 000 C 88 000 D 120 000 E 550 000

[©VCAA 2007]

21 A component for silicon wafer production bought for $61 400 is designed for 24 000 hours of operation, after which it will be written off with zero residual value. Calculate:a the accumulated depreciation after the first 8000 hours of operationb its book value after 16 000 hoursc the number of operating hours required for the machine to reach a book value of $400.

442

SUmmary

Simple interest

Simple interest is given by • IPRT=100where

I = interest ($)P = principal ($)R = rate of interest (% p.a.)T = term of interest (years)

The total amount is given by • A = P + I.When calculating simple interest, the interest earned is the same for each time period.•

Compound interest

Compound interest formula is given by • A Pr

n

= +

1100where

A = amount at the end of n compounding periods ($)P = principal ($)r = rate of interest per period (%)n = number of compounding periods

When dealing with compound interest, the interest is calculated on the principal as well as the interest over • each time period.Inflation is the measure of the rate at which prices increase.• The inflation rate is given as a percentage and is called the Consumer Price Index.• To estimate the cost of an item after a particular year or number of years, we use the compound interest • formula, using P as the original price, r as inflation rate and n as the number of years.Rare items such as collectables and memorabilia increase in value as time goes on at a rate that is usually • greater than inflation.If solving equations where the unknown is a power, take the logarithm to base 10 of both sides of the • equation; that is,

an = blog10 (a)n = log10 (b)

n × log10 (a) = log10 (b)

n = log ( )log ( )

10

10

ba

A capital gain occurs when the amount received for an asset is greater than the amount originally paid for the • asset.

Savings and credit card accounts

Two methods used by banks for calculating interest on savings accounts are:• (a) minimum monthly balance(b) minimum daily balance.The minimum daily balance offers the best rate of interest as it credits the customer for all monies in the • account.A credit card is a source of an instant loan to the cardholder.• The two options for Visa or MasterCard are:• (a) no annual fee and no interest-free period(b) an annual fee and a specific interest-free period.



The bank requires a minimum monthly payment that is usually the greater of a certain fixed amount or a • specific percentage of the closing balance.For all transactions made with a ‘no interest-free period’ credit card and for cash advances obtained with an • ‘up to 55 days interest-free period’ credit card, the interest is calculated from the date of the first purchase.For ‘interest-free period’ credit cards, if the closing balance is paid in full by the due date indicated on the • statement, no interest is incurred. Otherwise, interest is charged until the balance is repaid.

Purchase options

When purchasing goods with cash, the customer owns the goods outright and no further payments are • required.Credit cards and store cards are sources of an instant loan to the cardholder.• Credit cards and store cards are repaid monthly with a minimum payment.• There are many different types of credit cards: some have an interest-free period; others charge interest from • the date of purchase. This leads to varying interest rates.Hire-purchase allows customers to take goods home after paying an initial deposit. They are then required to • make regular payments. The goods belong to the store until all payments have been made.Flat rate is the simple rate of interest charged on the original sum borrowed.• Real or effective rate is the rate of interest being paid on the average principal outstanding.•

Ref = 2400

1I

P m( )+

= 21

Rnn +

where Ref = real or effective rate is the rate of interest (% p.a.)

I = total interest paid on the loan ($)P = total principal owed ($)m = number of monthly instalmentsR = flat rate of interest (% p.a.)n = total number of instalments to be made

The relationship between the flat rate of interest, • R, and the effective rate, Ref, is Ref ≈ 2R.The most common method of calculating interest on personal loans is the reducing-interest method.• A reducing-interest loan has interest calculated on the amount still owing.• Every time a payment is made, the amount of interest to be paid is reduced.•

Depreciation

Straight-line depreciationStraight-line depreciation allocates an equal amount of depreciation to each time period over the asset’s • useful life.The book value of an item at a specific time is defined as the difference between the purchase price and the • accumulated depreciation at that time.

The book value is given by • B V PRT

. . = −

1100

The depreciation rate is given by • RP SnP

= − ×100

whereP = purchase price ($)R = depreciation rate (% p.a.)T = period of depreciation (years)S = scrap value ($)n = number of time periods (years)

Straight-line depreciation may be illustrated graphically as a straight line with a negative gradient.•

444

Reducing-balance depreciationReducing-balance depreciation involves a fixed percentage applied to an amount (book value) changing at • the beginning of each time period.

The book value is given by • B V PR

n

. . = −

1100

whereP = purchase price ($)R = allowed annual rate of depreciation (% p.a.)n = number of depreciation periods (years)

Reducing-line depreciation may be illustrated graphically by an exponential curve.•

Unit cost depreciation

U.C.D. = P SE−

whereU.C.D. = unit cost depreciation ($/unit)

P = purchase price ($)S = scrap value ($)E = expected life of asset, number of production units



Chapter revieW

mUltiple ChoiCe

1 Anthony earned $1016 in simple interest when he invested $19 800 for 9 months. The rate of simple interest was:A 5.13% B 6.14%C 6.84% D 7.62%E 8.21%

2 With an interest rate of 4.11% p.a., the sum of $634 was earned in simple interest over 2 1

2 years. The

amount invested was close to:A $6170 B $4892C $7306 D $2937E $5607

3 The ‘Farmbrand’ dozen eggs (size 67) costs $2.40. With an inflation rate expected to average 3.5% over each of the next 4 years, the dozen eggs will then cost:A $2.50 B $2.60C $2.70 D $2.75E $2.80

4 An investment of $12 000 at 6.15% compounded quarterly to reach $15 000 will take close to:

A 3 years B 4 yearsC 5 years D 6 yearsE 7 years

5 The points on the graph below show the balance of an investment at the start of each quarter for a period of six years. The same rate of interest applied for these six years.

0 1 2 3Year

Bal

ance

(do

llars

)

4 5 6

In relation to this investment, which one of the following statements is true?A Interest is compounding annually and is

credited annually.B Interest is compounding annually and is

credited quarterly.C Interest is compounding quarterly and is

credited quarterly.D Simple interest is paid on the opening balance

and is credited annually.E Simple interest is paid on the opening balance

and is credited quarterly.

exam tip This question concerned the growth of an investment, but was formulated graphically rather than numerically. The aim was to test conceptual understanding of the different forms of investment growth. Only 37% of students were successful in answering this question. Teachers and students are reminded that graphical analysis of business and financial situations is a requirement of the study design.


[©VCAA 2006]

6 A sum of $5000 is invested for 2 years at the rate of 4.75% p.a., compounded quarterly. The value of the interest at the end of the term, to the nearest dollar, is:

A $495 B $5468C $4532 D $936E $391

7 A current household income is $61 800 p.a. and grows at the rate of 6% p.a. With this rate being constant, the number of years required for the household income to reach $100 000 is:

A 7 B 8C 9 D 10E 11

Questions 8, 9 and 10 relate to the following.The table below shows the transactions in Anastasia’s passbook for August.

Date Transaction Deposits Withdrawals Balance

3 August 2009 Salary $1862.50 $2084.72

10 August 2009 ATM — Ivanhoe $650.00

14 August 2009 Cheque 9652 $840.00

28 August 2009 Cheque 9654 $389.40

446

8 In the bank statement above, the minimum balance for the month is:

A $2084.72 B $1434.72C $594.72 D $222.22E $205.32

9 The interest earned on the minimum monthly balance if the bank pays 5.5% p.a. simple interest monthly is:

A $9.55 B $6.58C $2.73 D $1.02E $0.94

10 The total debits for the month of August are:

A $1490 B $1862.50C $1879.40 D $1229.40E $205.32

11 Ben has a store credit card with an outstanding balance of $850. If the interest rate is 24%, calculate the amount Ben will be charged if the balance is not paid by the due date.

A $646 B $204C $1054 D $826E $25

12 An electrical store offers its customers an 8% discount if goods are fully paid for in cash on the day. How much will Carlie pay for a mini DV camcorder with a marked price of $799 if she pays for it in cash?

A $63.92 B $735.08C $862.92 D $159.80E $639.20

13 A hire-purchase agreement on a loan of $3000 requires 24 monthly payments. The effective annual rate interest is 17%. The interest paid, in dollars, is close to:

A $289 B $361C $482 D $679E $531

14 A new cooktop worth $935 is bought under hire-purchase with a deposit of $100 and

18 monthly instalments of $50. The flat rate of interest is:

A 8.94% B 7.69%C 6.34% D 5.19%E 4.88%

15 Brad investigated the cost of buying a $720 washing machine under a hire-purchase agreement. A deposit of $180 is required and the balance will be paid in 24 equal monthly repayments. A flat interest rate of 12% per annum applies to the balance. Brad correctly calculated the monthly repayment to be:

A $22.50 B $25.20C $26.10 D $27.90E $29.70

exam tip This question was based on a 2-year hire-purchase agreement. With a success rate of 36%, this question was not well done. However, another 30% of students started correctly, but only took into account 1 year’s interest, not 2.


[©VCAA 2007]

16 A laser printer is purchased for $790. It has an expected lifetime of 6 years and zero residual (scrap) value. The amount of depreciation to be allowed per year, in dollars, assuming a straight-line depreciation, is:

A $790 B $212C $306.67 D $98.37E $131.67

17 Equipment worth $18 600 is bought for a snowboard workshop. It depreciates at 12.5% p.a. constant depreciation. With zero scrap value, the book value of the equipment after 5 years is expected to be:

A $11 625 B $6975C $8114 D $5678E $7316

18 A new computer costs $3490. If depreciation is calculated at 33% p.a. (reducing balance), then the computer’s book value at the end of 5 years will be close to:

A $471 B $509C $432 D $567E $288

19 A new kitchen in a restaurant cost $50 000. Its value is depreciated over time using the reducing balance method. The value of the kitchen in dollars



at the end of each year for 10 years is shown in the graph below.

0 1

10000

20000

30000

40000

50000

2 3

Year

Val

ue (

dolla

rs)

4 5 6 7 8 9 10

Which one of the following statements is true?A The kitchen depreciates by $4000 annually.B At the end of five years, the kitchen’s value is

less than $20 000.C The reducing balance depreciation rate is less

than 5% per annum.D The annual depreciation rate increases over

time.E The amount of depreciation each year decreases

over time. [©VCAA 2007]

20 A piece of equipment, originally worth $49 600, diminishes at a rate of 14.5% p.a. The owner decides to replace the equipment when its book value falls below $5000. The time passing before the next replacement is required is close to:

A 10 years B 12 yearsC 14 years D 15 yearsE 16 years

21 The unit cost depreciation for a machine bought for $11 340 with scrap value of $750 and designed for 48 000 hours of operation is close to:A 6 c B 12 cC 22 c D 35 cE 49 c

22 A company car purchased for $28 500 and depreciating at an average rate of 47.4 c/km has travelled 23 600 km in the first year and 20 400 km in the second year. In its third year, its book value will be:

A $20 856 B $11 186C $7644 D $9670E $18 631

Short anSWer

1 An investment in BT bonds of $18 000 over 3 12

years earned Rebecca the same amount as $21 600 invested at 7.5% in a term deposit for 5 years. Calculate the interest rate offered by BT bonds.

2 Find:a the simple interest on $2900 for 8 months at

5.85% p.a.b the annual rate of simple interest so that $1600

will amount to $2000 after 3 yearsc the number of years over which $5000 will

amount to $6000 at 6.5% simple interestd the sum invested at 8% p.a. simple interest if it

amounted to $4000 after 4 years. 3 Karen borrowed $1100 from a credit union at 8.55%

per annum. She repaid the loan in a lump sum when the principal and interest amounted to $1400. How long, to the nearest year, did she keep the money?

4 Find each of the following and compare with the results from question 2.a The amount owing on $2900 for 8 months at

5.85% p.a. compounded daily.b The annual rate of compound interest so that

$1600 will amount to $2000 after 3 years.c The number of years over which $5000 will

amount to $6000 at 6.5% p.a. compounded quarterly.

d The sum invested at 8% p.a. compound interest if it amounted to $4000 after 4 years.

5 It is estimated that inflation will average 2% per annum over the next 8 years. If a new machine costs $60 000 now, calculate the cost of a similar new machine in 8 years time, adjusted for inflation. Assume no other cost change. Write your answer correct to the nearest dollar.

[©VCAA 2006]

6 The company prepares for this expenditure by establishing three different investments.a $7 000 is invested at a simple interest rate of

6.25% per annum for 8 years. Determine the total value of this investment at the end of 8 years.

b $10 000 is invested at an interest rate of 6% per annum compounding quarterly for 8 years. Determine the total value of this investment at the end of 8 years. Write your answer correct to the nearest dollar. [©VCAA 2006]

exam tip Many students earned only 1 mark by showing the calculation but finding only the interest earned — $3500. [Assessment report 2006]

448

7 For the given statement below, calculate:a the total creditsb the total debitsc the balance after each transaction

d the final balance for the period of the statement.


2 April 2009 Cheque 2694 $128.00 $5684.29 CR

5 April 2009 Salary $1693.80

11 April 2009 ABWDL Northland $800.00

19 April 2009 Salary $1693.80

23 April 2009 ABWDL Jam Factory

$800.00

29 April 2009 Cheque 2695 $2652.95

8 For the given statement in question 7, calculate the interest the account will earn in April if the bank pays 5 1

3% p.a. simple interest

a on the minimum monthly balanceb on the minimum daily balance.

9 The minimum monthly repayment on a credit card is (a) 1.5% of the balance rounded down to the nearest dollar or (b) $25, whichever is greater. Calculate the minimum monthly repayment on a balance of:a $3500 b $1194.50 c $492.76d $150 e $1820.53

10 Study the credit card statement below.

Date Credit DebitBalance

(amount owing)

1 January 2010 $1548.50 DR

8 January 2010 $500 — Repayment

15 January 2010 $399 — Purchase

1 February 2010 ??? — Interest

8 February 2010 ??? — Repayment

1 March 2010 ??? — Interest

a If the interest rate is 16.5% p.a., calculate the daily rate of interest, correct to 4 decimal places (take 1 year = 365 days).

b Calculate the interest added to the account on 1 February.

c On 8 February the minimum repayment of 5% is made. Calculate the amount of this repayment.

d Calculate the outstanding balance on the account on 1 March (assuming a non-leap year).

11 A new refrigerator is bought under a hire-purchase agreement for $3300 with $300 deposit and 24 monthly instalments of $153 each to be made. Calculate:

a the amount of interest chargedb the flat rate of interestc the effective rate of interest.



12 Alvin borrows $2000 from a bank at 14.6% p.a. calculated on a reducing monthly balance with monthly repayments of $40.a Calculate the amount still owed after the fourth

repayment.b Calculate the amount of interest paid in

4 repayments.c Express the interest paid over the period of

4 months as a percentage of the total payments.

13 Machinery bought for $29 600 is expected to have a scrap value of $1200 after 10 years.a Assuming straight-line depreciation, find the

yearly depreciation change.

b Find the amount by which depreciation at the end of the second year exceeds the depreciation at the end of the sixth year.

14 What was the original cost of an item that has a book value of $6344 after 7 years if the depreciation rate is 18% p.a.?

15 A photocopier is purchased for $11 400 and depreciates at an average rate of 2.1 × 10−3 cents per copy.a Calculate the book value of the photocopier

after 4 × 108 copies were made.b After how many copies will the photocopier be

written off (that is, have a zero scrap value)?

extended reSponSe

1 The passbook page below shows all the transactions for the month of February.

Date Credit Debit Balance

2 Feb. $122.81 $877.35

7 Feb. $461.76

15 Feb. $931.99

21 Feb. $461.76

28 Feb. $819.89

a Calculate the balance after each transaction and enter the results in the table.b Calculate the total credits for the month of February.c Calculate the total debits for the month of February.d Find the interest that will be earned in February if the bank pays 5.25% p.a. simple interest on the

minimum monthly balance.

2 Use the passbook page from question 1 to find the interest that will be earned in February if the bank pays 5.25% p.a. simple interest on the minimum daily balance.

3 An ‘up to 55 days interest free period’ credit card was used for purchases which after the 30-day interval totalled $1825. Find the minimum amount payable, if the current credit limit on this card is $2000 and the bank requires the largest of $25 or 1.5% of the outstanding balance. Any excess above the card limit must also be included in the payment.

4 Chris bought an outboard motor for $5600 under hire purchase with $1000 deposit and 24 monthly instalments of $220. Calculate:a the amount of interest chargedb the flat rate of interestc the effective rate of interest.

5 A personal loan of $4770, borrowed at 10% per annum reducing monthly balance, is to be repaid in monthly instalments of $80. Calculate the amount still owing at the end of the second month.

6 Machinery bought for $70 000 has an expected scrap value of $5000 after 13 years. Assuming straight-line depreciation, find the yearly depreciation of the machinery.

7 A factory writes off 3.5% per annum for its building with the current book value of $450 000. If the adopted method to calculate depreciation is straight line, find the book value of the factory in 12 years’ time.

450

8 A road works company purchased machinery for $120 000. It depreciates at 10% per annum reducing balance. Calculate:a the book value of the machinery after 4 yearsb the total depreciation over the first 6 yearsc the number of years (to the nearest month) until the machinery has a value below $25 000d the number of years (to the nearest month) until the machinery reaches its scrap value of $10 000e a depreciation schedule for the first 5 years.

9 A company writes off 25% p.a. for reduced value depreciation of its machinery with a current book value of $90 000. Determine the purchase cost of the machinery 5 years ago.

10 A silver mining company purchased a new machine for $130 000. Calculate its book value after 5 years if it is depreciated at the rate of 17% per annum. Use the reducing balance method.

11 A machine cost $56 000 when new, and had an estimated trade-in value of $4000 and a useful life of 16 000 operating hours. Calculate:a the unit cost depreciationb the machine’s book value after it has been operating for 7200 hours.

12 Michelle is planning to travel as a backpacker around Australia on her summer holiday next year. She estimates that she needs to save $1400 over the next 12 months. She has already saved $1095 and is considering two options.Option Aa Buying government bonds paying 9.85% p.a. i How much interest will Michelle earn on this investment? ii How long will it take for Michelle to save the remaining amount of $305 required for the holiday?Option Bb Opening a ‘Super-Saver’ investment account that pays 9.55% p.a. compounded daily. i How much interest will Michelle earn on this investment? ii How long will it take for Michelle to save the remaining amount of $305 required for the holiday?c Michelle chose Option B. She also realised that she would not have enough money for the trip.

Therefore, she saved some extra money and combined this with the total calculated in part b and reinvested for another 12 months at the same rate. How much does she need to add, to ensure that she has $1400 at the end of 12 months?

13 a Ken bought for his business, ‘Ken’s Lawn Mowing’, a new mower for $4850. It depreciates at the rate of 18% of its value at the start of each year.

i Find the value of the equipment at the end of 6 years, correct to the nearest dollar.

ii Approximately how much time will pass before the equipment will be worth less than $500?

b Ken decided to buy a new car, which cost $19 890. He traded in his old car and small trailer for $2600 as a deposit. The balance was to be paid on hire-purchase over 4 years, in weekly payments of $115.

i How much will Ken pay over the period of 4 years? ii How much interest will Ken pay over that period? iii Calculate the flat rate of interest charged. iv Calculate the effective rate of interest.

14 Christopher wishes to purchase a notebook. Its retail value is $3599. Christopher’s first option for financing the purchase is using a hire-purchase plan. The terms offered by the department store are 10% deposit with fortnightly instalments over 2 years at an interest rate of 9.2% p.a.a How much will Christopher need to withdraw from his savings account to pay the deposit?b Calculate the fortnightly repayments and total interest charge.



c What is the total cost of the notebook?A personal loan is advertised at 13.5% p.a. For Christopher to compare the interest rate, he needs to convert the hire-purchase flat rate of interest to the effective interest rate. d Calculate the effective interest rate and comment on

how it compares with the personal loan rate.Another option is for Christopher to save up until he has the cash to pay for the notebook. He can place the balance of his savings account, $2268.50, into a term deposit offering 6.3% p.a. simple interest for a 2-year term.e Calculate the total value of his investment at the end

of 2 years.f If he uses the term deposit investment towards the

purchase of the notebook, what amount still needs to be paid?

g What extra fortnightly savings will be needed over the next 2 years to make up the balance of $3599?

h What is the main attraction of the hire-purchase option over the other two options?

15 Khan wants to buy some office furniture that is valued at $7000.a i A store requires 25% deposit. Calculate the deposit. The balance is to be paid in 24 equal monthly

instalments. No interest is charged. ii Determine the amount of each instalment. Write your answer in dollars and cents.b Another store offers the same $7000 office furniture for $500 deposit and 36 monthly instalments of

$220. i Determine the total amount paid for the

furniture at this store. ii Calculate the annual flat rate of interest

charged by this store. Write your answer as a percentage correct to 1 decimal place.

A third store has the office furniture marked at $7000 but will give 15% discount if payment is made in cash at the time of sale.

c Calculate the cash price paid for the furniture after the discount is applied.Khan decides to extend his home office and borrows $30 000 for building costs. Interest is charged on the loan at a rate of 9% per annum compounding monthly. Assume Khan will pay only the interest on the loan at the end of each month.d Calculate the amount of interest he will pay

each month.Khan paid $900 for a fax machine. This price includes 10% GST (goods and services tax).

e Determine the price of the fax machine before GST was added. Write your answer correct to the nearest cent.

f Khan will depreciate his $900 fax machine for taxation purposes. He considers two methods of depreciation.

Flat-rate depreciationUnder flat rate depreciation the fax machine will be valued at $300 after 5 years. i Calculate the annual depreciation in dollars.

exam tip This question was quite poorly answered and the deposit was often not allowed for. One mark was available for clearly identifying the interest of $1420 that was then to be used in the simple interest formula.


exam tip Many students were puzzled by not being given a period of time for this loan. A loan where the interest only is paid each month is not a reducing balance question as the balance does not decrease or increase. It becomes a simple interest calculation for one period (1 month in this case).


exam tip Reverse GST questions such as this continue to be a problem for many students. Answers of $818.20 were not accepted as accuracy correct to the nearest cent was required. Some students simply reduced $900 by 10% of $90 to get $810.


452

Unit cost depreciationSuppose Khan sends 250 faxes a year. The $900 fax machine is depreciated by 46 cents for each fax it sends. ii Determine the value of the fax machine after 5 years.The books in Khan’s offi ce are valued at $10 000.g Calculate the value of these books after 5 years if they are depreciated by 12% per annum using the

reducing balance method. Write your answer correct to the nearest dollar.Khan believes his books should be valued at $4000 after 5 years.h Determine the annual reducing balance depreciation rate that will produce this value. Write your answers

as a percentage correct to 1 decimal place. [©VCAA 2007]

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Digital docTest YourselfChapter 9



eBookpluseBookplus aCtivitieS

Chapter openerDigital doc

10 Quick Questions: Warm up with ten quick • questions on fi nancial arithmetic. (page 388)

9A Simple interestTutorial

We3 • int-0898: Watch how to calculate the amount of interest received for each payment and in total for an investment option. (page 391)

Digital docs

SkillSHEET 9.1: Practise time conversions — • simple interest. (page 393)Spreadsheet 118: Investigate simple interest. • (page 393)

9B Compound interest, inflation and appreciation

Interactivity

Compound interest • int-0810: Consolidate your understanding of compound interest. (page 394)

Tutorial

We4 • int-0899: Watch how to compare four different investment options. (page 398)

Digital docs

SkillSHEET 9.2: Practise conversions — compound • interest. (page 403)Spreadsheet 009: Investigate compound interest. • (page 403)WorkSHEET 9.1: Calculate simple and compound • interest amounts and compare interest options. (page 404)

9C Savings and credit card accountsTutorial

We 12 • int-0900: Watch how to calculate the minimum credit card repayments due for four different balances. (page 409)

9D Purchase optionsDigital docs

Spreadsheet 112: Investigate reducing-interest loans. • (page 428)WorkSHEET 9.2: Apply fi nancial arithmetic to • calculate credit card and loan and investment quantities. (page 428)

9E DepreciationTutorials

We21 • int- : Watch how to make calculations about the depreciation of a company car over a number of years. (page 429) We23 • int-0902: Watch how to calculate the book value of a photocopier given the scrap value and the expected and current number of uses. (page 436) We24 • int-0903: Watch how to calculate the book value of an industrial machine and estimate how many years it will take for the machine to be written off. (page 437)

Chapter reviewDigital doc

Test Yourself: Take the end-of-chapter test to test • your progress. (page 452)

To access eBookPLUS activities, log on to

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exam praCtiCe 3 ChapterS 8 to 9

mUltiple ChoiCe 15 minutes

There is 1 mark for each correct multiple-choice answer.

1 Pedro purchases office equipment for $1995. It depreciates at a flat rate of 8% each year. After 5 years the book value will be closest to:A $798 B $1197 C $1315D $1835 E $1955

2 Which one the following point does not lie within the region defined by the inequality? 2x + 5 ≤ 3yA (1, 4) B (2, 3) C (−1, 1)D (−3, −2) E (−4, −1)

3 A stereo system after GST of 10% is added is priced at $275. The cost of the stereo system before GST was added is closest to:A 248 B 250 C 265 D 285 E 303

4 Georgina purchases 3 bananas and 4 apples for $3.15. The following week she buys 4 bananas and 5 apples for $4.10. If the next week she buys 2 bananas and 3 apples she would pay, in dollars:A $0.65 B $0.95 C $2.20D $2.25 E $2.30

5 Garth purchases a stereo system for $1565.95. He pays a 10% deposit and pays the amount owing in regular monthly instalments of $137.40 for 12 months. The total amount he pays for the stereo system is closest to: A 1648 B 1649 C 1703D 1805 E 1806

6

DATE DEPOSIT WITHDRAWAL BALANCE

3 February $354 $984.02

7 February $510.00 $474.02

24 February $150.00 $634.02

3 March $250.00 $374.02

The monthly bank statement for Georgina’s account is shown above. Her account pays 3.05% per annum paid on the minimum monthly balance. The amount of interest her account is paid for the month of February is:A $0.95 B $1.20 C $1.61D $11.41 E $14.46

7 Sarah and Chester both invest $5000 in two different accounts for 1 year. Sarah invests hers in a savings account that pays simple interest at a rate of 5.65% per annum. Chester invests his in an account that pays compound interest, compounded quarterly, at 4.5% per annum. Which one of the following is correct?A Sarah’s account would have $53.67 more than

Chester’s account.B Sarah’s account would have $680.09 less than

Chester’s account. C Sarah’s account would have $57.50 more than

Chester’s account.D Sarah’s account would have $59.71 less than

Chester’s account.E Sarah’s account would have $59.71 more than

Chester’s account.The following information relates to Questions 8 and 9. Sid borrows $15850 for a car. He will fully pay the loan within eight years with equal monthly payments. Interest is charged at the rate of 8.75% per annum, calculated monthly, on the reducing balance.

8 The amount Sid pays each month is closest to:A 165 B 180 C 230 D 236 E 281

9 Sid decides to make regular monthly payments of $300. The number of payments he will need to make to pay off the loan in full will be closest to:A 6 B 45 C 53 D 57 E 67

Short anSWer 18 minutes

1 Garth purchases a ride on lawnmower for his gardening service. When new, the lawnmower is valued at $5000. It is depreciated according to the unit cost method at $4.60 per hour. The expected life of the lawnmower is 1000 hours. a Write an equation that determines the scrap

value, x, in dollars, of the lawnmower.b Hence, determine the scrap value, in dollars.c Determine the book value after the lawnmower

has been used for 450 hours. 2 + 2 + 2 = 6 marks

2 At the beginning of 2009, the cost of a DVD is $19.95. The average inflation is 3% per annum. a Determine the cost of the DVD at the

beginning of 2010, to the nearest cent.


455exam practice 3 455

b At the beginning of 2010 the infl ation increases to 3.5% per annum.

Determine the year when the DVD will cost $22.80 if infl ation remains at this constant rate. 2 + 3 = 5 marks

extended reSponSe 30 minutes

Georgina and Garth employ two workers, Rose and Lily for their garden nursery business. Let x represent the number of hours that Rose can work each day.Let y represent the number of hours that Lily can work each day.The working hours for both Rose and Lily can be represented by inequalities 1 to 5.Inequality 1: x + y ≤ 8Inequality 2: x ≤ 6Inequality 3: y ≤ 2x Inequality 4: x ≥ 0Inequality 5: y ≥ 0

1 In terms of the number of hours Rose and Lily can work, explain inequality 3. 2 marks 2 The inequalities 1 to 5 are represented in the graph below. The feasible region is the unshaded area.

The feasible region has vertices A, B, C and D.The coordinates for vertices A, C and D are: A (0, 0)C (6, 2)D (6, 0)a Determine the exact coordinates of vertex B.b Rose is paid $14.60 per hour and Lily is paid $14.10 per hour.

Write an equation for the maximum cost in wages, C, each day, in dollars.

c Hence, determine the maximum cost in wages, C each day, in dollars, correct to the nearest cent.

d i On one particular day, Rose asks if she can only work 2 hours and Lily asks if she can only work 5 hours. Georgina refuses this request. Explain, in the context of this problem, why Georgina is justifi ed in refusing the employees request.

ii If Rose is granted her request to only work 2 hours, what is the maximum number of hours Lily can work? 2 + 2 + 2 + 1 + 1 = 8 marks

3 Georgina’s garden nursery business was priced at $375 000. She pays a 20% deposit and borrows the rest. She will fully pay the loan within fifteen years with equal monthly payments. Interest is charged at a rate of 9.75% per annum, calculated monthly, on the reducing balance.a Determine the amount, in dollars, she pays in the deposit.b Determine the amount of money, Georgina needs to borrow.c i Determine the number of instalments Georgina needs to make to fully pay off the loan with fi fteen

years. ii Hence, determine her monthly repayments, correct to the nearest cent.After 2 years, the interest rate decreases to 9.25% per annum. Georgina decides to maintain her monthly repayments before the decrease.d i Determine the amount Georgina still owes on her loan after 2 years. ii Hence, determine total number of payments she will need to make to fully pay off the loan.

2 + 1 + (1 + 2) + 4 = 10 marks

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9A 9B 9C 9D 9E Depreciation Financial arithmetic · 2016. 7. 19. · 9E Depreciation 9 areaS oF StUdy • Applications of simple interest and compound interest formulas • Appreciation