Business pre-2006

BUSINESS, FUR1 2005 VCAA 5 MC

An investor borrows `$200\ 000` for five years to buy an apartment.
The interest rate is `text(8.5%)` per annum compounding monthly.
It is an interest only loan, that is, at the end of five years, the investor will still owe `$200\ 000`.
He is required to make monthly repayments.
Correct to the nearest cent, his monthly repayment will be

A. $666.67

B. $1416.67

C. $1757.67

D. $4103.31

E. $6789.95

Show Answers Only

`B`

Show Worked Solution

`text(Using)\ \ A = PR^n − (Q(R^n − 1))/(R − 1) \ \ \ … \ (1)`

`A`	`= 200\ 000`
`P`	`= 200\ 000`
`R`	`= 1 + 8.5/(12 xx 100)`
	`= 1.0070833`
`n`	`= 12 xx 5`
	`= 60`

`text(Sub above values into (1))`

`200\ 000`	`= 200\ 000(1.0070833)^(60) − (Q(1.0070833^(60) − 1))/(1.0070833 − 1)`
`(Q(1.0070833^(60) − 1))/(0.0070833)`	`= 200\ 000(1.0070833)^(60) − 200\ 000`
`(Q(1.0070833^(60) − 1))/(0.0070833)`	`= 200\ 000(1.0070833^60 − 1)`
`Q`	`= 200\ 000 xx 0.0070833`
	`= $1416.66…`
	`≈ $1416.67`

`:.\ text(Monthly payments are $1416.67)`

`=> B`

BUSINESS, FUR1 2005 VCAA 4 MC

A machine that makes boxes costs `$45\ 000`.
Its value depreciates by five cents for every box it makes.
Each year it makes `120\ 000` boxes.
The depreciated value of this machine at the end of two years is

A. $33 000

B. $38 000

C. $39 000

D. $45 000

E. $115 000

Show Answers Only

`A`

Show Worked Solution

`text(Depreciation)`	`= (120\ 000 xx 2) xx 0.05`
	`= 12\ 000`

`text(Depreciated value)`	`= 45\ 000 − 12\ 000`
	`= 33\ 000`

`=> A`

BUSINESS, FUR1 2005 VCAA 2 MC

The price of a book is `$22`. It includes a Goods and Services Tax (GST) of `text(10%)`.
The price before this Goods and Services Tax (GST) is added is

A. $19.80

B. $20.00

C. $24.00

D. $24.20

E. $24.40

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ x = text(original price before GST)`

`x + 0.1x`	`= 22`
`1.1x`	`= 22`
`x`	`= 20`

`:.\ text(Price before GST) = $20`

`=> B`

BUSINESS, FUR1 2005 VCAA 1 MC

An investment of `$1200` for one year returns `$140.38` interest.
The simple interest rate per annum for this investment is closest to

A. 0.88%

B. 3.21%

C. 4.38%

D. 11.70%

E. 15.99%

Show Answers Only

`D`

Show Worked Solution

`text(Using)\ \ I`	`= (PrT)/100`
`r`	`= (100 xx I)/(PT)`
	`= (100 xx 140.38)/(1200 xx 1)`
	`= 11.698…`
	`≈ 11.70`

`=> D`

BUSINESS, FUR1 2004 VCAA 9 MC

An amount of `$130\ 000` is borrowed at an interest rate of `text(7.5%)` per annum, compounding monthly. The loan is fully repaid over ten years with equal monthly repayments.
Which of the following statements is not true?

A. The monthly interest rate is 0.625%.

B. No money will be owed after 10 years.

C. The total number of repayments is 120.

D. A monthly repayment of $1500 will reduce the length of the loan.

E. At the end of five years, the amount of the principal still owing will exceed $65 000.

Show Answers Only

`D`

Show Worked Solution

`text(Consider)\ A`

`text(Monthly interest rate)`	`= text(7.5%)/12`
	`= text(0.625%)`

`:.\ text(True, eliminate)\ A`

`text(Consider)\ B`

`text(If loan is fully repaid after 10 years, no money)`

`text(will be owed then.)`

`:.\ text(True, eliminate)\ B`

`text(Consider)\ C`

`text(Total number of repayments)`	`= 12 xx 10`
	`= 120`

`:.\ text(True, eliminate)\ C`

`text(Consider)\ D`

`text(Using)\ \ A`	`= PR^n − (Q (R^n − 1))/(R − 1)`
`text(Now)\ A`	`= 0 text( after 10 years)`
`0`	`= PR^n − (Q (R^n − 1))/(R − 1)`
`(Q (R^n − 1))/(R − 1)`	`= PR^n`
`Q`	`= (PR^n (R − 1))/(R^n − 1)`

`text(Now)\ \ P = 130\ 000, R = 1.00625, n = 120`

`:.\ Q`	`= (130\ 000(1.00625^(120))(1.00625 − 1))/(1.00625^(120) − 1)`
	`= $1543.12…`

`:.\ `	`text(Monthly repayments of $1543.12 needed to`
	`text(repay loan in 10 years. Hence repayments of)`
	`text($1500 will lengthen the loan.)`

`:.\ text(False)`

`text(Similarly)\ E\ text(can be eliminated.)`

`=> D`

BUSINESS, FUR1 2004 VCAA 8 MC

A machine is purchased for `$15\ 000`. Using the reducing balance method of depreciation, its book value after six years will be `$5000`.
The graph that best represents the value of the machine at the end of each year over the six-year period is

Show Answers Only

`A`

Show Worked Solution

`text(As a reducing balance method of depreciation)`

`text(is used, book value will decrease every year)`

`text(and it will decrease by the according % of the)`

`text(value each year.)`

`text{(Depreciates less in terms of cost each year.)}`

`=> A`

BUSINESS, FUR1 2004 VCAA 7 MC

Binnie invests `$12 000` for `5` years at an interest rate of `text(3.6%)` per annum, compounding annually.
The amount of interest she earns during the third year of the investment is closest to

A. $463.66

B. $470.41

C. $480.36

D. $1 343.22

E. $1 823.57

Show Answers Only

`A`

Show Worked Solution

`text(Using)\ \ A = PR^n`

`text(After 2 years,)\ \ P = 12000, R = 1.036, n = 2`

`:.\ A`	`= (12000)(1.036)^2`
	`= $12879.552`

`:.\ text(Interest in 3)^text(rd)\ text(year)`	`= 12879.552 xx text(3.6%)`
	`= $463.664…`
	`~~ $463.66`

`=> A`

BUSINESS, FUR1 2004 VCAA 6 MC

A loan of `$250\ 000` is to be paid back over a period of `20` years at an interest rate of `text(7.4%)` per annum, compounding monthly.
To the nearest dollar, the monthly repayment is closest to

A. $1 963

B. $1 999

C. $2 998

D. $4 343

E. $13 326

Show Answers Only

`B`

Show Worked Solution

`text(Using)\ \ A`	`= PR^n − (Q (R^n − 1))/(R − 1)`
`A`	`= 0`
`P`	`= 250\ 000`
`R`	`= 1 + 7.4/(100 xx 12)`
	`= 1.006167`
`n`	`= 20 xx 12`
	`= 240`

`text(Sub above into equation)`

`0`	`= 250000(1.006167^(240)) − (Q(1.006167^(240) − 1))/(1.006167 − 1)`
`(Q(1.006167^(240) − 1))/0.006167`	`= 250000(1.006167^(240))`
`Q`	`= (250000(1.006167^(240))(0.006167))/(1.006167^(240) − 1)`
	`= $1998.7849…`
	`≈ $1999`

`:.\ text(Monthly repayments are $1999.)`

`=> B`

BUSINESS, FUR1 2004 VCAA 4-5 MC

Chen buys a new refrigerator. The advertised price was `$2 700`. He chooses to pay a deposit of `$500` and monthly repayments of `$115` over `2` years.

Part 1

Under this arrangement, the total cost of the refrigerator is

A. $2 200

B. $2 760

C. $3 200

D. $3 260

E. $4 260

Part 2

The annual flat rate of interest paid is closest to

A. 10.4%

B. 12.2%

C. 12.7%

D. 20.7%

E. 25.4%

Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Total cost)`	`= 500 + (115 xx 24)`
	`= $3260`

`=> D`

`text(Part 2)`

`text(Interest amount)`	`= 3260 − 2700 − 500`
	`= $560`

`text(For interest rate, using)\ \ I = (PrT)/100`

`text(Now)\ P`	`= 2700 − 500`	`($500\ text(deposit paid))`
	`= 2200`

`:.\ 560`	`= ((2200)(r)(2))/100`
`r`	`= (560 xx 100)/(2200 xx 2)`
	`= 12.7272…`
	`≈ 12.7`

`:.\ text(Flat interest rate is 12.7%)`

`=> C`

BUSINESS, FUR1 2004 VCAA 3 MC

Leonard charged `$36` per hour for tutoring. In September 2004, he increased his fees by `text(15%)`.
For two hours of tutoring, Leonard now charges

A. $10.80

B. $41.40

C. $47.60

D. $72.00

E. $82.80

Show Answers Only

`E`

Show Worked Solution

`text(New charge per hour)`	`= 36 xx (1 + text(15%))`
	`= 36 xx 1.15`
	`= $41.40`

`:.\ text(For 2 hours)`	`= 41.40 xx 2`
	`= 82.80`

`=> E`

BUSINESS, FUR1 2004 VCAA 2 MC

Ardy invests `$150\ 000` for `6` years at an interest rate of `text(3.5%)` per annum, compounding annually.
The value of the investment at the end of the `6` years is

A. $31 500.00

B. $34 388.30

C. $178 107.00

D. $181 500.00

E. $184 388.30

Show Answers Only

`E`

Show Worked Solution

`text(Using)\ \ A`	`= PR^n`
`A`	`= 150\ 000 xx (1 + 3.5/100)^6`
	`= 150\ 000 xx 1.035^6`
	`= $184\ 388.299…`
	`≈ $184\ 388.30`

`:.\ text(Investment after 6 years is $184 388.30.)`

`=> E`

BUSINESS, FUR1 2004 VCAA 1 MC

Sarah invests `$37\ 000` at a simple interest rate of `text(4%)` per annum.
The total amount of interest earned in two years is

A. $1 480

B. $2 960

C. $5 920

D. $38 480

E. $39 960

Show Answers Only

`B`

Show Worked Solution

`text(Using)\ \ I`	`= (PrT)/100`
`I`	`= (37\ 000 xx 4 xx 2)/100`
	`= $2960`

`=> B`

BUSINESS, FUR1 2003 VCAA 9 MC

Peter borrows `$80\ 000` for `10` years at `text(5.6%)` per annum, compounding monthly, with monthly repayments of `$555`.
Which one of the following statements is true?

A. The loan will be fully paid out in ten years.

B. At the end of five years, the balance of the loan will be `$40\ 000`.

C. The amount of interest paid each month during the loan increases.

D. Weekly repayments of `$132` compounding weekly would reduce the period of the loan.

E. If one extra payment of `$2000` is to be made, it would be better to make it at the end of year eight than at the end of year two.

Show Answers Only

`D`

Show Worked Solution

`text(Consider)\ A`

`text($ paid after 10 years)`

`= 555 xx 12 xx 10`

`= $66\ 600`

`text{(less than loan of $80 000)}`

`:.\ text(False, eliminate)\ A`

`text(Consider)\ B`

`text(Total repaid after 5 years)`

`= $66\ 600 -: 2\ \ text{(from}\ A\ text{above)}`

`= $33\ 300`

`text(Principle owing)`

`= (80\ 000 − 33\ 300) + text(interest on loan)`

`= $46\ 700 + text(interest on loan)`

`:.\ text(False, eliminate)\ B`

`text(Consider)\ C`

`text(Monthly repayments are higher than)`

`text(monthly interest, so balance owed)`

`text(decreases monthly.)`

`text(Hence, monthly interest decreases.)`

`:.\ text(False, eliminate)\ C`

`text(Consider)\ D`

`text(Weekly payment of $132 is the same)`

`text(as monthly payments of $572, and there)`

`text(there is also extra benefit of compounding)`

`text(weekly to reduce the period of the loan.)`

`:.\ text(True)`

`text(Similarly)\ E\ text(can be eliminated)`

`=> D`

BUSINESS, FUR1 2003 VCAA 8 MC

The following is an extract from a bank account showing all transactions for the period 1 January to 30 June, 2003.

Interest on this account is calculated at a rate of `text(0.25%)` per month on the minimum monthly balance and paid into the account quarterly. Interest for the June period (April to June) is paid on 30 June.
The balance in the account after interest is paid on 30 June 2003 is

A. $6039.64

B. $6038.49

C. $6024.76

D. $6023.51

E. $6022.36

Show Answers Only

`B`

Show Worked Solution

`text(In April,)`

`text(Interest)`	`= 6452.4 xx text(0.25%)`
	`= 16.131`

`text(In May,)`

`text(Interest)`	`= 5992.4 xx text(0.25%)`
	`= 14.981`

`text(In June,)`

`text(Interest)`	`= text(same as May)`
	`= 14.981`

`:.\ text(Balance on 30 June 2003)`

`= 5992.4 + 16.131 + 2(14.981)`

`= 6038.4930`

`=> B`

BUSINESS, FUR1 2003 VCAA 7 MC

Lim invested `$8000` in an investment account, earning `r text(%)` interest per annum, compounding quarterly.
The balance in dollars, after `5` years, is given by

A. `8000(1 + r/100)^5`

B. `8000(1 + r/100)^20`

C. `8000(1 + r/400)^5`

D. `8000(1 + r/400)^20`

E. `8000(1 + r/1200)^60`

Show Answers Only

`D`

Show Worked Solution

`text(Using)\ \ A = PR^n`

`P = 8000`

`R`	`= 1 + r/(4 xx 100)`
	`= 1 + r/400`

`n`	`= 5 xx 4`
	`= 20`

`:.\ A = 8000(1 + r/400)^20`

`=> D`

BUSINESS, FUR1 2003 VCAA 6 MC

Interest is paid monthly into an account at a rate of `text(3%)` per annum. Each month, immediately after the interest is paid, the account is debited `$5` in fees. No other transactions take place. The initial amount of money in the account is `$12\ 200`.
After all interest has been paid and fees debited, the balance in the account at the end of two months is

A. $12 251.06

B. $12 261.08

C. $12 271.09

D. $12 932.83

E. $12 953.13

Show Answers Only

`A`

Show Worked Solution

`text(Interest rate/month) = 3/12 % = text(0.25%)`

`text(Using)\ \ A = PR^n`

`text(After 1 month)`

`A_1`	`= 12200(1 + 0.25/100) − 5`
	`= 12200(1.0025) − 5`
	`= 12225.5`

`text(After 2 months)`

`A_2`	`= 12225.5(1.0025) − 5`
	`= 12251.0638`
	`= $12251.06`

`=> A`

BUSINESS, FUR1 2003 VCAA 5 MC

Zoltan is running a convenience store. He purchases equipment for `$6500`. It is anticipated that the equipment will last `5` years and have a depreciated value of `$2000`.
Assuming the straight line method of depreciation, the equipment is depreciated annually by

A. $400

B. $900

C. $1027

D. $1300

E. $4500

Show Answers Only

`B`

Show Worked Solution

`text(Depreciation)`	`= 6500 − 2000`
	`= 4500`

`text(Annual depreciation)`

`= 4500/5`

`= 900`

`=> B`

BUSINESS, FUR1 2003 VCAA 4 MC

Swee borrowed `$150\ 000` at `text(6.2%)` per annum compounding monthly. The repayments are `$1100` per month.
The balance of the loan at the end of five years is closest to

A. $0

B. $84 000

C. $127 000

D. $137 000

E. $148 000

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ A = PR^n`

`P = 150\ 000`

`R = 1 + ((6.2/12))/100 = 1.005167`

`n = 5 xx 12 = 60`

`:.\ A`	`= 150\ 000 xx 1.005167^60`
	`= 204\ 354.6788`

`text(Repayments) = (Q(R^n − 1))/(R − 1)`

`text(where)\ Q = 1100`

`:.\ text(Repayments)`	`= (1100(1.005167^60 − 1))/(1.005167 − 1)`
	`= 77\ 143.5993`

`text(Amount owing after 5 years)`

`= A − text(repayments)`

`= 204\ 354.6788 − 77\ 143.5993`

`= $127\ 211.0795`

`=> C`

BUSINESS, FUR1 2003 VCAA 3 MC

Heather invests `$45\ 000` at `text(4%)` per annum for `5` years compounding annually.
The total amount of interest earned is

A. $1 800

B. $2 100

C. $9 000

D. $9 750

E. $54 750

Show Answers Only

`D`

Show Worked Solution

`text(Using)\ \ A = PR^n`

`text(Sub)\ P = 45000, R = 1.04, n = 5\ text(into)`

`text(equation)`

`A`	`= 45000 xx 1.04^5`
	`= 54749.3806`

`text(Interest)`	`= text(Final Amount − Original Amount)`
`	`= 54749.3806 − 45000`
	`= 9749.3806`
	`~~ 9750`

`=> D`

BUSINESS, FUR1 2003 VCAA 2 MC

Derek invested `$26\ 000` for eighteen months and earned `$975` in simple interest.
The annual interest rate for the investment is

A. 0.025%

B. 0.0563%

C. 2.5%

D. 3.75%

E. 5.63%

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ I = (PrT)/100`

`text(Sub)\ \ I = 975, P = 26\ 000, T = 1.5`

`text(into the equation)`

`975`	`= (26\ 000 xx r xx 1.5)/100`
`r`	`= (975 xx 100)/(26\ 000 xx 1.5)`
	`= 2.5`

`:.\ text(Rate is 2.5%)`

`=> C`

BUSINESS, FUR1 2003 VCAA 1 MC

Under a hire purchase agreement, Sheng will pay a total of `$960` for a television set. He is required to pay a deposit of `$120` and to pay the balance in regular equal monthly payments over `6` months.
The monthly repayments are

A. $70

B. $80

C. $140

D. $160

E. $180

Show Answers Only

`C`

Show Worked Solution

`text(Balance owing) = 960 − 120 = 840`

`text(Payment over 6 months)`

`= 840/6`

`= 140`

`=> C`

BUSINESS, FUR1 2002 VCAA 9 MC

Roland wants to increase the price of his products by `text(5%)`. By mistake, he increases the price by `text(8%)`. Realising the mistake, he tries to correct his mistake by discounting the new prices by `text(3%)`.
As a result, the products have had a price increase (from the original price) of

A. 2.60%

B. 3.00%

C. 4.76%

D. 5.00%

E. 5.24%

Show Answers Only

`C`

Show Worked Solution

`text(Let original price be)\ x.`

`text(Price after 8% increase)`	`= 1.08 xx x`
	`= 1.08x`
`text(Price after 3% decrease)`	`= 1.08x xx (1 − 0.03)`
	`= 1.08x xx 0.97`
	`= 1.0476x`

`:.\ text(% increase from original price)\ x\ text(is 4.76%).`

`=> C`

BUSINESS, FUR1 2002 VCAA 8 MC

The following graph represents the growth of an investment over several years.

If `A` dollars is the value of the investment after `n` years, then a rule for describing the growth of this investment could be

A. `A = 2000 xx (1.06)^n`

B. `A = 2000 xx (0.06)^n`

C. `A = 2000 xx 1.06n`

D. `A = 2000 xx 0.06n`

E. `A = 2000 + (1.06)^n`

Show Answers Only

`A`

Show Worked Solution

`text(As investment value increases each year,)`

`text(in the multiple choices given, as)\ n`

`text(increases,)\ A\ text(increases.)`

`:.\ text(Eliminate)\ B, D.`

`text(As line that can be drawn on graph is not)`

`text(linear,)\ C\ text(is wrong.)`

`:.\ text(Eliminate)\ C.`

`text(For each year that passes, the value that is)`

`text(increased is proportional to the value of the)`

`text(previous year.)`

`:.\ text(Eliminate)\ E.`

`=> A`

BUSINESS, FUR1 2002 VCAA 7 MC

Ravi has a loan of `$135\ 000` at `text(7%)` per annum interest, compounding monthly. The loan is to be repaid monthly over `20` years. The scheduled repayments are `$1046.65` per month. However, he finds that he can afford to pay `$1200` per month and decides to do so for the duration of the loan.
The amount of time this will save in paying off the loan is closest to

A. 6 months

B. 1 year

C. 5 years

D. 10 years

E. 15 years

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ A`	`= PR^n − (Q (R^n − 1))/(R − 1) \ \ \ … \ (1)`
`A`	`= 0 text( after loan is repaid)`
`P`	`= 135\ 000`
`R`	`= 1 + 7/(12 xx 100)`
	`= 1.0058333…`
`Q`	`= 1200`

`text(Sub above into)\ (1)`

`0`	`= (135000)(1.0058333)^n − (1200(1.0058333^n − 1))/(1.0058333 − 1)`
`(1200(1.0058333^n − 1))/0.0058333`	`= (135000)(1.0058333^n)`
`1200(1.0058333^n) − 1200`	`= (135000)(1.0058333^n)(0.0058333)`
`1200 − (1200)/(1.0058333^n)`	`= (135000)(0.0058333)`
`− (1200)/(1.0058333^n)`	`= (135000)(0.0058333) − 1200`
`1.0058333^n`	`= (1200)/(1200 − (135000)(0.0058333))`
	`= 2.9090877…`
`log(1.0058333^n)`	`= log(2.9090877)`
`n`	`= 183.59`
	`≈ 184\ text(months)`
	`≈ 15.299\ text(years)`
`:.\ text(Time reduced)`	`= 20 − 15.299`
	`= 4.701`
	`≈ 5\ text(years)`

`=> C`

BUSINESS, FUR1 2002 VCAA 6 MC

A machine is purchased for `$36\ 000`. Its value depreciates at a rate of `$0.16` for each unit it produces. On average, the machine produces `24\ 000` units a year.
Using the unit cost method of depreciation, the value of the machine after six years of use is closest to

A. $3840

B. $7200

C. $12 960

D. $23 040

E. $32 160

Show Answers Only

`C`

Show Worked Solution

`text(Depreciation per year)`	`= 24\ 000 xx 0.16`
	`= $3840`
`text(Value after depreciation)`	`= 36\ 000 − (3840)(6)`
	`= $12\ 960`

`:.\ text(After 6 years, value is $12 960.)`

`=> C`

BUSINESS, FUR1 2002 VCAA 5 MC

A loan of `$18\ 000` is to be repaid by making `48` payments of `$480` per month.
For this loan, the effective rate of interest is closest to

A. 7.0% per annum

B. 13.7% per annum

C. 26.7% per annum

D. 28.0% per annum

E. 54.9% per annum

Show Answers Only

`B`

Show Worked Solution

`text(Total paid)`	`= 48 xx 480`
	`= $23\ 040`
`text(Interest)`	`= 23\ 040 − 18\ 000`
	`= 5040`

` text(Flat rate interest)`

`= 5040/(18\ 000) -: (48/12)`

`= 0.07`

`= text(7%)`

`text(Using effective rate of interest)`	`≈ (2n)/(n + 1) xx text(flat rate)`
`text(Effective rate of interest)`	`≈ (2(48))/(48 + 1) xx 7`
	`≈ 13.71…`
	`≈ text(13%)`

`=> B`

BUSINESS, FUR1 2002 VCAA 4 MC

Rho takes a `20`-year loan of `$172\ 000` at `text(6%)` per annum, compounding monthly and with monthly repayments.
To fully repay the loan in `20` years, the amount he must repay each month is

A. $716.67

B. $1216.54

C. $1232.26

D. $9058.63

E. $10 320.00

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ A = PR^n − (Q (R^n − 1))/(R − 1) \ \ \ … \ (1)`

`P`	`= 172\ 000`
`R`	`= 1 + 6/(12 xx 100)`
	`= 1.005`
`n`	`= 20 xx 12`
	`= 240`

` text(After 20 years,)\ A = 0`

`:.\ text{Sub above into (1)}`

`0 =`	`(172\ 000)(1.005)^240 − (Q(1.005^240− 1))/(1.005 − 1)`
`(Q(1.005^240− 1))/0.005 =`	`(172\ 000)(1.005)^240`
`Q =`	`((172\ 000)(1.005^240)(0.005))/(1.005^240 − 1)`
`=`	`$1232.261…`

`:.\ text(Monthly repayments are $1232.26)`

`=> C`

BUSINESS, FUR1 2002 VCAA 3 MC

Kuong invests `$8500` at `text(6%)` interest per annum, compounding quarterly.
The amount of interest he earns during the fourth year of the investment is

A. $521.59

B. $623.62

C. $2231.05

D. $2286.38

E. $4489.32

Show Answers Only

`B`

Show Worked Solution

`text(Using)\ \ A = PR^n`

`text(After 3 years,)\ P`	`= 8500`
`R`	`= 1 + 6/(4 xx 100)`
	`= 1.015`
`n`	`= 3 xx 4`
	`= 12`

`A`	`= 8500 xx 1.015^(12)`
	`= $10\ 162.754…`

` text(After 4 years,)\ n = 4 xx 4 = 16`

`A`	`= 8500 xx 1.015^(16)`
	`= $10\ 786.377…`

`text(Interest in)\ 4^text(th)\ text(year)`	`= 10\ 786.377 − 10\ 162.754`
	`= $623.62`

`=> B`

BUSINESS, FUR1 2002 VCAA 2 MC

An investment of `$16\ 000` is made at `4%` interest per annum, compounding yearly.
The value of the investment at the end of the second year is

A. $17 280.71

B. $17 305.60

C. $17 325.71

D. $21 120.00

E. $21 897.10

Show Answers Only

`B`

Show Worked Solution

`text(Using)\ \ A`	`= PR^n`
`A`	`= 16\ 000 xx (1.04)^2`
	`= $17\ 305.6`

`:.\ text(After 2 years, investment is $17 305.60.)`

`=> B`