smc-603-65-CAS (2 step)

Recursion, GEN2 2024 NHT 9

Cleo took out a loan of $35 000 to pay for an overseas holiday.

Interest is charged at the rate of 10% per annum compounding quarterly.

For the first year of this loan, Cleo made quarterly repayments of $1722.

Let $V_n$ be the balance of Cleo's loan, in dollars, after $n$ quarters.
Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$ that can model the value of the loan from quarter to quarter for the first year. (1 mark)

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For the second year of the loan, Cleo increased her quarterly repayments to $2000.

Determine the total amount of interest Cleo paid in the first two years of the loan.
Round your answer to the nearest cent. (2 marks)

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Show Answers Only

a. $V_0=35\,000\ \qquad V_{n+1}=1.025 \times V_n-1722 $

b. $\text{Total interest}\ = 14\,888-8553.89=\$6334.11$

Show Worked Solution

a. $\text{Interest per quarter}\ =\dfrac{10\%}{4}=2.5\% $

$\text{Multiplication factor}\ = 1+0.025=1.025$

$\text{Recurrence relation:}$

$V_0=35\,000\ \qquad V_{n+1}=1.025 \times V_n-1722 $

b. $\text{Find \(PV$ when $N=4$ (by TVM Solver):}\)

$N=4$

$I\%=10.0\%$

$PV=35\,000$

$PMT = -1722$

$FV= \text{SOLVE} = 31\,482.82$

$P/Y = C/Y = 4$

$\text{Find \(PV$ when $N=4$ (by TVM Solver):}\)

$N=4$

$I\%=10.0\%$

$PV=31\,482.82$

$PMT = -2000$

$FV= \text{SOLVE} = 26\,446.11$

$P/Y = C/Y = 4$

$\text{Total repayments}\ =(4 \times 1722) + (4 \times 2000) = \$14\,888 $

$\text{Principal paid}\ = 35\,000-26\,446.11=\$8553.89 $

$\text{Total interest}\ = 14\,888-8553.89=\$6334.11$

Financial Maths, GEN2 2024 VCAA 8

Emi takes out a reducing balance loan of $500 000.

The interest rate is 5.3% per annum, compounding monthly.

Emi makes regular monthly repayments of $3071.63 for the duration of the loan, with only the final repayment amount being slightly different from all the other repayments.

Determine the total cost of Emi's loan, rounding your answer to the nearest cent, and state the number of payments required to fully repay the loan. (2 marks)

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$\text{Total cost}\ =$884\,633.62\ ,\ 288\ \text{payments}$

Show Worked Solution

$\text{Using CAS: Find}\ N\ \text{when}\ FV=0$

♦♦ Mean mark 29%.

$\text{Using CAS: Find amount still owing if 288 payments are made.}$

$\text{Final payment}\ =$3071.63+$4.18 =$3075.81$

$\text{Total cost of loan}\ =3071.63\times 287 + 3075.81=$884\,633.62$

$\text{Total number of payments}\ =288$

Financial Maths, GEN2 2023 VCAA 7

Arthur takes out a new loan of $60 000 to pay for an overseas holiday.

Interest on this loan compounds weekly.

The balance of the loan, in dollars, after $n$ weeks, $V_n$, can be determined using a recurrence relation of the form

$V_0=60\ 000, \quad V_{n+1}=1.0015\,V_n-d$

Show that the interest rate for this loan is 7.8% per annum. (1 mark)

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Determine the value of $d$ in the recurrence relation if
1. Arthur makes interest-only repayments (1 mark)
  
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2. Arthur fully repays the loan in five years. Round your answer to the nearest cent. (1 mark)
  
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Arthur decides that the value of $d$ will be 300 for the first year of repayments.
If Arthur fully repays the loan with exactly three more years of repayments, what new value of $d$ will apply for these three years? Round your answer to the nearest cent. (1 mark)

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For what value of $d$ does the recurrence relation generate a geometric sequence? (1 mark)

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Show Answers Only

a. $ 7.8\%$

b.i. $d=90$

b.ii. $d= $278.86$

c. $ d= $350.01$

d. $d=0$

Show Worked Solution

a. $\text{Weekly interest rate factor}\ = 1.0015-1 = 0.0015 = 0.15\% $

$I\%(\text{annual}) = 52 \times 0.15 = 7.8\%$

♦♦ Mean mark (a) 32%.

b.i. $d= 0.15\% \times 60\ 000 = \dfrac{0.15}{100} \times 60\ 000 = 90$

b.ii. $\text{By TVM solver:} $

$N$	$=5 \times 52 = 260$
$I\%$	$=7.8$
$PV$	$= -60\ 000$
$PMT$	$= ?$
$FV$	$=0$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow PMT = d= $278.86$

♦ Mean mark (b)(i) 47%.
♦ Mean mark (b)(ii) 40%.

c. $d=300\ \text{for the 1st 52 weeks.}$

$\text{Find}\ FV\ \text{after 52 weeks:}$

$N$	$=52$
$I\%$	$=7.8$
$PV$	$= -60\ 000$
$PMT$	$= 300$
$FV$	$=?$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow FV = $48\ 651.67$

$\text{Find}\ PMT\ \text{given}\ FV=0\ \text{after 156 more weeks:}$

$N$	$=156$
$I\%$	$=7.8$
$PV$	$= -48\ 651.67$
$PMT$	$= ?$
$FV$	$=0$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow PMT = d= $350.01$

♦♦ Mean mark (c) 32%.

d. $\text{Geometric sequence when}\ \ d=0$

$V_0=60\ 000, V_1=60\ 000(1.0015), V_2=60\ 000(1.0015)^2, …$

♦♦♦ Mean mark (d) 11%.

CORE, FUR2 2021 VCAA 8

For renovations to the coffee shop, Sienna took out a reducing balance loan of $570 000 with interest calculated fortnightly.

The balance of the loan, in dollars, after `n` fortnights, `S_n` can be modelled by the recurrence relation

`S_0 = 570 \ 000,` `S_{n+1} = 1.001 S_n - 1193`

Calculate the balance of this loan after the first fortnightly repayment is made. (1 mark)

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Show that the compound interest rate for this loan is 2.6% per annum. (1 mark)

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For the loan to be fully repaid, to the nearest cent, Sienna's final repayment will be a larger amount.
Determine this final repayment amount.
Round your answer to the nearest cent. (1 mark)

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Show Answers Only

`$ 569 \ 377`
`2.6text(%)`
`$ 1198.56`

Show Worked Solution

a.	`S_1`	`= 1.001 xx 570 \ 000-1193`
		`= $ 569 \ 377`

b. `text{Fortnights in 1 year} = 26`

`text{Rate per fortnight} = (1.001-1) xx 100text(%) = 0.1text(%)`

`:.\ text(Annual compound rate)`	`= 26 xx 0.1text(%)`
	`= 2.6text(%)`

c. `text{Find}\ N \ text{by TVM Solver:}`

`N`	`= ?`
`I text{(%)}`	`= 2.6`
`PV`	`= 570 \ 000`
`PMT`	`= -1193`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 26`

`=> N = 650.0046 …`

`text{Find} \ FV \ text{after exactly 650 payments:}`

`N`	`= 650`
`Itext{(%)}`	`= 2.6`
`PV`	`= 570 \ 000`
`PMT`	`= -1193`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 26`

`=> FV = -5.59`

`:. \ text{Final repayment}`	`= 1193 + 5.59`
	`= $ 1198.59`

CORE, FUR2 2020 VCAA 11

Samuel took out a new reducing balance loan.

The interest rate for this loan was 4.1% per annum, compounding monthly.

The balance of the loan after four years of monthly repayments was $329 587.25

The balance of the loan after seven years of monthly repayments was $280 875.15

Samuel will continue to make the same monthly repayment.

To ensure the loan is fully repaid, to the nearest cent, the required final repayment will be lower.

In the first seven years, Samuel made 84 monthly repayments.

From this point on, how many more monthly repayments will Samuel make to fully repay the loan? (2 marks)

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Show Answers Only

`text(150 more monthly payments.)`

Show Worked Solution

`text(Find monthly payment)`

♦♦ Mean mark 21%.

`text(By TVM Solver:)`

`N`	`= 36`
`I(%)`	`= 4.1`
`PV`	`= 329\ 587.25`
`PMT`	`= ?`
`FV`	`= -280\ 875.15`
`text(PY)`	`= text(CY) = 12`

`=> PMT = -2400.00`

`text(Find)\ N\ text(when)\ FV = 0`

`text(By TVM Solver:)`

`N`	`= ?`
`I(%)`	`= 4.1`
`PV`	`= 280\ 875.15`
`PMT`	`= -2400`
`FV`	`= 0`
`text(PY)`	`= text(CY) = 12`

`=> N = 149.67…`

`:.\ text(After 7 years, 150 more monthly payments are required.)`

Financial Maths, GEN1 2019 NHT 22-23 MC

Armand borrowed $12 000 to pay for a holiday.

He will be charged interest at the rate of 6.12% per annum, compounding monthly.

This loan will be repaid with monthly repayments of $500.

Part 1

After four months, the total interest that Armand will have paid is closest to

$231
$245
$255
$734
$1796

Part 2

After eight repayments, Armand decided to increase the value of his monthly repayments.

He will make a number of monthly repayments of $850 and then one final repayment that will have a smaller value.

This final repayment has a value closest to

$168
$169
$180
$586
$681

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ B`

Show Worked Solution

`text(Part 1)`

`text(By TVM Solver:)`

`N`	`= 4`
`Itext(%)`	`= 6.12`
`PV`	`= 12\ 000`
`PMT`	`= −500`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=>\ text(FV) = 10\ 231.33`

`:.\ text(Total interest)`	`=\ text(total payments − reduction in principle)`
	`= (4 xx 500) – (12\ 000 – 10\ 231.33)`
	`= 231.33`

`=>\ A`

`text(Part 2)`

`text(Loan after 8 repayments (by TVM Solver)):`

`N`	`= 8`
`I(text(%))`	`= 6.12`
`PV`	`= 12\ 000`
`PMT`	`= −500`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = 8426.30`

`text(Number of repayments required at $850)`

`N`	`= ?`
`I(text(%))`	`= 6.12`
`PV`	`= 8426.30`
`PMT`	`= −500`
`FV`	`= 0`
`text(P/Y)`	`= 12`
`text(C/Y)`	`= 12`

`=> N = 10.2`

`text(Balance after 10 repayments = $168.29)`

`:.\ text(Final repayment)`	`= 168.29 xx (1 + 6.12/12)`
	`= 169.15`

`=>\ B`

CORE, FUR2 2016 VCAA 7

Ken has borrowed $70 000 to buy a new caravan.

He will be charged interest at the rate of 6.9% per annum, compounding monthly.

For the first year (12 months), Ken will make monthly repayments of $800.
1. Find the amount that Ken will owe on his loan after he has made 12 repayments. (1 mark)
  
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2. What is the total interest that Ken will have paid after 12 repayments? (1 mark)
  
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After three years, Ken will make a lump sum payment of $L in order to reduce the balance of his loan.
This lump sum payment will ensure that Ken’s loan is fully repaid in a further three years.
Ken’s repayment amount remains at $800 per month and the interest rate remains at 6.9% per annum, compounding monthly.
What is the value of Ken’s lump sum payment, $L?
Round your answer to the nearest dollar. (2 marks)

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Show Answers Only

a.i.`$65\ 076.22`

a.ii.`$4676.22`

b. `$28\ 204.02`

Show Worked Solution

a.i. `text(By TVM Solver,)`

♦ Mean mark 46%.

`N`	`= 12`
`I(%)`	`= 6.9`
`PV`	`= 70\ 000`
`PMT`	`= -800`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = -65\ 076.219…`

`:.\ text(Ken will owe $65 076.22)`

♦♦ Mean mark 26%.

a.ii.	`text(Total interest paid)`	`= text(Total repayments) – text(reduction in principal)`
		`= (12 xx 800) – (70\ 000 – 65\ 076.22)`
		`= $4676.22`

b. `text(Find the loan balance after 3 years)`

♦♦♦ Mean mark 20%.
MARKER’S COMMENT: Correct input tables allowed for a method mark if answers were calculated incorrectly.

`N`	`= 12 xx 3 = 36`
`I(%)`	`= 6.9`
`PV`	`= 70\ 000`
`PMT`	`= -800`
`FV`	`= ?`
`text(P/Y)`	`=text(C/Y)=12`

`=> FV = 54\ 151.60`

`text(Find the loan amount that can be)`

`text(fully repaid by monthly payments)`

`text(of $800 over 3 years.)`

`N`	`= 36`
`I(%)`	`= 6.9`
`PV`	`= ?`
`PMT`	`= -800`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y)= 0`

`=> PV = 25\ 947.58`

`:.\ $L`	`= 54\ 151.60 -25\ 947.58`
	`= $28\ 204.02`

CORE*, FUR2 2008 VCAA 5

Michelle took a reducing balance loan for $15 000 to purchase her car. Interest is calculated monthly at a rate of 9.4% per annum.

In order to repay the loan Michelle will make a number of equal monthly payments of $350.

The final repayment will be less than $350.

How many equal monthly payments of $350 will Michelle need to make? (1 mark)

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How much of the principal does Michelle have left to pay immediately after she makes her final $350 payment? Find this amount correct to the nearest dollar. (1 mark)

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Exactly one year after Michelle established her loan the interest rate increased to 9.7% per annum. Michelle decided to increase her monthly payment so that the loan would be fully paid in three years (exactly four years from the date the loan was established).

What is the new monthly payment Michelle will make? Write your answer correct to the nearest cent. (2 marks)

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`52`
`$147`
`$388.30`

Show Worked Solution

a. `text(By TVM Solver,)`

♦♦ Mean mark of all parts (combined) was 23%.
MARKER’S COMMENT: This part was not well done. An area where students consistently struggle each year.

`N`	`= ?`
`I(text(%))`	`= 9.4`
`PV`	`= 15\ 000`
`PMT`	`=-350`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>N = 52.422…`

`:. 52\ text(payments of $350 will be required.)`

b. `text(Find principal left after 52 repayments.)`

`text(By TVM Solver,)`

`N`	`= 52`
`I(text(%))`	`= 9.4`
`PV`	`= 15\ 000`
`PMT`	`=-350`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`FV =-147.056…`

`:.\ text(Principal left to pay) = $147\ \ text{(nearest dollar)}`

c. `text(Find principal left after 12 months.)`

`text(By TVM Solver,)`

`N`	`= 12`
`I(text(%))`	`= 9.4`
`PV`	`= 15\ 000`
`PMT`	`=-350`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=>FV = -12\ 086.602…`

`text(If loan repaid over the next 3 years,)`

`text(By TVM Solver,)`

`N`	`= 3 xx 12 = 46`
`I(text(%))`	`= 9.7`
`PV`	`= 12\ 086.602…`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT = 388.300…`

`:.\ text(New monthly payments is $388.30.)`

CORE*, FUR2 2015 VCAA 5

Jane and Michael borrow $50 000 to expand their business.

Interest on the unpaid balance is charged to the loan account monthly.

The $50 000 is to be fully repaid in equal monthly repayments of $485.60 for 12 years.

Determine the annual compounding rate of interest.

Write your answer correct to two decimal places. (1 mark)

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Calculate the amount that will be paid off the principal at the end of the first year.

Write your answer correct to the nearest dollar. (1 mark)

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Halfway through the term of the loan, at the end of the sixth year, Jane and Michael make an additional one-off payment of $3500.

Assume no other changes are made to their loan conditions.

Determine how much time Jane and Michael will save in repaying their loan.

Give your answer correct to the nearest number of months. (2 marks)

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Show Answers Only

`text(5.91%)`
`$2951`
`10\ text(months)`

Show Worked Solution

a. `text(By TVM Solver, after 12 years,)`

`N`	`=12 xx 12 = 144`
`I (%)`	`=?`
`PV`	`= 50\ 000`
`PMT`	`=-485.60`
`FV`	`=0`
`text(P/Y)`	`=text(C/Y) = 12`

`=> I=5.9100…`

`:.\ text(The annual compounding interest rate = 5.91%)`

b. `text(By TVM Solver,)`

MARKER’S COMMENT: Showing solver “input” allowed a possible error mark in part (b) even if part (a) was incorrect.

`N`	`=12`
`I (%)`	`=5.91`
`PV`	`= 50\ 000`
`PMT`	`= -485.60`
`FV`	`=?`
`text(P/Y)`	`=12`
`text(C/Y)`	`=12`

`=> FV=-47\ 048.7077…`

`:.\ text(Paid off)`	`= 50\ 000-47\ 048.7077…`
	`= $2951.29…`
	`=$2951\ \ text{(nearest dollar)}`

c. `text(By TVM Solver, after 6 years,)`

`N`	`=12 xx 6=72`
`I (%)`	`=5.91`
`PV`	`= 50\ 000`
`PMT`	`=-485.60`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV=-29\ 376.045…`

`:.\ text(New)\ PV = 29\ 376.05-3500 = 25\ 086.05`

`text(Find)\ \ N\ \ text(such that)\ \ FV=0,`

`N`	`=?`
`I (%)`	`=5.91`
`PV`	`= 25\ 086.05`
`PMT`	`=-485.60`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N=61.96…`

`:.\ text(Time saving)`	`= 72-61.96…`
	`= 10.03…`
	`=10\ text(months)`

CORE*, FUR1 2015 VCAA 8 MC

Cindy took out a reducing balance loan of $8400 to finance an overseas holiday.

Interest was charged at a rate of 9% per annum, compounding quarterly.

Her loan is to be fully repaid in six years, with equal quarterly payments.

After three years, Cindy will have reduced the balance of her loan by approximately

A. 9%

B. 35%

C. 43%

D. 50%

E. 57%

Show Answers Only

`C`

Show Worked Solution

`text(By TVM Solver,)`

♦♦ Mean mark 33%.

`N`	`=6 xx 4 = 24`
`I (%)`	`=9`
`PV`	`= – 8400`
`PMT`	`=?`
`FV`	`=0`
`text(P/Y)`	`= text(C/Y) = 4`

`:.\ text(Quarterly payment)\ = $456.793…`

`text(By TVM Solver, after 3 years)`

`N`	`=3 xx 4 = 12`
`I (%)`	`=9`
`PV`	`= – 8400`
`PMT`	`=456.793…`
`FV`	`=?`
`text(P/Y)`	`= text(C/Y) = 4`

`:.\ FV\ text{((balance of loan)} = 4757.4076…`

`:.\ text(% reduction)`	`=(8400 – 4757.41)/8400`
	`=0.4336…`
	`=43.36…%`

`=> C`

CORE*, FUR1 2009 VCAA 9 MC

To purchase a house Sam has borrowed $250 000 at an interest rate of 4.45% per annum, fixed for ten years.

Interest is calculated monthly on the reducing balance of the loan. Monthly repayments are set at $1382.50.

After 10 years, Sam renegotiates the conditions for the balance of his loan. The new interest rate will be 4.25% per annum. He will pay $1750 per month.

The total time it will take him to pay out the loan fully is closest to

A. 17 years.

B. 20 years.

C. 21 years.

D. 22 years.

E. 23 years.

Show Answers Only

`C`

Show Worked Solution

`text(By TVM solver:)`

♦ Mean mark 45%.

`N`	`= 10 × 12 = 120`
`I`	`= 4.45text(%)`
`PV`	`= – 250\ 000`
`PMT`	`= 1382.50`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`FV = 181\ 324.43…`

`:.\ text(Loan remaining after 10 years = $181 324.43…)`

`text(From 10 years on …)`

`N`	`= ?`
`I`	`= 4.25text(%)`
`PV`	`= 181\ 324.4`
`PMT`	`= 1750`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N`	`= 129.33text( months)`
	`= 10.77…text( years)`

`:.\ text(Total time to pay off loan)`

`= 10 + 10.77…\ text(years)`

`= 20.77…\ text( years)`

`=> C`

CORE*, FUR1 2011 VCAA 9 MC

Xavier borrows $45 000 from the bank to buy a car.

He is offered a reducing balance loan for three years with an interest rate of 9.75% per annum, compounding monthly.

He can repay this loan by making 36 equal monthly payments.

Instead, Xavier decides to repay the loan in 18 equal monthly payments.

If there are no penalties for repaying the loan early, the amount he will save is closest to

A. $2697

B. $3530

C. $3553

D. $6581

E. $7083

Show Answers Only

`B`

Show Worked Solution

`text(By TVM solver:)`

♦ Mean mark 44%.

`N`	`= 3 xx 12=36`
`I (%)`	`= 9.75text(%)`
`PV`	`= – 45\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PMT = $1446.75`

`N`	`= 18`
`I (%)`	`= 9.75text(%)`
`PV`	`= – 45\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PMT = $2697.39`

`:.\ text(Savings)`	`=(36 xx 1446.75) – (18 xx 2697.39)`
	`=52\ 083 – 48\ 553.02`
	`=$3529.98`

`=> B`

CORE*, FUR1 2013 VCAA 9 MC

The following information relates to the repayment of a home loan of $300 000.

• The loan is to be repaid fully with monthly payments of $2500.
• Interest compounds monthly.
• After the first monthly payment has been made, the amount owing on the loan is $299 000.

Which one of the following statements is true?

After two months, $297 995 is still owing on the loan.
$1000 of interest has been paid in the first month.
The loan will be fully repaid in less than 15 years.
Halfway through the term of the loan, the amount still owing will be $150 000.
Payments of $2750 rather than $2500 per month will reduce the time to repay the loan fully by more than three years.

Show Answers Only

`A`

Show Worked Solution

`text(Find the interest rate using a TVM Solver,)`

♦♦ Mean mark 29%.

`N`	`= 1`
`I (%)`	`= ?`
`PV`	`= -300\ 000`
`PMT`	`= 2500`
`FV`	`= 299\ 000`
`text(P/Y)`	`= text(C/Y) = 12`

`:.I =6(%)`

`text(Test each statement by TVM solver:)`

`text(Consider A,)`

`N`	`= 2`
`I (%)`	`= 6`
`PV`	`= -300\ 000`
`PMT`	`= 2500`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`:.\ FV = $297\ 995`

`=> A`

CORE*, FUR1 2006 VCAA 9 MC

Jenny borrowed $18 000. She will fully repay the loan in five years with equal monthly payments.

Interest is charged at the rate of 9.2% per annum, calculated monthly, on the reducing balance.

The amount Jenny has paid off the principal immediately following the tenth repayment is

A. $1876.77

B. $2457.60

C. $3276.00

D. $3600.44

E. $3754.00

Show Answers Only

`B`

Show Worked Solution

`text(Find monthly repayments,)`

♦♦ Mean mark 27%.

`text(By TVM Solver:)`

`N`	`= 5 xx 12 = 60`
`I(%)`	`= 9.2text(%)`
`PV`	`= − 18\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. \ PMT = $375.40`

`text(Find the loan value after the 10th repayment,)`

`text(By TVM Solver:)`

`N`	`= 10`
`I(%)`	`= 9.2text(%)`
`PV`	`= − 18\ 000`
`PMT`	`= 375.40`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`:. \ FV= 15\ 542.40`

`:.\ text(Amount paid off)`	`= 18\ 000 − 15\ 542.40`
	`= $2457.60`

`=>B`

\(N\)	\(=5 \times 52 = 260\)
\(I\%\)	\(=7.8\)
\(PV\)	\(= -60\ 000\)
\(PMT\)	\(= ?\)
\(FV\)	\(=0\)
\(\text{P/Y}\)	\(=\ \text{C/Y}\ = 52\)

\(N\)	\(=52\)
\(I\%\)	\(=7.8\)
\(PV\)	\(= -60\ 000\)
\(PMT\)	\(= 300\)
\(FV\)	\(=?\)
\(\text{P/Y}\)	\(=\ \text{C/Y}\ = 52\)

\(N\)	\(=156\)
\(I\%\)	\(=7.8\)
\(PV\)	\(= -48\ 651.67\)
\(PMT\)	\(= ?\)
\(FV\)	\(=0\)
\(\text{P/Y}\)	\(=\ \text{C/Y}\ = 52\)

Recursion, GEN2 2024 NHT 9

Financial Maths, GEN2 2024 VCAA 8

Financial Maths, GEN2 2023 VCAA 7

CORE, FUR2 2021 VCAA 8

CORE, FUR1 2021 VCAA 24 MC

CORE, FUR2 2020 VCAA 11

Financial Maths, GEN1 2019 NHT 22-23 MC

CORE, FUR1 2017 VCAA 24 MC

CORE, FUR2 2016 VCAA 7

CORE*, FUR2 2008 VCAA 5

CORE*, FUR2 2015 VCAA 5

CORE*, FUR1 2015 VCAA 8 MC

CORE*, FUR1 2009 VCAA 9 MC

CORE*, FUR1 2011 VCAA 9 MC

CORE*, FUR1 2013 VCAA 9 MC

CORE*, FUR1 2006 VCAA 9 MC

`=> I(%)`	`= 4.1427…`
	`= 4.14text{% (2 d.p.)}`