smc-603-60-CAS (1 step)

Financial Maths, GEN2 2025 VCAA 9

Declan takes out a new loan of $50 000 to promote his new film.

Interest on this loan compounds weekly and Declan makes weekly repayments of $75.

With these weekly repayments of $75, suppose the balance of Declan’s loan does not change over time.
Determine the weekly interest rate. (1 mark)

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With these weekly repayments of $75, suppose the balance of Declan's loan now reduces over time.
The balance of the loan, in dollars, after $n$ weeks, $L_n$, can be determined using a recurrence relation of the form
1. $L_0=50\,000, \quad L_{n+1}=R L_n-75$
Assume there are exactly 52 weeks in a year
After one year, Declan owes $49 565.34
Determine
1. the per annum interest rate, compounding weekly, as a percentage, rounded to two decimal places. (1 mark)
  
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2. the value of $R$ rounded to four decimal places. (1 mark)
  
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Show Answers Only

a. $\text{Weekly interest rate}=0.15 \%$

b.i. $\text{Interest rate}=6.96 \% \ \text{(2 d.p)}$

b.ii. $R=1.0013$

Show Worked Solution

a. $\text{Weekly interest rate}=\dfrac{75}{50\,000} \times 100=0.15 \%$

b.i. $L_0=50\,000 \quad L_{n+1}=R L_n-75$

$\text{When} \ \ N=52,\ \text{balance}=\$ 49\,565.34$

$\text{Solve for \(I$ (by CAS):}\)

$N$	$=52$
$I(\%)$	$=\boldsymbol{6.96000}$
$PV$	$=50\,000$
$PMT$	$=-75$
$FV$	$=-49\,565.34$
$PY$	$=CY=52$

$\text{Interest rate}=6.96 \% \ \text{(2 d.p)}$

b.ii. $R=1+\dfrac{6.96}{100 \times 52}=1.0013$

♦ Mean mark (a) 40%.
♦ Mean mark (b.i) 40%
♦♦ Mean mark (b.ii) 30%

Financial Maths, GEN2 2025 VCAA 7

Declan is a filmmaker and content creator.

He has taken out a reducing balance loan to fund a new production.

Interest is calculated monthly and Declan makes monthly repayments.

Three rows of the amortisation table for Declan’s loan are shown below.

\begin{array}{|c|c|c|c|c|}
\hline
\hline \rule{0pt}{2.5ex}\quad \textbf{Payment} \quad & \quad\textbf{Payment} \quad & \quad\textbf{Interest} \quad& \textbf{Principal} & \quad\textbf{Balance}\quad\\
\textbf{number} & \textbf{(\$)} \rule[-1ex]{0pt}{0pt}& \textbf{(\$)} & \quad\textbf{reduction (\$)} \quad& \textbf{(\$)}\\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 0.00 & 0.00 & 0.00 & 850\,000.00 \\
\hline \hline \rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & 15\,730.88 & 2975.00 & 12\,755.88 & 837\,244.12 \\
\hline \hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & 2930.35 & 12\,800.53 & 824\,443.59 \\
\hline
\end{array}

What amount, in dollars, did Declan borrow? (1 mark)
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Why is the interest associated with payment 2 lower than the interest associated with payment 1? (1 mark)
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The interest rate on Declan’s loan is 4.2% per annum, compounding monthly.
Using the values in the table, complete the table below.
Round all values to the nearest cent. (1 mark)

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The last payment required to fully repay the loan is $15 730.71, correct to the nearest cent.
How many payments of $15 730.88 did Declan make before this final payment? (1 mark)

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a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

d. $\text{59 payments made before the final payment.}$

Show Worked Solution

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

♦ Mean mark (c) 40%.

d. $\text{Solve for \(N$ (by CAS):}\)

$N$	$=\boldsymbol{59.99 \ldots}$
$I(\%)$	$=4.2$
$PV$	$=850\,000$
$PMT$	$=-15\,730.88$
$FV$	$=0$
$PY$	$=CY=12$

$\therefore \ \text{59 payments made before the final payment.}$

♦♦♦ Mean mark (d) 21%.

Financial Maths, GEN1 2024 VCAA 22-23 MC

Stewart takes out a reducing balance loan of \$240 000, with interest calculated monthly.

Stewart makes regular monthly repayments.

Three lines of the amortisation table are shown below.

\begin{array}{|c|r|r|r|r|}
\hline
\rule{0pt}{2.5ex} \textbf{Payment} & \textbf {Payment} & \textbf {Interest} &\textbf{Principal reduction} & \textbf{Balance}\\
\rule[-1ex]{0pt}{0pt}\textbf{number} & \textbf{($$$)}\ \ \ \ \ & \textbf{($$$)}\ \ \ \ \ & \textbf{($$$)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ & \textbf{($$$)}\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} 0 & 0.00 & 0.00 & 0.00\ \ \ \ \ \ \ \ \ \ & 240\,000.00 \\
\hline
\rule{0pt}{2.5ex} 1 & 2741.05 & 960.00 & 1781.05\ \ \ \ \ \ \ \ \ \ & 238\,218.95 \\
\hline
\rule{0pt}{2.5ex} 2 & 2741.05 & & & \\
\hline
\end{array}

Part 1

The principal reduction associated with Payment number 2 is closest to

$1773.93
$1781.05
$1788.17
$2741.05

Part 2

The number of years that it will take Stewart to repay the loan in full is closest to

Show Answers Only

Part 1: $C$

Part 2: $A$

Show Worked Solution

Part 1

$\text{Interest rate (monthly)}\ =\dfrac{960}{240\,000} = 0.004$

$\text{Interest repaid (payment 2)}\ =\dfrac{960}{240\,000}\times 238\,218.95=$952.88$

$\text{Principal Reduction (payment 2)}\ =2741.05-952.88=$1788.17$

$\Rightarrow C$

Part 2

$\text{Using CAS:}$

$\text{Annual interest rate}\ = 12 \times 0.004 = 4.8\%$

$\text{Number of repayments }= 108\ \text{months} = 9\ \text{years}$

$\Rightarrow A$

♦ Mean mark (Part 2) 54%.

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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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%.

Financial Maths, GEN2 2023 VCAA 5

Arthur borrowed $30 000 to buy a new motorcycle.

Interest on this loan is charged at the rate of 6.4% per annum, compounding quarterly.

Arthur will repay the loan in full with quarterly repayments over six years.

How many repayments, in total, will Arthur make? (1 mark)

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The balance of the loan, in dollars, after $n$ quarters, $A_n$, can be modelled by the recurrence relation

$A_0=30\ 000, \quad A_{n+1}=1.016 A_n-1515.18$

Showing recursive calculations, determine the balance of the loan after two quarters.
Round your answer to the nearest cent. (1 mark)

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The final repayment required will differ slightly from all the earlier repayments of $1515.18
Determine the value of the final repayment.
Round your answer to the nearest cent. (1 mark)

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a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

c. $ \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Show Worked Solution

a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

Mean mark (b) 51%.
♦♦ Mean mark (c) 25%

c. $\text{Solve for}\ N\ \text{using TMV calculator:}$

$N$	$=24$
$I\%$	$=6.4$
$PV$	$= -30\ 000$
$PMT$	$= 1515.18$
$FV$	$=?$
$\text{P/Y}$	$=\ \text{C/Y}\ = 4$

$FV = -0.14$

$\therefore \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Financial Maths, GEN1 2022 VCAA 18-19 MC

The balance of a loan, $V_n$, in dollars, after $n$ months is modelled by the recurrence relation

$V_0=400\ 000,\ \ \ V_{n+1}=1.003\,V_n-2024$

Question 18

The balance of the loan first falls below $398 000 after how many months?

Question 19

With a small change to the final payment, the loan is expected to be repaid in full in

25 years.
26 years.
28 years.
29 years.
30 years.

Show Answers Only

$\text{Question 18: C}$

$\text{Question 19: A}$

Show Worked Solution

$\text{Question 18}$

$V_1$	$=1.003 \times 400\ 000-2024 = $399 176$
$V_2$	$=1.003 \times 399\ 176-2024= $398 349.528$
$V_3$	$=1.003 \times 398\ 349.528-2024= $397 520.58$

$\Rightarrow C$

$\text{Question 19}$

$\text{Monthly interest rate = 0.3%}$

$\text{Annual interest rate}\ (r) = 12 \times 0.3 = 3.6\%$

$\text{Solve for}\ N\ \text{using TMV calculator:}$

$I\%$	$=3.6$
$PV$	$= 400\ 000$
$PMT$	$= -2024$
$FV$	$=0$
$\text{P/Y}$	$=12$
$\text{C/Y}$	$=12$

$N = 300.002\ \text{months}\ \approx \dfrac{300}{12} \approx 25\ \text{years}$

$\Rightarrow A$

♦ Mean mark (Q19) 47%.

CORE, FUR1 2021 VCAA 23 MC

Bimal has a reducing balance loan.

The balance, in dollars, of the loan from month, `B_n`, is modelled by the recurrence relation below.

`B_o = 450\ 000, \ B_{n+1} = RB_n - 2633`

Given that the loan will be fully repaid in 20 years, the value of `R` is closest to

1.003
1.0036
1.03
1.036
1.36

Show Answers Only

`A`

Show Worked Solution

`text{By TVM Solver:}`

♦♦ Mean mark 31%.

`N`	`= 240`
`Itext{(%)}`	`= ?`
`PV`	`= 450\ 000`
`PMT`	`= -2633`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. Itext{(%)}= 3.5999 …`

`r_text{monthly} = 3.6/12 = 0.3`

`R`	`= 1 + r/100`
	`= 1 + 0.3/100`
	`=1.003`

`=> A`

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 2019 VCAA 9

Phil would like to purchase a block of land.

He will borrow $350 000 to make this purchase.

Interest on this loan will be charged at the rate of 4.9% per annum, compounding fortnightly.

After three years of equal fortnightly repayments, the balance of Phil’s loan will be $262 332.33.

What is the value of each fortnightly repayment Phil will make?

Round your answer to the nearest cent. (1 mark)

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What is the total interest Phil will have paid after three years?

Round your answer to the nearest cent. (1 mark)

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Over the next four years of his loan, Phil will make monthly repayments of $3517.28 and will be charged interest at the rate of 4.8% per annum, compounding monthly.

Let `B_n` be the balance of the loan `n` months after these changes apply.

Write down a recurrence relation, in terms of `B_0, B_(n + 1)` and `B_n`, that could be used to model the balance of the loan over these four years. (2 marks)

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`$1704.03`
`$45\ 246.67`
`B_0=262\ 332.33,\ \ \ B_(n+1) = 1.004 B_n – 3517.28`

Show Worked Solution

a. `text(Find)\ PMT \ text{(by TVM solver)}:`

`N`	`= 3 xx 26 = 78`
`I(%)`	`= 4.9`
`PV`	`= 350\ 000`
`PMT`	`= ?`
`FV`	`= -262332.33`
`text(P/Y)`	`= text(C/Y) = 26`

`=> PMT = -1704.030`

`:.\ text(Fortnightly payment) = $1704.03`

b.	`text(Total interest)`	`= text(Payments) – text(decrease in principal)`
		`= 78 xx 1704.03-(300\ 000-262\ 332.33)`
		`= $45\ 246.67`

c. `B_0 = 262\ 332.33`

`text(Monthly interest) = 4.8/12 = 0.4%`

`text(Monthly payment) = 3517.28`

`:.\ text(Recurrence relation:)`

`B_0=262\ 332.33,\ \ \ B_(n+1) = 1.004 B_n-3517.28`

CORE, FUR1 2018 VCAA 22 MC

Adam has a home loan with a present value of $175 260.56

The interest rate for Adam’s loan is 3.72% per annum, compounding monthly.

His monthly repayment is $3200.

The loan is to be fully repaid after five years.

Adam knows that the loan cannot be exactly repaid with 60 repayments of $3200.

To solve this problem, Adam will make 59 repayments of $3200. He will then adjust the value of the final repayment so that the loan is fully repaid with the 60th repayment.

The value of the 60th repayment will be closest to

$368.12
$2831.88
$3200.56
$3557.09
$3568.12

Show Answers Only

`E`

Show Worked Solution

`text(By TVM Solver:)`

♦♦ Mean mark 29%.

`N`	`= ?`
`I(%)`	`= 3.72`
`PV`	`= 175\ 260.56`
`PMT`	`= −3200`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = −368.12`

`:.\ text(Final payment)`	`= 3200 + 368.12`
	`= $3568.12`

`=> E`

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 2006 VCAA 4

A company anticipates that it will need to borrow $20 000 to pay for a new machine.

It expects to take out a reducing balance loan with interest calculated monthly at a rate of 10% per annum.

The loan will be fully repaid with 24 equal monthly instalments.

Determine the total amount of interest that will be paid on this loan.

Write your answer to the nearest dollar. (2 marks)

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Show Answers Only

`$2150\ \ text{(nearest $)}`

Show Worked Solution

`text(Find monthly payment by TVM Solver,)`

♦♦ Mean mark 21%.

`N`	`= 2 xx 12 = 24`
`I(text(%))`	`= 10`
`PV`	`= 20\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT = -922.898…`

`text(Total repayments)`

`= 24 xx 922.898…`

`= $22\ 149.65…`

`:.\ text(Total interest paid)`

`= 22\ 149.56…-20\ 000`

`= 2149.56…`

`= $2150\ \ text{(nearest $)}`

CORE*, FUR2 2007 VCAA 2

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.

Calculate the amount of interest he will pay each month. (1 mark)

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Suppose the interest rate remains at 9% per annum compounding monthly and Khan pays $400 each month for five years.

Determine the amount of the loan that is outstanding at the end of five years.

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

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Khan decides to repay the $30 000 loan fully in equal monthly instalments over five years.

The interest rate is 9% per annum compounding monthly.

Determine the amount of each monthly instalment. Write your answer correct to the nearest cent. (1 mark)
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Show Answers Only

`$225`
`$16\ 801`
`$622.75`

Show Worked Solution

a. `text(Interest paid each month)`

♦ Mean mark of all parts (combined) was 37%.
MARKER’S COMMENT: Many students were confused by the fact that part (a) did not have a given loan period.

`= 1/12 xx 0.09 xx 30\ 000`

`= $225`

b. `text(Find principal left after 5 years.)`

`text(By TVM Solver:)`

`N`	`= 5 xx 12 = 60`
`I(text(%))`	`= 9`
`PV`	`= 30\ 000`
`PMT`	`= −400`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = −16\ 800.77…`

`:. $16\ 801\ text(is outstanding after 5 years.)`

c. `text(By TVM Solver,)`

`N`	`= 5 xx 12 = 60`
`I(text(%))`	`= 9`
`PV`	`= 30\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT = −622.750…`

`:.\ text(Monthly installment is $622.75)`

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)

--- 3 WORK AREA LINES (style=lined) ---
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)

--- 5 WORK AREA LINES (style=lined) ---

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)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`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 2009 VCAA 5

In order to drought-proof the course, the golf club will borrow $200 000 to develop a water treatment facility.

The club will establish a reducing balance loan and pay interest monthly at the rate of 4.65% per annum.

$1500 per month will be paid on this loan.

How much of the principal will be left to pay after five years?

Write your answer in dollars correct to the nearest cent. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Determine the total interest paid on the loan over the five-year period.

Write your answer in dollars correct to the nearest cent. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
When the amount outstanding on the loan has reduced to $95 200, the interest rate increases to 5.65% per annum.

Calculate the new monthly repayment that will fully repay this amount in 60 equal instalments.

Write your answer in dollars correct to the nearest cent. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`$151\ 133.38`
`$41\ 133.38`
`$1825.03`

Show Worked Solution

a. `text(Find principal after 5 years.)`

♦♦ Mean mark of all parts (combined) was 28%.

`text(By TVM Solver,)`

`N`	`= 12 xx 5 = 60`
`I(text(%))`	`= 4.65`
`PV`	`= 200\ 000`
`PMT`	`=-1500`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=>FV =-151\ 133.38…`

`:. text(Principal left) = $151\ 133.38`

b. `text(Total repayments)`

`= 60 xx 1500`

`= $90\ 000`

`text(Principal paid off after 5 years)`

`= 200\ 000-151\ 133.38`

`= $48\ 866.62`

`:.\ text(Total interest paid)`

`= 90\ 000-48\ 866.62`

`= $41\ 133.38`

c. `text(By TVM Solver,)`

`N`	`= 60`
`Itext(%)`	`= 5.65`
`PV`	`= 95\ 200`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT =-1825.029…`

`:. text(New monthly repayment) = $1825.03.`

CORE*, FUR2 2010 VCAA 4

A home buyer takes out a reducing balance loan of $250 000 to purchase an apartment.

Interest on the loan will be calculated and paid monthly at the rate of 6.25% per annum.

The loan will be fully repaid in equal monthly instalments over 20 years.
1. Find the monthly repayment, in dollars, correct to the nearest cent. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
2. Calculate the total interest that will be paid over the 20 year term of the loan.
3. Write your answer correct to the nearest dollar. (2 marks)
  
  --- 4 WORK AREA LINES (style=lined) ---
After 60 monthly repayments have been made, what will be the outstanding principal on the loan?
Write your answer correct to the nearest dollar. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

By making a lump sum payment after nine years, the home buyer is able to reduce the principal on his loan to $100 000. At this time, his monthly repayment changes to $1250. The interest rate remains at 6.25% per annum, compounding monthly.

With these changes, how many months, in total, will it take the home buyer to fully repay the $250 000 loan? (1 mark)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

i. `$1827.32`
ii. `$188\ 557\ \ text{(nearest $)}`
`$213\ 118`
`212\ text(months)`

Show Worked Solution

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

♦♦ This question (all parts) were generally poorly answered, although exact data unavailable.

`N`	`= 20 xx 12 = 240`
`I(text(%))`	`= 6.25`
`PV`	`= 250\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT = -1827.320…`

`:. text(Monthly repayment) = $1827.32`

a.ii. `text(Total of all repayments)`

`= 240 xx 1827.32`

`= $438\ 556.80`

`:.\ text(Total interest paid)`

`= 438\ 556.80-250\ 000`

`= 188\ 556.80`

`= $188\ 557\ \ text{(nearest $)}`

b. `text(By TVM Solver:)`

`N`	`= 60`
`I(text(%))`	`= 6.25`
`PV`	`= 250\ 000`
`PMT`	`=-1827.32`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=>FV =-213\ 117.807…`

`:.\ text(After 60 repayments, the loan balance is $213 118.)`

c. `text(By TVM Solver:)`

`N`	`= ?`
`I(text(%))`	`= 6.25`
`PV`	`= 100\ 000`
`PMT`	`=-1250`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 103.75…`

MARKER’S COMMENT: A common error was to forget to add the initial 108 months.

`:.\ text(Total time to repay loan)`

`= 104 + 9 xx 12`

`= 212\ text(months)`

CORE*, FUR2 2011 VCAA 4

Tania takes out a reducing balance loan of $265 000 to pay for her house.

Her monthly repayments will be $1980.

Interest on the loan will be calculated and paid monthly at the rate of 7.62% per annum.

1. How many monthly repayments are required to repay the loan? Write your answer to the nearest month. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
2. Determine the amount that is paid off the principal of this loan in the first year. Write your answer to the nearest cent. (1 mark)
  
  --- 4 WORK AREA LINES (style=lined) ---

Immediately after Tania made her twelfth payment, the interest rate on her loan increased to 8.2% per annum, compounding monthly.

Tania decided to increase her monthly repayment so that the loan would be repaid in a further nineteen years.

Determine the new monthly repayment.
Write your answer to the nearest cent. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

i. `300`
ii. `$3694.25\ \ text{(nearest cent)}`
`$2265.04`

Show Worked Solution

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

`N`	`= ?`
`I(text(%))`	`= 7.62`
`PV`	`= 265\ 000`
`PMT`	`= -1980`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 299.573…`

`:.\ text(After 300 months, the loan will be repaid.)`

a.ii. `text(After 12 months, by TVM Solver:)`

`N`	`= 12`
`I(text(%))`	`= 7.62`
`PV`	`= 265\ 000`
`PMT`	`= -1980`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = -261\ 305.74…`

`:.\ text(Amount paid off after 1 year)`

`= 265\ 000-261\ 305.747…`

`= 3694.252…`

`= $3694.25\ \ text{(nearest cent)}`

b. `text(By TVM Solver:)`

`N`	`= 19 xx 12 = 228`
`Itext(%)`	`= 8.2`
`PV`	`= 261\ 305.75`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PMT = -2265.043`

`:.\ text(New monthly repayment is $2265.04)`

CORE*, FUR2 2012 VCAA 3

An area of a club needs to be refurbished.

$40 000 is borrowed at an interest rate of 7.8% per annum.

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

Suppose the $40 000 loan is to be fully repaid in equal monthly instalments over five years.

Determine the monthly payment, correct to the nearest cent. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
If, instead, the monthly payment was $1000, how many months will it take to fully repay the $40 000? (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
Suppose no payments are made on the loan in the first 12 months.
1. Write down a calculation that shows that the balance of the loan account after the first 12 months will be $43 234 correct to the nearest dollar. (1 mark)
  
  --- 5 WORK AREA LINES (style=lined) ---
2. After the first 12 months, only the interest on the loan is paid each month.
3. Determine the monthly interest payment, correct to the nearest cent. (1 mark)
  
  --- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`$807.23`
`text(47 months)`
i. `text(See Worked Solutions)`
ii. `$281.02\ \ text{(nearest cent)}`

Show Worked Solution

a. `text(Find monthly payment by TVM solver:)`

♦♦ Mean mark of all parts (combined) 28%.

`N`	`= 5 xx 12 = 60`
`I(text(%))`	`= 7.8`
`PV`	`= 40\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PMT = -807.232…`

`:.\ text{Monthly payment is $807.23 (nearest cent)}`

b. `text(Find)\ n\ text(when loan fully paid:)`

`N`	`= ?`
`I(text(%))`	`= 7.8`
`PV`	`= 40\ 000`
`PMT`	`= -1000`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 46.47…`

`:.\ text(Loan will be fully repaid after 47 months.)`

c.i. `text(Loan balance after 12 months)`

`= 40\ 000 xx (1 + (7.8)/(12 xx 100))^12`

`= 43\ 233.99…`

`= $43\ 234\ \ text{(nearest $) … as required}`

c.ii. `text(Interest paid each month)`

`= 43\ 234 xx (7.8)/(12 xx 100)`

`= 281.021`

`= $281.02\ \ text{(nearest cent)}`

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)

--- 2 WORK AREA LINES (style=lined) ---
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)

--- 4 WORK AREA LINES (style=lined) ---
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)

--- 5 WORK AREA LINES (style=lined) ---

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 9 MC

Ravi borrowed $160 000 at an interest rate of 6.18% per annum.

Interest is calculated monthly on the reducing balance of the loan.

The loan will be fully repaid with monthly payments of $1950.

Which one of the following statements is not true?

A. His first payment reduces the loan by less than $1950.

B. His second payment reduces the loan by more than the first payment.

C. Repaying more than $1950 per month will reduce the term of the loan.

D. His final payment will be less than $1760.

E. His final payment includes interest.

Show Answers Only

`D`

Show Worked Solution

`text(By TVM solver,)`

♦ Mean mark 37%.
STRATEGY: Eliminating A, B and C is clear, and once a decimal month is confirmed for full payment, the last payment must contain interest and E can be eliminated.

`N`	`=?`
`I (%)`	`=6.18`
`PV`	`=-160\ 000`
`PMT`	`=1950`
`FV`	`=0`
`text(P/Y)`	`= text(C/Y) = 12`

`:.\ text(Number of monthly payments)\ = 106.906…`

`text{After 106 payments,}`

`text{FV (balance of loan)} = $1759.88…`

`:.\ text(Final payment)`	`= 1759.88…xx (1 + 6.18/(12 xx 100))`
	`= $1768.94…`

`=> D`

CORE*, FUR1 2008 VCAA 8 MC

A loan of $300 000 is taken out to finance a new business venture.

The loan is to be repaid fully over twenty years with quarterly payments of $6727.80.

Interest is calculated quarterly on the reducing balance.

The annual interest rate for this loan is closest to

A. 4.1%

B. 6.5%

C. 7.3%

D. 19.5%

E. 26.7%

Show Answers Only

`B`

Show Worked Solution

`text(Using a TVM Solver,)`

`N`	`=4 xx 20=80`
`I (%)`	`= ?`
`PV`	`= – 300\ 000`
`PMT`	`= 6727.80`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 4`

`:.I = 6.50text(%)`

`=> B`

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 2009 VCAA 8 MC

Robin takes out a reducing balance loan of $100 000 with quarterly repayments of $2150.

After seven years of quarterly repayments, Robin still owes $80 000.

Correct to one decimal place, the interest rate per annum for this loan is

A. 6.3%

B. 8.2%

C. 12.9%

D. 18.9%

E. 24.7%

Show Answers Only

`A`

Show Worked Solution

`text(Using TVM solver,)`

♦ Mean mark 45%.

`N`	`= 7 × 4 = 28`
`I(%)`	`= ?`
`PV`	`= – 100\ 000`
`PMT`	`= 2150`
`FV`	`= 80\ 000`
`text(P/Y)`	`= text(C/Y) = 4`

`:. I(%) = 6.305… text(%)`

`=> A`

CORE*, FUR1 2009 VCAA 7 MC

A loan of $17 500 is to be paid back over four years at an interest rate of 6.25% per annum on a reducing monthly balance.

The monthly repayment, correct to the nearest cent, will be

A. $364.58

B. $413.00

C. $802.08

D. $1156.77

E. $5079.29

Show Answers Only

`B`

Show Worked Solution

`text(By TVM solver:)`

`N`	`= 4 × 12 = 48`
`I(%)`	`= 6.25text(%)`
`PV`	`= – 17\ 500`
`PMT`	`= ?`
`FV`	`=0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. PMT = $412.99…`

`=> B`

CORE*, FUR1 2010 VCAA 7 MC

A loan of $300 000 is to be repaid over a period of 20 years. Interest is charged at the rate of 7.25% per annum compounding quarterly.

The quarterly repayment to the nearest cent is

A. $2371.13

B. $5511.46

C. $7113.39

D. $7132.42

E. $7156.45

Show Answers Only

`D`

Show Worked Solution

`text(By TVM Solver:)`

`N`	`= 4 xx 20=80`
`I (%)`	`= 7.25text(%)`
`PV`	`= – 300\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 4`

`:.PMT = $ 7132.41…`

`=> D`

CORE, FUR1 2011 VCAA 8 MC

Teresa borrowed $120 000 at an interest rate of 7.67% per annum, compounding monthly.

The loan is to be repaid with equal monthly payments.

She decides to repay the loan by making monthly payments of $1430.

Which of the following statements is true?

She will pay out the loan fully in less than ten years.
The amount of interest that she pays on the loan will increase each year.
After four years the amount that she owes on the loan will be less than $80 000.
Every monthly payment that she makes reduces the amount that she owes on the loan by the same amount.
Monthly payments of $1560 (instead of $1430) will enable her to repay this loan in less than nine years.

Show Answers Only

`E`

Show Worked Solution

`text(Consider E,)`

`text(By TVM solver:)`

`N`	`= ?`
`I (%)`	`= 7.67text(%)`
`PV`	`= – 120\ 000`
`PMT`	`= 1560`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 106.2`

`text{Loan will be paid when}\ \ n<108\ text {months (9 years).}`

`:.\ text(E is true.)`

`text(All other options can be shown to be false.)`

`=> E`

CORE*, FUR1 2012 VCAA 5 MC

A second-hand car is purchased for $9000.

A deposit of $2500 is paid.

Interest is calculated at the rate of 14.95% per annum on the reducing monthly balance.

The balance and interest will be repaid over two years with equal monthly payments.

The monthly payment is closest to

A. $315

B. $415

C. $436

D. $575

E. $587

Show Answers Only

`A`

Show Worked Solution

`text(Balance owing after deposit)`

`= $ 9000 – 2500`

`= $ 6500`

`text(By TVM Solver:)`

`N`	`= 2 xx 12=24`
`I (%)`	`= 14.95`
`PV`	`= – 6500`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:.PMT = $315.01`

`=> A`

CORE*, FUR1 2007 VCAA 9 MC

Petra borrowed $250 000 to buy a home. The interest rate is 7% per annum, calculated monthly on the reducing balance over the life of the loan. She will fully repay the loan over 20 years with equal monthly instalments.

The total amount of interest she will pay on the loan is closest to

A. $215 000

B. $266 000

C. $281 000

D. $350 000

E. $465 000

Show Answers Only

`A`

Show Worked Solution

`text(By TVM Solver:)`

♦ Mean mark 37%.

`N`	`= 20 xx 12 = 240`
`I(%)`	`= 7text(%)`
`PV`	`= − 250\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. \ PMT = $1938.25`

`:.\ text(Total interest paid)`

`= 240 xx PMT – 250\ 000`

`= 240 xx $1938.25 – 250\ 000`

`= $215\ 180`

`=> A`

CORE, FUR1 2014 VCAA 5 MC

A bank approves a $90 000 loan for a customer.

The loan is to be repaid fully over 20 years in equal monthly payments.

Interest is charged at a rate of 6.95% per annum on the reducing monthly balance.

To the nearest dollar, the monthly payment will be

A. $478

B. $692

C. $695

D. $1409

E. $1579

Show Answers Only

`C`

Show Worked Solution

`text(Monthly payments)\ = 20 xx 12 = 240`

`text(Annual interest rate)\ = 6.95text(%)`

`text(By TVM Solver:)`

`N`	`=20 xx 12=240`
`I(%)`	`= 6.95`
`PV`	`= -90000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. PMT = $695.07`

`=> C`

CORE*, FUR1 2014 VCAA 9 MC

Leslie borrowed $35 000 from a bank.

Interest is charged at the rate of 4.75% on the reducing monthly balance.

The loan is to be repaid with 47 monthly payments of $802.00 and a final payment that is to be adjusted so that the loan will be fully repaid after exactly 48 monthly payments.

Correct to the nearest cent, the amount of the final payment will be

A. $0.39

B. $3.57

C. $802.00

D. $802.39

E. $805.57

Show Answers Only

`E`

Show Worked Solution

`text(Calculate the FV of the loan after 47 payments)`

♦♦♦ Mean mark 18%.
MARKERS’ COMMENT: A majority of students found the amount still owed after the second last payment but failed to add interest during the last month.

`text(By TVM Solver:)`

`N`	`= 47`
`I(%)`	`= 4.75text(%)`
`PV`	`= -35\ 000`
`PMT`	`= 802`
`FV`	`= text(?)`
`text(P/Y)`	`= text(C/Y) = 12`

`:.\ text(Balance owing after 47 payments)\ =$802.39…`

`text(Final payment)`

`=\ text(Balance after 47 payments + month’s interest)`

`=802.39… + (802.39… xx 4.75/12 xx 100text{%})`

`=$805.57`

`=> E`

\(N\)	\(=52\)
\(I(\%)\)	\(=\boldsymbol{6.96000}\)
\(PV\)	\(=50\,000\)
\(PMT\)	\(=-75\)
\(FV\)	\(=-49\,565.34\)
\(PY\)	\(=CY=52\)

\(N\)	\(=\boldsymbol{59.99 \ldots}\)
\(I(\%)\)	\(=4.2\)
\(PV\)	\(=850\,000\)
\(PMT\)	\(=-15\,730.88\)
\(FV\)	\(=0\)
\(PY\)	\(=CY=12\)

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

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

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