smc-6926-30-Other Loan Tables

Financial Maths, STD2 EQ-Bank 32

Patricia borrows $20 000 as a reducing balance loan at an interest rate of 6% per annum. She is comparing two repayment options, each repaying the loan over 2 years.

Option A: equal monthly repayments
Option B: equal quarterly repayments

A table of present value interest factors for an annuity of $1 is shown.

Using the table, calculate Patricia's repayment under each option. (2 marks)
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Determine which option costs less in total interest, and by how much. (3 marks)
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a. $\text{Monthly} = \$886.41,\ \ \text{Quarterly} = \$2671.69$

b. $\text{Monthly repayments cost}\ \$99.68\ \text{less in interest.}$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{6}{12 \times 100}\ \ \Rightarrow\ \ r=0.005$

$\text{Quarterly interest rate} = \dfrac{6}{4 \times 100}\ \ \Rightarrow\ \ r=0.015$

$\text{Using Repayment} = \dfrac{\text{loan amount}}{\text{PV factor}}:$

$\text{Monthly repayment}= \dfrac{20\,000}{22.5629} = \$886.41$

$\text{Quarterly repayment}= \dfrac{20\,000}{7.4859} = \$2671.69$

b. $\text{Total paid} = \text{repayment} \times \text{number of repayments}$

$\text{Monthly:}\ 886.41 \times 24 = \$21\,273.84$

$\text{Quarterly:}\ 2671.69 \times 8 = \$21\,373.52$

$\text{Interest (monthly)} = 21\,273.84-20\,000 = \$1273.84$

$\text{Interest (quarterly)} = 21\,373.52-20\,000 = \$1373.52$

$\text{Difference} = 1373.52-1273.84 = \$99.68$

$\therefore\ \text{Monthly repayments cost}\ \$99.68\ \text{less in interest.}$

Financial Maths, STD2 EQ-Bank 29

Shown below is part of the output from a spreadsheet used to model a 25-year reducing balance loan with equal monthly repayments and a constant interest rate, $r$, expressed as a decimal.

Note that many rows and values have been removed from the output shown.

Show that the amount owing at the end of month 120 is $407 860.99. (4 marks)

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$\text{See worked solution}$

Show Worked Solution

$\text{Find } r:$

$\text{Interest}$	$= r \times \text{amount owing}$
$3500$	$= r \times 500\,000$
$r$	$= 0.007$

$\text{Find monthly repayment using PVIF table:}$

$\text{At } N = 300\ \text{ and }\ r = 0.007,\ \text{ PVIF} = 125.23492$

$\text{Monthly repayment}= \dfrac{500\,000}{125.23492}= \$3992.50$

$\text{Amount owing at start of month 120:}$

$\text{Amount owing at end of month 119} = \$408\,990.56$

$\therefore\ \text{Interest charged in month 120}$	$= 0.007 \times \$408\,990.56$
	$= \$2862.93$

$\text{Amount owing at end of month 120:}$

$= \$408\,990.56+\$2862.93-\$3992.50$

$= \$407\,860.99 \quad \checkmark$

Financial Maths, STD2 F4 2023 HSC 29

The table shows monthly repayments for each $1000 borrowed.

A couple borrows $520 000 to buy a house at 8% per annum over 25 years.
How much does the couple repay in total for this loan? (3 marks)

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Chris borrows some money at 7% per annum. Chris will repay the loan over 15 years, paying $3596 per month.
How much money does Chris borrow? (1 mark)

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a. `$1\ 204\ 320`

b. `400\ 000`

Show Worked Solution

a. `text{8.0% interest over a 25 year loan}`

`text{Monthly repayments to borrow $1000 = $7.72}`

`text{Total months}\ = 25 xx 12 = 300`

`text{Monthly repayments}=520 xx 7.72=$4014.40`

`:.\ text{Total repayments}= 4014.40 xx 300=$1\ 204\ 320`

♦ Mean mark (a) 50%.

b. `text{7.0% interest over a 15 year loan}`

`text{Monthly repayments to borrow $1000 = $8.99}`

`:.\ text{Amount borrowed}=3596/8.99 xx 1000=$400\ 000`

♦♦♦ Mean mark (b) 11%.

Financial Maths, STD2 F4 2006 HSC 27a

Liliana wants to borrow money to buy a house. The bank sent her an email with the following table attached.

Liliana decides that she can afford $1000 per month on repayments.

What is the maximum amount she can borrow, and how many years will she have to repay the loan? (1 mark)

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Zali intends to borrow $160 000 over 15 years from the same bank.

If she chooses to borrow $160 000 over 20 years instead, how much more interest will she pay? (2 marks)

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a. `text{$130 000 (over 30 years)}`

b. `$45\ 964.80`

Show Worked Solution

a. `text(From table:)`

`text{$130 000 (over 30 years)}`

b. `text(Total repayments over 15 years)`

`=1529.04 xx 180 = $275\ 227.20`

`text(Total repayments over 20 years)`

`=1338.30 xx 240 = $321\ 192.00`

`:.\ text(Extra interest over 20 years)`

`= 321\ 192.00-275\ 227.20 = $45\ 964.80`

Financial Maths, STD2 F4 2004 HSC 27a

Aaron decides to borrow $150 000 over a period of 20 years at a rate of 7.0% per annum.

Using the Monthly Repayment Table, calculate Aaron’s monthly repayment. (2 marks)

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How much interest does he pay over the 20 years? (2 marks)

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Aaron calculates that if he repays the loan over 15 years, his total repayments would be `$242\ 730`.

How much interest would he save by repaying the loan over 15 years instead of 20 years? (2 marks)

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a. `$1162.50`

b. `$129\ 000`

c. `$36\ 270`

Show Worked Solution

a. `text(Using the table:)`

`text(Monthly repayment on $1000 at 7.0% over 20 years = $7.75)`

`:.\ text(Monthly repayment on $150 000 loan)`

`= 150 xx 7.75= $1162.50`

b. `text(Total repayments over 20 years)`

`= 20 xx 12 xx 1162.50= $279\ 000`

`:.\ text(Interest paid over 20 years)`

`= 279\ 000-150\ 000= $129\ 000`

c. `text(Savings)`	`=\ text{Total paid (20 years) – Total paid (15 years)`
	`= 279\ 000-242\ 730= $36\ 270`

Financial Maths, STD2 F4 2005 HSC 10 MC

The table is used to calculate monthly loan repayments.

Samantha has borrowed $70 000 at 8% per annum for 15 years.

What is her monthly loan repayment?

$143.40
$669.20
$8030.40
$10 038.00

Show Answers Only

`B`

Show Worked Solution

`text(Monthly repayment of $1000 at 8% for 15 years)= $9.56`

`:.\ text(Monthly repayment of $70 000)= 70 × $9.56= $669.20`

`=> B`

Financial Maths, STD2 F4 2010 HSC 25b

William wants to buy a car. He takes out a loan for $28 000 at 7% per annum interest for four years.

Monthly repayments for loans at different interest rates are shown in the spreadsheet.

How much interest does William pay over the term of this loan? (2 marks)

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`$4183.52`

Show Worked Solution

♦ Mean mark 42%
MARKER’S COMMENT: An incorrect table value used correctly in the following calculations received half-marks here. Show your working!

`text(Loan) = $28\ 000,\ \ \ \ r =\ text(7% p.a.)`

`text(Monthly repayment = $670.49`

`text(# Repayments) = 4 xx 12 = 48`

`text(Total repaid)= 48 xx 670.49= $32\ 183.52`

`:.\ text(Interest paid)`	`=32\ 183.52-28\ 000`
	`=$4183.52`

Financial Maths, STD2 F4 2010 HSC 28a

The table shows monthly home loan repayments with interest rate changes from February to October 2009.

What is the change in monthly repayments on a $250 000 loan from February 2009 to April 2009? (1 mark)

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Xiang wants to borrow $307 000 to buy a house.

Xiang’s bank approves loans for customers if their loan repayments are no more than 30% of their monthly gross salary.

Xiang’s monthly gross salary is $6500.

If she had applied for the loan in October 2009, would her bank have approved her loan?

Justify your answer with suitable calculations. (3 marks)

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Jack took out a loan at the same time and for the same amount as Xiang.

Graphs of their loan balances are shown.

Identify TWO differences between the graphs and provide a possible explanation for each difference, making reference to interest rates and/or loan repayments. (2 marks)

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a. `text(Monthly repayments decrease by $15)`

b. `text(S)text(ince repayments of 1987.29 > 1950, the loan)`

`text(would not have been approved.)`

c. `text(Differences)`

`text(1. Jack’s loan balance falls more sharply for first 12 years)`

`text(2. Jack’s loan balance falls less sharply between years 12-30.)`

`text(Explanation)`

`text(1. Jack made larger repayments for first 12 years, OR)`

`text(Jack made the same repayments but had a lower interest rate)`

`text(for the first 12 years.)`

`text(2. Jack made smaller repayments in years 12 – 30.)`

Show Worked Solution

a.	`text(Repayment)\ text{(Feb 09)}`	`= 1588`
	`text(Repayment)\ text{(Apr 09)}`	`= 1573`

`text(Difference) = 1588-1573 = 15`

`:.\ text(Monthly repayments decrease by $15)`

♦ Mean mark 39%
MARKER’S COMMENT: Borrowing $307,000 can be achieved by borrowing $300,000, and then `7` times the table repayment value for borrowing $1000.

b. `text(Loan) = $307\ 000`

`text{Repayments (Oct 09)}`	`= 1942 + (7 xx 6.47)= 1942 + 45.29`
	`= $1987.29\ text(per month)`

`text(30% Gross salary)= 6500 xx\ text(30%)= $1950\ text(per month)`

`:.\ text(S)text(ince repayments of $1987.29 > $1950, the loan)`

`text(would not have been approved.)`

c. `text(Differences)`

`text(1. Jack’s loan balance falls more sharply for first 12 years.)`

`text(2. Jack’s loan balance falls less sharply between years 12-30.)`

`text{Explanation(s)}`

♦ Mean mark 36%
MARKER’S COMMENT: Explanations were generally poor and many failed to refer directly to the graphs shown, or reference Xiang or Jack directly.

`text(1. Jack made larger repayments for 1st 12 years, OR)`

`text(Jack made the same repayments but had a)`

`text(lower interest rate for the first 12 years.)`

`text(2. Jack made smaller repayments in years 12-30.)`

\(\text{Interest}\)	\(= r \times \text{amount owing}\)
\(3500\)	\(= r \times 500\,000\)
\(r\)	\(= 0.007\)