Band 4

Financial Maths, STD2 EO-Bank 28

Leon opens a superannuation account to build up savings for retirement. At the end of each year he pays in $4000, and the account earns 5% per annum, compounded annually.

The spreadsheet below models the first 4 years of the account.

Write down the formula used in cell C9, using appropriate grid references. (1 mark)

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Determine the value that belongs in cell C9. (1 mark)
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Starting from the end of year 4, Leon lifts his yearly payment from $4000 to $7000. Find the balance in the account at the end of year 7, and state how much larger this is than if he had stayed with $4000 payments. (3 marks)
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a. $\text{=E8*B3}$

b. $\text{C9}=\$410.00$

c. $\text{Balance at end of year 7}=\$42\,025.54$

$\text{Leon has}\ \$9457.50\ \text{more than at the standard contribution.}$

Show Worked Solution

a. $\text{Formula: =E8*B3}$

b. $\text{C9 (Year 3 interest)}=\text{balance at start}\times\text{rate}$

$\text{C9}=8200\times 0.05=\$410.00$

c. $\text{Using}\ \ P+I+C\ \ \text{from end of year 4 balance}\ \$17\,240.50:$

$\text{Increased contributions of}\ \$7000\ \text{from year 5:}$

$\text{Year 5:}\ 17\,240.50+17\,240.50\times 0.05+7000=\$25\,102.53$

$\text{Year 6:}\ 25\,102.53+25\,102.53\times 0.05+7000=\$33\,357.66$

$\text{Year 7:}\ 33\,357.66+33\,357.66\times 0.05+7000=\$42\,025.54$

$\text{Standard contributions of}\ \$4000\ \text{from year 5:}$

$\text{Year 5:}\ 17\,240.50+17\,240.50\times 0.05+4000=\$22\,102.53$

$\text{Year 6:}\ 22\,102.53+22\,102.53\times 0.05+4000=\$27\,207.66$

$\text{Year 7:}\ 27\,207.66+27\,207.66\times 0.05+4000=\$32\,568.04$

$\text{Difference}=42\,025.54-32\,568.04=\$9457.50$

$\therefore\ \text{Leon has}\ \$9457.50\ \text{more by increasing his contributions.}$

Financial Maths, STD2 EO-Bank 19

The table shows the present value of an annuity with a contribution of $1.

Rina and Owen each set up an annuity, depositing a fixed amount at the end of every year.

Rina pays $2500 each year for 5 years into an annuity earning 3% per annum, compounded annually. Using the table, find the present value of Rina’s annuity. (1 mark)
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Owen pays $4000 each year for 3 years into an annuity earning 5% per annum, compounded annually. Whose annuity has the greater present value? Justify your answer with calculations. (2 marks)
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a. $\$11\,449.25$

b. $\text{Rina’s annuity is greater: }\$11\,449.25>\$10\,892.80$

Show Worked Solution

a. $\text{Table factor when}\ n=5,\ r=3\%\ \Rightarrow\ 4.5797$

$\therefore\ PVA\ \text{(Rina)}=2500\times 4.5797=\$11\,449.25$

b. $\text{Table factor when}\ n=3,\ r=5\%\ \Rightarrow\ 2.7232$

$ PVA\ \text{(Owen)}=4000\times 2.7232=\$10\,892.80$

$\text{Rina’s annuity is greater: }\$11\,449.25>\$10\,892.80$

Financial Maths, STD2 EO-Bank 5 MC

Omar purchased a laptop for $1680 on 15 May using a credit card. Compound interest was charged daily at a rate of 18.98% per annum. There were no other purchases on this credit card account.

There was no interest-free period. The period for which interest was charged included the date of purchase and the date of payment.

What amount was paid when the account was paid in full on 21 June?

$1711.74
$1712.63
$1713.52
$1714.41

Show Answers Only

$C$

Show Worked Solution

$\text{Days for interest}\ (n)=17+21=38$

$\text{Daily interest rate}\ (r)=\dfrac{0.1898}{365}=0.00052$

$\text{Total paid}\ (FV)$	$=PV(1+r)^n$
	$=1680(1.00052)^{38}$
	$=\$1713.52$

$\Rightarrow C$

Financial Maths, STD2 EO-Bank 26

Kiara wants to purchase a used motorbike selling for $2500 and will be in a position to settle the whole debt in a single payment 40 days after the purchase.

Two methods of financing the purchase are open to her.

Using a credit card, where interest of 20.5% per annum is compounded daily. There is no interest-free period, so interest is charged from the day after the purchase.
Taking out a 40-day personal loan, where simple interest is charged at 12% per annum.

Determine the interest that would build up under each method across the 40 days. (2 marks)
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A friend insists that the personal loan will leave Kiara better off by more than $25 compared with the credit card. Decide whether the friend is correct, justifying your answer with calculations. (2 marks)
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a. $\text{Credit card: }\$56.78\ \text{, Personal loan: }\$32.88$

b. $\text{Difference}=\$23.90$

$\text{As }\$23.90\lt\$25,\text{ the friend’s claim is incorrect.}$

Show Worked Solution

a. $\text{Credit card option (compounding daily):}$

$\text{Amount owing}$	$=2500\left(1+\dfrac{0.205}{365}\right)^{40}$
	$=2556.784\ldots$
	$=\$2556.78\ \text{(nearest cent)}$

$\therefore\ \text{Interest}=2556.78-2500=\$56.78$

$\text{Personal loan option (simple interest):}$

$I=Prn$	$=2500\times 0.12\times\dfrac{40}{365}$
	$=32.876\ldots$
	$=\$32.88\ \text{(nearest cent)}$

b. $\text{Difference}=56.78-32.88=\$23.90$

$\text{As }\$23.90\lt\$25,\text{ the friend’s claim is incorrect.}$

Financial Maths, STD2 2025 HSC 27 (Adapted)

Noor buys a lounge suite for $650 on 3 April using a credit card. The card has an interest-free period of 30 days from and including the date of purchase. Interest is charged on purchases, compounding daily at a rate of 16.8% per annum, from and including the day following the interest-free period.

No other purchases were made on this credit card.

The account was paid in full on 20 May.

What was the total interest charged when the account was paid in full? (3 marks)

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$\text{Interest charged}=\$5.41$

Show Worked Solution

$\text{Total days (3 April to 20 May)}=28+20=48$

$\text{Interest-free days}=30$

$\text{Days accruing interest}\ (n)=48-30=18$

$\text{Daily interest rate}\ (r)=\dfrac{0.168}{365}=0.00046027\ldots$

$\text{Amount owing}=650(1+0.00046027\ldots)^{18}=\$655.41$

$\therefore\ \text{Interest charged}=655.41-650=\$5.41$

Financial Maths, STD2 EO-Bank 6 MC

Grace repays her credit card gradually over several months. She notices that, although the interest rate on her card stays the same, the amount of interest charged changes from month to month.

This occurs because a credit card is an example of a reducing balance loan.

Which statement best explains why a credit card is an example of a reducing balance loan?

Interest is calculated on the outstanding balance, which reduces as repayments are made.
The interest rate decreases each time a repayment is made.
The same amount of interest is charged each month, regardless of the balance owing.
A fixed portion of the balance is repaid each month, with no interest charged.

Show Answers Only

$A$

Show Worked Solution

A is correct: interest is charged on the outstanding balance, which reduces as repayments are made – the defining feature of a reducing balance loan.

Other options:

B is incorrect: the interest rate stays fixed; it is the balance that reduces, not the rate.
C is incorrect: the interest charged is not fixed – it depends on the balance owing.
D is incorrect: interest is charged on a credit card, and repayments are not a fixed portion of the principal.

$\Rightarrow A$

Financial Maths, STD2 2020 HSC 22 (Adapted)

Diego pays a $600 car repair bill using his credit card. The card charges interest at 18.6% per annum, compounded daily, and has no interest-free period.

Twenty days after the purchase, Diego makes a part-payment of $300.

Determine how much Diego still owes after making the $300 part-payment. (3 marks)

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$\text{Amount owing}=\$306.14$

Show Worked Solution

$\text{Days of interest}\ (n)=20$

$\text{Daily interest rate}\ (r)=\dfrac{18.6\%}{365}=\dfrac{0.186}{365}=0.00050959\ldots$

$\text{Amount owing}\ (FV)$	$=PV(1+r)^n-300$
	$=600(1+0.00050959\ldots)^{20}-300$
	$=606.14-300$
	$=\$306.14$

v1 Measurement, STD2 M1 2008 HSC 28b*

A tunnel is excavated with a cross-section as shown.

Find an expression for the area of the cross-section using the Trapezoidal rule. (2 marks)

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The area of the cross-section must be 600 m². The tunnel is 80 m wide.

If the value of `a` increases by 2 metres, by how much will `b` change? (2 marks)

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a. `h(2a + b)`

b. `b\ text(decreases by 4.)`

Show Worked Solution

`A`	`~~ h/2[0 + 2(a + b + a) + 0]`
	`~~ h/2(4a + 2b)`
	`~~ h(2a + b)`

b. `A = 600\ text(m²)`

`text(If tunnel is 80 metres wide)`

`4h=80\ \ =>\ \ h=20`

`text{Using part (a):}`

`600`	`=20(2a+b)`
`2a + b`	`= 30`
`b`	`= 30-2a`

`:.\ text(If)\ a\ text(increases by 2,)\ b\ text(must decrease by 4.)`

v1 Measurement, STD2 2012 HSC 26g

Milly purchases a new container of Gutho's Fertilizer which holds 25 kg.

The container comes with a small cup that holds 350 grams of fertilizer when full.

Milly has 5 pot plants that each receive 1.5 cups of fertilizer on the 1st day of the month and 1 cup each in the middle of the month.

Milly begins using this new container on the first day of October.

In what month will her fertilizer run out? Justify your answer with calculations. (3 marks)

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`text(March)`

Show Worked Solution

`text(Cups used per month)= 5 xx (1.5 + 1)=12.5`

`text(Fertiliser used per month)= 12.5 xx 350 = 4375\ text(g)`

`text(Total fertilizer available)\ = 35\ text(kg) = 35\ 000\ text(grams)`

`text(Time it will last)= (25\ 000)/4375= 5.71…\ text{months}`

`:.\ text(The container will run out in March.)`

Financial Maths, STD2 EQ-Bank 21

Part of George Sample's electricity bill is shown below.

Calculate the supply charge for the billing period. (1 mark)

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Show that the average daily usage is 12 kWh per day. (1 mark)

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Assuming George used exactly 12 kWh of electricity each day, calculate his Tier 1 and Tier 2 usage over the period and hence the total charge for his electricity usage. (2 marks)

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a. $\$99.00$

b. $\text{See worked solution}$

c. $\$315.00$

Show Worked Solution

a. $\text{Supply charge} = 90 \times \$1.10 = \$99.00$

b. $\text{Average daily usage} = \dfrac{1080}{90} = 12 \ \text{kWh per day}$

c. $\text{Tier 1 usage} = 90 \times 10 = 900 \ \text{kWh} \ @ \ \$0.28 = \$252.00$

$\text{Tier 2 usage} = 1080-900 = 180 \ \text{kWh} \ @ \ \$0.35 = \$63.00$

$\text{Total usage charge} = \$252.00 + \$63.00 = \$315.00$

Financial Maths, STD2 EQ-Bank 29

Mei sets up a superannuation account to save for retirement. She contributes $5000 at the end of each year into the account which earns interest at 6% per annum, compounded annually.

The spreadsheet shown models the first 4 years of the account.

Calculate the value in cell C9. (1 mark)
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From the end of year 4, Mei increases her annual contribution from $5000 to $8000. Calculate the balance in her superannuation account at the end of year 7, and determine how much more this is than if she had continued contributing $5000. (3 marks)
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a. $\text{C9} = \$618.00$

b. $\text{Balance at end of year 7} = \$51\,519.99$

$\text{Mei has}\ \$9550.80\ \text{more than at the standard contribution.}$

Show Worked Solution

a. $\text{C9 (Year 3 interest)} = \text{balance at start} \times \text{rate}$

$\text{C9} = 10\,300 \times 0.06 = \$618.00$

b. $\text{Using}\ \ P+I+C\ \ \text{from end of year 4 balance}\ \$21\,873.08:$

$\text{Increased contributions of}\ \$8000\ \text{from year 5:}$

$\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+8000=\$31\,185.46$

$\text{Year 6:}\ 31\,185.46+31\,185.46 \times 0.06+8000=\$41\,056.59$

$\text{Year 7:}\ 41\,056.59+41\,056.59 \times 0.06+8000=\$51\,519.99$

$\text{Standard contributions of}\ \$5000\ \text{from year 5:}$

$\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+5000=\$28\,185.46$

$\text{Year 6:}\ 28\,185.46+28\,185.46 \times 0.06+5000=\$34\,876.59$

$\text{Year 7:}\ 34\,876.59+34\,876.59 \times 0.06+5000=\$41\,969.19$

$\text{Difference}=51\,519.99-41\,969.19=\$9550.80$

$\therefore\ \text{Mei has}\ \$9550.80\ \text{more by increasing her contributions.}$

Financial Maths, STD2 EQ-Bank 27

Haruki opens a savings account. He deposits $2000 at the end of each year into an account earning 3% per annum, compounded annually.

Complete the table below to show the growth of Haruki's savings over the first 4 years. (3 marks)

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Hence, calculate the total interest Haruki earns over the 4 years. (1 mark)

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b. $\$367.25$

Show Worked Solution

a. $\text{Year 2:}$

$\text{Interest} = 2000 \times 0.03 = \$60.00$

$\text{Closing balance} = 2000+60+2000 = \$4060.00$

$\text{Year 3:}$

$\text{Interest} = 4060 \times 0.03 = \$121.80$

$\text{Closing balance} = 4060+121.80+2000 = \$6181.80$

$\text{Year 4:}$

$\text{Interest} = 6181.80 \times 0.03 = 185.454\ldots = \$185.45$

$\text{Closing balance} = 6181.80+185.45+2000 = \$8367.25$

b. $\text{Total deposits} = 4 \times 2000 = \$8000.00$

$\text{Total interest} = 8367.25-8000 = \$367.25$

Financial Maths, STD2 EQ-Bank 25

Marcus takes out a reducing balance loan of $\$5000$ at an interest rate of 12% per annum, compounded monthly. He repays the loan with monthly repayments of $\$700$.

Complete the table below to show the first 3 months of the loan. (3 marks)
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$\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{10pt} \text{Month} & \text{Balance at start} & \text{Interest} & \text{Repayment} & \text{Balance at end} \\[8pt] \hline
\rule{0pt}{10pt} 1 & \$5000.00 & \$50.00 & \$700.00 & \$4350.00 \\[8pt] \hline
\rule{0pt}{10pt} 2 & & & \$700.00 & \\[8pt] \hline
\rule[-6pt]{0pt}{16pt} 3 & & & \$700.00 & \\[8pt] \hline
\end{array}$
Calculate the total interest Marcus pays over the first 3 months. (1 mark)
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$\begin{array}{|c|c|c|c|c|} \hline \text{Month} & \text{Start} & \text{Interest} & \text{Repay} & \text{End} \\ \hline 1 & \$5000.00 & \$50.00 & \$700.00 & \$4350.00 \\ \hline 2 & \$4350.00 & \$43.50 & \$700.00 & \$3693.50 \\ \hline 3 & \$3693.50 & \$36.94 & \$700.00 & \$3030.44 \\ \hline \end{array}$

b. $\$130.44$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{12\%}{12} = 1\%$

$\text{Each month: Interest} = \text{balance at start} \times 0.01$

$\text{Balance at end} = \text{start} + \text{interest}-700$

$\text{Month 2:}$

$\text{Interest} = 4350 \times 0.01 = \$43.50$

$\text{End} = 4350+43.50-700 = \$3693.50$

$\text{Month 3:}$

$\text{Interest} = 3693.50 \times 0.01 = \$36.94$

$\text{End} = 3693.50+36.94-700 = \$3030.44$

b. $\text{Total interest} = 50.00+43.50+36.94 = \$130.44$

Financial Maths, STD2 EO-Bank 29

Dao takes out a reducing balance loan of $5000. The loan has an interest rate of 12% per annum, compounded monthly, and Dao makes monthly repayments of $900.

The spreadsheet shown models the first 4 months of the loan.

Calculate the value in cell C9. (1 mark)
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By continuing the spreadsheet, determine the number of months it takes Dao to repay the loan in full, and calculate the value of the final repayment. (3 marks)
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a. $\text{C9} = \$41.50$

b. $\text{The loan is repaid in}\ 6\ \text{months.}$

$\text{Final repayment} = \$670.78$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{12\%}{12} = 1\%$

$\text{C9 (Month 2 interest)} = 4150 \times 0.01 = \$41.50$

b. $\text{Continue the schedule using}\ \ P+I-R\ \ \text{each month:}$

$\text{Month 5:}\ 1548.65+1548.65 \times 0.01-900=\$664.14$

$\text{Month 6:}\ 664.14+664.14 \times 0.01=\$670.78$

$\text{The Month 6 balance owing}\ (\$670.78)\ \text{is less than the}$

$\text{usual}\ \$900\ \text{repayment, so this final repayment clears the loan.}$

$\therefore\ \text{The loan is repaid in}\ 6\ \text{months, with a final}$

$\text{repayment of}\ \$670.78.$

Financial Maths, STD2 EO-Bank 28

Talia takes out a reducing balance loan of $12 000 to buy a boat. The loan has an interest rate of 7.2% per annum and Talia makes monthly repayments of $300.

The spreadsheet shown models the first 3 months of the loan.

Calculate the value in cell C9. (1 mark)
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From the end of month 3, Talia increases her monthly repayment from $300 to $500. Calculate the balance owing at the end of month 6, and determine how much less Talia owes compared to keeping repayments at $300. (3 marks)
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a. $\text{C9} = \$70.63$

b. $\text{Balance at end of month 6} = \$10\,007.71$

$\text{Talia owes}\ \$603.61\ \text{less than at the standard repayment.}$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{7.2\%}{12} = 0.6\%$

$\text{C9 (Month 2 interest)} = 11\,772 \times 0.006 = \$70.63$

b. $\text{Using}\ \ P+I-R\ \ \text{from end of month 3 balance}\ \$11\,311.89:$

$\text{Increased repayments of}\ \$500\ \text{from month 4:}$

$\text{Month 4:}\ 11\,311.89+11\,311.89 \times 0.006-500=\$10\,879.76$

$\text{Month 5:}\ 10\,879.76+10\,879.76 \times 0.006-500=\$10\,445.04$

$\text{Month 6:}\ 10\,445.04+10\,445.04 \times 0.006-500=\$10\,007.71$

$\text{Standard repayments of}\ \$300\ \text{from month 4:}$

$\text{Month 4:}\ 11\,311.89+11\,311.89\times 0.006-300=\$11\,079.76$

$\text{Month 5:}\ 11\,079.76+11\,079.76\times 0.006-300=\$10\,846.24$

$\text{Month 6:}\ 10\,846.24+10\,846.24\times 0.006-300=\$10\,611.32$

$\text{Difference}=10\,611.32-10\,007.71=\$603.61$

$\therefore\ \text{Talia owes}\ \$603.61\ \text{less by increasing her repayments.}$

Financial Maths, STD2 EO-Bank 20

Noah takes out a reducing balance loan of $10 000 to renovate his bathroom. The loan has an interest rate of 6% per annum and Noah makes monthly repayments of $400.

The spreadsheet shown models the first 4 months of the loan.

Complete the missing values for cells C10, B11 and E11. (3 marks)
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Calculate the total interest Noah pays over the first 4 months of the loan. (1 mark)
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a. $\text{C10} = \$46.49$

$\text{B11} = \$8944.74$

$\text{E11} = \$8589.46$

b. $\$189.46$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{6\%}{12} = 0.5\%$

$\text{C10 (Month 3 interest)} =9298.25 \times 0.005 = \$46.49$

$\text{B11 (Month 4 start)}=9298.25+46.49-400=\$8944.74$

$\text{E11 (Month 4 end)}=8944.74+44.72-400=\$8589.46$

b. $\text{Total interest} = 50+48.25+46.49+44.72=\$189.46$

Financial Maths, STD2 EO-Bank 24

Ravi takes out a short-term loan to buy a laptop with a cash price of $1800.

The loan has the following terms:

Establishment fee: $110 charged when the loan starts
Monthly account-keeping fee: $40
Weekly repayments of $75 over 6 months (26 weeks)

Calculate the total amount Ravi will pay back over the term of the loan. (2 marks)

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How much more than the cash price does Ravi pay for the laptop? (1 mark)

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a. $\$2300$

b. $\$500$

Show Worked Solution

a. $\text{Account-keeping fees} = 40 \times 6 = \$240$

$\text{Weekly repayments} = 75 \times 26 = \$1950$

$\text{Total paid} = 110+240+1950= \$2300$

b. $\text{Extra paid} = 2300-1800 = \$500$

Financial Maths, STD2 EO-Bank 22

Tariq uses a buy now, pay later service to purchase headphones for $360.

He pays a $60 deposit upfront, with the remaining balance split into 4 equal fortnightly payments.

No interest is charged if all payments are made on time.

Tariq misses his final payment. He is charged a late fee of $12, plus a further $6 per week until the missed payment is made. Tariq pays the missed payment 3 weeks late.

What is the amount of each fortnightly payment? (1 mark)

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What is the total amount Tariq pays for the headphones? (2 marks)

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a. $\$75$

b. $\$390$

Show Worked Solution

a. $\text{Remaining balance} = \$360-\$60 = \$300$

$\text{Each payment} = \dfrac{\$300}{4} = \$75$

b. $\text{Late fee} = \$12$

$\text{Weekly fees} = 3 \times \$6 = \$18$

$\text{Total} = \$60 + \$300 + \$12 + \$18 = \$390$

Financial Maths, STD2 EQ-Bank 20_3

Mei uses a buy now, pay later payment option to make a purchase of $1600. Her repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Mei misses her final payment and is charged a late fee of $68. Mei's payment schedule is shown, with her balance totalling $468.

Find the total amount Mei pays for her purchase if repaying in full on 28 September 2026. (1 mark)
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Mei's bank offers short-term loans where simple interest is charged at 18% per annum.
Suppose Mei had borrowed $1600 from the bank to make this purchase on 3 August 2026 and repaid it in full 9 weeks later.
How much would Mei have saved using this approach instead of the buy now, pay later option? (2 marks)
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a. $$1668$

b. $$18.29$

Show Worked Solution

a. $\text{If total owing paid on 28 September:}$

$\text{Total paid} = 400+400+400+468=$1668$

b. $r=18\%=0.18,\ \ n=\dfrac{9 \times 7}{365} = \dfrac{63}{365}$

$I=Prn=1600 \times 0.18 \times \dfrac{63}{365} = 49.709… = $49.71 $

$\text{Amount saved} = 68-49.71=$18.29$

Financial Maths, STD2 EQ-Bank 20_2

Tane uses a buy now, pay later payment option to make a purchase of $200. His repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Tane misses his final payment and is charged a late fee of $22. Tane's payment schedule is shown, with his balance totalling $72.

Find the total amount Tane pays for his purchase if repaying in full on 1 June 2026. (1 mark)
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Tane's bank offers short-term loans where simple interest is charged at 14% per annum.
Suppose Tane had borrowed $200 from the bank to make this purchase on 6 April 2026 and repaid it in full 7 weeks later.
How much would Tane have saved using this approach instead of the buy now, pay later option? (2 marks)
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a. $$222$

b. $$18.24$

Show Worked Solution

a. $\text{If total owing paid on 1 June:}$

$\text{Total paid} = 50+50+50+72=$222$

b. $r=14\%=0.14,\ \ n=\dfrac{7 \times 7}{365} = \dfrac{49}{365}$

$I=Prn=200 \times 0.14 \times \dfrac{49}{365} = 3.758… = $3.76 $

$\text{Amount saved} = 22-3.76=$18.24$

Financial Maths, STD2 EQ-Bank 20_4

Idris uses a buy now, pay later payment option to make a purchase of $440. His repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Idris misses his final payment and is charged a late fee of $26. Idris's payment schedule is shown, with his balance totalling $136.

Find the total amount Idris pays for his purchase if repaying in full on 30 November 2026. (1 mark)
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Idris's bank offers short-term loans where simple interest is charged at 12% per annum.
Suppose Idris had borrowed $440 from the bank to make this purchase on 5 October 2026 and repaid it in full 8 weeks later.
How much would Idris have saved using this approach instead of the buy now, pay later option? (2 marks)
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a. $$466$

b. $$17.90$

Show Worked Solution

a. $\text{If total owing paid on 30 November:}$

$\text{Total paid} = 110+110+110+136=$466$

b. $r=12\%=0.12,\ \ n=\dfrac{8 \times 7}{365} = \dfrac{56}{365}$

$I=Prn=440 \times 0.12 \times \dfrac{56}{365} = 8.100… = $8.10 $

$\text{Amount saved} = 26-8.10=$17.90$

v1 Financial Maths, STD2 F4 2023 HSC 28

A graphic designer purchases a computer system valued at $45 000.

The salvage value of the system after a number of years can be calculated using either of the two methods of depreciation shown in the table.

\begin{array} {|l|l|} \hline \rule{0pt}{2.5ex} \text{Method of depreciation} \rule[-1ex]{0pt}{0pt} & \text{Rate of depreciation} \\ \hline \rule{0pt}{2.5ex} \text{Straight-line method} \rule[-1ex]{0pt}{0pt} & \text{\$4000 per annum} \\ \hline \rule{0pt}{2.5ex} \text{Declining balance method} \rule[-1ex]{0pt}{0pt} & \text{10% per annum} \\ \hline \end{array}

Under which method of depreciation would the salvage value of the equipment be lower at the end of 4 years? Justify your answer with appropriate mathematical calculations. (3 marks)

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`text{Straight-line method:}`

`S=V_0-Dn=45\ 000-4000×4=$29\ 000`

`text{Declining-balance method:}`

`S`	`=V_0(1-r)^n`
	`=45\ 000(1-0.10)^4`
	`=$29\ 524.50`

`text{Salvage value is lower for the straight-line method.}`

Show Worked Solution

`text{Straight-line method:}`

`S=V_0-Dn=45\ 000-4000×4=$29\ 000`

`text{Declining-balance method:}`

`S`	`=V_0(1-r)^n`
	`=45\ 000(1-0.10)^4`
	`=$29\ 524.50`

`text{Salvage value is lower for the straight-line method.}`

v1 Financial Maths, STD2 F1 2007 HSC 26bii

Isla is in her third year working as a freelance photographer.

Isla purchased photography equipment for $4200.

Isla earns $760 per week. Calculate her taxable income for this year if the only allowable deduction is the amount of depreciation of her photography equipment in the third year of use. (1 mark)

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Use this tax table to calculate Isla’s tax payable. (2 marks)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Taxable income} \rule[-1ex]{0pt}{0pt}& \textit{Tax on this income} \\
\hline
\rule{0pt}{2.5ex}0 - \$18\,200 \rule[-1ex]{0pt}{0pt}& \text{Nil} \\
\hline
\rule{0pt}{2.5ex}\$18 \, 201 - \$45\,000 \rule[-1ex]{0pt}{0pt}& \text{16 cents for each \$1 over \$18 200} \\
\hline
\rule{0pt}{2.5ex}\$45\,001 - \$135\,000 \rule[-1ex]{0pt}{0pt}& \$4288 \text{ plus 30 cents for each \$1 over \$45 000} \\
\hline
\rule{0pt}{2.5ex}\$135\,001 - \$190\,000 \rule[-1ex]{0pt}{0pt}& \$31 \, 288 \text{ plus 37 cents for each \$1 over \$135 000} \\
\hline
\rule{0pt}{2.5ex}\$190\,001 \text{ and over} \rule[-1ex]{0pt}{0pt}& \$51 \, 638 \text{ plus 45 cents for each \$1 over \$190 000} \\
\hline
\end{array}

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a. $\$38\,790$

b. $\$3294.40$

Show Worked Solution

a. $\text{Income per year}=52 \ 760= $39\,520$

$\text{Taxable income}= 39\ 520-350= $38\,790$

b.	$\text{Tax payable}$	$= 0.16 \times (38\,790-18\,200)$
		$= 0.16 \times 20\,590$
		$= $3294.40$

Calculus, EXT1 EQ-Bank 16

Find the equation of the tangent to the curve $y=\tan ^{-1}\left(x^2\right)$ at the point on the curve where $x=1$. (3 marks)

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$y=x-1+\dfrac{\pi}{4}$

Show Worked Solution

$y=\tan ^{-1}\left(x^2\right) \ \Rightarrow \ \dfrac{dy}{dx}=\dfrac{1}{1+\left(x^2\right)^2} \times 2 x=\dfrac{2 x}{1+x^4}$

$\text{At} \ \ x=1:$

$y=\tan ^{-1}\left(1^2\right)=\dfrac{\pi}{4}$

$\dfrac{dy}{dx}=\dfrac{2}{1+1^4}=1$

$\text{Find equation of tangent,} \ \ m=1 \ \ \text {through}\ \left(1, \dfrac{\pi}{4}\right):$

$y-\dfrac{\pi}{4}$	$=1(x-1)$
$y$	$=x-1+\dfrac{\pi}{4}$

Calculus, EXT1 EQ-Bank 19

The function $y=f(x)$ has an inverse function $y=f^{-1}(x)$.

The tangent to $y=f(x)$ at the point $(2,3)$, $\ell$, has a gradient of 1.

Show that the tangent to $y=f^{-1}(x)$ at the point $(3,2)$ is parallel to $\ell$. (3 marks)

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$\text{Since tangent to}\ f(x) \ \text {touches at}\ (2,3):$

$f(2)=3 \ \ \Rightarrow\ \ f^{-1}(3)=2$

$\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}$

$\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}$

$m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1$

$\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:$

$(f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1$

$\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell$.

Show Worked Solution

$\text{Since tangent to}\ f(x) \ \text {touches at}\ (2,3):$

$f(2)=3 \ \ \Rightarrow\ \ f^{-1}(3)=2$

$\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}$

$\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}$

$m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1$

$\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:$

$(f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1$

$\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell$.

Financial Maths, STD2 EQ-Bank 28

Sam needs to borrow $3000 and plans to repay it in full after 30 days. He is comparing two options.

Credit card: compound interest charged daily at 19.9% per annum, with no interest-free period.
Personal loan: simple interest charged at 11.5% per annum.

Calculate the interest Sam would be charged on the credit card over the 30 days. (2 marks)

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How much does Sam save by choosing the personal loan. (2 marks)

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a. $\$49.46$

b. $\text{Saving =}\ \$21.10$

Show Worked Solution

a. $\text{Daily interest rate} = \dfrac{0.199}{365}$

$\text{Amount owing}$	$= 3000\left(1+\dfrac{0.199}{365}\right)^{30}$
	$= 3049.458\ldots$
	$= \$3049.46\ \text{(nearest cent)}$

$\therefore \text{Interest} = 3049.46-3000 = \$49.46$

b. $\text{Personal loan (simple interest):}$

$I = Prn$	$=3000 \times 0.115 \times \dfrac{30}{365}$
	$= 28.356\ldots$
	$= \$28.36\ \text{(nearest cent)}$

$\therefore \text{Saving} = 49.46-28.36 = \$21.10$

Financial Maths, STD2 EQ-Bank 27

A credit card account has no interest-free period. Compound interest is calculated daily at a rate of 18.5% per annum and charged to the account up to and including the statement date.

A single purchase of $1500 was made and interest accrued on the purchase for 28 days, including the statement date. There were no other purchases on the account.

Calculate the closing balance on the statement date. (2 marks)

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The minimum payment is calculated as $30 or 3% of the closing balance, whichever is greater. Calculate the minimum payment due. (2 marks)

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a. $\$1521.43$

b. $\$45.64$

Show Worked Solution

a. $\text{Daily interest rate} = \dfrac{0.185}{365}$

$\text{Closing balance}$	$= 1500\left(1+\dfrac{0.185}{365}\right)^{28}$
	$= 1521.433\ldots$
	$= \$1521.43$

b. $3\% \text{ of closing balance} = 0.03 \times 1521.43= \$45.64$

$\text{Since } \$45.64 > \$30, \text{ the minimum payment due is } \$45.64.$

Financial Maths, STD2 EQ-Bank 5 MC

A credit card is an example of a reducing balance loan.

Which statement best explains why this is the case?

Interest is charged at a fixed amount each month, regardless of the balance owing.
Interest is charged on the outstanding balance, which decreases as repayments are made.
The interest rate reduces each month as the balance is repaid.
A fixed portion of the amount borrowed is repaid each month, with no interest charged.

Show Answers Only

$B$

Show Worked Solution

B is correct: Like any reducing balance loan, interest is charged on the outstanding balance, which decreases as repayments are made.

Other options:

A is incorrect: the interest is not a fixed amount; it depends on the balance.
C is incorrect: it is the balance that reduces, not the interest rate, which stays fixed.
D is incorrect: interest is charged on a credit card, and repayments are not a fixed portion of the principal.

$\Rightarrow B$

Networks, STD2 EQ-Bank 28

Lena is opening a new café. The project requires the completion of 7 activities, $A$ to $G$. The project is due to be completed in 15 days.

The directed network diagram shows these activities with their completion times in days.

Use the information from the network diagram to complete the Gantt chart, including the critical path and all other activities required for the project. (3 marks)
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List all activities, not on the critical path, which could be occurring at midday on day 7. (1 mark)
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b. $B\text{, }D\text{ and }E$

Show Worked Solution

b. $\text{By inspection of the Gantt chart.}$

$\text{Activities which could be occurring at }\approx 6.5\ \text{on chart (midday day 7):}$

$B\text{, }D\text{ and }E$

Networks, STD2 EQ-Bank 27

A project requires the completion of 8 activities, $A$ to $H$. The project is due to be completed in 16 days.

The directed network diagram shows these activities with their completion times in days.

Use the information from the network diagram to complete the Gantt chart, including the critical path and all other activities required for the project. (3 marks)
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List all activities, not on the critical path, which could be occurring at midday on day 9. (1 mark)
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b. $C, D\ \text{and}\ F$

Show Worked Solution

b. $\text{By inspection of the Gantt chart.}$

$\text{Activities which could be occurring at}\ \approx 8.5\ \text{on chart (midday day 9):}$

$C, D\ \text{and}\ F$

Networks, STD2 EQ-Bank 6 MC

The Gantt chart below shows the activities involved in organising a community market day. Critical path activities are shown as solid bars and non-critical activities show available float as dashed extensions.

Activity $D$ is delayed by 7 hours and activity $B$ is delayed by 1 hour. What is the new minimum completion time for the project?

14 hours
15 hours
16 hours
22 hours

Show Answers Only

$C$

Show Worked Solution

$\text{From the Gantt chart:}$

$\text{D has float of 5 hours. Delay of 7 exceeds float by 2.}$

$\text{B has float of 3 hours. Delay of 1 is within float.}$

$\text{Only the delay to D affects completion time.}$

$\text{New minimum} = 14+2 = 16 \text{ hours}$

$\Rightarrow C$

Networks, STD2 EQ-Bank 25

The construction of a new reptile exhibit is a project involving nine activities, $A$ to $I$. The network diagram below shows the activities and their completion times in weeks. Some values are missing.

The Gantt chart below has been created for this project.

Using the Gantt chart, identify the critical path. (1 mark)
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Use the Gantt chart to determine the missing values in the network diagram. (2 marks)
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Activity $E$ is delayed by 8 weeks. Using the Gantt chart, explain whether this will affect the minimum completion time of the project. (2 marks)
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a. $ACDFGI$

b. $\text{B} = 5 \text{ weeks, H} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Show Worked Solution

a. $\text{The critical path is the continuous solid bar on row 1 of the Gantt chart.}$

$\text{Critical path:}\ ACDFGI$

b. $\text{Activities B and H are not labelled in the network diagram.}$

$\text{From the Gantt chart:}$

$\text{B starts at week 0, ends at week 5} \to \text{duration} = 5 \text{ weeks}$

$\text{H starts at week 7, ends at week 14} \to \text{duration} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Networks, STD2 EQ-Bank 18

A Gantt chart for a project with activities $A, B, C, D, E, F, G$ and $H$ has been created. The Gantt chart can be used to complete the missing information on the edges in the network diagram.

Use the Gantt chart to determine the three missing values in the network diagram. (3 marks)
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State the minimum completion time for the project. (1 mark)
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a. $\text{C} = 4 \text{ hours, D} = 5 \text{ hours, E} = 7 \text{ hours}$

b. $26 \text{ hours}$

Show Worked Solution

a. $\text{Using the Gantt chart}$

$\text{C (between A and F):} $

$\Rightarrow\ \text{Starts hour 7, ends hour 11 = 4 hours duration}$

$\text{D (between B and G):}$

$\Rightarrow\ \text{Starts hour 3, ends hour 8 = 5 hours duration}$

$\text{E (between B and H):}$

$\Rightarrow\ \text{Starts hour 3, ends hour 10 = 7 hours duration}$

b. $\text{From the Gantt chart, the project ends at hour 26.}$

$\text{Minimum completion time} = 26 \text{ hours}$

Measurement, STD2 EQ-Bank 30

A 2500-watt air-conditioning system is turned on for 3 hours each day. Electricity is charged at 27 cents per kWh.

What is the cost of electricity for using the air-conditioning system over a seven-day period? (2 marks)

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`$14.18`

Show Worked Solution

`text{Daily usage}`	`=2500 xx 3`
	`= 7500\ text{Wh}`
	`=7.5\ text{kWh (1000 Wh = 1 kWh)}`

`text{Cost}`	`= 7 xx 7.5 xx 0.27`
	`= 14.175=$14.18\ \ text{(nearest cent)}`

♦♦ Mean mark 39%.

Measurement, STD2 EQ-Bank 25

The table compares the fuel costs of a petrol car with an electric car.

\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Petrol car} & \textit{Electric car}\\
\hline
\rule{0pt}{2.5ex}\text{Fuel consumption}\rule[-1ex]{0pt}{0pt} & \text{8.6 L/100 km} & \text{18 kWh/100 km} \\ \hline \rule{0pt}{2.5ex}\text{Cost of fuel}\rule[-1ex]{0pt}{0pt} & \text{\$1.87/L} & \text{\$0.25/kWh} \\ \hline \end{array}

Jun travels on average 35 000 km per year.

How much will he save on fuel costs in a year by using an electric car? (3 marks)

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$$4053.70$

Show Worked Solution

$\text{Petrol car fuel costs}$

$\text{Litres used}=\dfrac{8.6\times 35\ 000}{100}=3010\text{ L}$

$\text{Cost of fuel}=3010\times $1.87 = $5628.70$

$\text{Electric car power costs}$

$\text{kWh}=\dfrac{18\times 35\ 000}{100}=6300\text{ kWh}$

$\text{Cost of power}=6300\times $0.25 = $1575$

$\text{Saving}=$5628.70-$1575=$4053.70$

Statistics, EXT1 EQ-Bank 6 MC

The lifetime of a certain brand of batteries has a mean lifetime of 20 hours and a standard deviation of 2 hours. A random sample of 40 batteries is selected.

The probability that the mean lifetime of this sample of 40 batteries exceeds 19.5 hours is closest to

0.0571
0.5987
0.8944
0.9429

Show Answers Only

$D$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}$

$\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 20, \dfrac{4}{40}\right)$

$\sigma_{\bar{X}} = \dfrac{1}{\sqrt{10}} $

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{X}-20}{\frac{1}{\sqrt{10}}}\sim N(0,1)$

$Z=\dfrac{19.5-20}{\frac{1}{\sqrt{10}}}=-1.58 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$\Pr(\overline{X}>19.5)=\Pr(Z>-1.58)$

$\phantom{\Pr(\overline{X}>19.5)}=0.9429$

$\Rightarrow D$

Statistics, EXT1 EQ-Bank 29

The waiting time, $T$ hours, to see a particular doctor at a clinic has a mean of 0.5 hours and a standard deviation of 0.3 hours.

A sample of 35 waiting times is chosen at random.

Use the standard normal distribution table (provided) to find the probability that the average waiting time of the sample of 35 patients is between 0.43 hours and 0.50 hours. Give your answer as a percentage correct to 1 decimal place. (3 marks)

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$\Pr(0.43<\overline{T}<0.50) \approx 41.6\%$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}$

$\text{The sample mean,}\ \overline{T}, \text{for random samples of size 35 is}$

$\text{approximately normally distributed, where:}$

$\mu=0.5\ \ \text{and}\ \ \sigma=\dfrac{0.3}{\sqrt{n}}=\dfrac{0.3}{\sqrt{35}} \approx 0.0507$

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{T}-0.5}{\frac{0.3}{\sqrt{35}}} \sim N(0,1)$

$\Pr(0.43<\overline{T}<0.50)$	$=\Pr\left(\dfrac{0.43-0.50}{0.0507}<Z<\dfrac{0.50-0.50}{0.0507}\right)$
	$=\Pr(-1.38<Z<0)$
	$=\Pr(0<Z<1.38)$
	$=0.9162-0.5000$
	$=0.4162 = 41.6\%\ \text{(1 d.p.)}$

Statistics, EXT1 EQ-Bank 22

The gestation period of cats has a mean of 66 days and a variance of 9 days$^2$.

A sample of 32 cats is chosen at random.

Using the Normal Distribution Table of Values, determine the probability that the sample has an average gestation period greater than 65 days. (3 marks)

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$0.9706$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}$

$\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 66, \dfrac{9}{32}\right)$

$\sigma_{\bar{X}} = \dfrac{3}{\sqrt{32}} $

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{X}-66}{\frac{3}{\sqrt{32}}}\sim N(0,1)$

$Z=\dfrac{65-66}{\frac{3}{\sqrt{32}}}=-1.89 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$P(\overline{X}>65)$	$=P(Z>-1.89)$
	$=1-P(Z\leq -1.89)$
	$=1-0.0294$
	$=0.9706$

Statistics, EXT1 EQ-Bank 26

A research team is investigating the amount of sleep obtained by Year 12 students. Previous studies indicate that the sleep time of Year 12 students has a population mean of 7.4 hours and a population standard deviation of 2.42 hours.

A random sample of 100 Year 12 students is selected.

Using the normal distribution table (included), determine the probability that the mean sleep time of the sample is less than 7 hours. Give your answer correct to four decimal places. (3 marks)

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$4.94 \%$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}$

$\text{The sample mean,}\ \overline{X}, \text{for random samples of size 100 is}$

$\text{approximately normally distributed, where:}$

$\mu=7.4\ \ \text{and}\ \ \sigma=\dfrac{2.42}{\sqrt{n}}=\dfrac{2.42}{\sqrt{100}}$

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{X}-7.4}{\frac{2.42}{\sqrt{100}}} \sim N(0,1)$

$Z=\dfrac{7-7.4}{\frac{2.42}{\sqrt{100}}}=-1.65 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$P(\overline{X} < 7)=P(Z < -1.65)=0.0494=4.94 \%$

Probability, STD2 EQ-Bank 31

A survey of 60 people found the following information about their exercise habits.

35 enjoy hiking $(H)$
28 enjoy cycling $(C)$
8 enjoy neither hiking nor cycling

Draw a Venn diagram to represent this information. (2 marks)

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One person is selected at random. What is the probability that the person enjoys hiking or cycling? (1 mark)

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b. $\dfrac{13}{15}$

Show Worked Solution

a. $\text{Number in either }H\text{ or }C = 60-8 = 52$

$\text{Number in both }H\text{ and }C = 35+28-52 = 11$

$H\text{ only} = 35-11 = 24$

$C\text{ only} = 28-11 = 17$

b. $P(H \text{ or } C) = \dfrac{24+11+17}{60} = \dfrac{52}{60} = \dfrac{13}{15}$

$\text{OR, using the complement:}$

$P(H \text{ or } C) = 1-\dfrac{8}{60} = \dfrac{13}{15}$

Probability, STD2 EQ-Bank 5 MC

The Venn diagram shows information about 40 people surveyed about whether they own a dog $(D)$ or a cat $(C)$.

Which two-way table correctly represents the information in the Venn diagram?

A.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 12 & 8 & 20 \\ \hline \text{Not }C & 5 & 15 & 20 \\ \hline \text{Total} & 17 & 23 & 40 \\ \hline \end{array}$	B.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 8 & 12 & 20 \\ \hline \text{Not }C & 15 & 5 & 20 \\ \hline \text{Total} & 23 & 17 & 40 \\ \hline \end{array}$
C.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 15 & 12 & 27 \\ \hline \text{Not }C & 8 & 5 & 13 \\ \hline \text{Total} & 23 & 17 & 40 \\ \hline \end{array}$	D.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 5 & 12 & 17 \\ \hline \text{Not }C & 15 & 8 & 23 \\ \hline \text{Total} & 20 & 20 & 40 \\ \hline \end{array}$

Show Answers Only

$B$

Show Worked Solution

$\text{Reading the Venn diagram:}$

$\text{C and D (intersection)} = 8$

$\text{C only (not D)} = 12$

$\text{D only (not C)} = 15$

$\text{Neither} = 5$

$\text{In the two-way table:}$

$\text{Row }C\text{: }\ D\text{ column} = 8,\ \text{Not }D\text{ column} = 12,\ \text{Total} = 20$

$\text{Row Not }C\text{: }\ D\text{ column} = 15,\ \text{Not }D\text{ column} = 5,\ \text{Total} = 20$

$\Rightarrow B$

Algebra, STD2 A4 EQ-Bank 27

SunPower Solutions is a business that installs solar panels. Fixed costs are $1200. Each panel costs $150.00 to install and generates revenue of $350.00.

The spreadsheet below models the business's costs and revenue for different numbers of panels installed.

Calculate the spreadsheet values for the installation of 4 panels (cells B9, C9, D9) and 6 panels (cells B10, C10, D10). (2 marks)
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Using the spreadsheet, identify the break-even point and explain what it means for the business. (2 marks)
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SunPower Solutions has received a large order that will see them make a profit of $4200. Calculate the number of solar panels $(x)$ they will be installing. (2 marks)
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a. $\text{4 panels: TC (B9) = \$1800.00}$

$\text{Revenue (C9) = \$1400.00, Profit/Loss (D9) = }-\$400.00$

$\text{6 panels: TC (B10) = \$2100.00}$

$\text{Revenue (C10) = \$2100.00, Profit/Loss (D10) = \$0.00}$

b. $\text{Break-even = 6 panels, See worked solution}$

c. $27 \text{ panels}$

Show Worked Solution

a. $\text{4 panels:}$

$\text{Total Cost (B9)} = \$1200+4\times\$150 = \$1800.00$

$\text{Revenue (C9)} = 4\times\$350 = \$1400.00$

$\text{Profit/Loss (D9)} = \$1400.00-\$1800.00 = -\$400.00$

$\text{6 panels:}$

$\text{Total Cost (B10)} = \$1200+6\times\$150 = \$2100.00$

$\text{Revenue (C10)} = 6\times\$350 = \$2100.00$

$\text{Profit/Loss (D10)} = \$2100.00-\$2100.00 = \$0.00$

b. $\text{When 6 panels are installed:}$

$\text{Revenue = Total costs = \$2100.00 (breakeven)}$

$\text{This is the point at which the business covers all of its costs and}$

$\text{begins to make a profit.}$

$\text{OR}$

$\text{If fewer than 6 panels are installed the business will make a loss.}$

c. $\text{Let } x = \text{the number of solar panels to be installed.}$

$\text{Profit}$	$ = \text{Revenue}-\text{Total costs}$
$4200$	$ = 350x-(1200+150x)$
$4200$	$= 200x-1200$
$5400$	$=200x$
$x$	$=27$

$\text{SunPower Solutions will install 27 solar panels.}$

Algebra, STD2 A4 EQ-Bank 16

Harmony Arts Festival is a cultural event with fixed costs of $510. Each ticket costs $8.00 to provide and sells for $25.00.

The spreadsheet below models the festival's costs and revenue for different numbers of tickets sold.

Calculate the values for cells C12 and D12 in the spreadsheet. (2 marks)
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Using the spreadsheet, identify the break-even point and explain what it means for the festival. (2 marks)
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a. $\text{C12} = \$1250.00 \quad \text{D12} = \$340.00$

b. $\text{See worked solution}$

Show Worked Solution

a. $\text{C12: Revenue} = 50 \times \$25.00 = \$1250.00$

$\text{D12: Profit/Loss} = \$1250.00-\$910.00 = \$340.00$

b. $\text{When 30 tickets sold:}$

$\text{Revenue = Total costs = \$750.00 (Breakeven)}$

$\text{This is the point at which the festival covers all of its costs}$

$\text{and begins to make a profit.}$

$\text{OR}$

$\text{If less than 30 tickets are sold the festival will make a loss.}$

Complex Numbers, EXT2 EQ-Bank 15

Prove that for any complex numbers $z_1$ and $z_2$,
$\abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}$ (2 marks)
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Hence, or otherwise, show that if $\abs{z-1}+\abs{z+1} \leqslant 4$ for $z\in C,$
$\abs{z} \leqslant 2$ (2 marks)

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a. $\text{Proof (See Worked Solutions)}$

b. $\text{See Worked Solutions}$

Show Worked Solution

a. $\text {Prove}\ \ \abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}:$

$\abs{z_1+z_2}^2$	$=\left(z_1+z_2\right)\left(\overline{z}_1+\overline{z}_2\right)$
	$=\abs{z_1}^2+\abs{z_2}^2+z_1 \overline{z}_2+\overline{z}_1 z_2$
	$=\abs{z_1}^2+\abs{z_2}^2+2 \operatorname{Re}\left(z_1 \overline{z}_2\right)$

$\text{Since}\ \ \operatorname{Re}(w) \leqslant\abs{w}\ \ \text{for} \ \ w\in C,$

$\abs{z_1+z_2}^2$	$\leqslant\abs{z_1}^2+\abs{z_2}^2+2\abs{z_1 \overline{z}_2}$
	$\leqslant\abs{z_1}^2+2\abs{z_1}\abs{z_2}+\abs{z_2}^2$
	$\leqslant\left(\abs{z_1}+\abs{z_2}\right)^2$

$\therefore\abs{z_1+z_2} \leqslant\abs{z_1}+\abs{z_2}$

b. $|z-1|+|z+1| \leqslant 4 \ \text{(given)}\ …\ (1)$

$\text {Using triangle inequality:}$

$|(z-1)+(z+1)| \leqslant|z-1|+|z+1|$

$\text{Since}\ \ (z-1)(z+1)=2 z:$

$\abs{2z}$	$\leqslant\abs{z-1}+\abs{z+1}$
$\abs{2z}$	$\leqslant 4\ \ \text{(using (1) above)}$
$2\abs{z}$	$\leqslant 4$
$\abs{z}$	$\leqslant 2$

Calculus, EXT2 C1 2020 HSC 13d*

By expanding `(text{cis}\theta + text{cis}(-theta))^4` show that
`cos^4 theta = frac{1}{8} ( cos (4 theta) + 4 cos (2 theta) + 3 )`. (3 marks)

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Hence, or otherwise, find `int_0^(frac{pi}{2}) cos^4 theta\ d theta`. (2 marks)

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a. `text{See Worked Solution}`

b. `frac{3 pi}{16}`

Show Worked Solution

a. `text{cis}\theta + text{cis}(-theta) = 2 cos theta\ \ …\ (1)`

`(text{cis}\theta + text{cis}(-theta))^4= 16 cos^4(4theta)`

`text{Expand LHS:}`

`(text{cis}\theta + text{cis}(-theta))^4`

`= text{cis}(4theta)+4text{cis}(2theta)+6+4text{cis}(-2theta)+text{cis}(4theta)`

`= 2text{cos}(4theta)+8text{cos}(2theta)+6\ \ text{(using (1) above)}`

`text{Equating sides:}`

`16 cos^4 theta`	`= 2 cos (4 theta) + 8 cos (2 theta) + 6`
`cos^4 theta`	`= frac{1}{8} cos(4 theta) + 1/2 cos(2 theta) + 3/8`
`cos^4 theta`	`= frac{1}{8} (cos(4 theta) + 4 cos(2 theta) + 3)`

b.	`int_0^(frac{pi}{2}) cos^4 theta\ d theta`	`= frac{1}{8} int_0^(frac{pi}{2}) cos(4 theta) + 4 cos(2 theta) + 3\ d theta`
		`= frac{1}{8} [ frac{1}{4} sin(4 theta) + 2 sin (2 theta) + 3 theta ]_0^(frac{pi}{2}`
		`= frac{1}{8} [( frac{1}{4} sin (2 pi) + 2 sin pi + frac{3 pi}{2}) – 0 ]`
		`= frac{1}{8} ( frac{3 pi}{2})`
		`= frac{3 pi}{16}`

Mechanics, EXT2 EQ-Bank 36

An experimental rocket is at a height of 5000 m, ascending at a speed of $50\sqrt{2}$ m s$^{-1}$ at an angle of 45° to the horizontal, when its engine stops. The rocket is then subject to gravity and to air resistance proportional to its velocity. Take $g$ = 10 m s$^{-2}$.

The velocity vector of the rocket, $t$ seconds after the engine stops, is

$\mathbf{v}(t) = 50e^{-0.2t}\,\mathbf{i} + (100e^{-0.2t}-50)\mathbf{j}.$ (Do NOT prove this.)

Show that the rocket reaches its greatest height when $t =5\ln 2$ seconds, and calculate its greatest height. (3 marks)

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The pilot can only operate the ejection seat while the rocket is descending at an angle between 45° and 60° to the horizontal. Find the earliest and latest times at which the pilot can eject. (3 marks)

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As the rocket continues to fall, its speed approaches a limiting value. Find this terminal speed, justifying your answer. (1 mark)

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a. $\text{See Worked Solutions}$

b. $\text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{Terminal speed}=50 \ \text{ms}^{-1}$

Show Worked Solution

a. $\mathbf{v}(t)=50 e^{-0.2 t}\,\mathbf{i}+\left(100 e^{-0.2 t}-50\right)\mathbf{j}$

$\text{Max height occurs when} \ \ \dot{y}=0:$

$100 e^{-0.2 t}-50$	$=0$
$e^{-0.2 t}$	$=\dfrac{1}{2}$
$-0.2 t$	$=-\ln 2$
$t$	$=5 \ln 2$

$\text{Find} \ y \ \text{when}\ \ t=5 \ln 2:$

$y(t)=\displaystyle \int 100 e^{-0.2 t}-50\, d t=-500 e^{-0.2 t}-50 t+c$

$\text{When} \ \ t=0, y=5000:$

$5000=-500 e^{\circ}+c \ \Rightarrow \ c=5500$

$y=5500-500 e^{-0.2 t}-50 t$

$\text{At} \ \ t=5\ln 2:$

$y=5500-500 e^{-\ln 2}-50 \times 5 \ln 2=5076.71 \ldots=5077 \ \text{m}$.

b. $\text {On descent,} \ \ \dot{y}<0.$

$\text{Let} \ \ \theta=\text{angle below the horizontal}$

$\tan \theta=\dfrac{\abs{\dot{y}}}{\dot{x}}$

$\text{Let}\ \ a=e^{-0.2 t}$

$\tan \theta=\dfrac{50-100 a}{50 a}=\dfrac{1}{a}-2 \ \Rightarrow \ \theta=\tan ^{-1}\left(\dfrac{1}{a}-2\right)$

$\text{When} \ 45^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=1 \ \Rightarrow \ a=3$

$e^{-0.2 t}=\dfrac{1}{3} \ \Rightarrow \ t=\dfrac{\ln 3}{0.2} \approx 5.5 \ \text{s (1 d.p.)}$

$\text{When} \ 60^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=\sqrt{3} \ \Rightarrow \ a=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}$

$e^{-0.2 t}$	$=2-\sqrt{3}$
$-0.2 t$	$=\ln (2-\sqrt{3})$
$t$	$=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s (1 d.p.)}$

$\therefore \ \text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{As} \ \ t \rightarrow \infty:$

$e^{-0.2 t} \rightarrow 0\ \ \Rightarrow\ \ \dot{x} \rightarrow 0, \ \ \dot{y} \rightarrow -50$

$\therefore \ \text{Terminal speed}=\sqrt{0^2+50^2}=50 \ \text{ms}^{-1}$

Mechanics, EXT2 EQ-Bank 31

A particle is projected from the origin with speed, $V$, at an angle of $\alpha$ above the horizontal. It is subject to both gravity and an air resistance proportional to its velocity, so that its horizontal and vertical components of acceleration while it is rising are given by

$\ddot{x}=-k\dot{x}$ and $\ddot{y} = -g-k\dot{y}$

Show that $\dot{x} = V\cos\,\alpha\ e^{-kt}$ and $\dot{y} = \left( \dfrac{g}{k} + V\sin\,\alpha\right)e^{-kt}-\dfrac{g}{k} $ (2 marks)

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Show that when the particle reaches its greatest height, it has travelled a horizontal distance of
$\dfrac{V^2\sin\,2\alpha}{2(g+Vk\,\sin\,\alpha)}$. (3 marks)

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Show Answers Only

a. $\text{See Worked Solutions}$

b. $\text{See Worked Solutions}$

Show Worked Solution

a. $\ddot{x}=-k \dot{x} \ \Rightarrow \ \dfrac{d \dot{x}}{d t}=-k \dot{x}$

$\text{Solving 1st order differential equation:}$

$\dot{x}=A e^{-k t}$

$\text{Since}\ \ \dot{x}=V \cos \alpha\ \ \text{when} \ \ t=0\ \ \Rightarrow\ \ A=V \cos \alpha$

$\dot{x}=V \cos \alpha e^{-k t}$

$\ddot{y}=-g-k \dot{y}\ \Rightarrow \ \dfrac{d \dot{y}}{d t}+k \dot{y}=-g$

$\text{Solving 1st order differential equation:}$

$\dot{y}=B e^{-k t}-\dfrac{g}{k}$

$\text{Since} \ \ \dot{y}=V \sin \alpha \ \ \text{when} \ \ t=0:$

$V \sin \alpha=B e^{-k t}-\dfrac{g}{k} \ \Rightarrow \ B=\dfrac{g}{k}+V \sin \alpha$

$\dot{y}=\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}$

b. $\text{At max height,} \ \ \dot{y}=0$

$\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}=0 \ \Rightarrow \ e^{-k t}=\dfrac{g}{g+V k \, \sin \alpha}\ \ldots\ (1)$

$\text{Find horizontal distance} \ (x):$

$x$	$=\displaystyle \int_0^t V \cos \alpha\, e^{-k t}\, d t$
	$=-\dfrac{V \cos \alpha}{k}\big[e^{-k t}\big]_0^t$
	$=\dfrac{V \cos \alpha}{k}\left(1-e^{-k t}\right)$
	$=\dfrac{V \cos \alpha}{k}\left(1-\dfrac{g}{g+Vk\, \sin \alpha}\right)\ \ \ \text{(using (1) above)}$
	$=\dfrac{V \cos \alpha}{k}\left(\dfrac{g+Vk\, \sin \alpha-g}{g+Vk\, \sin \alpha}\right)$
	$=\dfrac{V^2 \sin \alpha\, \cos \alpha}{g+Vk\, \sin \alpha}$
	$=\dfrac{V^2 \sin 2 \alpha}{2(g+Vk\, \sin \alpha)}$

Proof, EXT2 EQ-Bank 21

Prove that $\lim\limits_{x \to 0}x^2\, \sin \left(\dfrac{1}{x}\right)=0$. (2 marks)

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$-1 \leqslant \sin \left(\dfrac{1}{x}\right) \leqslant 1$

$-x^2 \leqslant x^2\, \sin \left(\dfrac{1}{x}\right) \leqslant x^2$

$\text{By squeeze theorem:}$

$\text{Since} \ \ \lim\limits_{x \to 0}-x^2=0\ \ \text{and}\ \ \lim\limits _{x \rightarrow 0} x^2=0$

$\Rightarrow \lim\limits _{x \rightarrow 0} x^2\, \sin \left(\dfrac{1}{x}\right)=0$

Show Worked Solution

$-1 \leqslant \sin \left(\dfrac{1}{x}\right) \leqslant 1$

$-x^2 \leqslant x^2\, \sin \left(\dfrac{1}{x}\right) \leqslant x^2$

$\text{By squeeze theorem:}$

$\text{Since} \ \ \lim\limits_{x \to 0}-x^2=0\ \ \text{and}\ \ \lim\limits _{x \rightarrow 0} x^2=0$

$\Rightarrow \lim\limits _{x \rightarrow 0} x^2\, \sin \left(\dfrac{1}{x}\right)=0$

Vectors, EXT2 EQ-Bank 26

Let $S$ be a sphere with equation

\begin{align*}
\left|r-\left(\begin{array}{c} 2 \\ -3 \\ 3
\end{array}\right)\right|=15
\end{align*}

Show the point $P(1,-1,2)$ lies outside the sphere $S$. (2 marks)

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$\text{Centre of circle} \ (C)=(2,-3,3)$

$\text{Radius}=15 \ \text{(given)}$

$\text{Find distance from} \ C \text { to } P(1,-1,2):$

$\operatorname{dist}=\sqrt{(1-2)^2+(-1+3)^2+(2-3)^2}=\sqrt{6}$

$\text{Since \(\ \sqrt{6}<15, P$ lies inside sphere.}\)

Show Worked Solution

$\text{Centre of circle} \ (C)=(2,-3,3)$

$\text{Radius}=15 \ \text{(given)}$

$\text{Find distance from} \ C \text { to } P(1,-1,2):$

$\operatorname{dist}=\sqrt{(1-2)^2+(-1+3)^2+(2-3)^2}=\sqrt{6}$

$\text{Since \(\ \sqrt{6}<15, P$ lies inside sphere.}\)

Proof, EXT2 EQ-Bank 34

Consider a sequence of rectangles with side lengths $a_{ n }$ and $b_{ n }$.

The first rectangle has $a_1=2$ and $b_1=1$.

For integers $n \geq 1,\ \ a_{n+1}=\dfrac{a_n+b_n}{2}$ and $b_{n+1}=\dfrac{2}{a_{n+1}}.$

Show that every rectangle in the sequence has an area of 2 square units. (1 mark)

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Use the relationship between the arithmetic mean and the geometric mean to prove that $a_n \geq \sqrt{2}$ for any integer $n \geq 1$. (2 marks)

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Use mathematical induction to prove that $a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2})$ for any integer $n \geq 1$. (4 marks)

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Use the squeeze theorem to show that the rectangles approach a square as $n$ approaches infinity. (2 marks)

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Show Worked Solution

a. $\text{Since}\ \ a_1 b_1=2\ \ \text{and}\ \ a_{n+1} b_{n+1}=a_{n+1} \times \dfrac{2}{a_{n+1}}=2$

$\Rightarrow\ \text{Each rectangle has area 2.}$

b. $a_1=2\ \ \Rightarrow\ \ a_1 \geq \sqrt{2}\ \ \text{(true for 1st rectangle)}$

$\text{AM/GM inequality:}\ \ \dfrac{a_n+b_n}{2} \geq \sqrt{a_nb_n} $

$a_nb_n=2\ \ \text{(from part (a))}$

$\dfrac{a_n+b_n}{2} \geq \sqrt{2}\ …\ (1)$

$\text{Since}\ \ a_{n+1}=\dfrac{a_n+b_n}{2}:$

$\ a_{n+1} \geq \sqrt{2}\ \ \ \text{(using (1) above)}$

$\therefore a_n \geq \sqrt{2}$

c. $\text{Prove}\ \ a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}),\ \ \text{for}\ \ n \geq 1$

$\text{If}\ \ n=1:$

$\ a_1-\sqrt{2}=2-\sqrt{2} \leq \dfrac{1}{2^{0}}(2-\sqrt{2})$.

$\Rightarrow\ \text{True for}\ \ n=1.$

$\text{Assume true for}\ \ n=k:$

$a_k-\sqrt{2} \leq \dfrac{1}{2^{k-1}}(2-\sqrt{2})\ …\ (1) $

$\text{Prove true for}\ \ n=k+1:$

$\text{i.e.}\ \ a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\text{By definition,} \ \ a_{k+1}=\dfrac{1}{2}\left(a_k+b_k\right)$

$a_{k+1}-\sqrt{2}$	$=\dfrac{1}{2}\left(a_k-\sqrt{2}\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) $
	$\leq \dfrac{1}{2}\left(\dfrac{1}{2^{k-1}}(2-\sqrt{2})\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right)\ \ \text{(see (1) above)}\ $

$\text{Since}\ \ a_k \geq \sqrt{2}\ \ \text{and}\ \ a_kb_k=2:$

$\ b_k \leq \sqrt{2}\ \ \text{and}\ \ b_k-\sqrt{2} \leq 0$

$a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\Rightarrow\ \text{True for}\ \ n=k+1.$

$\therefore\ \text{Since true for}\ \ n=1,\ \text{by PMI, true for integers}\ \ n \geq 1.$

d. $\text {Combining parts (b) and (c):}$

$0 \leq a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}).$

$\text{As}\ \ n \rightarrow \infty, \ \dfrac{1}{2^{n-1}} \rightarrow 0$

$\Rightarrow a_n-\sqrt{2} \rightarrow 0\ \ \text{(by squeeze theorem)}$

$\text{Since rectangles have an area = 2:}$

$\text{As}\ \ n \rightarrow \infty,\ b_n \rightarrow \sqrt{2}$

$\text{i.e. rectangles approach a square.}$

Mechanics, EXT2 EQ-Bank 30

Luggage at an airport is delivered to its owners via a ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.

By resolving forces parallel to the ramp, show that the magnitude of the acceleration, `ddot{x}\ text(ms)^(-2)`, of the suitcase down the ramp is given by `ddot{x} = (g-2v)/2`. (2 marks)

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Using 9.8 `text(ms)^(-2)` as the acceleration due to gravity, find the distance `x` metres that the suitcase has slid as a function of `v`. Give your answer in the form `x = bv + c\ log_e(c/(c-v))`, where `b, c in R`. (3 marks)

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a. `text(Proof)\ \ text{(See Worked Solutions)}`

b. `x = -v + 4.9 ln ((4.9)/(4.9-v))`

Show Worked Solution

`sumF`	`=20g sin 30^@-20v`
`m ddot{x}`	`=10g-20v`
`ddot{x}`	`= (g-2v)/2`

b. `text{Using}\ \ ddot{x}=v *(dv)/(dx):`

`(dv)/(dx)`	`= (g-2v)/(2v)`
`(dx)/(dv)`	`= (2v)/(g-2v)`
`(dx)/(dv)`	`= -(2v)/(2v-g)=-((2v-g + g))/(2v-g)= -1-g/(2v-g)`

`text{Find the distance travelled:}`

`x`	`= int_0^v-1-g/(2v-g)\ dv`
	`= int_0^v-1-g/2 (2/(2v-g))\ dv`
	`= [-v-4.9 xx ln\ \|2v-g\|]_0^v`

`text(When)\ \ x=0, v=0:`

`2v-g < 0\ \ =>\ \ |2v-g| = g-2v`

`x`	`= [-v-4.9 ln (g-2v)]_0^v`
	`= -v-4.9 ln (g-2v)-(0-4.9 ln (g))`
	`= -v + 4.9 ln (g/(g-2v))`
	`=-v + 4.9 ln(9.8/(9.8-2v))`
	`= -v + 4.9 ln ((4.9)/(4.9-v))`

Complex Numbers, EXT2 N2 2023 14a*

Let $z$ be the complex number $z=\text{cis}\dfrac{\pi}{6} $ and $w$ be the complex number $w=\text{cis}\dfrac{3\pi}{4} $.

By first writing $z$ and $w$ in Cartesian form, or otherwise, show that
$|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}$. (3 marks)

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The complex numbers $z, w$ and $z+w$ are represented in the complex plane by the vectors $\overrightarrow{O A},\overrightarrow{O B}$ and $\overrightarrow{O C}$ respectively, where $O$ is the origin.
Show that $\angle A O C=\dfrac{7 \pi}{24}$. (2 marks)

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Deduce that $\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}$. (1 mark)

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i. $\text{See Worked Solutions}$

ii. $\text{See Worked Solutions}$

iii. $\text{See Worked Solutions}$

Show Worked Solution

i. $z= \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i $

$w= \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} $

$\|z+w\|^2$	$=\Bigg{\|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{\|}$
	$=\Bigg{\|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{\|}$
	$=\Bigg{\|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{\|}$
	$= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}$
	$= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}$
	$=\dfrac{16-4\sqrt6+4\sqrt2}{8} $
	$=\dfrac{4-\sqrt6+\sqrt2}{2} $

ii.

$\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} $

$ |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} $

$\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC $

$\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB $

$\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} $

iii. $\text{In}\ \triangle AOC: $

$ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} $

$\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. $

$\text{Using the cos rule in}\ \triangle AOC: $

$\cos\,\dfrac{7\pi}{24}$	$=\dfrac{\|z\|^2+\|z+w\|^2-\|w\|^2}{2\|z\|\|z+w\|}$
	$=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1 \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} $
	$=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} $
	$=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} $
	$=\dfrac{8-2\sqrt6+2\sqrt2}{4} $

♦♦ Mean mark (iii) 26%.

Mechanics, EXT2 EQ-Bank 19

A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and $m$ kg attached to the ends of the string.

The acceleration due to gravity is 9.8 m s$^{-2}$.

If the acceleration of the 1 kg particle is 4.9 ms$^{-2}$ upwards, then determine the value of $m$. (2 marks)

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$m=3\ \text{kg}$

Show Worked Solution

$\text{Resolving the forces:}$

$\text{Using}\ \ \Sigma F = m \ddot{x},\ \text{consider forces on 1 kg mass:}$

$T-(9.8 \times 1) = 4.9 \times 1\ \ \Rightarrow\ \ T=14.7$

$\text{Consider forces on}\ m\ \text{kg mass:}$

$m \times 9.8-T$	$=m \times 4.9$
$4.9m$	$= 14.7$
$:. m$	$= \dfrac{14.7}{4.9}=3\ \text{kg}$

Mechanics, EXT2 EQ-Bank 16

Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest and with acceleration due to gravity equal to 9.8 m s$^{-2}$.

Assuming the system remains connected, determine the speed of the 5 kg mass after two seconds. (3 marks)

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$v= 4.9\ \text{m s}^{-1}$

Show Worked Solution

$\Sigma F=5g-3g=2 \times 9.8 = 19.6\ \text{N}$

$\text{Find acceleration, using}\ \ \Sigma F=m \ddot{x}:$

$19.6$	$=(5+3) \ddot{x}$
$\ddot{x}$	$=\dfrac{19.6}{8}=2.45\ \text{m s}^{-2}$

$v= \displaystyle \int \ddot{x}\,dt=\int 2.45\,dt=2.45t+c$

$v=0\ \ \text{at}\ \ t=0\ \ \Rightarrow\ \ c=0$

$\text{At}\ \ t=2:$

$v=2 \times 2.45 = 4.9\ \text{m s}^{-1}$

Mechanics, EXT2 2019 SPEC1 9

A light inextensible string is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs in equilibrium from a smooth ring on the string, as shown in the diagram below. The string makes an angle `alpha` with the ceiling.

Express the tension, `T` newtons, in the string in terms of `m`, `g` and `alpha`. (1 mark)

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A different light inextensible sting is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs from a smooth ring on the string. A horizontal force of `F` newtons is applied to the ring. The tension in the sting has a constant magnitude and the system is in equilibrium. At one end the string makes an angle `beta` with the ceiling and at the other end the string makes an angle `2beta` with the ceiling, as shown in the diagram below.

Show that `F = mg((1-cos(beta))/(sin(beta)))`. (3 marks)

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a. `T = (mg)/(2sinalpha)`

b. `text(See Worked Solutions)`

Show Worked Solution

`2 xx Tsinalpha`	`= mg`
`:.T`	`= (mg)/(2sinalpha)`

`text(Resolving forces vertically:)`

`Tsin(beta) + Tsin(2beta)`	`= mg`
`T`	`= (mg)/(sin(beta) + sin(2beta))`

`text(Resolving forces horizontally:)`

`F + Tcos(2beta)`	`= Tcos(beta)`
`F`	`= Tcos(beta)-Tcos(2beta)`
	`= T(cos(beta)-cos(2beta))`
	`= T[cos(beta)-(2cos^2beta-1)]`
	`= T(−2cos^2(beta) + cos(beta) + 1)`
	`= T(−2cos(beta)-1)(cos(beta)-1)`
	`= (mg(1 -cos(beta))(2cosbeta + 1))/(sin(beta) + 2sin(beta)cos(beta))`
	`= (mg(1-cos(beta))(2cos(beta) + 1))/(sin(beta)(1+2cos(beta)))`
	`= mg((1-cos(beta))/(sin(beta)))`

Mechanics, EXT2 EQ-Bank 25

The diagram below shows objects of mass 5 kg and $M$ kg attached to the ends of a light, inextensible string that passes over a smooth pulley.

The 5 kg object is accelerating upwards at a rate of 4.9 m/s$^2$. Let the tension in the string be $T$ newtons.

Using 9.8 m s$^{-2}$ as the acceleration due to gravity, find the value of $T$ and hence determine the value of $M$. (3 marks)

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$M=15\ \text{kg}$

Show Worked Solution

$\text{Using} \ \ F=m \ddot{x}:$

$\text{Consider the 5 kg mass:}$

$T-5 g$	$=m \ddot{x}$
$T-5(9.8)$	$=5(4.9)$
$T$	$=73.5 \ N$

$\text{Consider the \(M$ kg mass:}\)

$\text{Mass is accelerating downward at 4.9 ms\(^{-2}$}\)

$M \times 9.8-73.5$	$=M \times 4.9$
$4.9 M$	$=73.5$
$M$	$=\dfrac{73-5}{49}=15 \ \text{kg}$

Mechanics, EXT2 EQ-Bank 20

A 6-kilogram mass is placed on a frictionless plane inclined 30° to the horizontal.

The mass is connected to another 10-kilogram mass by a light, inextensible string via a pulley, as shown in the diagram

The 6-kilogram mass is released from rest and slides up the plane.

Given the acceleration due to gravity is $g$ m s$^{-2}$, express the velocity, $v$, of the 6 kilogram mass in terms of $t$ as it moves up the plane. (3 marks)

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$v=\dfrac{7}{16}gt$

Show Worked Solution

$\text{Resolving forces on the 6 kg mass:}$

$\text{Let} \ \ F_d=\text{force down slope:}$

$\sin 30^{\circ}$	$=\dfrac{F_d}{6 g}$
$F_d$	$=6 g\, \sin 30=3 g$

$\text{Using} \ \ \sum F=m a$:

$T-3 g=6 a\ \ldots\ (1)$

$\text{Consider 10 kg mass:}$

$10 g-T=10 a\ \ldots\ (2)$

$\text{Add (1) + (2):}$

$7 g=16 a \ \Rightarrow \ a=\dfrac{7}{16} g$

$v=\displaystyle \int \dfrac{7}{16}g \, dt=\dfrac{7}{16} g t+c$

$\text{When} \ \ t=0 \quad v=0 \ \Rightarrow \ c=0$

$\therefore v=\dfrac{7}{16}gt$

Mechanics, EXT2 EQ-Bank 22

A 12 kilogram object is suspended from a horizontal ceiling by two light, inextensible strings at angles of 30° and 45°, as shown in the diagram.

The acceleration due to gravity is $g$ m s$^{-2}$ and the tensions in the strings are $T_1$ newtons and $T_2$ newtons.

Show that $T_2=\sqrt{\dfrac{3}{2}} T_1$ (2 marks)

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Determine the tensions, in newtons, of $T_1$ and $T_2.$ (2 marks)

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a. $\text{See Worked Solutions}$

b. $T_1=12(\sqrt{3}-1)g$

$T_2=6 \sqrt{6}(\sqrt{3}-1)g\ \ \ (=6 \sqrt{2}(3-\sqrt{3}))$

Show Worked Solution

a. $\text{Resolve forces into horizontal / vertical components:}$

$\text{Horizontal forces are equal.}$

$T_1\cos 30^{\circ}$	$=T_2 \cos 45^{\circ}$
$T_1 \times \dfrac{\sqrt{3}}{2}$	$=T_2 \times \dfrac{1}{\sqrt{2}}$
$T_2$	$=\dfrac{\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \, T_1=\sqrt{\dfrac{3}{2}}\, T_1$

b. $\text{Vertical forces are equal.}$

$T_1 \sin 30+T_2 \sin 45$	$=12 g$
$T_1 \times \dfrac{1}{2}+T_2 \times \dfrac{1}{\sqrt{2}}$	$=12 g$

$\text{Substitute} \ \ T_2=\sqrt{\dfrac{3}{2}}\, T_1:$

$T_1 \times \dfrac{1}{2}+T_1 \times \sqrt{\dfrac{3}{2}} \times \dfrac{1}{\sqrt{2}}$	$=12 g$
$T_1\left(\dfrac{\sqrt{3}+1}{2}\right)$	$=12 g$

$T_1=\dfrac{24g}{\sqrt{3}+1} \times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}=12(\sqrt{3}-1)g$

$T_2$	$=\sqrt{\dfrac{3}{2}} \times 12(\sqrt{3}-1)g$
	$=6 \sqrt{6}(\sqrt{3}-1)g\ \ \ (=6 \sqrt{2}(3-\sqrt{3}))$

Mechanics, EXT2 EQ-Bank 17

Two light inextensible strings are attached to a horizontal surface and suspended a 10-kilogram object as shown in the diagram below

The tension in the strings are $T_1$ newtons and $T_2$ newtons.

Express $T_1$ in terms of $T_2$. (2 marks)

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If the acceleration due to gravity is 9.8 ms$^{-2}$, determine the exact values of $T_1$ and $T_2$, in newtons. (2 marks)

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a. $\text{See Worked Solutions}$

b. $T_1=49 \sqrt{3}, T_2=49$

Show Worked Solution

a. $\text{Resolve forces into horizontal/vertical components:}$

$\text{Horizontal forces are equal.}$

$T_1 \cos 60^{\circ}$	$=T_2 \cos 30^{\circ}$
$T_1 \times \dfrac{1}{2}$	$=T_2 \times \dfrac{\sqrt{3}}{2}$
$T_1$	$=\sqrt{3}\, T_2$

b. $\text{Vertical forces are equal.}$

$T_1 \sin 60^{\circ}+T_2 \sin 30^{\circ}$	$=10 \times 9.8$
$T_1 \times \dfrac{\sqrt{3}}{2}+T_2 \times \frac{1}{2}$	$=98$

$\text {Substitute} \ \ T_1=\sqrt{3}\, T_2 :$

$\sqrt{3}\, T_2 \times \dfrac{\sqrt{3}}{2}+T_2 \times \dfrac{1}{2}$	$=98$
$2\, T_2$	$=98$
$T_2$	$=49 \ \text{newtons}$

$\therefore T_1=49 \sqrt{3}, \ T_2=49$

Mechanics, EXT2 EQ-Bank 30

In a circus act, an 8 kg cannon ball is projected from the origin into the air with an initial velocity of 26 m s$^{-1}$ and at an angle of 67.4° to the horizontal. The ball is caught at the top of its trajectory by a performer who is at the position $(A, B)$.

The velocity vector, $\mathbf{v} (t)$, of the ball at time $t$ seconds after launch is given by

$\mathbf{v}(t)=10 e^{-0.8 t} \mathbf{i} +\left[36.5 e^{-0.8 t}-12.5\right] \mathbf{j}$. (Do NOT Prove this.)

Show that the ball reaches the performer at $t=1.339$ (to three decimal places). (2 marks)

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Find the values of $A$ and $B$. (3 marks)

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What is the speed of the ball when it reaches the performer? (1 mark)

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What is the magnitude of the force on the ball when it reaches the performer? (2 marks)

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a. $t=1.339 \ \text{s}$

b. $A=8.217\ \text{m},\ \ B=13.257\ \text{m}$

c. $\text{Speed}\ = 3.426\ \text{m s}^{-1}$

d. $82.951 \ \text{N}$

Show Worked Solution

a. $\text{At top of trajectory:}$

$36.5 e^{-0.8 t}-12.5$	$=0$
$36.5 e^{-0.8 t}$	$=12.5$
$e^{-0.8 t}$	$=\dfrac{12.5}{36.5}$
$-0.8 t$	$=\ln \dfrac{12.5}{36.5}$
$ t$	$=\dfrac{\ln\frac{12.5}{36.5}}{-0.8}=1.339 \ \text{s (3 d.p.)}$

b. $\text{Horizontal velocity}\ =10 e^{-0.8 t}$

$x=\displaystyle \int 10 e^{-0.8 t}\,d t=\dfrac{10}{-0.8} e^{-0.8 t}+c_1=-12.5 e^{-0.8 t}+c_1$

$\text{When} \ \ t=0, \ x=0:$

$0=-12.5 e^0+c_1\ \ \Rightarrow\ \ c_1=12.5$

$x=-12.5 e^{-0.8 t}+12.5$

$\text{When } t=1.339, \ x=A:$

$A=-12.5 e^{-0.8 \times 1.339}+12.5=8.217\ \text{m (3 d.p.)}$

$\text {Vertical velocity}\ =36.5 e^{-0.8 t}-12.5$

$y=\displaystyle \int\left(36.5 e^{-0.8 t}-12.5\right)\,d t=\dfrac{36.5}{-0.8} e^{-0.8 t}-12.5 t+c_2=-45.625 e^{-0.8 t}-12.5 t+c_2$

$\text{When} \ \ t=0, \ y=0:$

$0=-45.625+c_2\ \ \Rightarrow\ \ c_2=45.625$

$y=-45.625 e^{-0.8 t}-12.5 t+45.625$

$\text{When}\ \ t=1.339, \ y=B:$

$B=-45.625 e^{-0.8 \times 1.339}-12.5 \times 1.339+45.625=13.257\ \text{m (3 d.p.)}$

c. $\text{At top of trajectory,}\ \mathbf{j} \text{-component of velocity = 0.}$

$\Rightarrow\ \text{Speed at top is the}\ \mathbf{i} \text{-component of velocity at}\ t=1.339:$

$\text{Speed}\ =10e^{-0.8 \times 1.339} = 3.426\ \text{m s}^{-1}\ \text{(3 d.p.)}$

d. $\text{Using}\ \ F=m \ddot{x}:$

$\mathbf{v}(t)=10 e^{-0.8 t}\mathbf{i} +\left(36.5 e^{-0.8 t}-12.5\right) \mathbf{j}$

$\mathbf{a} =\dfrac{d v }{d t}=-8 e^{-0.8 t} \mathbf{i} -29.2 e^{-0.8 t} \mathbf{j}$

$\text{When} \ \ t=1.339:$

$\mathbf{a}=-8 e^{-0.8 \times 1.339}\mathbf{i}-29.2 e^{-0.8 \times 1.339}\mathbf{j}=-2.74 1\,\mathbf{i} -10\, \mathbf{j}$

$\text{Magnitude of acceleration}$

$=\sqrt{(2.741)^2+(10)^2}=10.36885 \ldots\ \text{ms}^{-2}$

$\therefore \ \text{Magnitude of force}$

$=8 \times 10.36885 \ldots =82.951 \ \text{N (3 d.p.)}$

\(\text{Total paid}\ (FV)\)	\(=PV(1+r)^n\)
	\(=1680(1.00052)^{38}\)
	\(=\$1713.52\)

\(\text{Amount owing}\)	\(=2500\left(1+\dfrac{0.205}{365}\right)^{40}\)
	\(=2556.784\ldots\)
	\(=\$2556.78\ \text{(nearest cent)}\)

\(I=Prn\)	\(=2500\times 0.12\times\dfrac{40}{365}\)
	\(=32.876\ldots\)
	\(=\$32.88\ \text{(nearest cent)}\)

\(\text{Amount owing}\ (FV)\)	\(=PV(1+r)^n-300\)
	\(=600(1+0.00050959\ldots)^{20}-300\)
	\(=606.14-300\)
	\(=\$306.14\)

b.	\(\text{Tax payable}\)	\(= 0.16 \times (38\,790-18\,200)\)
		\(= 0.16 \times 20\,590\)
		\(= $3294.40\)

\(y-\dfrac{\pi}{4}\)	\(=1(x-1)\)
\(y\)	\(=x-1+\dfrac{\pi}{4}\)

\(\text{Closing balance}\)	\(= 1500\left(1+\dfrac{0.185}{365}\right)^{28}\)
	\(= 1521.433\ldots\)
	\(= \$1521.43\)

\(\Pr(0.43<\overline{T}<0.50)\)	\(=\Pr\left(\dfrac{0.43-0.50}{0.0507}<Z<\dfrac{0.50-0.50}{0.0507}\right)\)
	\(=\Pr(-1.38<Z<0)\)
	\(=\Pr(0<Z<1.38)\)
	\(=0.9162-0.5000\)
	\(=0.4162 = 41.6\%\ \text{(1 d.p.)}\)

\(P(\overline{X}>65)\)	\(=P(Z>-1.89)\)
	\(=1-P(Z\leq -1.89)\)
	\(=1-0.0294\)
	\(=0.9706\)