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Financial Maths, STD2 2020 HSC 37 (Adapted)

Estelle deposits a single lump sum into an account that earns 3% per annum compound interest.

Present value interest factors for an annuity of $1 for various interest rates \((r)\) and numbers of periods \((N)\) are shown in the table.

  

From this account, Estelle plans to make the following withdrawals.

  • $2000 at the end of each year for the first 15 years (the first withdrawal is one year after the deposit).
  • $5000 at the end of each year for a further 10 years, that is, in years 16 to 25.

Find the smallest lump sum Estelle must deposit so that both sets of withdrawals can be made.   (3 marks)

Show Answers Only

\(\text{Lump sum required}=\$51\,251\)

Show Worked Solution

\(\text{Annuity 1:}\ \ PV\ \text{of}\ \ \$2000\ \text{annuity for 15 years at}\ r=0.03\)

\(\Rightarrow PV\ \text{factor}=11.938\)

\(\therefore\ PV\ \text{Annuity 1}=11.938\times 2000=\$23\,876\)
  

\(\text{Annuity 2:}\ \ PV\ \text{of}\ \ \$5000\ \text{annuity for years 16−25 at}\ r=0.03\)

\(PV\ \text{Annuity 2}\) \(=PV(25\ \text{years})-PV(15\ \text{years})\)  
  \(=5000\times 17.413-5000\times 11.938\)  
  \(=5000\times(17.413-11.938)\)  
  \(=\$27\,375\)  

 
\(\therefore\ \text{Lump sum required}=23\,876+27\,375=\$51\,251\)

Financial Maths, STD2 2025 HSC 34 (Adapted)

The table shows future value interest factors for an annuity of $1.
   

Larry invests a single amount of $18 000 for 5 years at 9% per annum, compounding monthly.

Tobias wants to end up with the same amount as Larry by using an annuity. He will pay a fixed sum into an account at the end of each month for 5 years, with the account also paying 9% per annum, compounding monthly.

Using the table, work out how much Tobias must deposit each month.   (3 marks)

Show Answers Only

\(\$373.65\)

Show Worked Solution

\(r=\dfrac{0.09}{12}=0.0075,\ \ n=12\times 5=60\)

\(\text{Larry’s investment:}\)

\(FV=18\,000(1+0.0075)^{60}=28\,182.26\)
  

\(\text{Tobias’s investment:}\)

\(\text{Annuity factor:}\ 75.42414\)

\(\text{Annuity}\times 75.42414\) \(=\$28\,182.26\)
\(\text{Annuity}\) \(=\dfrac{28\,182.26}{75.42414}=\$373.65\)

Financial Maths, STD2 2023 HSC 25 (Adapted)

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

  
 

  1. Sue wants to save $180 000 over the next 5 years. The account pays 8% per annum, compounding annually.
  2. Using the table, find the amount Sue should contribute each year, to the nearest dollar.   (2 marks)
  3. Instead, Sue decides to contribute $7500 every three months for 5 years into an account paying 8% per annum, compounding quarterly.
  4. Using the table, find how much Sue will have at the end of 5 years.   (3 marks)
Show Answers Only

a.    \(\$30\,680\)

b.    \(\$182\,227.50\)

Show Worked Solution

a.    \(\text{Applicable interest rate}=8\%\)

\(\text{Compounding periods}=5\times 1=5\)

\(\Rightarrow\ \text{Factor}=5.867\)

\(\therefore\ \text{Contribution (annual)}=\dfrac{180\,000}{5.867}=\$30\,680\ \text{(nearest dollar)}\)
 

b.    \(\text{Applicable interest rate}=\dfrac{8\%}{4}=2\%\ \text{per quarter}\)

\(\text{Compounding periods}=5\times 4=20\)

\(\Rightarrow\ \text{Factor}=24.297\)

\(\therefore\ \text{Total (after 5 years)}=7500\times 24.297=\$182\,227.50\)

Financial Maths, STD2 2025 HSC 18 (Adapted)

The table shows future value interest factors for an annuity of $1.
  

A scholarship fund receives a contribution of $4000 at the end of each year for 15 years. The fund earns 5% per annum, compounded annually.

Using the table, calculate the value of the fund at the end of 15 years.   (2 marks)

Show Answers Only

\(\$86\,316\)

Show Worked Solution

\(r=5\%\ \text{annually}\)

\(\text{Compounding periods}=15\)

\(\text{Annuity factor}=21.579\)

\(\therefore\ FV\ \text{(annuity)}=4000\times 21.579=\$86\,316\)

Financial Maths, STD2 2018 HSC 26c (Adapted)

Sofia contributes $175 to an annuity at the end of every month and plans to keep this up for 3 years.

Ignoring interest, how much will Sofia have paid into the annuity in total over the 3 years?   (1 mark)

Show Answers Only

\(\text{Total contributed}=\$6300\)

Show Worked Solution

\(\text{Total contributed}=3\times 12\times 175=\$6300\)

Financial Maths, STD2 2013 23 MC (Adapted)

Elias opens a savings account that pays 4% per annum, compounded quarterly. At the end of every quarter he deposits $800, beginning one quarter after the account is opened, and continues for the following 18 months.

How much is in Elias’s account at the end of the 18 months?

  1. $4800.00
  2. $4921.61
  3. $4970.83
  4. $5095.30
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Interest: 4% p.a.}\ \ \Rightarrow\ \  \text{1% per quarter}\)

\(\text{18 months}=6\ \text{end-of-quarter deposits}\)

\(\text{Value of 1st deposit}=800(1.01)^5=840.81\)

\(\text{Value of 2nd deposit}=800(1.01)^4=832.48\)

\(\text{Value of 3rd deposit}=800(1.01)^3=824.24\)

\(\text{Value of 4th deposit}=800(1.01)^2=816.08\)

\(\text{Value of 5th deposit}=800(1.01)^1=808.00\)

\(\text{Value of 6th deposit}=800\)
   

\(\therefore\ \text{Amount in account}\)

\(=840.81+832.48+824.24+816.08+808.00+800.00\)

\(=\$4921.61\)

\(\Rightarrow B\)

Financial Maths, STD2 2016 HSC 17 MC (Adapted)

A credit card balance of $920 is outstanding for 20 days. Compound interest is charged at 0.041% per day, with no interest-free period.

Which calculation gives the amount of interest charged on this balance?

  1. \(920\times 0.00041\times 20\)
  2. \(920(1.00041)^{20}\)
  3. \(920(1.00041)^{20}-920\)
  4. \(920(1+0.00041\times 20)\)
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: compound interest \(=\) total owing \(-\) principal \(=920(1.00041)^{20}-920\).

Other options:

  • A is incorrect: this is simple interest \((Prn)\), not compound interest.
  • B is incorrect: this gives the total amount owing, not the interest.
  • D is incorrect: this gives the total owing using simple interest.

\(\Rightarrow C\)

Financial Maths, STD2 2019 HSC 27 (Adapted)

Bianca has a credit card that offers no interest-free period. At the end of each month, interest is added to the account at 19.5% per annum, compounded daily. Interest is worked out from the day a purchase is made (included) through to the last day of the month (included).

Part of Bianca’s June statement is shown below, with two figures left blank.
  

 

For this account, the minimum payment is set at 3% of the closing balance on 30 June.

Find the minimum payment, giving your answer to the nearest cent.   (3 marks)

Show Answers Only

\(\text{Minimum payment}=\$78.42\)

Show Worked Solution

\(\text{Days of interest}\ (n)=\ \text{21 June – 30 June}=10\)

\(\text{Daily interest rate}\ (r)=\dfrac{0.195}{365}=0.0005342\ldots\)

\(\text{Closing balance}=2600(1+0.0005342\ldots)^{10}=\$2613.92\)

\(\text{Minimum payment}\) \(=\dfrac{3}{100}\times 2613.92\)
  \(=78.4176\ldots\)
  \(=\$78.42\ \text{(nearest cent)}\)

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)

Show Answers Only

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

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 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)

Show Answers Only

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

Show Worked Solution

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

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

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

Measurement, STD2 M1 2008 HSC 28b* (Adapted)

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

 

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

  2. 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)

Show Answers Only

a.    `h(2a + b)`

b.    `b\ text(decreases by 4.)`

Show Worked Solution
a.    
`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.)`

Financial Maths, STD2 F4 2023 HSC 28 (Adapted)

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)

Show Answers Only

`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.}`

Financial Maths, STD1 F3 2025 HSC 24 (Adapted)

A used car has a sale price of \(\$18\,600\). In addition to the sale price, the following costs are charged:

  • transfer of registration $50
  • stamp duty which is calculated at $3 for every $100, or part thereof, of the sale price.

Tahlia borrows the total amount to be paid for the car, including transfer of registration and stamp duty. Simple interest at the rate of 7.2% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 4 years.

Calculate Tahlia's monthly repayment, correct to the nearest cent.   (5 marks)

Show Answers Only

\(\$515.41\)

Show Worked Solution

\(\text{Stamp duty}=\dfrac{18\,600}{100}\times 3=\$558\)

\(\text{Total borrowed}=18\,600+50+558=\$19\,208\)

\(\text{Interest}=Prn=19\,208\times 0.072\times 4=\$5531.904\)

\(\text{Total to repay}=19\,208+5531.904=\$24\,739.904\)

 
\(\therefore\ \text{Monthly repayment}=\dfrac{24\,739.904}{48}=515.4146…\approx \$515.41\)

Financial Maths, STD1 F1 2025 HSC 19 (Adapted)

At the end of the 2024-2025 financial year, Hannah had a gross annual salary of \(\$78\,500\). She had allowable tax deductions totalling $2,300 for work-related expenses.

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


The Medicare levy is 2% of taxable income. During the year, Hannah paid $1,250 per month in Pay As You Go (PAYG) tax.

Determine whether Hannah will receive a tax refund or owe money to the Australian Taxation Office. Justify your answer with calculations.    (4 marks)

Show Answers Only

\(\text{Hannah owes}\ \$172\ \text{to the ATO.}\)

Show Worked Solution

\(\text{Taxable income}=78\,500-2300=\$76\,200\)

\(\text{Tax payable}\) \(=4288+0.30\times (76\,200-45\,000)\)
  \(=4288+0.30\times 31\,200\)
  \(=4288+9360\)
  \(=\$13\,648\)

 
\(\text{Medicare levy}=0.02\times 76\,200=\$1524\)

\(\text{Total tax + Medicare}=13\,648+1524=\$15\,172\)

\(\text{Total PAYG paid}=1250\times 12=\$15\,000\)

\(\therefore\ \text{Hannah owes}\ \ 15\,172-15\,000=\$172\ \ \text{to the ATO.}\)

Financial Maths, STD1 F1 2025 HSC 19 (Adapted)

At the end of the 2024-2025 financial year, Priya's taxable income was $92 400.

  1. The table shows the income tax rate for Australian residents for the 2024-2025 financial year.

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

  1. Using the table, calculate Priya's tax payable.   (3 marks)

  2. The Medicare levy is 2% of taxable income.
  3. Calculate the Medicare levy payable by Priya.   (1 mark)

Show Answers Only

a.    \(\$18\,508\)

b.    \(\$1848\)

Show Worked Solution

a.    \(\text{Calculate tax payable:}\)

\(\text{Tax payable}\) \(=4288+0.30\times (92\,400-45\,000)\)
  \(=4288+0.30\times 47\,400\)
  \(=4288+14\,220\)
  \(=\$18\,508\)

 

b.    \(\text{Calculate Medicare levy:}\)

\(\text{Medicare levy}=0.02\times 92\,400=\$1848\)

Financial Maths, STD1 F1 2024 HSC 4 MC (Adapted)

Liam works at a cafe in Newcastle and earns $24 per hour. His hourly pay rate increases by 3%.

How much will he earn for a 5-hour shift with this increase?

  1. $3.60
  2. $24.72
  3. $120.00
  4. $123.60
Show Answers Only

\(D\)

Show Worked Solution

\(\text{New hourly rate}=24\times 1.03=\$24.72\)

\(\therefore\ \text{Shift earnings}=24.72\times 5=\$123.60\)

\(\Rightarrow D\)

Financial Maths, STD1 F1 2022 HSC 21 (Adapted)

A real estate agent's commission for selling houses is 3% for the first \(\$600\,000\) of the sale price and 1.8% for any amount over \(\$600\,000\).

Calculate the commission earned in selling a house for \(\$950\,000\).   (2 marks)

Show Answers Only

\(\$24\,300\)

Show Worked Solution

\(\text{Commission on first}\ \$600\,000=3\%\times 600\,000=\$18\,000\)

\(\text{Amount over}\ \$600\,000= 950\,000- 600\,000=\$350\,000\)

\(\text{Commission on remainder}=1.8\%\times 350\,000=\$6300\)

\(\text{Total commission}= 18\,000+ 6300=\$24\,300\)

Financial Maths, STD1 F1 2021 HSC 12 (Adapted)

A coding bootcamp runs a school holiday course and charges $200 per student.

The costs of running the course are:

  • Instructor: $130 per hour
  • Venue hire: $60 per hour plus 10% GST.

The course runs for 3 hours a day for 5 days.

What profit does the coding bootcamp make if 18 students pay for the course?   (3 marks)

Show Answers Only

\(\$660\)

Show Worked Solution

\(\text{Total hours}=3\times 5=15\ \text{hours}\)

\(\text{Revenue}=18\times 200=\$3600\)

\(\text{Instructor cost}=15\times 130=\$1950\)

\(\text{Venue hire}=15\times 60=\$900\)

\(\text{GST on venue hire}=10\%\times 900=\$90\)

\(\text{Total venue hire}= 900+ 90=\$990\)

\(\text{Total costs}= 1950+ 990=\$2940\)

\(\text{Profit}= 3600- 2940=\$660\)

Financial Maths, STD1 F1 2019 HSC 26 (Adapted)

Harper has a weekly net income of $720. She has created a budget where she allocates this income to rent, food, phone, subscriptions and the rest to savings.

Her budget is shown below, with some details missing.

Item Weekly amount
Rent $300
Food ?
Phone $25
Subscriptions $35
Savings ?

 
Harper allocates 25% of her weekly net income to food.

How many weeks will it take Harper to save $4860?   (3 marks)

Show Answers Only

\(\text{27 weeks}\)

Show Worked Solution

\(\text{Food}=25\%\times 720=0.25\times 720=\$180\)

\(\text{Savings}= 720- 300- 180- 25- 35=\$180\)

\(\text{Weeks}=\dfrac{4860}{180}=27\ \text{weeks}\)

Financial Maths, STD1 F1 2019 HSC 11 (Adapted)

Jack works as a casual lifeguard and earns $30 per hour. He is also paid a $20 wet-weather allowance for each shift in which he works in the rain.

In one week, Jack worked three shifts of 5 hours each. It rained during two of these shifts.

How much did Jack earn in total this week?   (2 marks)

Show Answers Only

\(\$490\)

Show Worked Solution

\(\text{Total hours}=3\times5=15\ \text{hours}\)

\(\text{Wages}=15\times 30=\$450\)

\(\text{Wet-weather allowances}=2\times 20=\$40\)

\(\text{Total earnings}=450+40=\$490\)

Financial Maths, STD1 F1 2022 HSC 7 MC (Adapted)

Sienna works a 40-hour week at a Bunnings warehouse and is paid at an hourly rate of $25. Any overtime hours worked are paid at time-and-a-half.

In a particular week, she earned $1450.

How many hours in total did Sienna work in this week to earn this amount?

  1. 12
  2. 38.6
  3. 52
  4. 58
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Normal pay}=40\times 25=\$1000\)

\(\text{Overtime pay earned}= 1450-1000=\$450\)

\(\text{Overtime rate}= 25\times1.5=\$37.50\ \text{per hour}\)

\(\text{Overtime hours}=\dfrac{450}{37.50}=12\ \text{hours}\)

\(\text{Total hours}=40+12=52\ \text{hours}\)

\(\Rightarrow C\)

Financial Maths, STD1 F1 2021 HSC 19 (Adapted)

Maya purchased a motorbike for $12 000. The value of the motorbike decreases according to a linear model. The graph shows the value of the motorbike, $\(V\), against the time, \(t\) months, since it was purchased.
 

  1. By how much does the value of the motorbike decrease every 10 months?   (1 mark)

  2. Find the value of the motorbike after 4 years.   (1 mark)

  3. Identify ONE problem with using this model to determine the value of Maya's motorbike over time.   (1 mark)

Show Answers Only

a.     \(\$1000\)

b.     \(\$7200\)

c.     \(\text{Motorbike will have a negative value after 120 months.}\)

Show Worked Solution

a.    \(\text{Total decrease over 100 months}= 12\,000-2000= \$10\,000\)

\(\text{Decrease per 10 months}= \dfrac{10\,000}{10}= \$1000\)
 

b.    \(\text{4 years} = 4 \times 12 = 48\ \text{months}\)

\(\text{Depreciation rate}= \$100\ \text{per month}\)

\(V = 12\,000-(100 \times 48)=\$7200\)
 

c.    \(\text{Model limitations:}\)

\(\text{The linear model will eventually predict a value of \$0 (at 120 months) and negative}\)

\(\text{values beyond that, which is unrealistic.}\)

Financial Maths, STD1 F1 2023 HSC 6 MC (Adapted)

A courier van was valued at $48 000 when new. The value of the van depreciates at a rate of 18 cents per kilometre travelled.

What is the value of the van after it has travelled a total distance of 95 400 km?

  1. $17 172
  2. $30 828
  3. $37 440
  4. $65 172
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Depreciation} = 95\,400\times\dfrac{18}{100}=\$17\,172\)

\(\text{Value} = 48\,000-17\,172=\$30\,828\)

\(\Rightarrow B\)

Financial Maths, STD2 F1 2010 HSC 5 MC (Adapted)

Noah saw the following advertisement at his local bank in Newcastle:

Noah invests $6000 for a term of 9 months.

How much interest will Noah earn at the end of the term?

  1. $216
  2. $259
  3. $288
  4. $298
Show Answers Only

\(A\)

Show Worked Solution

\(P=\$6000,\ r=4.8\%=0.048,\ n=\dfrac{9}{12}\ \text{years}\)

\(I=Prn=6000 \times 0.048 \times \dfrac{9}{12}=\$216\)

\(\Rightarrow A\)

Financial Maths, STD2 F1 2010 HSC 5 MC (Adapted)

Hayden invests $3000 for 1 year and 8 months. Simple interest is paid on the investment at a rate of 5% per annum.

What is the total value of the investment at the end of this period?

  1. $3150
  2. $3250
  3. $3270
  4. $3500
Show Answers Only

\(B\)

Show Worked Solution

\(\text{1 year and 8 months = 20 months}\)

\(I=Prn=3000 \times 0.05 \times \dfrac{20}{12}=\$250\)

\(\therefore\ \text{Value of Investment} =3000+250=\$3250\)

\(\Rightarrow B\)

Financial Maths, STD1 F1 2023 HSC 2 MC (Adapted)

Mia deposits $4000 into a savings account at the Commonwealth Bank that pays simple interest at a rate of 4% per annum.

How much interest will she earn in the first three years?

  1. $120
  2. $160
  3. $480
  4. $4480
Show Answers Only

\(C\)

Show Worked Solution

\(P=\$4000,\ r=4\%=0.04,\ n=3\ \text{years}\)

\(I=Prn=4000 \times 0.04 \times 3=\$480\)

\(\Rightarrow C\)

Calculus, 2ADV C4 2011 HSC 4d (Adapted)

  1. Differentiate  `y=sqrt(16 -x^2)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, find  `int (8x)/sqrt(16 -x^2)\ dx`.    (2 marks)

Show Answers Only

a.    `- x/sqrt(16-x^2)`

b.    `-8 sqrt(16-x^2) + C`

Show Worked Solution

a.    `y= sqrt(16-x^2)= (16-x^2)^(1/2)`

`dy/dx` `=1/2 xx (16-x^2)^(-1/2) xx d/dx (16-x^2)`
  `= 1/2 xx (16-x^2)^(-1/2) xx -2x`
  `= – x/sqrt(16-x^2)`

 

b.    `int (8x)/sqrt(16-x^2)\ dx` `= -8 int (-x)/sqrt(16-x^2)\ dx`
    `= -8 (sqrt(16-x^2)) + C`
    `= -8 sqrt(16-x^2) + C`

Calculus, 2ADV C4 2024 HSC 5 MC (Adapted)

What is \( {\displaystyle \int x(3 x^2+1)^4 d x} \) ?

  1. \( \dfrac{1}{30}(3x^2+1)^5+C \)
  2. \( \dfrac{1}{5}(3x^2+1)^5+C \)
  3. \( \dfrac{5}{6}(3x^2+1)^5+C \)
  4. \( \dfrac{6}{5}(3x^2+1)^5+C \)
Show Answers Only

\( A \)

Show Worked Solution
\[ \int x(3x^2+1)^4 dx\] \(=\dfrac{1}{5} \cdot \dfrac{1}{6x}x(3x^2+1)^5+C\)  
  \(=\dfrac{1}{30}(3x^2+1)^5+C\)  

 
\( \Rightarrow A \)

NOTE: Integrating by the reverse chain rule only works if some form of the derivative is already present outside of the brackets.

Calculus, 2ADV C1 2014 HSC 13c (Adapted)

The displacement of a particle moving along the  `x`-axis is given by

 `x =2t -3/sqrt(t+1)`,

where  `x`  is the displacement from the origin in metres,  `t`  is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.   (2 marks)

  2. What value does the velocity approach as  `t`  increases indefinitely?   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2`
Show Worked Solution
a.    `x` `=2t -3/sqrt(t+1)`
    `=2t -3(t+1)^(-1/2)`

 

`dot x` `= 2\ -3(-1/2) (t+1)^(-3/2)`
  `= 2 + 3/(2(t+1)^(3/2))`

 

`ddot x` `= -(9/4)(t+1)^(-5/2)`
  `= – 9/(4sqrt((t+1)^5))`

 
`text(S)text(ince)\ \ t >= 0,`

`=> 1/sqrt((t+1)^5) > 0`

`=> – 9/(4sqrt((t+1)^5)) < 0`
 

`:.\ text(Acceleration is always negative.)`

 

b.    `text(Velocity)\ (dot x) = 2 + 3/(2(t+1)^(3/2))`

 
`text(As)\ t -> oo,\ 3/(2(t+1)^(3/2)) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 2`

Calculus, 2ADV C1 2018 HSC 12d (Adapted)

The displacement of a particle moving along the `x`-axis is given by

`x = 1/4t^4 -t^3 -1/2t^2 +3t,`

where `x` is the displacement from the origin in metres and `t` is the time in seconds, for `t >= 0`.

  1. What is the initial velocity of the particle?   (1 mark)

  2. At which times is the particle stationary?   (2 marks)

  3. Find the position of the particle when the acceleration is `4/3`.   (2 marks)

Show Answers Only

a.    `3\ text(ms)^(-1)`

b.    `t = 1 or 3\ text(seconds)`

c.    `-329/324\ text(m)`

Show Worked Solution

a.    `x = 1/4t^4 -t^3 -1/2t^2 +3t`

`v = (dx)/(dt) = t^3-3t^2-t+3`
 

`text(Find)\ \ v\ \ text(when)\ \ t = 0:`

`v` `= 0 -3(0) -0+ 3`
  `= 3\ text(ms)^(-1)`

 

b.  `text(Particle is stationary when)\ \ v = 0`

`t^3-3t^2-t+3` `=0`  
`t^2(t-3)-1(t-3)` `=0`  
`(t-3)(t^2-1)` `=0`  
`(t-3)(t-1)(t+1)` `=0`  

 

`t = 1 or 3\ text(seconds), t >= 0`
 

c.  `a = (dv)/(dt) = 3t^2-6t-1`
 

`text(Find)\ \ t\ \ text(when)\ \ a = 4/3`

`3t^2-6t-1` `= 4/3`
`3t^2-6t-7/3` `= 0`
`9t^2-18t-7` `=0`
`(3t-7)(3t+1)` `=0`
`t` `=7/3`, `t >= 0`
`x(7/3)` `= 1/4(7/3)^4 -(7/3)^3 -1/2(7/3)^2 +3(7/3)`
  `= -329/324\ text(m)`

Calculus, 2ADV C1 2019 HSC 14d (Adapted)

The equation of the tangent to the curve  `y = ae^(2x)+bx`  at the point where  `x = 0`  is  `y = 3x +2`.

Find the values of  `a`  and  `b`.   (3 marks)

Show Answers Only

`b = -1,\ \ a = 2`

Show Worked Solution
`y ` `= ae^(2x)+bx`
`(dy)/(dx)` `= 2ae^(2x)+b`

 
`text(When)\ \ x = 0,\ \ (dy)/(dx) = 3`

`2a + b` `= 3\ …\ (1)`

 
`text(The point)\ (0, 2)\ text(lies on)\ y:`

`a(1) +b(0)` `=2`
`a` `= 2\ …\ (2)`

  

`text(Substitute into)\ (1)`

`2(2)+b` `= 3`
`b` `= -1`

Calculus, 2ADV C1 2023 HSC 14 (Adapted)

Find the equation of the tangent to the curve  `y=x(3x+2)^2`  at the point `(1,25)`.   (3 marks)

Show Answers Only

`y=55x-30`

Show Worked Solution
`y` `=x(3x+2)^2`  
`dy/dx` `=6x(3x+2) + (3x+2)^2`  
  `=(3x+2)(9x+2)`  

 
`text{At}\ x=1\ \ =>\ \ dy/dx=(3(1)+2)(9(1)+2)=55`
 

`text{Find equation of line}\ \ m=55,\ text{through}\ (1,25)`

`y-y_1` `=m(x-x_1)`  
`y-25` `=55(x-1)`  
`y` `=55x-30`  

Calculus, 2ADV C1 2009 HSC 1d (Adapted)

Find the gradient of the tangent to the curve `y = 2x^3-5x^2 + 4` at the point `(2, 0)`.   (2 marks)

Show Answers Only

`text(Gradient = 2.)`

Show Worked Solution
`y` `= 2x^3-5x^2 + 4`  
`dy/dx` `= 6x^2-10x`  
     

`text(At)\ x = 2:`

`dy/dx= 6(2)^2-10(2)=24-20=4`

`:.\ text(Gradient of tangent at)\ (2, 0) = 4.`

Calculus, 2ADV C1 2013 HSC 11b (Adapted)

Evaluate  `lim_(x->1) ((x-1)(x+2)^2)/(x^2+x-2)`.   (2 marks)

Show Answers Only

 `3`

Show Worked Solution

`lim_(x ->1) ((x-1)(x+2)^2)/(x^2+x-2)`

COMMENT: This question has been simplified as students no longer need to factorise the difference between 2 cubes (`x^3-2^3`).

`=lim_(x->1) ( (x -1)(x+2)^2)/( (x-1)(x+2)`

`=lim_(x->1) (x+2)`

`=3`

Calculus, 2ADV C1 2019 HSC 11c (Adapted)

Differentiate  `(4x + 3)/(3x-4)`.   2 marks)

Show Answers Only

`-25/(3x-4)^2`

Show Worked Solution

`text(Using quotient rule:)`

`u=4x+3,`     `v=3x-4`  
`u^{′} = 4,`     `v^{′} = 3`  
     
`y^{′}` `= (u^{′} v-v^{′} u)/v^2`
  `= (4(3x-4)-3(4x+3))/(3x-4)^2`
  `= (12x-16-12x-9)/(3x-4)^2`
  `= -25/(3x-4)^2`

Calculus, 2ADV C1 2015 HSC 12c (Adapted)

Find  `f^{′}(x)`, where  `f(x) = (2x^2-3x)/(2-x).`   (2 marks)

Show Answers Only

`((x-3) (x + 1))/(x-1)^2`

Show Worked Solution

`f(x) = (2x^2-3x)/(2-x)`

`text(Using the quotient rule:)`

`u` `= 2x^2-3x` `\ \ \ \ \ \ v` `= 2-x`
`u^{′}` `= 4x-3` `\ \ \ \ \ \ v^{′}` `= -1`
`f^{′}(x)` `= (u^{′} v-uv^{′})/v^2`
  `= ((4x-3)(2-x)-(2x^2-3x) xx -1)/(2-x)^2`
  `= (-2x^2 + 8x-6)/(x-2)^2`
  `= (-2(x^2-4x+3)/(x-2)^2`
  `= (-2(x-3) (x-1))/(x-2)^2`

Calculus, 2ADV EO-Bank 9 MC

 Let  `f^(')(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

  1. `f(x)=2sqrt(2x-3)`
  2. `f(x)=sqrt(2x-3)-2`
  3. `f(x)=2sqrt(2x-3)-2`
  4. `f(x)=sqrt(2x-3)+2`
Show Answers Only

`=>C`

Show Worked Solution
`f^{‘}(x)` `=2/(sqrt(2x-3))`  
`f(x)` `=2 int(2x-3)^{- 1/2}`  
  `=2*1/2*2(2x-3)^{1/2}+c`  
  `=2sqrt(2x-3)+c`  

 
`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3)-2`

`=>C`

Calculus, 2ADV C1 EO-Bank 11 MC (Adapted)

Two functions, \(f\) and \(g\), are continuous and differentiable for all  \(x\in R\). It is given that  \(f(-1)=7,\ g(-1)=5\)  and  \(f^{′}(-1)=-4,\ g^{′}(-1)=-2\).

The gradient of the graph  \(y=\dfrac{f(x)}{g(x)}\)  at the point where  \(x=-1\)  is

  1. \(-\dfrac{6}{49}\)
  2. \(\dfrac{6}{49}\)
  3. \(\dfrac{6}{25}\)
  4. \(-\dfrac{6}{25}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Using the Quotient Rule when}\ \ x=-1:\)

\(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)\) \(=\dfrac{g(x)f^{′}(x)-f(x)g^{′}(x)}{g(x)^2}\)
  \(=\dfrac{g(-1)f^{′}(-1)-f(-1)g^{′}(-1)}{g(-1)^2}\)
  \(=\dfrac{5 \times -4-7 \times -2}{5^2}\)
  \(=-\dfrac{6}{25}\)

 
\(\Rightarrow D\)

Calculus, 2ADV C1 2023 HSC 7 MC (Adapted)

It is given that  \(y=f(g(x))\), where  \(f(2)=5\), \(f^{′}(2)=3\), \(g(4)=2\)  and  \(g^{′}(4)=-2\).

What is the value of \(y^{′}\) at  \(x=4\)?

  1. \(-6\)
  2. \(-2\)
  3. \(3\)
  4. \(6\)
Show Answers Only

\(A\)

Show Worked Solution
\(y\) \(=f(g(x))\)  
\(y^{′}\) \(=f^{′}(g(4)) \times g^{′}(4)\)  
  \(=f^{′}(2) \times -2\)  
  \(=3 \times -2\)  
  \(=-6\)  

 
\(\Rightarrow A\)

Calculus, 2ADV C1 EO-Bank 1 MC (Adapted)

The derivative of  \((n^2-1) x^{3n-2}\)  can be expressed as

  1. \(3(n-1)(n^2-1) x^{3n-2}\)
  2. \(3(n-1)(n^2-1) x^{3(n-1)}\)
  3. \((3n-2) (n^2-1) x^{3(n-1)}\)
  4. \((3n-2) (n^2-1) x^{3n-2}\)
Show Answers Only

\(C\)

Show Worked Solution
\(y\) \(=(n^2-1) x^{3n-2}\)  
\(y^{′}\) \(=(3n-2) (n^2-1) x^{3n-2-1)}\)  
  \(=(3n-2) (n^2-1) x^{3(n-1)}\)  

 
\(\Rightarrow C\)

Measurement, STD2 M7 2018 HSC 27a (Adapted)

Alex used a cloud storage service for one month.

The plan has a base monthly cost of $25. The service also charges 45 cents per GB uploaded and 12 cents per GB downloaded.

During the month, Alex uploaded 180 GB and downloaded 350 GB.

What was the total bill for the month?   (2 marks)

Show Answers Only

`$148.00`

Show Worked Solution

`text(Upload charge = 180 × 45c = $81.00)`

`text(Download charge = 350 × 12c = $42.00)`

`:.\ text(Total bill)` `= 25 + 81 + 42`
  `= $148.00`

Measurement, STD2 M7 2016 HSC 15 MC (Adapted)

Parking in a city car park is charged at the rate of $3.40 per 20 minutes, or part thereof.

What is the cost of parking for 1 hour and 8 minutes?

  1. $10.20
  2. $13.60
  3. $17.00
  4. $20.40
Show Answers Only

`=> B`

Show Worked Solution

`68 -: 20 = 3.4 \ \Rightarrow\ 4\ \text{blocks}`

♦ Mean mark 48%.
`:.\ \text(Cost)` `= 4 xx 3.40`
  `= $13.60`

`=> B`

Measurement, STD2 M7 2023 HSC 5 MC (Adapted)

Four mobile data packs are shown, each with the data included and its cost.

Which one represents the best value?

Show Answers Only

`C`

Show Worked Solution

`text{Calculate cost per GB of each option}`

`text{Option}\ A:\ $22.50 -: 10= $2.25\ text{per GB}`

`text{Option}\ B:\ $25.60 -: 12= $2.133\ldots\ \text{per GB}`

`text{Option}\ C:\ $27.30 -: 14= $1.95\ \text{per GB}`

`text{Option}\ D:\ $30.00 -: 15= $2.00\ \text{per GB}`

`=>C`

Algebra, STD2 A2 2014 HSC 22 MC (Adapted)

Maya’s hybrid uses fuel at the rate of 7.4 L per 100 km for highway driving and 10.2 L per 100 km for urban driving.

She drove a total of 720 km, of which 120 km were urban driving.

Approximately how much fuel did Maya’s car use on the journey?

  1. 55 L
  2. 57 L
  3. 60 L
  4. 65 L
Show Answers Only

`B`

Show Worked Solution

`text(Fuel used in urban driving)`

`= 120/100 xx 10.2\ text(L) = 12.24\ text(L)`

`text(Fuel used on highway)`

`= 600/100 xx 7.4\ text(L) = 44.4\ text(L)`

`:.\ text(Total Fuel)` `= 12.24 + 44.4`
  `= 56.64\ text(L) ~~ \approx 57\ text(L)`

`=> B`

Measurement, STD2 M7 2022 HSC 38 (Adapted)

A 5.0 L container is filled with a mixture of milk and coffee in the ratio 3:2.

After removing 1.0 L of the mixture, pure milk is added to refill the container.

What is the ratio of milk to coffee in the final mixture?   (3 marks)

Show Answers Only

`17:8`

Show Worked Solution

`text{Original mixture amounts}`

`text{Milk : Coffee = 3000 mL : 2000 mL}`

`text{Removing 1.0 L (1000 mL) removes}`

`text{→ }(3/5)×1000=600\text{ mL milk and }(2/5)×1000=400\text{ mL coffee}`
 

`text{Milk after removal}`

`=3000-600=2400\text{ mL}`
 

`text{Coffee after removal}`

`=2000-400=1600\text{ mL}`
 

`text{After refilling with 1000 mL milk:}`

`text{Milk }=2400+1000=3400\text{ mL}`

`text{Coffee }=1600\text{ mL}`
 

`:.\ \text{Final ratio } \text{Milk : Coffee}`

`=3400:1600`

`=17:8`


♦♦ Mean mark 34%.

Measurement, STD2 M7 2021 HSC 25 (Adapted)

A rectangular sportsground has been drawn to scale on a 1-cm grid as shown. The scale used is `1:2000`.
 

Johnny took 12 minutes to walk around the perimeter of this sportsground.

What was Johnny's average speed in kilometres per hour?   (4 marks)

Show Answers Only

`3.4 \ text{km/hr}`

Show Worked Solution
`text{Distance walked on map} \ = \ 2 times  (10 + 7) = 34 \ text{cm}`
Mean mark 54%.

 

`text{Actual distance}` ` =34 times 2000`  
  `= 68\ 000 \ text{cm}`  
  `=680 \ text{m}`  
  `= 0.68 \ text{km}`  

 
`12 \ text{minutes}\ = 12/60 = 0.2 \ text{hours}`
 

`text{Speed}` `= text{distance}/text{time}`  
  `= 0.68/0.2`  
  `= 3.4 \ text{km/hr}`  

Measurement, STD1 M5 2021 HSC 26 (Adapted)

The diagrams show two similar shapes. The dimensions of the small shape are enlarged by a scale factor of 1.5 to produce the large shape.
 

Calculate the area of the large shape.   (3 marks)

Show Answers Only

`94.5\ text(cm)^2`

Show Worked Solution

`text(Dimension of larger shape:)`

♦♦ Mean mark 32%.

`text(Width) = 6 xx 1.5 = 9\ text(cm)`

`text(Height) = 8 xx 1.5 = 12 \ text(cm)`

`text(Triangle height) = 2 xx 1.5 = 3\ text(cm)`

`:.\ text(Area)` `= 9 xx (12-3) + 1/2 xx 9 xx 3`
  `= 94.5\ text(cm)^2`

Measurement, STD2 M7 2012 HSC 27c (Adapted)

A topographic map has a scale of 1 : 250 000.

  1. Two lookouts are 3.6 cm apart on the map.

     

    What is the actual distance between the two lookouts, in kilometres?   (1 mark)

  2. Two towns are 42.5 km apart. How far apart are the two towns on the map, in centimetres?   (1 mark)

Show Answers Only

a.    `text(9 km)`

b.    `text(17 cm)`

Show Worked Solution
♦ Mean mark 41%
MARKER’S COMMENT: Better responses converted between centimetres, metres and kilometres systematically using the map scale (1 unit on the map represents 250 000 of the same unit in reality).
a.     `text{Actual distance (3.6 cm)}` `= 3.6 xx 250\ 000`
    `= 900\ 000\ text(cm)`
    `= 9\ 000\ text(m)`
    `= 9\ text(km)`

`:.\ text(The 2 lookouts are 9 km apart.)`

♦ Mean mark 45%

b.    `text(Towns are 42.5 km apart.)`

`text{From the scale, } 1\ \text{cm} = 250\ 000\ \text{cm} = 2\ 500\ \text{m} = 2.5\ \text{km}`

`=>\ \text{On the map, } 42.5\ \text{km} = 42.5/2.5 = 17\ \text{cm}`

`:.\ \text{Distance on the map is 17 cm.}`

Measurement, STD2 M7 2012 HSC 26f (Adapted)

The capture-recapture technique was used to estimate a population of turtles in 2015.

• 80 turtles were caught, tagged and released.

• Later, 200 turtles were caught at random.

• 40 of these 200 turtles had been tagged.

The estimated population of turtles in 2015 was 25% greater than the estimated population for 2010.

What was the estimated population for 2010?   (2 marks)

Show Answers Only

`text{320 turtles}`

Show Worked Solution

`text(Let population in 2015 =)\ P(2015)`

`text(Capture)`

`⇒ 80/{P(2015)}`

`text(Recapture)`

`=> 40/200 = 1/5`

`80/{P(2015)}` `=1/5`
`:. P(2015)` `= 80 xx 5 =400`

 

`text(We know)\ P(2015)\ text(is 25% greater than P(2010))`

`text{(100% + 25%)} xx P(2010)` `= 400`
`125% xxP(2010)` `=400`
`:. P(2010)` `=400/1.25`
  `= 320`

 

`:.\ text{2010 population estimate = 320 turtles}`

Measurement, STD2 M7 2020 HSC 23 (Adapted)

In a tropical punch, the ratio of passionfruit juice to guava juice to lime juice is 12 : 10 : 6 .

  1. How much lime juice is needed if the punch is to contain 2.4 litres of passionfruit juice?   (2 marks)

  2. The internal dimensions of a drink container, in the shape of a rectangular prism, are shown.
     

    To completely fill the container with the punch, how many litres of guava juice are required?   (3 marks)

Show Answers Only
  1. `1.2 text{L}`
  2. `10  text{L}`
Show Worked Solution
a.    `text(12 parts)` `= 2.4\ text(L)`
  `text(1 part)` `= 2.4/12`
    `= 0.2\ text(L)`

 

`:.\ text(6 parts)` `= 6 xx 0.2`  
  `= 1.2\ text(L)`  
b.     `text{Volume of container}` `= 40 xx 20 xx 35`
    `= 28\ 000 \ text{cm}^3`

`1 \ text{mL} \ to \ 1 \ text{cm}^3`

`⇒ \ 28\ 000 \ text{mL of punch}`

`therefore \ text{Guava juice required}` `= text{Guava parts}/text{Total parts} \ xx \ 28\ 000`
  `= \frac{10}{28} \ xx \ 28\ 000`
  `= 10\ 000 \ text{mL}`
  `= 10 \ text{L}`

Measurement, STD2 M7 2019 HSC 18 (Adapted)

Dana, Eden and Flynn are the first three players in the school basketball team. In a recent game, Dana scored 24 points, Eden scored 18 points and Flynn scored 30 points.

  1. What is the ratio of Dana's to Eden's to Flynn's points scored, in simplest form?   (2 marks)

  2. In this game, the ratio of the total number of points scored by Dana, Eden and Flynn to the total number of points scored by the whole team is `9:20`.
  3. How many points were scored by the whole team?   (2 marks)

Show Answers Only
  1. `4:3:5`
  2. `160\ text(points)`
Show Worked Solution
a.     `D:E:F` `= 24:18:30`
    `= 4:3:5`

 

b.     `text(Total points by)\ D,E,F` `= 24 + 18 + 30`
    `= 72`

 

`text(Let)\ P` `=\ text(team points)`
`P/72` `= 20/9`
`:. P` `= (20 xx 72)/9`
  `= 160\ text(points)`

Algebra, STD2 A4 2021 HSC 35 (Adapted)

A toy store releases a limited edition LEGO set for $20 each. At this price, 3000 LEGO sets are sold each week and the revenue is  `3000 xx 20=$60\ 000`.

The toy store considers increasing the price. For every dollar price increase, 15 fewer LEGO sets will be sold.

If the toy store charges `(20+x)` dollars for each LEGO set, a quadratic model for the revenue raised, `R`, from selling them is

`R=-15x^2+2700x+60\ 000`

 


 

  1. What price should be charged per LEGO set to maximise the revenue?   (2 marks)

  2. How many LEGO sets are sold when the revenue is maximised?   (2 marks)

  3. Find the value of the intercept of the parabola with the vertical axis.   (1 mark) 

Show Answers Only

a.   `$110`

b.    `1650`

c.   `$60\ 000`

Show Worked Solution

a.   `text{Highest revenue}\ (R_text{max})\ text(occurs halfway between)\ \ x= -20 and x=200.`

`text{Midpoint}\ =(-20 + 200)/2 = 90`

`:.\ text(Price of LEGO set for)\ R_text(max)`

`=90 + 20`

`=$110`
 

b.  `text{LEGO sets sold when}\ R_{max}`

`=3000-(90 xx 15)`

`=1650`
 

c.   `ytext(-intercept → find)\ R\ text(when)\ \ x=0:`

`R` `= -15(0)^2 + 2700(0) + 60\ 000`
  `=$60\ 000`

Algebra, STD2 A4 2017 HSC 28e (Adapted)

Sage brings 60 cartons of unpasteurised milk to the market each week. Each carton currently sells for $4 and at this price, all 60 cartons are sold each weekend.

Sage considers increasing the price to see if the total income can be increased.

It is assumed that for each $1 increase in price, 6 fewer cartons will be sold.

A graph showing the relationship between the increase in price per carton and the income is shown below.

 


 

  1. What price per carton should be charged to maximise the income?   (1 mark)

  2. What is the number of cartons sold when the income is maximised?   (1 mark)

  3. The cost of running the market stall is $40 plus $1.50 per carton sold.

    Calculate Sage's profit when the income earned from a day selling at the market is maximised.   (2 marks)

Show Answers Only

a.   `$7`

b.   `42`

c.   `$191`

Show Worked Solution

a.   `text(Graph is highest when increase = $3)`

`:.\ text(Carton price)\ = 4 + 3= $7`
 

b.   `text(Cartons sold)\ =60-(3 xx 6)=42`
  

c.   `text{Cost}\ = 42 xx 1.50 + 40 = $103`

`:.\ text(Profit when income is maximised)`

`= (42 xx 7)-103`

`= $191`

Measurement, STD2 M7 2008 HSC 20 MC (Adapted)

A point `P` lies between a lamp post, 1.5 metres high, and a building, 7 metres high. `P` is 2.5 metres away from the base of the post.

From `P`, the angles of elevation to the top of the lamp post and to the top of the building are equal.
 

What is the distance, `x`, from `P` to the top of the tower?

  1. 10.60
  2. 12.50
  3. 13.60
  4. 14.55
Show Answers Only

`C`

Show Worked Solution

`text(Triangles are similar)\ \ text{(equiangular)}`

`text(In smaller triangle:)`

`h^2` `= 1.5^2 + 2.5^2`
  `= 8.5`
`h` `= sqrt 8.5`
   
`x/sqrt8.5` `= 7/1.5 \ \ \ text{(sides of similar Δs in same ratio)}`
`x` `= (7 sqrt 8.5)/1.5`
  `= 13.60…`

 
`=>  C`