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Calculus, 2ADV C4 EQ-Bank 25

Let \(g(x)\) be a function defined so that  \(g^{\prime}(x)=\dfrac{1}{2 x+3}\)  and  \(g(1)=0, \ x \in \left(-\frac{3}{2}, \infty\right).\)

Find \(g(x)\).   (3 marks)

Show Answers Only

\(g(x)=\dfrac{1}{2} \ln \left(\dfrac{2 x+3}{5}\right)\)

Show Worked Solution
\(g^{\prime}(x)\) \(=\dfrac{1}{2 x+3}\)
\(g(x)\) \(=\displaystyle \int \frac{1}{2 x+3}\,dx=\frac{1}{2} \int \frac{2}{2 x+3} \, dx=\frac{1}{2} \ln (2 x+3)+c\)

 

\(\text{Given} \ \ g(1)=0:\)

\(\dfrac{1}{2} \ln(2+3)+c=0 \ \ \Rightarrow \ \ c=-\dfrac{1}{2} \ln 5\)

\(g(x)=\dfrac{1}{2} \ln (2 x+3)-\dfrac{1}{2} \ln 5=\dfrac{1}{2} \ln \left(\dfrac{2 x+3}{5}\right)\)

Calculus, MET1 2025 VCAA 2

Let \(g(x)\) be a function defined for  \(x>-\dfrac{3}{2}\)  so that  \(g^{\prime}(x)=\dfrac{1}{2 x+3}\)  and  \(g(1)=0\).

Find \(g(x)\).   (2 marks)

Show Answers Only

\(g(x)=\dfrac{1}{2} \ln \left(\dfrac{2 x+3}{5}\right)\)

Show Worked Solution
\(g^{\prime}(x)\) \(=\dfrac{1}{2 x+3}\)
\(g(x)\) \(=\displaystyle \int \frac{1}{2 x+3}\,dx=\frac{1}{2} \int \frac{2}{2 x+3} \, dx=\frac{1}{2} \ln (2 x+3)+c\)

 

\(\text{Given} \ \ g(1)=0:\)

\(\dfrac{1}{2} \ln(2+3)+c=0 \ \ \Rightarrow \ \ c=-\dfrac{1}{2} \ln 5\)

\(g(x)=\dfrac{1}{2} \ln (2 x+3)-\dfrac{1}{2} \ln 5=\dfrac{1}{2} \ln \left(\dfrac{2 x+3}{5}\right)\)

Calculus, 2ADV C1 EQ-Bank 18

Let  \(f(x)=6 \sqrt{x+1}+5\).

Find the gradient of the tangent to  \(y=f(x)\) at  \(x=8\).   (2 marks)

Show Answers Only

\(\text{Gradient of tangent }=1\)

Show Worked Solution
\(f(x)\) \(=6 \sqrt{x+1}+5\)
\(f^{\prime}(x)\) \(=6 \times \dfrac{1}{2} \times(x+1)^{-\tfrac{1}{2}}=\dfrac{3}{\sqrt{x+1}}\)

\(\text{At} \ \ x=8:\)

\(f^{\prime}(x)=\dfrac{3}{\sqrt{8+1}}=1\)

\(\therefore \ \text{Gradient of tangent }=1\)

Calculus, MET1 2025 VCAA 1b

Let  \(f(x)=6 \sqrt{x+1}+5\).

Find the gradient of the tangent to  \(y=f(x)\) at  \(x=8\).   (2 marks)

Show Answers Only

\(\text{Gradient of tangent }=1\)

Show Worked Solution
\(f(x)\) \(=6 \sqrt{x+1}+5\)
\(f^{\prime}(x)\) \(=6 \times \dfrac{1}{2} \times(x+1)^{-\tfrac{1}{2}}=\dfrac{3}{\sqrt{x+1}}\)

\(\text{At} \ \ x=8:\)

\(f^{\prime}(x)=\dfrac{3}{\sqrt{8+1}}=1\)

\(\therefore \ \text{Gradient of tangent }=1\)

Statistics, STD2 S1 EQ-Bank 23

A boxplot for the sale prices of a sample of 203 homes is shown.
 

  1. Calculate the range of the sale price data in the boxplot.   (1 mark)

  2. Calculate the upper fence for any outliers within the sale price data of the boxplot.   (2 marks)

Show Answers Only

a.    \(\text{Range}=900\,000\)

b.    \(1\,350\,000\)

Show Worked Solution

a.    \(\text{Range}=1\,300\,000-400\,000=900\,000\)
 

b.    \(IQR=900\,000-600\,000=300\,000\)

\(Q_3=900\,000\)

\(\text{Upper Fence}\) \(=Q_3+1.5 \times IQR\)
  \(=900\,000+1.5 \times 300\,000\)
  \(=1\,350\,000\)

Matrices, GEN2 2025 VCAA 14

An early learning centre runs seven different activities during its 40-day holiday program.

The activities are cooking \((C)\), drama \((D)\), gardening \((G)\), lunch \((L)\), music \((M)\), reading \((R)\) and sport \((S)\).

The timetabled order of the activities for day one of the holiday program is shown below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \text{9 am} \ \ \rule[-1ex]{0pt}{0pt} & \text{10 am} \rule[-1ex]{0pt}{0pt} & \text{11 am} \rule[-1ex]{0pt}{0pt} & \text{12 pm} \rule[-1ex]{0pt}{0pt} & \ \ \text{1 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{2 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{3 pm} \ \ \\
\hline
\rule{0pt}{2.5ex} \textit{C} \rule[-1ex]{0pt}{0pt} & \textit{D} \rule[-1ex]{0pt}{0pt} & \textit{G} \rule[-1ex]{0pt}{0pt} & \textit{L} \rule[-1ex]{0pt}{0pt} & \textit{M} \rule[-1ex]{0pt}{0pt} & \textit{R} \rule[-1ex]{0pt}{0pt} & \textit{S}\\
\hline
\end{array}

The timetabled order of the activities for day one is also shown in matrix \(X\) below.

\begin{aligned} 
X = & \begin{bmatrix}
C  \\
D \\
G \\
L \\
M \\
R\\
S \\
\end{bmatrix}
\end{aligned}

Matrix \(P\), shown below, is a permutation matrix used to determine the timetabled order of activities from one day to the next.

\begin{aligned} 
P = & \begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{aligned}

A column matrix containing the timetabled order of activities on one day is multiplied by matrix \(P\) to determine the timetabled order of activities for the next day.

  1. State the activities that are always held at the same time on each day of the program.  (1 mark)
  2. Determine the timetabled order of the seven activities on day three of the program.  (1 mark)
  3. \(P^4\) is an identity matrix.
  4. Explain what this means for the timetabled order of the activities over the 40-day holiday program.  (1 mark)
Show Answers Only

a.    \(\text{Gardening, lunch, music.}\)

b.    \(\text{Drama, cooking, gardening, lunch, music, sport, reading.}\)

c.    \(\text{Order of activities rotate on a 4 day cycle.}\)

\(\text{Over 40 days, there will be 10 cycles of activities}\)

Show Worked Solution

a.    \(\text{Activities held af the same time:}\)

\(\Rightarrow \ \text{correspond to 1’s in leading diagonal}\)

\(\text{Gardening, lunch, music.}\)
 

b. 

\begin{aligned}X_2=P \times X=\begin{bmatrix}
R \\
S \\
G \\
L \\
M \\
D \\
C
\end{bmatrix}, \quad X_3=P \times X_2=\begin{bmatrix}
D \\
C \\
G \\
L \\
M \\
S \\
R
\end{bmatrix}
\end{aligned}

\(\text{Order on day 3:}\)

\(\text{Drama, cooking, gardening, lunch, music, sport, reading.}\)

♦♦ Mean mark (b) 27%.

c.    \(P^4 \ \Rightarrow \ \text{identity matrix}\)

\(\text{Order of activities rotate on a 4 day cycle.}\)

\(\text{Over 40 days, there will be 10 cycles of activities}\)

♦♦♦ Mean mark (c) 11%.

Matrices, GEN2 2025 VCAA 13

An early learning centre offers a 10-week activity program for four-year-old children. There are 27 children enrolled in the program. They participate in three different activities over the 10 weeks. The activities are cooking \((C)\), gardening \((G)\) and music \((M)\).

The transition matrix \(K\), shown below, gives the expected proportion of children in the program who will change activities from one week to the next.

\begin{aligned}
& \quad \quad \ \ \ \textit{this week} \\
& \quad  \ \ C \quad \quad G \ \quad \ \ M \\
K = & \begin{bmatrix}
0  & 0.76   & 0.36 \\
0.55  & 0  & 0.64 \\
0.45  & 0.24  & 0\\
\end{bmatrix}\begin{array}{l}
C\\
G\\
M
\end{array} \ \textit{next week} \\
\end{aligned}

  1. What do the values on the leading diagonal in matrix \(K\) indicate?   (1 mark)
  2. In Week 1 of the program, all 27 children participate in cooking \((C)\).
    1. Calculate the expected percentage of children who will participate in cooking in Week 10 of the program. Round your answer to one decimal place.   (1 mark)
    2. Find the expected number of children who will participate in gardening \((G)\) in Week 3 of the program and then move across to music \((M)\) in Week 4 of the program. Round your answer to the nearest whole number.   (2 marks)
Show Answers Only

a.    \(\text{No child does the same activity in two consecutive weeks.}\)

b.i.  \(37.4\%\)

b.ii. \(2 \ \text{children}\)

Show Worked Solution

a.    \(\text{Leading diagonal made up of 0’s:}\)

\(\text{No child does the same activity in two consecutive weeks.}\)
 

b.i.  \(\text{Find number of children cooking in week 10:}\)

\begin{aligned}K^9\begin{bmatrix}
27 \\
0 \\
0
\end{bmatrix}=\begin{bmatrix}
10.08 \ldots\\
9.96 \ldots \\
6.94 \ldots
\end{bmatrix}
\end{aligned}

\(\text{% children cooking}=\dfrac{10.08 \ldots}{27} \times 100=37.35 \ldots=37.4\% \ \text{(1 d.p.)}\)

♦♦♦ Mean mark (b.i) 9%.

b.ii.  \(\text{At the start of week 3:}\)

\begin{aligned}K^2\begin{bmatrix}
27 \\
0 \\
0
\end{bmatrix}=\begin{bmatrix}
15.66 \\
7.776 \\
3.564
\end{bmatrix}
\end{aligned}

\(\text{Children who move from \(G\) (week 3) to \(M\) (week 4)}\)

\(=7.776 \times 0.24=1.886 \ldots=2 \ \text{children}\)

♦♦♦ Mean mark (b.ii) 21%.

Financial Maths, STD2 F4 2025 GEN1 7

Declan is a filmmaker and content creator.

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

Interest is calculated monthly and Declan makes monthly repayments.

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

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

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

\begin{array}{|c|c|c|c|c|}
\hline
\hline \rule{0pt}{2.5ex}\quad \textbf{Payment} \quad & \quad\textbf{Payment} \quad & \quad\textbf{Interest} \quad& \textbf{Principal} & \quad\textbf{Balance}\quad\\
\textbf{number} & \textbf{(\$)}  \rule[-1ex]{0pt}{0pt}& \textbf{(\$)} & \quad\textbf{reduction (\$)} \quad& \textbf{(\$)}\\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 0.00 & 0.00 & 0.00 & 850\,000.00 \\
\hline \hline \rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & 15\,730.88 & 2975.00 & 12\,755.88 & 837\,244.12 \\
\hline \hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & 2930.35 & 12\,800.53 & 824\,443.59 \\
\hline \hline \rule{0pt}{2.5ex}3 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & & & \\
\hline
\end{array}

Show Answers Only

a.    \(\$ 850\,000\)

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

\(\text{a reduced balance.}\)

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

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

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

Show Worked Solution

a.    \(\$ 850\,000\)
 

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

\(\text{a reduced balance.}\)
 

c.    \(\text{Monthly interest}=\dfrac{4.2}{12}=0.35\%\)

\(\text{Calculating missing values in the table:}\)

\(\text{Interest}=\dfrac{0.35}{100} \times 824\, 443.59=\$ 2885.55\)

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

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

♦ Mean mark (c) 40%.

Matrices, GEN2 2025 VCAA 11

An early learning centre contains three rooms, Nursery \((N)\), Toddler \((T)\) and Pre-kinder \((P)\).

The Nursery and Toddler rooms each have capacity for eight children and the Pre-kinder room has capacity for 20 children, as shown in matrix \(C\) below.

\begin{align*}
C=\left[\begin{array}{c}
8 \\
8 \\
20
\end{array}\right] \begin{aligned}
& N \\
& T \\
& P
\end{aligned}
\end{align*}

Matrix \(E\) shows enrolment numbers for each room for one week, Monday to Friday.

\begin{aligned}
& \quad \ \ \ Mon \quad Tue\quad Wed \ \ \ Thu \ \ \ Fri \\
E&=\begin{bmatrix}
6 & \quad 8 & \quad 8 & \quad 8 & \quad 5 \\
7 & \quad 8 & \quad 7 & \quad 8 & \quad 6 \\
18 & \ \ \ 18 &\ \ \ 17 &\ \ \ 15 &\ \ \ 13
\end{bmatrix}\begin{array}{l}
N \\
T \\
P 
\end{array}\\
\end{aligned}

  1. State the order of matrix \(E\).   (1 mark)
  3. The following matrix multiplication has been completed to determine a new matrix, \(W\).
    1. \begin{align*}
      \begin{bmatrix}
      1 & 1 & 1
      \end{bmatrix} \times E=W
      \end{align*}
  4. What information does matrix \(W\) provide regarding enrolments at the early learning centre?   (1 mark)

  5. It has been decided that the capacity of the Nursery room will be increased by 25% and the capacity of the Toddler room will be increased by 50%. The capacity of the Pre-kinder room will be reduced by 10%.
  6. The new capacities for the three rooms \((C_{\text {new}})\) can be determined from the matrix product
    1. \(C_{\text {new }}=F C\)
  7. where \(F\) is a diagonal matrix.
  8. Write down the matrix \(F\).   (1 mark)   
Show Answers Only

a.    \(\text{Order of matrix} \ \ E=3 \times 5\)

b.    \(\text { Total enrolments for each day of the week.}\)

c.

\(F=\begin{bmatrix}1.25 & 0 & 0 \\ 0 & 1.5 & 0 \\ 0 & 0 & 0.9\end{bmatrix}\)

Show Worked Solution

a.    \(\text{Order of matrix} \ \ E=3 \times 5\)
 

b.    \(\text { Total enrolments for each day of the week.}\)
  

c.

\(F=\begin{bmatrix}1.25 & 0 & 0 \\ 0 & 1.5 & 0 \\ 0 & 0 & 0.9\end{bmatrix}\)

♦ Mean mark (c) 44%.

Financial Maths, GEN2 2025 VCAA 8

Declan depreciates the value of his lighting equipment using flat rate depreciation.

The graph below shows the value, in dollars, of the lighting equipment, \(V_n\), after \(n\) years.
 

  1. The value of the lighting equipment could be modelled by either a recurrence relation or a rule.
    1. Write a recurrence relation in terms of \(V_0, V_{n+1}\) and \(V_n\).   (1 mark)

    2. Write a rule for \(V_n\) in terms of \(n\).   (1 mark)

  2. What is the annual flat rate depreciation percentage applied to the lighting equipment?   (1 mark)
Show Answers Only

a.i.   \(V_0=40\,000 \quad V_{n+1}=V_n-8000\)

a.ii.  \(V_n=40\,000-8000 \times n\)

b.    \(\text{Depreciation}=20\%\)

Show Worked Solution

a.i.   \(\text{Recurrence relation:}\)

\(V_0=40\,000 \quad V_{n+1}=V_n-8000\)
 

a.ii.  \(V_n=40\,000-8000 \times n\)
 

b.    \(\text{Depreciation %}=\dfrac{8000}{40\,000} \times 100=20\%\)

♦ Mean mark (a.ii) 50%.

Financial Maths, GEN2 2025 VCAA 7

Declan is a filmmaker and content creator.

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

Interest is calculated monthly and Declan makes monthly repayments.

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

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

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

\begin{array}{|c|c|c|c|c|}
\hline
\hline \rule{0pt}{2.5ex}\quad \textbf{Payment} \quad & \quad\textbf{Payment} \quad & \quad\textbf{Interest} \quad& \textbf{Principal} & \quad\textbf{Balance}\quad\\
\textbf{number} & \textbf{(\$)}  \rule[-1ex]{0pt}{0pt}& \textbf{(\$)} & \quad\textbf{reduction (\$)} \quad& \textbf{(\$)}\\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 0.00 & 0.00 & 0.00 & 850\,000.00 \\
\hline \hline \rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & 15\,730.88 & 2975.00 & 12\,755.88 & 837\,244.12 \\
\hline \hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & 2930.35 & 12\,800.53 & 824\,443.59 \\
\hline \hline \rule{0pt}{2.5ex}3 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & & & \\
\hline
\end{array}

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

Show Answers Only

a.    \(\$ 850\,000\)

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

\(\text{a reduced balance.}\)

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

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

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

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

Show Worked Solution

a.    \(\$ 850\,000\)
 

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

\(\text{a reduced balance.}\)
 

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

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

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

♦ Mean mark (c) 40%.

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

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

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

♦♦♦ Mean mark (d) 21%.

Data Analysis, GEN2 2025 VCAA 6

The time series plot below shows the number of homes sold in a town each month over a four‑year period.

Month 1 is January 2016 and month 48 is December 2019.
 

  1. Excluding any possible outliers, identify two qualitative features of the time series plot.   (2 marks)

  2. The total number of sales in each of the four years is given in the table below.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \textbf{Year} \quad \rule[-1ex]{0pt}{0pt}& \textbf{Total number of sales}\\
\hline \rule{0pt}{2.5ex}2016 \rule[-1ex]{0pt}{0pt}& 361 \\
\hline \rule{0pt}{2.5ex}2017 \rule[-1ex]{0pt}{0pt}& 354 \\
\hline \rule{0pt}{2.5ex}2018 \rule[-1ex]{0pt}{0pt}& 358 \\
\hline \rule{0pt}{2.5ex}2019 \rule[-1ex]{0pt}{0pt}& 357 \\
\hline
\end{array}

  1. A seasonal index can be calculated for each month based on the four-year period.
  2. Calculate this seasonal index for September, the ninth month in the calendar year. Round your answer to three decimal places.   (2 marks)

Show Answers Only

a.    \(\text{Seasonality, irregular fluctuations.}\)

b.    \(\text{Seasonal index(Sep)} =0.587\)

Show Worked Solution

a.    \(\text{Qualitative features of time series plot:}\)

\(\text{Seasonality}\)

\(\text{Irregular fluctuations}\)
 

b.    \(\text{Sept avg}=\dfrac{15+20+20+15}{4}=17.5\)

\(\text {Monthly avg}=\dfrac{361+354+358+357}{48}=29.79 \ldots\)

\(\text{Seasonal index (Sep)}=\dfrac{17.5}{29.79 \ldots}=0.5874 \ldots=0.587\ \text{(3 d.p.)}\)

Data Analysis, GEN2 2025 VCAA 4

The scatterplot below shows the sale price of a home, in dollars, against the distance of the home from the city centre of Melbourne, in kilometres, distance from city centre.

The sample consists of three‑bedroom homes sold between 2016 and 2018
 

The equation of the least squares line for the data in the scatterplot is

sale price\(=1\,765\,353-35\,054 \times\)distance from city centre

The coefficient of determination is 0.0806

  1. Identify the explanatory variable in the least squares equation.   (1 mark)

  2. Calculate the value of the correlation coefficient \(r\). Round your answer to three decimal places.   (1 mark)

  3. Use the equation of the least squares line to predict the sale price for a three-bedroom home, located in the city centre of Melbourne, sold between 2016 and 2018.   (1 mark)

  4. Jocelyn wants to sell her three-bedroom home located two kilometres from the city centre of Melbourne.
  5. Would the predicted sale price be an example of interpolation or extrapolation?
  6. Briefly explain your answer.   (1 mark)

  7. Describe the linear association between sale price and distance from city centre in terms of its strength and direction. Answer in the table below.  (2 marks)

\begin{array}{|l|l|}
\hline \rule{0pt}{2.5ex}\text {strength} \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad \quad \\
\hline \rule{0pt}{2.5ex}\text {direction} \rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

  1. A residual plot associated with the least squares line is shown below.
  2. It is missing one point.
     

  1. The residual associated with the home that is furthest from the city centre of Melbourne is missing from the residual plot. The home is 15.5 km from the city centre and sold for $1 250 000.
    1. Show that the value of the missing residual is 27 984.   (1 mark)

    2. Plot the residual from part i by placing an \(\text{X}\) on the residual plot above.   (1 mark)

Show Answers Only

a.    \(\text{Distance from city centre}\)

b.   \(r=-0.284\)

c.    \(\text{sale price}=1\,765\,353\)

d.    \(\text{Extrapolation as 2 km lies outside the explanatory variable data range.}\)

e.    \(\text{Strength: weak. Direction: negative}\)

f.i.  \(\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016\)

\(\text{Actual sale price}=1\,250\,000\)

\(\text{Residual}=1\,250\,000-1\,222\,016=27\,984\)

f.ii.

       

Show Worked Solution

a.    \(\text{Distance from city centre}\)
 

b.   \(\text{Since slope is negative}\)

\(r=-\sqrt{0.0806}=-0.284\)

♦♦♦ Mean mark (b) 21%.

c.    \(\text{Find sale price when distance fran city centre}=0:\)

\(\text{sale price}=1\,765\,353\)
 

d.    \(\text{Extrapolation as 2 km lies outside the explanatory variable data range.}\)

\(\text{(Note: interpolation/extrapolation should be referenced to the}\)

\(\text{explanatory variable range).}\)

♦♦ Mean mark (d) 27%.

e.    \(\text{Strength: weak}\)

\(\text{Direction: negative}\)
 

f.i.  \(\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016\)

\(\text{Actual sale price}=1\,250\,000\)

\(\text{Residual}=1\,250\,000-1\,222\,016=27\,984\)

♦ Mean mark (f.i) 41%.
♦ Mean mark (f.ii) 45%

f.ii.

       

Data Analysis, GEN2 2025 VCAA 3

The sale prices for homes in another suburb are normally distributed with a mean of $1 400 000.

A home in this suburb that sold for $952 000 has a standardised score of  \(z=-1.60\)

Using the 68-95-99.7% rule, calculate the percentage of homes sold in this suburb with a sale price between $560 000 and $1 680 000.   (2 marks)

Show Answers Only

\(83.85 \%\)

Show Worked Solution

\(\bar{x}=1\,400\,000, z\text{-score}\  (952\,000)=-1.6\)

\(\text {Using} \ \ z=\dfrac{x-\bar{x}}{s}:\)

\(-1.6=\dfrac{952\,000-1\,400\,000}{s} \ \Rightarrow \ s=280\,000\)

\(z \text{-score }(560,000)\) \(=\dfrac{560\,000-1\,400\,000}{280\,000}=-3\)
\(z \text{-score }(1\,680\,000)\) \(=1\)

 
\(\therefore \% \ \text{sale prices}=\dfrac{99.7 \%}{2}+\dfrac{68.0 \%}{2}=83.85 \%\)

Mean mark 54%.

Data Analysis, GEN2 2025 VCAA 1

Table 1, below, shows the prices in dollars, price, for a sample of 20 homes sold in an inner Melbourne suburb during 2017.

The type of home sold is either an apartment or a house.

Table 1

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \ \textit{Price(\$)}\quad \ \rule[-1ex]{0pt}{0pt}& \textit{Type} \\
\hline \rule{0pt}{2.5ex}350\,000 \rule[-1ex]{0pt}{0pt}& \ \ \text{apartment}\ \ \\
\hline \rule{0pt}{2.5ex}490\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}500\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}620\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}720\,000 \rule[-1ex]{0pt}{0pt}\rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}830\,000 & \text{apartment}\\
\hline \rule{0pt}{2.5ex}875\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}995\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,100\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,520\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}800\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}840\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,010\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,263\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,398\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,460\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline
\end{array}

  1. Find the median, in dollars, of the variable price.   (1 mark)

  2. State whether the variable type is numerical, nominal or ordinal.   (1 mark)

    1. Complete the table below by finding the standard deviation, to the nearest whole number, for the sale price of apartments in the sample.   (1 mark)

    2. Table 2
      \begin{array}{|c|c|}
      \hline \rule{0pt}{2.5ex} \quad \ \ \textbf{Type of home} \quad \ \ & \quad \textbf{Standard deviation of} \quad \\
      & \rule[-1ex]{0pt}{0pt}\textbf{sale price (\$)}\\
      \hline \rule{0pt}{2.5ex}\text{house} \rule[-1ex]{0pt}{0pt}& 300\,911 \\
      \hline \rule{0pt}{2.5ex}\text{apartment} \rule[-1ex]{0pt}{0pt}& \\
      \hline
      \end{array}
    3. Using the information in Table 2, comment on the relative spread in the distribution of the sale prices of houses compared with apartments in this sample.   (1 mark)

  4. Table 3, below, shows the percentage of houses and apartments with prices in the given ranges. Some information is missing.
  5. Use the data from Table 1 to complete Table 3.
  6. Table 3 
Show Answers Only

a.    \(\text{Median}=920\,000\)

b.    \(\text{Variable is nominal}\)

c.i.  \(\text{Std deviation = \$346 466}\)

c.ii.  \(\text {Apartment sale prices have a higher spread (standard deviation) than the}\)

\(\text{spread of house sale prices.}\)

d.

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Median}=\dfrac{10^{\text{th}}+11^{\text{th}}}{2}=920\,000\)
 

b.    \(\text{Type is qualitative and cannot be ordered}\)

\(\Rightarrow \ \text{Variable is nominal}\)
 

c.i.  \(\text{By calculator,}\)

\(\text{Std deviation = \$346 466}\)
 

c.ii.  \(\text {Apartment sale prices have a higher spread (standard deviation) than the}\)

\(\text{spread of house sale prices.}\)

♦ Mean mark (c.ii) 47%.

d.

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Networks, STD2 N3 EQ-Bank 38

The precedence table below shows the 12 activities required to complete a project. The duration in days and immediate predecessors are shown.

\begin{array}{|c|c|c|}
\hline \rule{0pt}{2.5ex}\ \ \textbf{Activity} \ \ & \ \ \textbf{Duration} \ \ & \textbf{Immediate predecessors} \\
\hline \rule{0pt}{2.5ex}A \rule[-1ex]{0pt}{0pt}& 4 & - \\
\hline \rule{0pt}{2.5ex}B \rule[-1ex]{0pt}{0pt}& 6 & A \\
\hline \rule{0pt}{2.5ex}C \rule[-1ex]{0pt}{0pt}& 8 & A \\
\hline \rule{0pt}{2.5ex}D \rule[-1ex]{0pt}{0pt}& 3 & A \\
\hline \rule{0pt}{2.5ex}E \rule[-1ex]{0pt}{0pt}& 9 & B\\
\hline \rule{0pt}{2.5ex}F \rule[-1ex]{0pt}{0pt}& 6 & C \\
\hline \rule{0pt}{2.5ex}G \rule[-1ex]{0pt}{0pt}& 7 & B, D, F \\
\hline \rule{0pt}{2.5ex}H \rule[-1ex]{0pt}{0pt}& 12 & C \\
\hline \rule{0pt}{2.5ex}I \rule[-1ex]{0pt}{0pt}& 6 & G, H \\
\hline \rule{0pt}{2.5ex}J \rule[-1ex]{0pt}{0pt}& 4 & E, I \\
\hline \rule{0pt}{2.5ex}K \rule[-1ex]{0pt}{0pt}& 3 & G, H \\
\hline \rule{0pt}{2.5ex}L \rule[-1ex]{0pt}{0pt}& 9 & J \\
\hline
\end{array}

The project is to be completed in minimum time.

  1. Sketch the network, identifying each activity and its duration, including any dummy activities.   (3 marks)
  2. Determine the critical path of the network.   (1 mark)
Show Answers Only

a.    \(\text{Sketch the network:}\) 
 

b.    \(\text {Critical Path:}\ ACFGIJL\)

Show Worked Solution

a.    \(\text{Sketch the network:}\) 
 

 
b.   \(\text{Critical path only requires the forward scanning in the above network.}\)

\(\text {Critical Path:}\ ACFGIJL\)

Networks, GEN1 2025 VCAA 39 MC

Four contractors, \(\text{P, Q, R}\) and \(\text{S}\), have been hired to complete work at one of four sites.

Each contractor will be allocated to a different site.

The cost for each contractor to complete the works, in dollars ($), is shown in the table below
 

The Hungarian algorithm is used to determine the minimum cost to complete the works.

Martha, the project manager, completes two steps of the Hungarian algorithm, as follows.

  • First, she subtracts the minimum entry in each row from each element in that row to obtain a new table of values.
  • Then, using this new table, she subtracts the minimum entry in each column from each element in that column.

Which one of the following tables correctly displays the results after these two steps are completed?
 

Show Answers Only

\(B\)

Show Worked Solution

\(\text{Using two steps of the Hungarian algorithm:}\)

\(\text{Step 1}\)

\(\text{Subtract minimum entry in each row from each element in that row.}\)

\(\text{This produces table in option A (eliminate A).}\)
 

\(\text{Step 2}\)

\(\text{Using the resultant table from step 1:}\)

\(\text{Subtract the minimum entry in each column from each element in that column.}\)

\(\text{This produces table in opton B.}\)

\(\text{Following step 2 only produces option C (eliminate C).}\)

\(\text{Step 2 followed by step 1 produces option D (eliminate D).}\)

\(\Rightarrow B\)

Networks, GEN1 2025 VCAA 36-37 MC

The network below shows the flow of water, in litres per minute, in a system of pipes connecting the source to the sink.
 

Part 1

What is the capacity of Cut 1?

  1. 27
  2. 31
  3. 33
  4. 37

 
Part 2

What is the maximum flow of water, in litres per minute, from source to sink?

  1. 16
  2. 17
  3. 26
  4. 30
Show Answers Only

\(\text{Part 1:}\ A\)

\(\text{Part 2:}\ B\)

Show Worked Solution

\(\text{Part 1}\)

\(\text{Capacity (Cut 1)}=12+5+3+7=27\)

\(\Rightarrow A\)
 

\(\text{Part 2}\)

\(\text{Minimum Cut}=6+4+7=17\)

\(\Rightarrow B\)

Networks, STD2 N2 2025 GEN1 6 MC

Consider the following weighted graph.
 

The minimum spanning tree for this graph contains the edge with weight \(w\).

The length of this minimum spanning tree is

  1. \(37+w\)
  2. \(44+w\)
  3. \(48+w\)
  4. \(49+w\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Minimum spanning tree:}\)
 

\(5+6+4+w+2+4+5+3+6+5+4=44+w\)

\(\Rightarrow B\)

Matrices, GEN1 2025 VCAA 32 MC

Kyle \((K)\), Lian \((L)\), Maggie \((M)\), Neil \((N)\) and Ophelia \((O)\) took part in a round-robin chess tournament in which each person played each of the others once. In each game there was a winner and a loser.

The winner of the tournament was determined by \(T=D+D^2\) where \(D\) and \(D^2\) are, respectively, the one-step and two-step dominance matrices.

Some of the individual match results were not recorded.

An incomplete matrix \(D\) is shown below.

The ' 1 ' in row \(K\), column \(M\) indicates that Kyle defeated Maggie.
 

\begin{aligned}
& \quad \ \ \  \ K \ \ \ \ \ L \ \ \ \ \ M \ \ \ \ N \ \ \  O \\
D=\ & \begin{array}{l}
K \\
L \\
M \\
N \\
O
\end{array}\begin{bmatrix}
0 & \ldots & 1 & \ldots & 0 \\
\ldots & 0 & 0 & \ldots & 0 \\
0 & 1 & 0 & 1 & 1 \\
\ldots & \ldots & 0 & 0 & 0 \\
1 & 1 & 0 & 1 & 0
\end{bmatrix}\\
\end{aligned}

The following information is known.

  • Maggie and Ophelia each won three of their four games
  • Kyle won two of his four games.
  • Lian and Neil each won one of their four games.
  • Kyle defeated Neil.

Which one of the following is matrix \(T\)?

\begin{aligned}
& \quad \ \ \  K \ \ \  L \ \  \ M \ \ \ N \ \  O \\
\textbf{A.}\ \ \ & \begin{array}{l}
K \\
L \\
M \\
N \\
O
\end{array}\begin{bmatrix}
0 & 2 & 1 & 2 & 1 \\
1 & 0 & 0 & 1 & 0 \\
2 & 2 & 0 & 3 & 1 \\
1 & 1 & 1 & 0 & 0 \\
2 & 2 & 1 & 2 & 0
\end{bmatrix}\\
\end{aligned}		 \begin{aligned}
& \quad \ \ \  K \ \ \  L \ \  \ M \ \ \ N \ \  O \\
\quad \quad \quad \quad \textbf{B.}\ \ \ & \begin{array}{l}
K \\
L \\
M \\
N \\
O
\end{array}\begin{bmatrix}
0 & 2 & 1 & 2 & 1 \\
1 & 0 & 1 & 1 & 0 \\
2 & 3 & 0 & 2 & 1 \\
1 & 1 & 0 & 0 & 0 \\
2 & 2 & 1 & 2 & 0
\end{bmatrix}\\
\end{aligned}
\begin{aligned}
& \quad \ \ \  K \ \ \  L \ \  \ M \ \ \ N \ \  O \\
\textbf{C.}\ \ \ & \begin{array}{l}
K \\
L \\
M \\
N \\
O
\end{array}\begin{bmatrix}
0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 1 & 0
\end{bmatrix}\\
\end{aligned}		 \begin{aligned}
& \quad \ \ \  K \ \ \  L \ \  \ M \ \ \ N \ \  O \\
\quad \quad \quad \quad \textbf{D.}\ \ \ & \begin{array}{l}
K \\
L \\
M \\
N \\
O
\end{array}\begin{bmatrix}
0 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 1 \\
0 & 1 & 0 & 0 & 0 \\
1 & 1 & 0 & 1 & 0
\end{bmatrix}\\
\end{aligned}
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Complete matrix \(D\) using information given:}\)

\(\text{Kyle had 2 wins → Maggie and Neil.}\)

\(\text{Kyle loses to Lian and Ophelia.}\)

\(\text{Lian must lose to Maggie, Neil, and Ophelia}\)

\(\text{Neil must lose to Kyle, Maggie, and Ophelia.}\)
 

\begin{aligned}
& \quad \ \ \  K \ \ \ L \ \ \ M \ \ N \ \  O \\
D=\ & \begin{array}{l}
K \\
L \\
M \\
N \\
O
\end{array}\begin{bmatrix}
0 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 1 \\
0 & 1 & 0 & 0 & 0 \\
1 & 1 & 0 & 1 & 0\end{bmatrix}\\
\end{aligned}

\(D+D^2 \ \ \text{ produces matrix in option B.}\)

\(\Rightarrow B\)

Matrices, GEN1 2025 VCAA 31 MC

Consider the following four matrices.

\begin{align*}
A=\begin{bmatrix}
4 & 6 \\
-2 & -3
\end{bmatrix}\quad B=\begin{bmatrix}
4 \\
2 \\
8
\end{bmatrix} \quad C=\begin{bmatrix}
5 & 0 \\
2 & 5 \\
3 & 1
\end{bmatrix}\quad D=\begin{bmatrix}
3 & -2 & 5
\end{bmatrix}
\end{align*}

Which one of the following computations is defined?

  1. \(D(A+B)\)
  2. \(C^T B D\)
  3. \(C+B D\)
  4. \((D A)^T+D\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Option A: \( A+B\) is not defined (eliminate)}\)

\(\text{Option B: \(C^{\text{T}}BD\) is defined (see below)}\)

\(C^{\text{T}}=(2 \times 3)\ \text{matrix. }\  C^{\text{T}} B=(2 \times 3)(3 \times 1)=(2 \times 1)\)

\(C^{\text{T}} B D=(2 \times 1)(1 \times 3)=(2 \times 3)\)

\(\text{Option C: \(B D=(3 \times 3)\) matrix, \(C+B D\) is not defined (eliminate)}\)

\(\text{Option D: \(DA=(1 \times 3)(2 \times 2) \Rightarrow\) not defined (eliminate)}\)

\(\Rightarrow B\)

Matrices, GEN1 2025 VCAA 30 MC

\(F\) is a \(4 \times 4\) matrix.

The element in row \(i\) and column \(j\) of \(F\) is \(f_{i j}\).

The elements are determined by the rule \(f_{i j}=i^2-j\).

How many of the elements in \(F\) will be negative?

  1. 2
  2. 3
  3. 5
  4. 16
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Constructing matrix} \ F:\)

\begin{aligned}\begin{bmatrix}
1^2-1 & 1^2-2 & 1^2-3 & 1^2-4 \\
2^2-1 & 2^2-2 & 2^2-3 & 2^2-4 \\
3^2-1 & 3^2-2 & 3^2-3 & 3^2-4 \\
4^2-1 & 4^2-2 & 4^2-3 & 4^2-4
\end{bmatrix}=\begin{bmatrix}
0 & -1 & -2 & -3 \\
3 & 2 & 1 & 0 \\
8 & 7 & 6 & 5 \\
15 & 14 & 13 & 12
\end{bmatrix}
\end{aligned}

\(\Rightarrow B\)

Financial Maths, GEN1 2025 VCAA 22 MC

Rita opens a savings account with an initial deposit of $4000.

The account earns interest compounding weekly. After the interest is added each week, Rita deposits an additional $50 into the account.

Assume there are exactly 52 weeks in one year.

The annual interest rate, compounding weekly, that is required to achieve a balance of $14 000 after three years is closest to

  1. 8.4%
  2. 14.2%
  3. 14.6%
  4. 17.2%
Show Answers Only

\(A\)

Show Worked Solution

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

\(N\) \(=3 \times 52=156\)
\(I(\%)\) \(=\boldsymbol{8.372…}\)
\(PV\) \(=-4000\)
\(PMT\) \(=-50\)
\(FV\) \(=14\,000\)
\(PY\) \(=CY=52\)

  

\(\Rightarrow A\)

Financial Maths, GEN1 2025 VCAA 21 MC

Donald invests $250 000 into a perpetuity at an interest rate of 5% per annum.

Donald receives a payment at the end of each year.

When will the sum of all annual payments to Donald first exceed $250 000?

  1. at the end of year 13
  2. at the end of year 17
  3. at the end of year 21
  4. at the end of year 25
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Initial payment}=\dfrac{5.0}{100} \times 250\, 000=\$ 12\,500\)

\(\text{Let}\ n=\text{number of payments}\)

\(\text{Find} \ n \ \text {when:}\)

\(n \times 12\,500=250\,000 \ \Rightarrow \ n=\dfrac{250\,000}{12\,500}=20 \ \text {payments}\)

\(\therefore \ \text {At end of 21 years, total payments}>\$250\,000\)

\(\Rightarrow C\)

Financial Maths, GEN1 2025 VCAA 20 MC

Steve owns gardening equipment that had an initial value of $12 000.

He will use the unit cost method to depreciate the value of this equipment per hour used.

Steve uses the gardening equipment for 960 hours per year.

The value of the gardening equipment is $7680 after two years.

By what amount does the gardening equipment depreciate per hour used?

  1. $2.25
  2. $4.00
  3. $4.17
  4. $4.50
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Value loss }=12\,000-7680=$4320\)

\(\text{Hours of use }=2 \times 960=1920\ \text{hours} \)

\(\text{Depreciation per hour }=\dfrac{4320}{1920}=\$2.25\)

\(\Rightarrow A\)

Mean mark 55%.

Data Analysis, GEN1 2025 VCAA 11-12 MC

The table below shows the life expectancy in years, life, and the number of doctors per 1000 people, doctors, for a sample of 10 countries in 2024. A scatterplot displaying the data is also shown.
 

 

Part 1

A logarithmic (base 10) transformation was applied to the variable life.

With \(\log _{10}(\textit{life})\) as the response variable, the equation of the least squares line fitted to the transformed data is closest to

  1. \(\log _{10}(life)=1.57+0.0123 \times doctors\) 
  2. \(\log _{10}(life)=1.63+0.0326 \times doctors\)
  3. \(\log _{10}(life)=1.79+0.0383 \times doctors\)
  4. \(\log _{10}(life)=1.85+0.0403 \times doctors\)

 
Part 2

A squared transformation was applied to the variable \(doctors\).

The equation of the least squares line fitted to this transformed data is of the form  \(\textit{life}=a+b \times(\textit{doctors})^2\).

Using this equation, the predicted \(life\), in years, for a country with two \(doctors\) per 1000 people is closest to

  1. 73.6
  2. 74.0
  3. 74.5
  4. 74.9
Show Answers Only

\(\text{Part 1:}\ C\)

\(\text{Part 2:}\ C\)

Show Worked Solution

\(\text{Transform data in table (use for parts 1 and 2):}\)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{doctors} \ \ \rule[-1ex]{0pt}{0pt} & \log_{10}(\textit{life}) & \textit{doctors}^2 & \quad \textit{life} \quad \\
\hline
\rule{0pt}{2.5ex} 0.21 \rule[-1ex]{0pt}{0pt} & 1.810 &0.044 &64.6\\
\hline
\rule{0pt}{2.5ex} 0.42 \rule[-1ex]{0pt}{0pt} & 1.809 &0.176 &64.4\\
\hline
\rule{0pt}{2.5ex} 0.59 \rule[-1ex]{0pt}{0pt} & 1.811 &0.348 &64.7\\
\hline
\rule{0pt}{2.5ex} 0.79 \rule[-1ex]{0pt}{0pt} & 1.814 &0.624 &65.1\\
\hline
\rule{0pt}{2.5ex} 1.12 \rule[-1ex]{0pt}{0pt} & 1.819 &1.254 &65.9\\
\hline
\rule{0pt}{2.5ex} 1.42 \rule[-1ex]{0pt}{0pt} & 1.824 &2.016 &66.7\\
\hline
\rule{0pt}{2.5ex} 1.72 \rule[-1ex]{0pt}{0pt} & 1.836 &2.958 &68.6\\
\hline
\rule{0pt}{2.5ex} 1.77 \rule[-1ex]{0pt}{0pt} & 1.851 &3.133 &70.9\\
\hline
\rule{0pt}{2.5ex} 1.94 \rule[-1ex]{0pt}{0pt} & 1.868 &3.764 &73.8\\
\hline
\rule{0pt}{2.5ex}2.05 \rule[-1ex]{0pt}{0pt} & 1.898 &4.203 &79.1\\
\hline
\end{array}

 
\(\text{Part 1}\)

\(\text{Calculate LSRL (by CAS):}\)

\(\log _{10}(\textit{life})=1.7879 \ldots+0.03829 \ldots \times \textit{doctors}\)

\(\Rightarrow C\)
 

\(\text{Part 2}\)

\(\text{Calculate LSRL (by CAS):}\)

\(\textit{life}=63.1165+2.8419 \times \textit{doctors}^2\)
 

\(\text{Find}\ \textit{life}\ \text{when} \ \textit{doctors}=2:\)

\(\textit{life}=63.1165+2.8419 \times 2^2=74.48\ldots \ \text{years}\)

\(\Rightarrow C\)

Data Analysis, GEN1 2025 VCAA 9-10 MC

In the scatterplot below, the games won are plotted against the goals against for each of the 12 Australian A-League men’s teams after 27 games of the 2022–2023 season. A least squares line has been fitted to the data.
 

Part 1

The equation for the least squares line is closest to

  1. games won\(=16.8-0.178 \times\)goals against
  2. games won\(=14.3-0.193 \times\)goals against
  3. games won\(=12.5-0.165 \times\)goals against
  4. games won\(=17.4+0.173 \times\)goals against

 
Part 2

The correlation coefficient between games won and goals against is  \(r=-0.466\)

Based on the correlation coefficient, it can be concluded that

  1. fewer goals scored against is associated with a smaller number of wins.
  2. fewer goals scored against causes a smaller number of wins.
  3. more goals scored against causes a smaller number of wins.
  4. more goals scored against is associated with a smaller number of wins.
Show Answers Only

\(\text{Part 1:}\ A\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\)

\(\text{By calculator, input 10 data points.}\)

\(\text{Gradient (by CAS)}=-0.1776 \ldots\)

\(\Rightarrow A\)
 

\(\text{Part 2}\)

\(\text{Eliminate B and C (correlation is not causation).}\)

\(\text{More goals scored against is associated with fewer wins.}\)

\(\Rightarrow D\)

Data Analysis, GEN1 2025 VCAA 1-2 MC

At a fruit shop, customers can buy avocados in bags.

The bag size ranges from one to six avocados per bag.

The histogram below shows the number of customers who bought each bag size on a particular day.
 

Part 1

The median bag size bought by customers on the day was

  1. 2
  2. 3
  3. 3.5
  4. 4


Part 2

The total number of avocados sold in bags was

  1. 39
  2. 134
  3. 136
  4. 138
Show Answers Only

\(\text{Part 1:}\ D\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\)

\(\text{Sum of columns}\ = 4+11+3+9+5+7=39\)

\(\text{Median = 20th value which is in the 4th column (bag size 4).}\)

\(\Rightarrow D\)
 

\(\text{Part 2}\)

\(\text{Total number of avocados}\)

\(=1 \times 4 + 2 \times 11+3 \times 3+4 \times 9+5 \times 5+6 \times 7 =138\)

\(\Rightarrow D\)

Calculus, 2ADV C1 EQ-Bank 12

A block of ice is melting. The mass \(M\) kilograms of the ice block remaining at time \(t\) hours after it begins to melt is given by  \(M(t)=50(12-3t)^2, 0 \leqslant t \leqslant 4\).

  1. Find the rate of change of the ice block's mass at any time \(t\).   (1 mark)

  2. How long does it take for the ice block to completely melt?   (1 mark)

  3. At what time is the ice melting at a rate of 2100 kilograms per hour?   (2 marks)

Show Answers Only

a.    \(\dfrac{dM}{dt}=-300(12-3t)\)

b.    \(4\ \text{hours}\)

c.    \(t=\dfrac{5}{3}\ \text{hours}\)

Show Worked Solution

a.    \(M(t)=50(12-3t)^2\)

\(\dfrac{dM}{dt}=50 \times 2 \times (-3) \times(12-3t)=-300(12-3t)\)
 

b.    \(\text{Find}\ t\ \text{when}\ \ M(t)=0:\)

\(50(12-3t)^2=0 \ \Rightarrow \ t=4\)

\(\text{Ice block is completely melted at} \ \ t=4 \ \ \text {hours}\)
 

c.    \(\text{Find}\ t \ \text{when}\ \ \dfrac{d M}{d t}=-2100:\)

\(-300(12-3t)\) \(=-2100\)
\(12-3t\) \(=7\)
\(-3t\) \(=-5\)
\(t\) \(=\dfrac{5}{3}\ \text{hours}\)

Calculus, 2ADV C1 EQ-Bank 13

A cylindrical water tank is being filled. The volume \(V\) litres of water in the tank at time \(t\) minutes after filling begins is given by  \(V(t)=500 \sqrt{(2 t+1)}, t \geqslant 0\).

  1. At what rate is water entering the tank at any time \(t\) ?   (1 mark)

  2. Find the rate at which the tank is being filled when \(t=12\) minutes.   (1 mark)

  3. At what time is the water flowing into the tank at a rate of 125 litres per minute?   (2 marks)

Show Answers Only

a.    \(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{2t+1}} \)

b.    \(100\ \text{L/min} \)

c.    \(\dfrac{15}{2}\ \text{mins}\)

Show Worked Solution

a.    \(V(t)=500(2t+1)^{\frac{1}{2}}\)

\(\dfrac{dV}{dt}=\dfrac{1}{2} \times 2 \times 500(2t+1)^{-\frac{1}{2}} = \dfrac{500}{\sqrt{2t+1}} \)
 

b.    \(\text{When}\ \ t=12:\)

\(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{25}} = 100\ \text{L/min} \)
 

c.    \(\text{Find}\ t\ \text{when}\ \dfrac{dV}{dt}=125:\)

\(125\) \(=\dfrac{500}{\sqrt{2t+1}}\)  
\(\sqrt{2t+1}\) \(=4\)  
\(2t+1\) \(=16\)  
\(t\) \(=\dfrac{15}{2}\ \text{mins}\)  

Networks, STD2 EQ-Bank 21

Five friends \((A, B, C, D, E)\) live in walking distance to each other's houses. The weighted network diagram below shows the walking time between the houses.
 

The table below summarises the same information. Fill in the missing table values in the shaded cells.   (3 marks)

Show Answers Only

Show Worked Solution

\(\text{Note the symmetry about the table diagonal.}\)

Financial Maths, STD2 EQ-Bank 29

Priya works as a sales representative and earns a base wage plus commission. A spreadsheet is used to calculate her weekly earnings.

Total weekly earnings = Base wage + Total sales \(\times\) Commission rate

A spreadsheet showing Priya's weekly earnings is shown.
  

 

  1. Write down the formula used in cell B9, using appropriate grid references.   (1 mark)

  2. In the following week, Priya earned total weekly earnings of $1437.50. Her base wage and commission rate remained unchanged. Calculate Priya's total sales for that week.   (2 marks)

  3. Priya's employer increases her commission rate to 4.2% but keeps her base wage the same. If Priya makes $22 000 in sales, calculate her new total weekly earnings.   (2 marks)

Show Answers Only

a.   \(=\text{B4}+\text{B5}^*\text{B6}/100\)

b.    \($22\, 500\)

c.    \($1574\)

Show Worked Solution

a.   \(\text{Total weekly earnings} = \text{Base wage}+\text{Total sales} \times \text{Commission rate}\)

\(\therefore\ \text{Formula:}\ =\text{B4}+\text{B5}^*\text{B6}/100\)
 

b.   \(\text{Total weekly earnings} = \text{Base wage}+\text{Total sales} \times \text{Commission rate}\)

\(\text{Let the Total sales}=S\)

\(1437.50\) \(=650+S\times \dfrac{3.5}{100}\)
\(787.50\) \(=S\times \dfrac{3.5}{100}\)
\(S\) \(=\dfrac{787.50\times 100}{3.5}=$22\,500\)

 
\(\text{The amount of Priya’s total sales was \$22 500.}\)
 

c.    \(\text{Base wage}=650\)

\(\text{Total sales}=$22\,000\)

\(\text{New commission rate}=4.2\%\)

\(\text{Total weekly earnings}\) \(=650+22\,000\times \dfrac{4.2}{100}\)
  \(=650+924=$1574\)

Financial Maths, STD2 EQ-Bank 20 MC

A company uses a spreadsheet to calculate employees' monthly salaries from their weekly salaries. The spreadsheet is shown below.
 

Which formula has been used in cell B7 to calculate the monthly salary?
 
  1. \(=\text{B4}^*4\)
  2. \(=\text{B4}^*52/12\)
  3. \(=\text{B4}^*12\)
  4. \(=\text{B4}/52^*12\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Monthly salary} = \dfrac{\text{weekly salary} \times 52}{12}\)

\(\therefore\ \text{Formula:}\ =\text{B4}^*52/12\)

\(\Rightarrow B\)

Financial Maths, STD2 EQ-Bank 21

Liam works in a factory assembling electronic components and is paid on a piecework basis. A spreadsheet is used to calculate his weekly earnings.

Weekly earnings = Number of units completed \(\times\) Rate per unit

A spreadsheet showing Liam's earnings for one week is shown.
  

  1. Write down the formula used in cell B8, using appropriate grid references.   (1 mark)

  2. In the following week, Liam earned $2036.25. The rate per unit remained at $3.75. Calculate the number of units Liam completed that week.   (2 marks)

Show Answers Only

a.   \(=\text{B4}^*\text{B5}\)

b.    \(\text{Number of units completed}=543\)

Show Worked Solution

a.   \(\text{Weekly earnings} = \text{Number of units completed} \times \text{Rate per unit}\)

\(\text{Formula:}\ =\text{B4}^*\text{B5}\)
 

b.   \(\text{Weekly earnings} = \text{Number of units completed} \times \text{Rate per unit}\)

\(\text{Let the Weekly earnings}=E\)

\(2036.25\) \(=E\times 3.75 \)
\(E\) \(=\dfrac{2036.25}{3.75}=543\)

Financial Maths, STD2 EQ-Bank 26

Maria works as a freelance writer and earns income through royalties. A spreadsheet is used to calculate her monthly royalty earnings.

Royalties = Number of books sold \( \times \) Royalty rate per book

A spreadsheet showing Maria's royalty earnings is shown.
  

  1. Write down the formula used in cell B8, using appropriate grid references.   (1 mark)

  2. In April, Maria's total royalty earnings were $8960. Her royalty rate remained at $2.85 per book. How many books did Maria sell in April?   (2 marks)

Show Answers Only

a.   \(=\text{B4}^*\text{B5}\)

b.    \(\text{Number of books}=3144\)

Show Worked Solution

a.   \(\text{Total royalty earnings} = \text{Number of books sold} \times \text{Royalty rate per book}\)

\(\text{Formula:}\ =\text{B4}^*\text{B5}\)
 

b.   \(\text{Total royalty earnings} = \text{Number of books sold} \times \text{Royalty rate per book}\)

\(\text{Let the Number of books sold}=N\)

\(8960.40\) \(=N\times 2.85 \)
\(N\) \(=\dfrac{8960.40}{2.85}=3144\)

Measurement, STD2 M1 2025 HSC 26*

A toy has a curved surface on the top which has been shaded as shown. The toy has a uniform cross-section and a rectangular base.
 

  1. Use two applications of the trapezoidal rule to find an approximate area of the cross-section of the toy.   (2 marks)

  2. The total surface area of the plastic toy is 1300 cm².
  3. What is the approximate area of the curved surface?   (2 marks)

Show Answers Only

a.   \(34.68 \ \text{cm}^2\)

b.   \(582.64 \ \text{cm}^2\)

Show Worked Solution

a.   \(\text{Solution 1}\)

\(A\) \(=\dfrac{5.1}{2}(6.0+3.8) + \dfrac{5.1}{2}(3.8+0) \)  
  \(=34.68\ \text{cm}^2\)  

 
\(\text{Solution 2}\)

\(\begin{array}{|c|c|c|c|}
\hline\rule{0pt}{2.5ex} \quad x \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 5.1 \quad & \quad 10.2 \quad\\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 6 & 3.8 & 0 \\
\hline
\end{array}\)

\(A\) \(\approx \dfrac{h}{2}\left(y_0+2y_1+y_2\right)\)
  \(\approx \dfrac{5.1}{2}\left(6+2 \times 3.8+0\right)\)
  \(\approx 34.68 \ \text{cm}^2\)

 

b.    \(\text{Toy has 5 sides.}\)

\(\text{Area of base}=10.2 \times 40=408 \ \text{cm}^2\)

\(\text{Area of rectangle}=6.0 \times 40=240 \ \text{cm}^2\)

\(\text{Approximated areas}=2 \times 34.68=69.36 \ \text{cm}^2\)

\(\therefore \ \text{Area of curved surface}\) \(=1300-(408+240+69.36)=582.64 \ \text{cm}^2\)
♦ Mean mark (b) 48%.

Algebra, STD2 EQ-Bank 28

Fried's formula for determining the medicine dosage for children aged 1 - 2 years is:

\(\text{Dosage}=\dfrac{\text{Age of infant (months)}\  \times \ \text{adult dose}}{150}\)

The spreadsheet below is used as a calculator for determining an infant's medicine dosage according to Fried's formula.
 

Amber, a 12 month old child, is being discharged from hospital with two medications. Medicine A has an adult dosage of 325 milligrams and she is to take 26 milligrams. She must also take 9.6 milligrams of Medicine B but the equivalent adult dosage has been left off the spreadsheet.

  1. By using appropriate grid references, write down a formula that could appear in cell B10.   (2 marks)

  2. Calculate the equivalent adult dosage for Medicine B (cell B7) using the information in the spreadsheet.    (2 marks)

Show Answers Only

a.    \(=\text{B5}^*\text{B6}/150\)

b.    \(120 \ \text{milligrams}\)

Show Worked Solution

a.     \(=\text{B5}^*\text{B6}/150\)
 

b.    \(\text{Let} \ A= \text{Adult dose}\)

\(\text{Using given formula:}\)

\(9.6\) \(=\dfrac{12 \times A}{150}\)
\(A\) \(=\dfrac{9.6 \times 150}{12}=120 \ \text{milligrams}\)

Financial Maths, STD1 EQ-Bank 24

A company uses the spreadsheet below to calculate the fortnightly pay, after tax, of its employees.
 

  1. Write down the formula that was used in cell D5, using appropriate grid references.   (1 mark)
  2. Hence, calculate Kim's fortnightly pay.   (2 marks)
Show Answers Only

a.  \(=\text{B6}-\text{C6}\)

b.    \(\$3140.85\)

Show Worked Solution

a.    \(\text{Formula for Kim’s Salary after tax:}\)

\(=\text{B5}-\text{C5}\)
 

b.    \(\text{Kim’s salary after tax}\ = \$81\,662\)

\(\text{Kim’s fortnightly pay}\ = \dfrac{81\,662}{26}=\$3140.85\)

Financial Maths, STD2 EQ-Bank 17

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}\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}

A company's spreadsheet was created that calculates its employees' after tax fortnightly pay, based on the table and excluding the Medicare levy.

  1. Write down the formula that was used in cell D6, using appropriate grid references.   (1 mark)
  2. Write down the formula that was used in cell C5, using appropriate grid references.   (1 mark)
  3. Hence, calculate Greg's fortnightly pay.   (2 marks)
Show Answers Only

a.  \(=\text{B6}-\text{C6}\)

b.  \(=4288+(\text{B5}-45000)^*0.30 \)

c.    \(\$3140.85\)

Show Worked Solution

a.    \(\text{Formula for Ian’s Salary after tax:}\)

\(=\text{B6}-\text{C6}\)
 

b.    \(\text{Formula for Greg’s estimated tax:}\)

\(=4288+(\text{B5}-45000)^*0.30 \)
 

c.    \(\text{Greg’s estimated tax}\ =4288+(103\,500-45\,000) \times 0.30=\$21\,838\)

\(\text{Greg’s salary after tax}\ = 103\,500-21\,838=\$81\,662\)

\(\text{Greg’s fornightly pay}\ = \dfrac{81\,662}{26}=\$3140.85\)

Algebra, STD2 EQ-Bank 26

Fried's formula for determining the medicine dosage for children aged 1 - 2 years is:

\(\text{Dosage}=\dfrac{\text{Age of infant (months)}\  \times \ \text{adult dose}}{150}\)

The spreadsheet below is used as a calculator for determining an infant's medicine dosage according to Fried's formula.
 

  1. By using appropriate grid references, write down a formula that could appear in cell B9.   (2 marks)

  2. Another infant requiring the same medicine has been recommended a dosage of 2 millilitres. What is the age of the infant?   (2 marks)

Show Answers Only

a.    \(=\text{B5}^*\text{B6}/150\)

b.    \(15 \ \text{months}\)

Show Worked Solution

a.     \(=\text{B5}^*\text{B6}/150\)
 

b.    \(\text{Let} \ n= \text{age of infant}\)

\(\text{Using given formula:}\)

\(2\) \(=\dfrac{n \times 20}{150}\)
\(n\) \(=\dfrac{2 \times 150}{20}=15 \ \text{months}\)

Financial Maths, STD2 EQ-Bank 32

Nigel's weekly wages are calculated using the partially completed spreadsheet below.
 

  1. Calculate the total wages Nigel earned on Wednesday.   (2 marks)

  2. Determine the time Nigel finished work on Saturday, given he earned $196.35 on the day.   (2 marks)

Show Answers Only

a.    \(\text{Total wages}=119.00+35.70=\$ 154.70\)

b.    \(14:30\)

Show Worked Solution

a.    \(\text{Wednesday wages:}\)

\(\text{Regular hours:} \ 5 \times 23.80=\$ 119.00\)

\(\text{Time-and-a-half:} \ 1 \times 1.5 \times 23.80=\$35.70\)

\(\text{Total wages}=119.00+35.70=\$ 154.70\)
 

b.    \(\text{Let \(h=\) total hours worked:}\)

\(\text{Since Saturday wages are time-and-a-half rate:}\)

\(h \times 1.5 \times 23.80\) \(=196.35\)
\(h\) \(=\dfrac{196.35}{1.5 \times 23.80}=5.5\ \text{hours}\)

 

\(\text{Nigel’s shift started at 09:00 and lasted 5.5 hours.}\)

\(\therefore\ \text{Nigel finished work at 14:30.}\)

Financial Maths, STD2 EQ-Bank 33

Trust Us Realty has three salespeople, Ralph, Ritchie, and Fonzi.

Their June monthly wages include a base wage and commission earned, which is modelled in the spreadsheet below.

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

  2. Calculate Fonzi's total pay for the month of June.   (2 marks)

  3. Ralph's total pay for June is $5850. Determine Ralph's total sales for the month.   (2 marks)

Show Answers Only

a.    \(=0.01^* \text{B9}\)

b.    \(\$ 20\,200\)

c.    \(\$ 385\,000\)

Show Worked Solution

a.    \(=0.01^* \text{B9}\)
 

b.    \(\text{Fonzi’s Commission:}\)

\(\text{Sales}\ \$0-\$500\,000=0.01 \times 500\,000=\$ 5000\)

\(\text{Sales over} \ \$500\, 000=(2\,150\,000-500\,000) \times 0.008=\$13\,200\)

\(\text{Total June wages}=5000+13\,200+2000=\$ 20\,200\)
 

c.    \(\text{Ralph’s sales commission}\ =5850-2000=\$3850\)

\(\text {Since Ralph earned} \ \$3850 \ \text{in commission:}\)

\(\text{Sales} \times 0.01\) \(=3850\)
\(\text{Sales}\) \(=\dfrac{3850}{0.01}=\$ 385\,000\)

Algebra, STD2 EQ-Bank 31

Clark's formula for determining the medicine dosage for children is:

\(\text{Dosage}=\dfrac{\text{weight in kilograms}\  \times \ \text{adult dosage}}{70}\)

The spreadsheet below is used as a calculator for determining a child's medicine dosage according to Clark's formula.
 

  1. By using appropriate grid references, write down a formula that could appear in cell E5.   (2 marks)

  2. Another child requiring the same medicine has been recommended a dosage of 62.5 milligrams. How much does the child weigh?   (2 marks)

Show Answers Only

a.    \(=\text{B6}^*\text{B5}/70\)

b.    \(17.5 \ \text{kilograms}\)

Show Worked Solution

a.    \(=\text{B6}^*\text{B5}/70\)
 

b.    \(\text{Let} \ w= \text{weight of the child.}\)

\(\text{Using given formula:}\)

\(62.5\) \(=\dfrac{w \times 250}{70}\)
\(w\) \(=\dfrac{62.5 \times 70}{250}=17.5 \ \text{kilograms}\)

Financial Maths, STD2 EQ-Bank 28

The spreadsheet shows a casual employee's partially completed timesheet.
 

Calculate Jane's total earnings for the week.   (2 marks)

Show Answers Only

\(\text{Earnings:}\)

\(\text{Sunday:}\ 4 \times 20.70 \times 2 = $165.60\)

\(\text{Thursday:}\ 3 \times 20.70 = $62.10\)

\(\text{Friday:}\ (2 \times 20.70) + (2 \times 20.70 \times 1.5) = $103.50\)

\(\text{Saturday:}\ 4 \times 20.70 \times 1.5 = $124.20\)
 

\(\text{Total earnings}\ =165.60+62.10+103.50+124.20=$455.40\)

Show Worked Solution

\(\text{Earnings:}\)

\(\text{Sunday:}\ 4 \times 20.70 \times 2 = $165.60\)

\(\text{Thursday:}\ 3 \times 20.70 = $62.10\)

\(\text{Friday:}\ (2 \times 20.70) + (2 \times 20.70 \times 1.5) = $103.50\)

\(\text{Saturday:}\ 4 \times 20.70 \times 1.5 = $124.20\)
 

\(\text{Total earnings}\ =165.60+62.10+103.50+124.20=$455.40\)

Algebra, STD2 EQ-Bank 24

QuickPrint Copy Centre charges for printing services using the formula below.

Total cost = Setup fee + Cost per page \( \times \) Number of pages

A spreadsheet used to calculate the total cost is shown.

  1. Write down the formula used in cell E3, using appropriate grid references.   (1 mark)

  2. QuickPrint increases their cost per page to \$0.42, but keeps the setup fee unchanged. Aisha needs to print 120 pages.

    How much more will Aisha pay compared to the original pricing shown in the spreadsheet?   (2 marks)

Show Answers Only

a.    \(=\text{B3}+ \text{B4} \ ^* \ \text{B5}\)

b.    \($14.40\)

Show Worked Solution

a.   \(=\text{B3}+ \text{B4} \ ^* \ \text{B5}\)
 

b.   \(\text{Original cost from spreadsheet}:\ $44.50\)

\(\text{At increased rate}:\)

\(\text{Total cost}\ =8.50+0.42\times 120=$58.90\)

\(\text{Additional amount}\ =58.90-44.50=$14.40\)

Algebra, STD2 EQ-Bank 23

Green Thumb Landscaping charges for their lawn mowing service based on the size of the lawn.

They use the formula below to calculate the cost of each service.

Total cost = Call-out fee + Cost per square metre \( \times \) Area of lawn

The spreadsheet they provide to their clients is included below.

  1. Write down the formula that has been used in cell E4, using appropriate grid references.   (1 mark)

  2. Miguel has a lawn with a different area. The call-out fee and cost per square metre remain the same. When Miguel's lawn area is entered into the spreadsheet, the total cost shown in cell E4 becomes \$153.00.

    What is the area of Miguel's lawn?   (2 marks)

Show Answers Only

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)

b.    \(\text{60 m}^2\)

Show Worked Solution

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)
 

b.     \(153\) \(=45+1.8A\)
  \(108\) \(=1.8A\)
  \(A\) \(=\dfrac{108}{1.8}=60\)

 
\(\therefore\ \text{The area of Miguel’s lawn is 60 m}^2.\)

Algebra, STD2 EQ-Bank 25

Two quantities,  \(M\)  and  \(t\),  have a relationship such that  \(M\)  varies directly with  \(t\)  and   \(M=2.4\)   when  \(t=8\).

Find the value of  \(t\)  when  \(M=5.1\).   (2 marks)

Show Answers Only

\(t=17\)

Show Worked Solution

\(M \propto t \ \ \Rightarrow \ \ M=kt\)

\(\text{Given} \ \ M=2.4 \ \ \text{when} \ \ t=8:\)

\(2.4=8k \ \ \Rightarrow \ \ k=\dfrac{2.4}{8}=0.3\)

\(\text{When} \ \ M=5.1:\)

\(5.1\) \(=0.3 \times t\)  
\(t\) \(=\dfrac{5.1}{0.3}=17\)  

Algebra, STD2 EQ-Bank 26

\(F\) varies directly with  \(m\)  and  \(F=14\)  when  \(m=3.5\).

Find the value of  \(m\)  when  \(F=50\).   (2 marks)

Show Answers Only

\(m=12.5\)

Show Worked Solution

\(F \propto m \ \ \Rightarrow \ \ F=km\)

\(\text{Given} \ \ F=14 \ \ \text{when} \ \ m=3.5:\)

\(14=3.5k \ \ \Rightarrow \ \ k=4\)

\(\text{When} \ \ F=50:\)

\(50\) \(=4 \times m\)  
\(m\) \(=\dfrac{50}{4} = 12.5\)  

Algebra, STD2 EQ-Bank 15 MC

\(d\) varies directly with \(h\) and it is known that  \(d=1.2\)  when  \(h=5\).

The value of \(h\) when  \(d=48\)  is

  1. \(11.52\)
  2. \(57.6\)
  3. \(200\)
  4. \(240\)
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\(C\)

Show Worked Solution

\(d \propto h \ \ \Rightarrow \ \ d=k \times h\)

\(\text{Find \(k\) given \(\ d=1.2\)  when \(\ h=5\):}\)

\(1.2\) \(=5 \times k\)  
\(k\) \(=\dfrac{1.2}{5}=0.24\)  

 
\(\text{Find} \ h \ \text{when} \ \ d=48:\)

\(48\) \(=0.24 \times h\)  
\(h\) \(=\dfrac{48}{0.24}=200\)  

 
\(\Rightarrow C\)

Algebra, STD2 EQ-Bank 11 MC

\(Q\) varies directly with \(r\) and  \(Q=2\) when  \(r=16\).

The value of \(r\) when  \(Q=13\)  is

  1. \(1.625\)
  2. \(26\)
  3. \(76\)
  4. \(104\)
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\(D\)

Show Worked Solution

\(Q \propto r \ \Rightarrow \ Q=kr\)

\(\text{Find \(k\) given \(\ Q=2\) given \(\ r=16\):}\)

\(2=16 \times k \ \Rightarrow \ k=\dfrac{1}{8}\)

\(\text{Find \(r\) when \(\ Q=13:\)}\)

\(13\) \(=\dfrac{1}{8} \times r\)  
\(r\) \(=13 \times 8=104\)  

 
\(\Rightarrow D\)

Algebra, STD2 EQ-Bank 27

Dial-A-Lift Luxury Transport offers ride services in the local area of Maitland.

They currently use the formula below to calculate the cost of each fare.

   Total fare \(=\) Booking fee \(+\) Cost per kilometre \(\times\) Number of kilometres travelled

A spreadsheet used to calculate the total fare is shown.
 

  1. By using appropriate grid references, write down a formula that could have been used in cell E4.   (2 marks)

  2. A trip with a different number of kilometres travelled is entered, but the booking fee and cost per kilometre remain unchanged. As a result, the value in cell E4 changes to $47.50.
  3. Calculate the value entered in cell B6.   (2 marks)

Show Answers Only

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)

b.    \(\text{Cell B6 value = 13.}\)

Show Worked Solution

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)
 

b.     \(47.5\) \(=15+2.5 n\)
  \(32.5\) \(=2.5n\)
  \(n\) \(=\dfrac{32.5}{2.5}=13\)

 
\(\text{Cell B6 value = 13.}\)

Algebra, STD2 EQ-Bank 25

The following formula can be used to calculate the recommended dosage of a medicine for a child.

   Recommended dosage \(=\) base dosage \(+\) adjustment factor \(\times\) weight of child,

where the recommended dosage and base dosage are in milligrams and the weight of the child is in kilograms.

A spreadsheet used to calculate the recommended dosage is shown.
 

  1. By using appropriate grid references, write down a formula that could have been used in cell E5.   (2 marks)

  2. The weight of a different child is entered, but the base dosage and adjustment factor remain unchanged. As a result, the value in cell E5 changes to 70.
  3. Calculate the value entered in cell B7.   (2 marks)

Show Answers Only

a.    \(=\text{B5}+ \text{B6} \ ^* \ \text{B7}\)

b.    \(\text{Cell B7 value = 25.}\)

Show Worked Solution

a.    \(=\text{B5}+ \text{B6} \ ^* \ \text{B7}\)
 

b.    \(\text{Let \(w\) = weight of the child}\)

\(70\) \(=50+0.8 w\)
\(20\) \(=0.8 w\)
\(w\) \(=\dfrac{20}{0.8}=25\)

 
\(\text{Cell B7 value = 25.}\)

Algebra, STD2 EQ-Bank 24

The area of a semicircle is given by  \(A=\dfrac{1}{2}\pi r^2\)  where \(r\) is the radius of the semicircle.

If the area of a semicircle is 250 cm², find the radius, to 1 decimal place.   (3 marks)

Show Answers Only

\(12.6\ \text{cm}\)

Show Worked Solution

\(A=\dfrac{1}{2}\pi r^2\ \ \Rightarrow \ \ r^2=\dfrac{2A}{\pi}\)

   
\(\text{When}\ \ A = 250:\)

\(r^2\) \(=\dfrac{2\times 250}{\pi}=\dfrac{500}{\pi}=159.154…\)
\( r\) \(=\sqrt{159.154…}=12.615…=12.6\ \text{cm (to 1 d.p.)}\)

Algebra, STD2 EQ-Bank 20 MC

If  \(w=4y^2-5\), what is the value of \(y\) when  \(w=43\)?

  1. \(3\)
  2. \(-3\)
  3. \(\sqrt{3}\)
  4. \(-2\sqrt{3}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{When}\ w=43:\)

\(w\) \(=4y^2-5\)
\(43\) \(=4y^2-5\)
\(4y^2\) \(=48\)
\(y^2\) \(=\dfrac{48}{4}=12\)
\(y\) \(=\pm 2\sqrt{3}\)

 
\(y=-2\sqrt{3}\ \text{(only possibility given options)}\)

\(\Rightarrow D\)

Combinatorics, EXT1 A1 EQ-Bank 19

Find the term independent of \(x\) in the expansion of  \(\left(2 x^3+\dfrac{1}{x^4}\right)^7\).   (2 marks)

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\(560\)

Show Worked Solution

\(T_k=\ \text {General term of} \ \ \left(2 x^3+\dfrac{1}{x^4}\right)^7\)

\(T_k\) \(=\displaystyle \binom{7}{k}\left(2 x^3\right)^{7-k} \cdot\left(x^{-4}\right)^k\)
  \(=\displaystyle\binom{7}{k} \cdot 2^{7-k} \cdot x^{3(7-k)} \cdot x^{-4 k}\)
  \(=\displaystyle\binom{7}{k} \cdot 2^{7-k} \cdot x^{21-7 k}\)

 

\(\text{Independent term occurs when:}\)

\(x^{21-7 k}=x^0 \ \Rightarrow \ k=3\)

\(\therefore \text{Independent term}=\displaystyle \binom{7}{3} \cdot 2^4=560\)

Polynomials, EXT1 EQ-Bank 25

The polynomial  \(R(x)=2 x^4+a x^3+b x^2+c x+d\)  has a double zero at  \(x=1\), a zero at  \(x=-3\), and passes through the point \((0,-12)\).

Find the integer values of \(a, b, c, d\) and the fourth zero of the polynomial.   (4 marks)

Show Answers Only

\(a=-2, \ b=-14, \ c=26, \ d=-12\)

\(\text{Fourth zero:} \ \ x=2\)

Show Worked Solution

\(R(x)=2 x^4+a x^3+b x^2+c x+d\)

\(\text{Since leading coefficient is 2 with a double zero at 1 and a zero at }-3:\)

\(R(x)=2(x-1)^2(x+3)(x-k) \ \ \text{where} \ k \ \text{is the fourth zero.}\)

\(\text{The polynomial passes through}\ (0,-12):\)

\(R(0)=2(0-1)^2(0+3)(0-k)=-12\ \ \Rightarrow\ \ k=2\)
 

\(\text{Expanding}\ R(x):\)

\(R(x)\) \(=2(x-1)^2(x+3)(x-2)\)  
  \(=2\left(x^2-2 x+1\right)(x+3)(x-2) \)  
  \(=2(x^3+3 x^2-2 x^2-6 x+x+3)(x-2) \)  
  \(=2(x^3+x^2-5 x+3)(x-2) \)  
  \(=2(x^4+x^3-5 x^2+3 x-2 x^3-2 x^2+10 x-6) \)  
  \(=2 x^4-2 x^3-14 x^2+26 x-12\)  

 

\(\text{Equating coefficients:}\)

\(a=-2, \ b=-14, \ c=26, \ d=-12\)

\(\text{Fourth zero:} \ \ x=2\)

Trigonometry, 2ADV EQ-Bank 5 MC

A tower \(B T\) has height \(h\) metres.

From point \(A\), the angle of elevation to the top of the tower is 26° as shown.
 

Which of the following is the correct expression for the length of \(A B\) ?

  1. \(h\, \tan 26^{\circ}\)
  2. \(h\, \cot 26^{\circ}\)
  3. \(h\, \sin 26^{\circ}\)
  4. \(h\, \operatorname{cosec}\, 26^{\circ}\)
Show Answers Only

\(B\)

Show Worked Solution
\(\tan 26^{\circ}\) \(=\dfrac{h}{AB}\)  
\(AB\) \(=\dfrac{h}{\tan 26^{\circ}}\)  
\(AB\) \(=h\,\cot 26^{\circ}\)  

 
\(\Rightarrow B\)