Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 2 · Simple familiar

Question 14

[6 marks] Graphics calculator
The question

A wildlife sanctuary monitors the female population of a species of quoll. The table shows the initial female population and the rates for each age group.

Age (years)0–11–22–3
Female population400250150
Birth (breeding) rate01.51.2
Survival rate0.40.60
a)

Determine the Leslie matrix for this population.

[1 mark] b)

Determine the total female population after one year.

[1 mark] c)

Use your result from Question 14a) to determine the number of females in each age group after five years.

[2 marks]

A ranger notes that the total female population increased during the first year, and claims that the total female population will continue to increase in the long term.

d)

Evaluate the reasonableness of the ranger’s claim.

[2 marks]
Watch the situation first

One good year, then a long decline

Watch the table become a Leslie matrix. Then watch the matrix move the quolls forward one year at a time, and follow the total for 40 years.

Three age groups, so the Leslie matrix is 3 × 3 Row 1 holds the birth rates: they make next year’s newborns The survival rates go just below the main diagonal Survival rate 0 for the 2–3 group: there is no 3–4 group Start with 400, 250 and 150 females: 800 in total One year on: 555, 160 and 150, so 865 in total Multiply by L once for each year, up to year 5 After five years: about 390, 165 and 108 females The totals go up, down, up, then down Run it for 40 years: from year 3 the total falls every year Long term, each total is about 0.950 × the one before Age group (years) 0–1 1–2 2–3 1.5 1.2 0.4 0.6 survival rate 0: no 3–4 group, so it is not entered L = 0–1 1–2 2–3 3 × 3: one row and column per age group 0 1.5 1.2 0.4 0 0 0 0.6 0 0 100 200 300 400 500 600 Females 0–1 1–2 2–3 Age group (years) 400 250 150 After 0 years: X0 Total: 800 555 160 150 After 1 year: X1 Total: 865 420 222 96 After 2 years: X2 Total: 738 448.2 168 133.2 After 3 years: X3 Total: 749.4 411.84 179.28 100.8 After 4 years: X4 Total: 691.92 389.88 164.736 107.568 After 5 years: X5 Total: 662.184 1.5 × 250 + 1.2 × 150 = 555 0.4 × 400 = 160 0.6 × 250 = 150 Total 800 → 865: up in year 1 Xn+1 = LXn 389.88 → 390164.736 → 165107.568 → 108 0 200 400 600 800 0 5 10 15 20 25 30 35 40 Years after the start, n Total female population 865 738 749.4 513.394 308.271 185.149 111.202 175.947 ÷ 185.149 ≈ 0.950 About 5% fewer each year, towards zero: the ranger’s claim is not reasonable

Blue arcs are births, which feed the 0–1 group. White arrows are survivals, which move females up one age group. Each year is one more multiplication by $\boldsymbol{L}$.

Now for the mathematics

Step the population forward, one year at a time

Each step multiplies by $\boldsymbol{L}$. Find the age groups after five years for part c). Then keep going until you can describe the long-term behaviour of the total for part d).

Females Age group (years) Years after the start, n Total females
Females in each age group after $n$ years Total for each year
Compared with the year before

This is the starting population, $\boldsymbol{X}_0$, with 800 females in total. Move forward a year.

After one year the total is 865, up from 800. That answers part b), and it is the growth the ranger saw. Keep going.

The total drops to 738. The growth in the first year did not last.

The total rises again, to 749.4. Early totals go up and down because they depend on the starting age mix, so they cannot settle the question.

This is $\boldsymbol{L}^4\boldsymbol{X}_0$, only four years. If you count the initial year as “year 1”, you stop here and get 412, 179 and 101. One more year.

This is part c). $\boldsymbol{L}^5\boldsymbol{X}_0$ rounds to 390 females aged 0–1, 165 aged 1–2 and 108 aged 2–3. Now keep going for part d).

This is $\boldsymbol{L}^6\boldsymbol{X}_0$, one year too many for part c). It gives 376, 156 and 99. For part d), keep going.

From year 3 onwards the total falls every year. Watch the ratio of successive totals settle as you go further.

The ratio has settled at about 0.950. In the long term the total falls by about 5% each year, so the ranger’s claim is not reasonable.

First find the age groups after five years. Then go far enough to see the long-term behaviour.

You have part c). Now go far enough to see the long-term behaviour of the total.

You have the long-term behaviour. Now find the age groups after exactly five years.

You have found both. One year of growth told you nothing about the long term.

Marking guide · 6 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · part a)1 mark

Put the birth rates in row 1 and the survival rates directly below the main diagonal.

$\boldsymbol{L}=\begin{bmatrix}0&1.5&1.2\\0.4&0&0\\0&0.6&0\end{bmatrix}$
Why?

Row 1 makes next year’s newborns from each group’s birth rate, and a female who survives moves up one age group, so 0.4 sits in row 2, column 1 and 0.6 in row 3, column 2. The survival rate of 0 for the 2–3 group is not entered, because the model has no 3–4 group.

Marker: correctly determines the Leslie matrix

Step 2 · part b)1 mark

Multiply the initial population matrix on the left by $\boldsymbol{L}$, then add the age groups.

$$\begin{aligned}\boldsymbol{L}\begin{bmatrix}400\\250\\150\end{bmatrix}&=\begin{bmatrix}555\\160\\150\end{bmatrix}\\555+160+150&=865\end{aligned}$$

865 females after one year
Calculator use

Define [A] as $\boldsymbol{L}$ ($3\times3$) and [B] as the initial population ($3\times1$), then enter [A]×[B]. TI-84 Plus CE: 2nd MATRIX → EDIT to define, 2nd MATRIX → NAMES to paste. Casio fx-CG50: Run-Matrix, F3 (MAT/VCT) to define, then Mat A×Mat B, EXE. Output: 555, 160, 150.

Marker: correctly determines the total female population after one year

Step 3 · part c)1 mark

Each year multiplies by $\boldsymbol{L}$ once, so five years from the initial population uses $\boldsymbol{L}^5$ (not $\boldsymbol{L}^4$ or $\boldsymbol{L}^6$).

$$\boldsymbol{L}^{5}\begin{bmatrix}400\\250\\150\end{bmatrix}\approx\begin{bmatrix}389.88\\164.74\\107.57\end{bmatrix}$$

Calculator use

Enter [A]^5×[B] (Casio fx-CG50: Mat A^5×Mat B). Output: 389.88, 164.736, 107.568.

Marker: correctly uses the fifth power of the Leslie matrix with the initial population matrix

Step 4 · part c)1 mark

Round each entry to the nearest whole female, only at this last step.

After five years: 390 aged 0–1, 165 aged 1–2 and 108 aged 2–3

Marker: correctly determines the number of females in each age group

Step 5 · part d)1 mark

The claim is about the long term, so compare the totals for large $n$.

$$\begin{aligned}&\text{Totals: }800,\ 865,\ 738,\ \ldots\\&\frac{\text{total after 31 years}}{\text{total after 30 years}}\approx\frac{175.95}{185.15}\approx0.950\end{aligned}$$

In the long term, the total female population decreases by approximately 5% each year.

Calculator use

Define [C] as the $1\times3$ row matrix $\begin{bmatrix}1&1&1\end{bmatrix}$, which adds the age groups. Enter [C]×[A]^30×[B] (Casio fx-CG50: Mat C×Mat A^30×Mat B). Output: 185.149, a $1\times1$ matrix holding the total. Change 30 for other values of $n$.

Marking

Totals for several large values of $n$ that show a continuing decrease (such as 308 after 20 years and 185 after 30 years) also gain this mark. Comparing the totals after one and five years only is not enough evidence of long-term behaviour.

Marker: correctly determines the long-term behaviour of the total female population

Step 6 · part d)1 mark

Compare the long-term behaviour with the ranger’s claim and give a clear verdict.

The claim is not reasonable. After the first year the total female population declines, and in the long term it falls by about 5% per year.
Why?

One year of growth (800 to 865) describes the start only. The ranger generalised from one year of data.

Marker: evaluates the reasonableness of the claim using a suitable comment

Putting it all together

Each phrase of the question gave you something

“the initial female population and the rates for each age group”
Three age groups, so $\boldsymbol{L}$ is $3\times3$, and the starting population is $\boldsymbol{X}_0=\begin{bmatrix}400\\250\\150\end{bmatrix}$.
“Birth (breeding) rate”
Row 1 of $\boldsymbol{L}$: $\begin{bmatrix}0&1.5&1.2\end{bmatrix}$, because births make next year’s 0–1 group.
“Survival rate”
0.4 and 0.6 go on the sub-diagonal. The 0 for the 2–3 group is not entered, because there is no 3–4 group.
“the total female population after one year”
$\boldsymbol{X}_1=\boldsymbol{L}\boldsymbol{X}_0=\begin{bmatrix}555\\160\\150\end{bmatrix}$, then add: $555+160+150=865$.
“Use your result from Question 14a) … after five years”
$\boldsymbol{X}_5=\boldsymbol{L}^5\boldsymbol{X}_0$, which rounds to 390, 165 and 108 females.
“increased during the first year”
800 to 865 describes the start only. The totals then go up and down before falling every year from year 3.
“will continue to increase in the long term”
You need totals for large $n$. The ratio of successive totals settles at about 0.950, so the total falls by about 5% each year.
“Evaluate the reasonableness of the ranger’s claim”
An explicit verdict: the total does not keep increasing, so the claim is not reasonable.
Common errors

If I had to guess where you might lose marks…

  • Transposing the matrix. If you put the survival rates in row 1 and the birth rates down column 1, you get $\begin{bmatrix}100\\690\\480\end{bmatrix}$. Multiplying the row $\begin{bmatrix}400&250&150\end{bmatrix}$ by $\boldsymbol{L}$ on the right gives $\begin{bmatrix}100&690&480\end{bmatrix}$. Either way the one-year total is 1270 instead of 865.
  • The wrong power in c). Using $\boldsymbol{L}^4$ (counting the initial year as “year 1”) gives 412, 179 and 101. Using $\boldsymbol{L}^6$ gives 376, 156 and 99. The correct values are 390, 165 and 108.
  • Generalising from too few years in d). Extending the first-year growth factor $865\div800=1.08125$ predicts $800\times1.08125^5\approx1182.28$ females after five years, but the model gives 662.184. Comparing only years 2 and 3 ($749.4\div738\approx1.01545>1$) also wrongly suggests growth. Neither earns the first mark in d).
  • Comparing only years 1 and 5. Using 865 against 662.184 to conclude “decreasing” is not enough. This is not sufficient evidence of long-term behaviour.
  • Rounding before multiplying in c). If you round each $\boldsymbol{X}_n$ to whole numbers year by year, you get 390, 165 and 107, so the 2–3 group comes out as 107 instead of 108. Round only the final $\boldsymbol{X}_5$.

Only mark this done when you could do it without help. Reading the solution does not count.

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.

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