VCE General Mathematics 2025 Exam 2

a. [1 mark]

With 20 values the median is the mean of the 10th and 11th values in order. Sorting all 20 prices, the 10th and 11th are both $920 000, so the median is $920 000.

b. [1 mark]

Type names a category with no natural order, so it is nominal.

c. i. [1 mark]

Enter the 10 apartment prices as a list and use one-variable statistics: $s_x \approx 346\,466.29$, so the standard deviation is $346 466.

ii. [1 mark]

The house prices have the smaller standard deviation ($300 911 against $346 466), so house prices are less spread out than apartment prices.

d. [1 mark]

Count each type (10 homes each, so each home is 10%):

From the report. Part c.i (69%) needed careful data entry. In c.ii (47%) the question said to use Table 2, so answers based on the range or IQR were not appropriate; the comparison had to rest on the standard deviations.

a. [1 mark]. Range $=$ maximum $-$ minimum $= 1\,300\,000 - 400\,000 =$ $900 000.

b. [1 mark]. $\text{IQR} = 900\,000 - 600\,000 = 300\,000$, so

The upper fence is $1 350 000.

[2 marks]. First find the standard deviation from the $z$-score:

Now locate the two prices: $1\,400\,000 - 3 \times 280\,000 = 560\,000$ is $z = -3$, and $1\,400\,000 + 280\,000 = 1\,680\,000$ is $z = 1$.

Between $z = -3$ and the mean is half of 99.7%, which is 49.85%; between the mean and $z = 1$ is half of 68%, which is 34%. So the percentage is

From the report. Most students found the standard deviation, but only 47% earned both marks. The report stresses that 83.85% must not be rounded to 84%: the exam instructions say to round only when told to.

a. [1 mark]

The variable on the right-hand side of the equation is the explanatory variable: distance from city centre.

b. [1 mark]

$r = \pm\sqrt{0.0806} = \pm 0.2839\ldots$ The slope is negative, so $r$ is negative: $r = -0.284$.

c. [1 mark]

In the city centre, distance $= 0$, so the predicted sale price is the intercept, $1 765 353.

d. [1 mark]

Extrapolation, because a distance of 2 km lies outside the range of the explanatory variable (distance) in the data, which starts at about 5 km.

e. [2 marks]

Strength: weak ($|r| \approx 0.28$). Direction: negative (price tends to fall as distance increases).

f. i. [1 mark]

Predicted value $= 1\,765\,353 - 35\,054 \times 15.5 = 1\,222\,016$. Then

ii. [1 mark]. Place an X at distance $= 15.5$ km and residual $= 27\,984$, which is just above the zero line (about one-seventh of the way from 0 to 200 000), at the right-hand edge of the plot.

From the report. Part b was the weakest on the paper (21%): a large proportion gave $+0.284$, missing that the negative slope makes $r$ negative. Part d (27%) needed the reason to be about the explanatory variable, distance; saying the sale price was outside the data range was not appropriate. In f.i (41%) all working leading to the given value had to be shown, and in f.ii (45%) the point had to be placed precisely against the scale.

a. [2 marks]. Enter days as the explanatory ($x$) list and sale price as the response ($y$) list, then run linear regression: $a = 1\,050\,360.6\ldots$ and $b = -8050.18\ldots$ To four significant figures,

b. [1 mark]. $r^2 = (-0.866)^2 \approx 0.75$, so 75% of the variation in sale price can be explained by the variation in days on the market.

From the report. Only 28% scored both marks in part a. Significant figures remained a problem, and many students misread calculator output in exponent form (such as 1.05E6).

a. [2 marks]. Seasonality (the same pattern of highs and lows repeats each year) and irregular fluctuations. There is no increasing or decreasing trend.

b. [2 marks]. Read the September values from the plot: 15 (2016), 20 (2017), 20 (2018) and 15 (2019). Divide each by that year's monthly average (total $\div 12$):

The seasonal index is the mean of these:

The report also accepted a pooled method: average September value $= 17.5$, overall monthly average $= \frac{361 + 354 + 358 + 357}{48} = 29.79\ldots$, giving $17.5 \div 29.79\ldots \approx 0.587$.

From the report. Part a asks for the formal features named in the study design; the report notes irregular fluctuations are present in every time series. Part b was difficult (12% full marks).

a. [1 mark]

The opening balance: $850 000.

b. [1 mark]

Interest is charged on the balance owing, and the balance has reduced after payment 1, so less interest accrues.

c. [1 mark]

Monthly rate $= \dfrac{4.2}{12}\% = 0.35\%$.

• Interest $= 0.0035 \times 824\,443.59 = 2885.55$
• Principal reduction $= 15\,730.88 - 2885.55 = 12\,845.33$
• Balance $= 824\,443.59 - 12\,845.33 = 811\,598.26$

d. [1 mark]. Finance solver: $I\% = 4.2$, $PV = 850\,000$, $PMT = -15\,730.88$, $FV = 0$, $P/Y = C/Y = 12$, solve for $N$: $N = 59.9999\ldots \approx 60$. The 60th payment is the slightly smaller final payment, so 59 payments of $15 730.88 come before it.

From the report. Part c (40%) had to use the table values; using a finance solver gave a slightly different balance to the nearest cent, which was not accepted. Part d was answered correctly by 41%.

a. i. [1 mark]. The value falls by $\frac{40\,000 - 8000}{4} = 8000$ each year:

ii. [1 mark]. $V_n = 40\,000 - 8000n$.

b. [1 mark]. $\dfrac{8000}{40\,000} \times 100\% =$ 20% of the purchase price per year.

From the report. In a.ii only half the students scored the mark: many repeated the recurrence relation instead of giving a rule in terms of $n$.

a. [1 mark]. If the balance does not change, each repayment exactly covers the week's interest (an interest-only loan):

b. i. [1 mark]. Finance solver: $N = 52$, $PV = 50\,000$, $PMT = -75$, $FV = -49\,565.34$, $P/Y = C/Y = 52$, solve for $I\%$: $I\% = 6.96$. The rate is 6.96% per annum.

ii. [1 mark]. $R = 1 + \dfrac{6.96}{100 \times 52} = 1.001338\ldots \approx$ 1.0013.

From the report. All three parts were weak (37%, 40% and 30%).

[2 marks]. First find the quarterly payment for the whole annuity (it runs out after 10 years). Finance solver: $N = 40$, $I\% = 6.4$, $PV = -650\,000$, $FV = 0$, $P/Y = C/Y = 4$: $PMT = 22\,126.27$.

The balance at the halfway point is the future value after 20 quarters: $N = 20$, same other entries, gives $FV = 376\,159.43$.

(The multiplier checks out: $1 + \frac{6.4}{400} = 1.016$.)

From the report. Only 11% scored both marks. Some students found the payment correctly but entered it as positive, when a payment received reduces the annuity balance and must be subtracted.

a. [1 mark]. 3 rows (rooms) and 5 columns (days): $3 \times 5$.

b. [1 mark]. The row matrix of ones adds the three entries in each column, so $W$ is a $1 \times 5$ matrix giving the total number of enrolments at the centre on each day of the week:

c. [1 mark]. Each capacity is multiplied by its own growth factor, which sits on the leading diagonal:

From the report. In part b (57%) many answers suggested the rows had been summed rather than the columns. Part c (44%) showed a poor understanding of what a diagonal matrix is.

a. [1 mark]. Work backwards with the inverse: $S_{2023} = M^{-1}S_{2024}$.

Check: $0.25 \times 16 = 4$, $0.625 \times 16 + 0.25 \times 20 = 15$, $0.625 \times 20 + 0.1 \times 25 = 15$, and $0.125 \times 16 + 0.125 \times 20 + 0.9 \times 25 = 27$.

b. [1 mark]. Apply the new rule twice:

Children enrolled are those in N, T and P (not L): $15.25 + 15.9375 + 19.11875 = 50.30\ldots$, so 50 children.

From the report. Part b was answered correctly by only 13%. Many gave 115 by including the children who had left the centre.

a. [1 mark]. The leading diagonal entries are all 0, so no child does the same activity in two consecutive weeks.

b. i. [1 mark]. Week 1 is $S_1 = \begin{bmatrix} 27 & 0 & 0 \end{bmatrix}^T$, and Week 10 is nine transitions later:

ii. [2 marks]. Week 3 is two transitions after Week 1:

So 7.776 children are expected to do gardening in Week 3. The proportion moving from G to M is 0.24 (row M, column G):

From the report. Part b.i was answered correctly by only 9%. Round only at the final step; the expected number of children should not have been rounded to 10 before finding the percentage. In b.ii (17% full marks) working was needed, since a wrong bare answer earns nothing while shown working can earn a method mark.

a. [1 mark]. A 1 on the leading diagonal keeps an activity in place. Rows 3, 4 and 5 have their 1 on the diagonal, so gardening, lunch and music never move.

b. [1 mark]. $P$ sends slot 6 to slot 1, slot 7 to slot 2, slot 2 to slot 6 and slot 1 to slot 7.

• Day 2: $PX$ = reading, sport, gardening, lunch, music, drama, cooking.
• Day 3: $P^2X$ = drama, cooking, gardening, lunch, music, sport, reading.

c. [1 mark]. $P^4X = X$, so the timetable repeats every four days: day 5 has the same order as day 1. The 40-day program is exactly 10 complete four-day cycles.

From the report. Part b was answered correctly by 27%, and part c by only 11%: few students showed an understanding of the effect of the identity matrix.

a. [1 mark]. Degrees: gym 2, coffee shop 4, supermarket 2, pet shop 3, home 3, so the sum is $2 + 4 + 2 + 3 + 3 = 14$. (This equals twice the 7 edges, as it must.)

b. [1 mark]. $v = 5$ and $e = 7$, so $f = 7 + 2 - 5 = 4$ (three inner regions plus the outside):

c. [1 mark]. Each vertex is visited exactly once, starting and finishing at different vertices: a Hamiltonian path (for example gym, pet shop, coffee shop, supermarket, home).

d. [1 mark]. Any 4 of the original edges that connect all 5 vertices without forming a cycle. For example: gym to coffee shop, coffee shop to supermarket, coffee shop to pet shop, coffee shop to home.

From the report. In part d (81%) the tree could be drawn many ways, but some students included an edge that was not in the original graph.

[2 marks]. On the map there are four junctions: one on the entry-recovery passage (call it $J_1$), one on the change room-weights passage near the change room ($J_2$), one on that passage near weights ($J_3$) and one on the recovery-weights passage ($J_4$). Cross passages join $J_1$ to $J_2$ and $J_3$ to $J_4$. Count the routes that avoid other areas:

• E to R: E-$J_1$-R, and E-$J_1$-$J_2$-$J_3$-$J_4$-R. So $x = 2$.
• R to W: R-$J_4$-W, R-$J_4$-$J_3$-W, R-$J_1$-$J_2$-$J_3$-W and R-$J_1$-$J_2$-$J_3$-$J_4$-W. So $y = 4$.
• W to C: W-$J_3$-$J_2$-C and W-$J_4$-$J_3$-$J_2$-C. So $z = 2$.

The same counting reproduces the given entries, for example the loops giving 1 at R-R and W-W.

From the report. Only 20% scored both marks; $y$ was often given as 3.

a. [1 mark]. A route that uses every edge once and returns to its start is an Eulerian circuit, which needs every vertex to have even degree. Here C (degree 3) and E (degree 1) have odd degree, so some edges must be repeated.

b. [1 mark]. Total length of all walkways:

The cheapest way to fix the two odd vertices is to repeat the shortest path between C and E, which is C-A-E $= 14 + 7 = 21$ m. The minimum distance is $175 + 21 =$ 196 m.

From the report. Parts a and b were answered correctly by 40% and 27%.

a. [1 mark]. Forward scan (earliest start times): C and D can start at 5; E and F at 4; H and I at $\max(5 + 7, 4 + 3) = 12$; G at 12; J at $\max(12 + 6, 12 + 7) = 19$; K at $\max(12 + 4, 4 + 6) = 16$; L at $\max(19 + 1, 16 + 4) = 20$. The project takes $20 + 5 = 25$ days.

Both A-D-H-J-L and A-D-I-K-L take 25 days, so the common activities are A, D and L.

b. [1 mark]

Work backwards: L starts by 20, K by $20 - 4 = 16$, I by $16 - 4 = 12$. E must finish by the start of H and I, which is 12, so the latest start time of E is $12 - 3 =$ 9 days.

c. [1 mark]

Floats (latest start minus earliest start): B 5, C 1, E 5, F $(16 - 6) - 4 = 6$, G 1, and 0 for the critical activities. The longest float belongs to F (6 days).

d. [1 mark]

Both critical paths must drop to 22 days, and the path A-C-G-J-L (24 days) must not become longer than that.

• A is on both critical paths (and on A-C-G-J-L): $500 per day, at most 2 days.
• Shortening H by 1 and K by 1 together cuts both paths by 1 day for $400, but only once, since K has 1 day available.

Cheapest: H and K by 1 day each ($400), then A by 2 days ($1000). All paths are then at most 22 days (A-C-G-J-L is also 22). The minimum additional cost is $1400.

From the report. Part d was answered correctly by only 21%. In part c, the report notes that any extra information given (such as float times) had to be correct; only F was needed for the mark.