VCE General Mathematics 2024 Exam 2

a. i. [1 mark]

2.39 m (1996).

ii. [1 mark]

Eleven of the 22 heights are above 2.25 m (2.25 itself is not "greater than"), so $\tfrac{11}{22} = 50\%$.

b. [1 mark]

c. [2 marks]. In order, the 22 values run from 1.94 to 2.39. The median is the mean of the 11th and 12th values, $\tfrac{2.25 + 2.33}{2} = 2.29$. $Q_1$ is the median of the lower 11 values (the 6th, 2.12) and $Q_3$ the median of the upper 11 (the 17th, 2.36).

Five-number summary: minimum 1.94, $Q_1 = 2.12$, median 2.29, $Q_3 = 2.36$, maximum 2.39. The lower fence is $2.12 - 1.5 \times 0.24 = 1.76$, so there are no outliers and the whiskers run to 1.94 and 2.39.

d. [2 marks]. Linear regression with year as $x$ and Mgold as $y$ gives $a = -7.971\ldots$ and $b = 0.005161\ldots$:

e. [1 mark]. 85.7% of the variation in Mgold can be explained by the variation in year.

From the report. In a.ii, 54.54% (including 2.25) was the most common wrong answer; a bare 50% was enough, with no sentence needed. In d (39% full marks) many students could not read calculator output in exponent form (5.16E-3) or round to three significant figures. In e (52%) some did not mention variation in both variables, and some rounded to 86% when no rounding was asked for.

a. [1 mark]

The long lower whisker and the median close to $Q_3$ show a negatively skewed distribution (with no outliers).

b. [1 mark]

With 19 values, $Q_1$ is the 5th value. The minimum (1.67) is below 1.85, but the 2nd to 5th values could all equal 1.85. So the smallest possible number is 1.

c. i. [1 mark]

$\text{IQR} = 2.04 - 1.85 = 0.19$.

ii. [1 mark]. The minimum, 1.67, is above the lower fence of 1.565, and the maximum, 2.06, is below the upper fence of 2.325. Every value lies between the fences, so there are no outliers.

From the report. Part b was answered correctly by only 3% of students. Most wrote the height 1.67 m instead of a number of heights, missing that one value is enough to extend the whisker. In c.ii, answers that squeezed both fence conditions into one statement often became mathematically incorrect.

a. [1 mark]. Wbest (the variable on the left of the equation).

b. [1 mark]. Calculate two points far apart and join them with a ruler:

• at Wgold $= 1.90$: Wbest $= 0.300 + 0.860 \times 1.90 = 1.934$
• at Wgold $= 2.08$: Wbest $= 0.300 + 0.860 \times 2.08 = 2.0888$

Plot $(1.90, 1.934)$ and $(2.08, 2.089)$ precisely and rule the line between them.

c. [1 mark]

$r^2 = 0.9318^2 = 0.8682\ldots$, so 86.8%.

d. [1 mark]

Strength: strong. Direction: positive.

e. [1 mark]

On average, Wbest increases by 0.86 m for each 1 m increase in Wgold.

f. [2 marks]

g. i. [1 mark]

Plot an X at Wgold $= 2.02$ and residual $= 0.0328$, the highest point on the residual plot.

ii. [1 mark]

Yes. The residuals are randomly scattered with no clear pattern, which supports a linear model.

h. [1 mark]

It is extrapolation: Wgold $= 1.90$ lies outside the range of Wgold values (1.92 to 2.06) used to fit the line.

From the report. In b (51%) many lines were drawn by eye instead of through calculated points, and 1.934 had to be placed precisely between 1.93 and 1.94. In d some said "moderate". In e (54%) answers needed the change in both variables. In g.i (47%) many did not connect the scatterplot to the residual plot. In g.ii the number of points above and below zero was irrelevant, and some wrongly asserted the prediction was reliable. In h the reason had to be about the explanatory variable; saying 1964 or 1.934 was outside the data range was not accepted.

a. [1 mark]. Each smoothed value is the median of five consecutive values, centred on the year:

• 1968: median of 1.85, 1.90, 1.82, 1.92, 1.93 is 1.90
• 1972: median of 1.90, 1.82, 1.92, 1.93, 1.97 is 1.92
• 1976: median of 1.82, 1.92, 1.93, 1.97, 2.02 is 1.93
• 1980: median of 1.92, 1.93, 1.97, 2.02, 2.01 is 1.97

Plot these as crosses and join them with a dashed line from 1960 to 1980. (There are no smoothed values for the first two or last two years.)

b. [1 mark]. An increasing trend and irregular fluctuations.

From the report. Part b (24%) needed the study design's formal features. There was no evidence of an outlier, seasonality or structural change, and irregular fluctuations are present in every time series.

a. [1 mark]

$60 per week (the coefficient of $n$).

b. [1 mark]

Four years is $4 \times 52 = 208$ weeks: $V_{208} = 15\,000 - 60 \times 208 = 2520$, so $2520.

c. [1 mark]

$V_0 = 15\,000$, $V_{n+1} = V_n - 60$.

d. [1 mark]

Annual depreciation $= 60 \times 52 = 3120$, and $\dfrac{3120}{15\,000} \times 100\% = 20.8\%$.

From the report. In b, $14 760 was a common wrong answer from not converting four years into 208 weeks. In c some wrongly put an $n$ on the end of the recurrence relation. In d (42%) a weekly rate of 0.4% was a common wrong answer.

a. [1 mark]. Solve $\left(1 + \dfrac{r}{26}\right)^{26} = 1.0507$:

so the nominal rate is 4.95% per annum. (CAS: use the nominal-rate conversion with 26 compounding periods.)

b. [1 mark]. The nominal rate does not take the fortnightly compounding into account. The effective rate includes the interest earned on interest during the year, so it is higher.

From the report. Part b was answered well by only 26%. Many students could define the effective rate but could not explain simply why the nominal rate is lower.

a. [1 mark].

The balance is $297 477.40.

b. [1 mark]

Finance Solver: $I\% = 3.6$, $PV = -300\,000$, $PMT = 2159.41$, $FV = 0$, 12 periods per year gives $N = 180$ months, which is 15 years.

c. [1 mark]

$0.003 \times 12 \times 100\% = 3.6\%$ per annum.

d. [1 mark]

A perpetuity pays only the interest: $0.003 \times 300\,000 = 900$, so $900 per month.

From the report. In part a, $297 477.4 is not correct to the nearest cent; two decimal places were required.

[2 marks]. Finance Solver: $I\% = 5.3$, $PV = 500\,000$, $PMT = -3071.63$, $FV = 0$, 12 periods per year gives $N = 287.99\ldots$, so there are 288 payments.

To find the final payment, set $N = 288$ and solve for $FV$: $FV = -4.1773\ldots$ This means 288 full payments leave $4.18 still owing, so the final payment is $3071.63 + 4.18 = 3075.81$ (slightly different, as the question says).

Equivalently, $288 \times 3071.63 + 4.18 = 884\,633.62$. The total cost is $884 633.62.

From the report. Only 13% scored both marks. Many found 288 but left the total cost blank. Some used 289 payments, giving a final payment of $4.20, which is not "slightly different" from the others, and a total of $885 633.64.

a. [1 mark]. Rows become columns:

b. [1 mark].

c. [1 mark]. From the first element, $n \times 1 = 36$, so $n = 36$. Then $36p = 90$ gives $p = 2.5$.

a. [1 mark]. A row matrix has one row, so $i = 1$ and $j = 1, 2, \ldots, 5$:

b. [1 mark]. The last element is $x_{18} = (1 - 8)^2 + 2 \times 8 = 49 + 16 =$ 65.

From the report. Many students did not understand the row-column notation (46% scored the mark in part a), and some who got part a could not apply the same rule in part b.

a. i. [1 mark]. Row 1 holds birth rates and the subdiagonal holds survival rates. Draw, with arrows on every edge:

• a loop on 0 to 1 labelled 0.4, an arrow from 1 to 2 back to 0 to 1 labelled 0.75, and an arrow from 2 to 3 back to 0 to 1 labelled 0.4 (births)
• arrows 0 to 1 → 1 to 2 labelled 0.4, 1 to 2 → 2 to 3 labelled 0.7, and 2 to 3 → 3 to 4 labelled 0.5 (survival)

ii. [1 mark]. $R_1 = LR_0$:

For example, $0.4 \times 70 + 0.75 \times 80 + 0.4 \times 90 = 124$ and $0.5 \times 90 = 45$.

b. [1 mark]. The initial total is 280, so find when the total first drops below 140. The totals are 253, 190.2, 163.96, 150.98 and then 130.88 after year 5. The answer is 5 years.

From the report. Part a.i was poorly answered (24%): all edges needed arrows, or it is no longer a directed graph.

a. [1 mark]. Write the columns as "this year" (C, F, M, L) and rows as "next year":

$S_1 = TS_0 = \begin{bmatrix} 109 & 80 & 13 & 188 \end{bmatrix}^T$ and $S_2 = TS_1 = \begin{bmatrix} 48.7 & 43 & 19.9 & 278.4 \end{bmatrix}^T$. Foremen fall from 50 to 43:

b. [1 mark]

L is an absorbing state (it keeps 100% of its staff and nobody ever returns), while every other group loses staff each year. In the long term everyone ends up in L, so the number of staff on site is 0.

c. [1 mark]

$V_1 = RV_0 + Z = \begin{bmatrix} 332 & 146 & 13 & 89 \end{bmatrix}^T$. Staff on site in 2024 are C, F and M: $332 + 146 + 13 = 491$. In 2023 there were 390, so there are 101 more.

d. [1 mark]

Keep iterating: foremen are 146 (2024), 167.2 (2025), 187.48 (2026) and 200.54 (2027). The number first exceeds 200 in 2027.

From the report. Only 12% scored part c; some forgot to subtract the staff who had left and answered 190. In part d the fourth year was also accepted.

a. [1 mark]

Add vertex P with edges P-F and P-G.

b. i. [1 mark]

The Bakery's only neighbour is General, so B must be the last department visited: Bakery. For example, P, F, D, G, B.

ii. [1 mark]

Each vertex is visited exactly once without returning to the start: a Hamiltonian path.

c. [1 mark]

Place the departments one at a time using their neighbours:

• E borders only B and P: the top-right region beside the Bakery.
• P borders B, E, F and G: the strip under the Bakery and E.
• D borders B, F and G but not P: the top region to the left of the Bakery.
• F borders B, D, G and P: the block below D.
• G borders D, F and P but not B or E: the large L-shaped region down the left side and along the bottom.

From the report. In b.ii (75%) a significant number wrote just "path"; the full term Hamiltonian path was needed.

a. [1 mark]. Count only edges flowing from the M side to the S side: L-Q (13), O-S (18), P-R (6) and N-R (9). Q-O flows back across the cut, so it is not counted.

b. [1 mark]. Find a small cut and a flow that matches it. The cut separating $\{M, L, N, P, R\}$ from $\{O, Q, S\}$ crosses L-Q (13), L-O (5), P-O (11) and R-S (8), so its capacity is

A flow of 37 is achievable: 13 along M-L-Q-S, 5 along M-L-O-S, 11 along M-P-O-S and 8 along M-N-R-S. A flow equal to a cut's capacity must be the maximum, so the maximum flow is 37.

c. [1 mark]. R and S. R-S (8) is the bottleneck on the bottom route: up to 15 deliveries can reach R (6 from P, 9 from N), and raising R-S lets the extra 7 through. Raising any other single edge adds at most 2.

From the report. Part b was answered correctly by 29%, using either a minimum cut or an exhaustion of paths. Part c (22%) was not answered well, with many other pairs given.

a. [1 mark]

Forward scan: C and D start at 9, E and F at 11, H and G at 16, I at $\max(9 + 8, 11 + 4) = 17$, J at $\max(16 + 8, 17 + 6) = 24$, K at 19. Finish $= \max(16 + 12, 24 + 5, 19 + 8) = 29$. The critical path is A-C-H-J ($9 + 7 + 8 + 5 = 29$).

b. [1 mark]

Floats (latest start minus earliest start): B 2, D 1, E 3, F 2, G 1, I 1, K 2. Activity E can be delayed longest (3 weeks).

c. [1 mark]

The new times show that H and G must now wait for D and E as well as C: H's latest start of 17 equals D's latest finish ($9 + 8$) and E's latest finish ($13 + 4$). Draw a dummy activity (a dashed arrow, labelled "dummy") from the node where D and E end (the start of I) to the node where C ends (the start of H and G). With it, the backward scan reproduces every latest start time in the table.

d. [1 mark]

H and G can now start only after $\max(9 + 7, 9 + 8, 11 + 4) = 17$, so J starts at $\max(17 + 8, 17 + 6) = 25$ and the project takes $25 + 5 =$ 30 weeks (critical path A-D-dummy-H-J).

e. [1 mark]

Only activities of at least 8 weeks can be shortened: A (by 2), B (2), D (1), F (1), G (2), H (1) and K (1). J, C, E and I cannot.

• A-C-H-J (29) can drop by at most $2 + 1 = 3$ (A and H), to 26, so 26 weeks is the best possible.
• To reach 26: shorten A by 2 and H by 1 (A-D-dummy-H-J drops from 30 to 27), then D by 1 (it drops to 26).
• B-E-dummy-H-J is then $11 + 4 + 7 + 5 = 27$, so shorten B by 1. That also brings B-E-G and B-F-K to 26.

Total reduction $2 + 1 + 1 + 1 = 5$ weeks, costing $5 \times 10\,000 =$ $50 000.

From the report. Part c (10%) was poorly answered: the dummy was often drawn without an arrow or a label (dashed or solid lines were both accepted). Part e was answered correctly by only 7%.