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AusGrader has 270 VCAA General Mathematics questions on Discrete mathematics, all with instant AI grading and detailed marking feedback.

VCE General Mathematics — Discrete mathematics Questions & Answers

Q3

2023

SCSA

Paper 1

9 marks

Q3

From January 1, 2020, a company offered its employees an income package with a starting wage of ＄4000 per month, paid at the end of each month. Also, as an incentive to stay with the company, there was a monthly increase of ＄50 each month.

Q3a

2 marks

Determine a recursive rule for the monthly wage.

Reveal Answer

$T_{n+1} = T_n + 50, T_1 = 4000$

Marking Criteria

Descriptor	Marks
states correct recursive rule	1
states correct first term	1

Q3b

2 marks

Deduce a simplified rule for the $n$ th term of the monthly wage.

Reveal Answer

$T_n = 4000 + (n-1)(50)$
$T_n = 3950 + 50n$

Marking Criteria

Descriptor	Marks
uses correct arithmetic formula	1
gives correct simplified rule for the $n^{\text{th}}$ term	1

Q3c

2 marks

Determine the monthly wage for December 2020.

Reveal Answer

$T_{12} = 3950 + 50(12) = 4550$

Therefore, the monthly wage for December 2020 is ＄4550

Marking Criteria

Descriptor	Marks
correctly identifies term 12	1
correctly calculates the ＄4550	1

Q3d

3 marks

The company has decided to make the monthly increase ＄60 from the end of December 2023.

Calculate the monthly wage for March 2024.

Reveal Answer

$12 \times 4 = 48$
$T_{48} = 3950 + 50(48) = 6350$

Therefore, the monthly wage for March 2024 is $6350 + 60 + 60 + 60 = ＄6530$

Marking Criteria

Descriptor	Marks
correctly calculates $T_{48}$	1
calculates correct term for March 2024	1
states correct solution for wage	1

Q23

2025

VCAA

Paper 1

1 mark

Q23

1 mark

Virat invested ＄5000 into an account that earned interest compounding fortnightly.

The effective annual interest rate for Virat's investment was 4.51%.

Assume that there are exactly 26 fortnights in one year.

After five years, the amount of interest earned by Virat was closest to

A

＄1128

B

＄1234

C

＄1262

D

＄1264

Reveal Answer

A

＄1128

This incorrectly calculates simple interest instead of compound interest ( $5000 \times 0.0451 \times 5 = ＄1127.50$ ).

B

＄1234

Correct Answer

Using the effective annual rate, the total amount after 5 years is $5000(1 + 0.0451)^5 \approx ＄6234.20$ . The interest earned is the total amount minus the principal ( $＄6234.20 - ＄5000 = ＄1234.20$ ).

C

＄1262

This incorrectly treats 4.51% as the nominal annual rate compounded fortnightly, calculating $5000(1 + 0.0451/26)^{130} - 5000 \approx ＄1262$ .

D

＄1264

This incorrectly treats 4.51% as a nominal annual rate compounded daily, which would yield $5000(1 + 0.0451/365)^{1825} - 5000 \approx ＄1264$ .

Q16

2022

SCSA

Paper 2

7 marks

Q16

After paying a deposit for his new apartment, Declan obtains a bank loan for the remaining amount of ＄112 000 at 3.26% per annum compounded monthly. He can currently afford to repay ＄970 per month at the end of every month.

Q16a

3 marks

Calculate how much he would owe after the 40th repayment.

Reveal Answer

Using the financial app with
N = 40, I = 3.26, PV = -112 000, PMT = 970, P/Y = C/Y = 12
FV = 83 910.19

He would owe ＄83 910.19 after the 40th repayment

Marking Criteria

Descriptor	Marks
states at least 4 correct entries	1
states all correct entries	1
states correct answer	1

Q16b

4 marks

Declan decided to deposit a one-off extra amount of ＄1600, after the 16th repayment. Calculate the new amount he would owe after the 40th repayment.

Reveal Answer

Step 1. Using the financial app
N = 16, I = 3.26, PV = –112 000, PMT = 970, P/Y = C/Y = 12
FV = ＄101 128.46
New PV = ＄101 128.46 − 1600 = ＄99 528.46

Step 2. Using financial app
N = 24(40 − 16), I = 3.26, PV = −99 528.46, PMT = 970, P/Y = C/Y = 12
FV = ＄82 202.54

New amount owing after the 40th repayment is ＄82 202.54

Marking Criteria

Descriptor	Marks
states all correct entries in step 1	1
correctly subtracts 1600 from FV in step 1 to give new PV	1
states all correct entries in step 2	1
correctly determines new FV	1

Q14

2021

SCSA

Paper 2

11 marks

Q14

Patrick has retired and invested his lump sum superannuation payout of ＄717 850 at a rate of 5.7% per annum compounded monthly. He begins the investment strategy from 1 January.

Q14a

Patrick will receive ＄4500 at the end of each month for general living expenses and will also receive a further ＄4000 at the end of each year for an annual holiday.

Q14a (i)

1 mark

Identify this type of investment account.

Reveal Answer

Annuity

Marking Criteria

Descriptor	Marks
states correct answer	1

Q14a (ii)

4 marks

Determine the balance in the account at the end of the first year.

Reveal Answer

N = 12, I = 5.7, PV = $-717\ 850$ , PMT = 4500, P/Y = 12, C/Y = 12
FV = 704 420.20

Balance at end of year 1 = $704\ 420.20 - 4000 = ＄700\ 420.20$

Marking Criteria

Descriptor	Marks
uses at least 4 correct values for N, I, PV, PMT, P/Y, C/Y	1
uses all correct values for N, I, PV, PMT, P/Y, C/Y	1
determines correct value for FV	1
determines correct end of year balance	1

Q14a (iii)

3 marks

Determine the balance in the account at the end of the second year.

Reveal Answer

N = 12, I = 5.7, PV = $-700\ 420.20$ , PMT = 4500, P/Y = 12, C/Y = 12
FV = 685 970.53

Balance at end of year 2 = $685\ 970.53 - 4000 = ＄681\ 970.53$

Marking Criteria

Descriptor	Marks
uses correct value for PV	1
determines correct FV	1
determines correct end of year 2 balance	1

Q14b

3 marks

When Patrick retired, he also considered the option of setting up a perpetuity with his superannuation payout still at 5.7% per annum compounded monthly. Calculate the quarterly payments Patrick would have received with this perpetuity in place.

Reveal Answer

N = 2 (can be any value), I = 5.7, PV = $-717\ 850$ , FV = 717 850, P/Y = 4, C/Y = 12

Quarterly payments = ＄10 278.03

Marking Criteria

Descriptor	Marks
uses at least 4 correct values for N, I, PV, FV, P/Y, C/Y	1
uses all correct values for N, I, PV, FV, P/Y, C/Y	1
states correct quarterly payments	1

Q10

2024

SCSA

Paper 2

13 marks

Q10

Matt is saving up to purchase a new boat. He deposits ＄14 500 into a savings account which is compounded monthly. The account pays an annual interest rate of 4.8% and he also deposits ＄300 into the account at the end of each month.

Q10c

After four years, Matt makes a one-off deposit of ＄2500 into the savings account. His goal is to have a total of ＄50 000 by the end of the fifth year.

Q10a (i)

1 mark

Calculate the monthly interest rate.

Reveal Answer

4.8 ÷ 12 = 0.4%

Marking Criteria

Descriptor	Marks
calculates correct rate	1

Q10a (ii)

2 marks

Determine a recursive rule to model the balance of the savings account at the end of each month.

Reveal Answer

$T_{n+1} = 1.004T_n + 300, \quad T_0 = 14 500$

Marking Criteria

Descriptor	Marks
states correct rule	1
states correct initial value	1

Q10b

2 marks

After how many months will the balance of Matt's account first exceed ＄20 000?

Reveal Answer

$T_{14} = 19 644.42$
$T_{15} = 20 023.00$
Therefore after 15 months

Marking Criteria

Descriptor	Marks
correctly calculates 14th and 15th terms	1
correctly concludes it is 15 months	1

Q10c

5 marks

Determine the equal monthly deposits during the fifth year he will need to make to reach this amount.

Reveal Answer

$T_{48} = 33 402.99$

33 402.99 + 2500 = 35 902.99

N = 12, I = 4.8, PV = –35 902.99, FV = 50 000, P/Y = 12, C/Y = 12
PMT = –1005.52

Therefore, deposits of ＄1005.52 per month

Marking Criteria

Descriptor	Marks
correctly calculates balance after 4 years	1
adds 2500 to balance after 4 years	1
states correct PV	1
correctly states the remaining parameters	1
determines correct monthly deposit	1

Q10d

3 marks

Matt purchases his new boat, which costs him ＄47 500. He decides to take the remaining money and re-invest it in one of the following high-interest savings accounts.

Option 1: 5.52% per annum, compounded six-monthly.

Option 2: 5.5% per annum, compounded quarterly.

Determine which option Matt should choose, by calculating the effective annual rates of interest.

Reveal Answer

Option 1: $i_e = \left(1 + \frac{0.0552}{2}\right)^2 - 1 = 0.05596 \times 100 = 5.60$

Option 2: $i_e = \left(1 + \frac{0.055}{4}\right)^4 - 1 = 0.0561 \times 100 = 5.61$

Therefore option 2 is the better choice as it has a higher effective interest rate.

Marking Criteria

Descriptor	Marks
correctly calculates effective interest rate for option 1	1
correctly calculates effective interest rate for option 2	1
correctly states option 2 is the better choice	1

Q6

2025

QCAA

Paper 1

1 mark

Q6

1 mark

A ball is dropped from a height of 25.6 m. After each bounce, the ball rebounds to 75% of its previous height.

Which option shows the ball's height after the third bounce?

A

8.1 m

B

10.8 m

C

14.4 m

D

19.2 m

Reveal Answer

A

8.1 m

This is incorrect because it represents the height of the ball after the fourth bounce ( $25.6 \times 0.75^4 = 8.1$ m), not the third.

B

10.8 m

Correct Answer

This is correct. The height after the third bounce is found by multiplying the initial height by the rebound rate cubed: $25.6 \times 0.75^3 = 10.8$ m.

C

14.4 m

This is incorrect because it represents the height of the ball after the second bounce ( $25.6 \times 0.75^2 = 14.4$ m).

D

19.2 m

This is incorrect because it represents the height of the ball after the first bounce ( $25.6 \times 0.75 = 19.2$ m).

Q22

2025

QCAA

Paper 1

4 marks

Q22

A ＄50 000 perpetuity earning fortnightly interest at 4.94% p.a. provides a regular fortnightly payment.

Q22a

2 marks

Calculate the fortnightly payment.

Reveal Answer

Fortnightly payment = $\frac{0.0494}{26} \times 50 000 =＄95$

Marking Criteria

Descriptor	Marks
correctly provides mathematical reasoning or working to support the answer	1
calculates fortnightly payment	1

Q22b

2 marks

Calculate the perpetuity's effective annual rate of interest as a percentage.

Reveal Answer

\begin{align*} i_{\text{effective}} &= \left(1 + \frac{i}{n}\right)^n - 1\\ &= \left(1 + \frac{0.0494}{26}\right)^{26} - 1\\ &\approx 0.05059 \end{align*}

The effective annual rate of interest is 5.06% p.a.

Marking Criteria

Descriptor	Marks
correctly provides mathematical reasoning or working to support the answer	1
calculates effective interest rate as a percentage	1

Q12

2023

QCAA

Paper 1

1 mark

Q12

1 mark

A reducing balance loan with an initial balance of ＄6000 is modelled by the recurrence relation
$A_{n+1} = \left(1 + \frac{0.03}{12}\right)A_n - 400$ , where $n$ is the number of months.
The loan balance at the end of two months is closest to

A

＄5100

B

＄5200

C

＄5215

D

＄5230

Reveal Answer

A

＄5100

This value is incorrect. It is significantly lower than the calculated balance and likely results from a calculation error.

B

＄5200

This option incorrectly ignores the interest component. It simply subtracts two repayments from the principal: $6000 - 2(400) = 5200$ .

C

＄5215

This option fails to add interest for the second month. It takes the balance after one month ( $A_1 = 5615$ ) and subtracts the repayment without applying the interest factor: $5615 - 400 = 5215$ .

D

＄5230

Correct Answer

By applying the recurrence relation iteratively: $A_1 = 1.0025(6000) - 400 = 5615$ and $A_2 = 1.0025(5615) - 400 \approx 5229.04$ . This is closest to ＄5230.

Q7

2024

VCAA

Paper 2

4 marks

Q7

Emi decides to invest a ＄300000 inheritance into an annuity.

Let $E_n$ be the balance of Emi's annuity after $n$ months.

A recurrence relation that can model the value of this balance from month to month is

$E_0 = 300000, \quad E_{n+1} = 1.003E_n - 2159.41$

Q7a

1 mark

Showing recursive calculations, determine the balance of the annuity after two months.
Round your answer to the nearest cent.

Reveal Answer

$E_0 = 300\,000$

$E_1 = 1.003 \times 300\,000.00 - 2159.41 = 298\,740.59$

$E_2 = 1.003 \times 298\,740.59 - 2159.41 = 297\,477.401 = ＄297\,477.40$

Marking Criteria

Descriptor	Marks
Shows recursive calculations and determines the correct balance after two months, rounded to the nearest cent (＄297,477.40)	1

Q7b

1 mark

For how many years will Emi receive the regular payment?

Reveal Answer

15 years

Marking Criteria

Descriptor	Marks
Correctly determines the number of years Emi will receive the regular payment (15)	1

Q7c

1 mark

Calculate the annual compound interest rate for this annuity.

Reveal Answer

3.6%

Marking Criteria

Descriptor	Marks
Correctly calculates the annual compound interest rate (3.6%)	1

Q7d

1 mark

If Emi wanted the annuity to act as a perpetuity, what monthly payment, in dollars, would she receive?

Reveal Answer

＄900

Marking Criteria

Descriptor	Marks
Correctly determines the monthly payment for a perpetuity (＄900)	1

Q22

2025

VCAA

Paper 1

1 mark

Q22

1 mark

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 ＄14000 after three years is closest to

A

8.4%

B

14.2%

C

14.6%

D

17.2%

Reveal Answer

A

8.4%

Correct Answer

Correct. Setting up the annuity equation $14000 = 4000(1+\frac{r}{52})^{156} + 50\frac{(1+\frac{r}{52})^{156}-1}{\frac{r}{52}}$ or using a financial solver yields an annual rate of $r \approx 8.4\%$ .

B

14.2%

Incorrect. An annual interest rate of $14.2\%$ would result in a final balance of approximately ＄15,800, which exceeds the target of ＄14,000.

C

14.6%

Incorrect. An annual interest rate of $14.6\%$ would result in a final balance of approximately ＄16,100, which is significantly higher than the target of ＄14,000.

D

17.2%

Incorrect. An annual interest rate of $17.2\%$ would result in a final balance of approximately ＄17,700, which is much higher than the target balance of ＄14,000.

Q17

2025

VCAA

Paper 1

1 mark

Q17

1 mark

A sequence of numbers is generated by the recurrence relation

$T_0 = 30, \quad T_{n+1} = 0.5T_n - 5$

The first negative term in the sequence is

A

$T_1$

B

$T_2$

C

$T_3$

D

$T_4$

Reveal Answer

A

$T_1$

Calculating the first term gives $T_1 = 0.5(30) - 5 = 10$ , which is a positive number.

B

$T_2$

Calculating the second term gives $T_2 = 0.5(10) - 5 = 0$ , which is neither positive nor negative.

C

$T_3$

Correct Answer

Calculating the third term gives $T_3 = 0.5(0) - 5 = -5$ , making it the first negative term in the sequence.

D

$T_4$

While $T_4 = 0.5(-5) - 5 = -7.5$ is negative, it is not the first negative term since $T_3$ is already negative.

Q17

2022

QCAA

Paper 1

4 marks

Q17

An investment of ＄50 000 that compounds interest monthly is modelled by the recurrence relation

$A_{n+1} = 1.00375A_n$ where $A_0 = 50\,000$ .

Q17a

2 marks

What would be the advertised interest rate per annum, compounding monthly?

Reveal Answer

$r = 1 + \frac{i}{n}$
$1.00375 = 1 + \frac{i}{12}$
$0.00375 = \frac{i}{12}$
$i = 0.045$

Therefore, the annual interest rate is 4.5% p.a. compounding monthly.

Marking Criteria

Descriptor	Marks
Correctly substitutes into an appropriate rule	1
Calculates annual interest rate	1

Q17b

2 marks

How many months would it take for the value of the investment to exceed ＄51 000?

Reveal Answer

$A_0 = 50\ 000$
$A_1 = 50\ 187.50$
$A_2 = 50\ 375.70$
$A_3 = 50\ 564.61$
$A_4 = 50\ 754.23$
$A_5 = 50\ 944.56$
$A_6 = 51\ 135.60$
Therefore, the investment would exceed ＄51 000 at 6 months.

Marking Criteria

Descriptor	Marks
correctly uses an appropriate method	1
determines when the investment would exceed $51 000	1

Q17

2021

QCAA

Paper 1

4 marks

Q17

4 marks

Determine the monthly repayment on a reducing balance loan of ＄720 000 at 4.8% p.a. over 25 years.
Give your answer to the nearest dollar.

Reveal Answer

$A = 720\,000$

$M = ?$
$i = \frac{0.048}{12} = 0.004$
$n = 25 \times 12 = 300$

$A = M\left(\frac{1-(1+i)^{-n}}{i}\right)$

$A = M\left(\frac{1-(1+0.004)^{-300}}{0.004}\right)$

$720\,000 = M \times 174.520 ...$

$M = \frac{720\,000}{174.520 ...}$

$M = 4125.578 ...$

The monthly repayment will be ＄4126 each month for 25 years.

Marking Criteria

Descriptor	Marks
correctly determines the $i$ and $n$ values	1
substitutes into appropriate annuity rule	1
determines monthly repayment	1
states solution to the nearest dollar	1

Q7

2025

VCAA

Paper 2

4 marks

Q7

Maria invests ＄300 000 into an annuity.

Interest is calculated fortnightly and Maria receives fortnightly payments of ＄1570.58 from the annuity.

Three complete lines of the amortisation table are shown below.

Payment number	Payment (＄)	Interest (＄)	Principal reduction (＄)	Balance (＄)
0	0.00	0.00	0.00	300 000.00
1	1570.58	750.00	820.58	299 179.42
2	1570.58	747.95	822.63	298 356.79
3

Q7a

1 mark

Write a calculation to show that the interest rate on the annuity is 0.25% per fortnight.

Reveal Answer

$Rate = \frac{750}{300 000} \times 100 = 0.25\%$

Marking Criteria

Descriptor	Marks
Provides the correct calculation to show the interest rate: $\frac{750}{300 000} \times 100 = 0.25\%$	1

Q7b

1 mark

Determine the principal reduction associated with payment number 3.

Round your answer to the nearest cent.

Reveal Answer

＄824.69

Marking Criteria

Descriptor	Marks
Determines the correct principal reduction: ＄824.69	1

Q7c

1 mark

Let the balance of Maria's annuity, in dollars, after $n$ fortnights be represented by $V_n$ .

Write a recurrence relation, in terms of $V_0$ , $V_{n+1}$ and $V_n$ , that can model the value of the annuity from one fortnight to the next.

Reveal Answer

$V_0 = 300 000 \quad V_{n+1} = 1.0025 \times V_n - 1570.58$

Marking Criteria

Descriptor	Marks
Writes the correct recurrence relation: $V_0 = 300 000 \quad V_{n+1} = 1.0025 \times V_n - 1570.58$	1

Q7d

1 mark

How many fortnightly payments will Maria receive over the life of the annuity?

Reveal Answer

260

Marking Criteria

Descriptor	Marks
Determines the correct number of payments: 260	1

Q3

2021

QCAA

Paper 2

5 marks

Q3

5 marks

Jo contributes ＄2500 per quarter to an annuity earning 3.6% p.a. compounding quarterly.

At the end of 4 years, Jo makes a one-off extra contribution of ＄10 000 and continues with the regular quarterly contributions.

Determine the value of the annuity at the end of 6 years, to the nearest dollar.

Reveal Answer

Value of regular contributions
$M = 2500$
$i = \frac{3.6}{400} = 0.009$
$n = 6 \times 4 = 24$

$A = M \left( \frac{(1+i)^n - 1}{i} \right)$
$= 2500 \left( \frac{(1.009)^{24} - 1}{0.009} \right)$
$= 66\,639.94$

Value of extra payment
$P = 10\,000$
$i = \frac{3.6}{400} = 0.009$
$n = 2 \times 4 = 8$

$A = P(1+i)^n$
$= 10\,000(1.009)^8$
$= 10\,743.09$

Total value $= 66\,639.94 + 10\,743.09$
$= 77\,383.03$
$= ＄77\,383$

Marking Criteria

Descriptor	Marks
correctly determines the $i$ and $n$ values	1
substitutes into appropriate annuity rule	1
substitutes into appropriate rule	1
determines sum of two values	1
determines total value, rounded to the nearest dollar	1

VCAA General Mathematics Discrete mathematics

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