Category Archives: Year 12 Mathematics Applications

November 7, 2024 · 9:51 am

Mathematics Applications – Counting can be Tricky Sometimes (off by one error)

Sequences are part of the Year 12 Mathematics Applications course and sometimes it’s tricky to work out which terms the question requires.

For example, ATAR 2020 Question 11

Judith monitors the water quality in her garden pond at the same time everyday. She likes to maintain the concentration of algae between 200 and 250 unites per 100 litres (L). Her measurements show that the concentration increases daily according to the recursive rule

C_{n+1}=1.025C_n where C_1=200 units per 100 L (the minimum concentration)

When the concentration gets above the 250 units per 100 L limit, she treats the water to bring the concentration back to the minimum 200 units per 100 l.

(a) If Judith treated the water on Sunday 6 December 2020, determine

(i) the concentration on Wednesday, 9 December 2020.
(ii) the day when she next treated the water.

(b) During the first week of January 2021, Judith monitored the water and recorded the following readings

Day1234567
Concentration (C)200206212.28218.55225.10231.85238.81

(i) Determine the revised recursive rule.
(ii) If she treated the water on 10 January and went on holiday until 20 January, when she next treated the water, calculate the concentration of the water on her return. Assuming the recursive rule from (b)(i) is used.

(a)(i) If C_1 is the 6th of December, then what term is the 9th of January?

I find most students simply do 9-6=3 so C_3, but this means they are off by one.
It’s better to list them
6th C_1
7th C_2
8th C_3
9th C_4
Hence we want to find C_4



The concentration on Wednesday 9 December is 215.38 units per 100 L

a(ii) We need to find when the concentration is greater than 250


C_{11}=256.02, what day is C_{11}?
The 9th is C_4, 10th C_5, etc. 16th is C_{11}
Judith next treats the water on Wednesday 16 December

(b)
(i) r=\frac{206}{200}=1.03
C_{n+1}=1.03C_n where C_1=200
(ii) C_1 is the 10th of January, 20th of January is C_{11} (20-10+1)


The concentration of the water on Judith’s return is 268.78 units per 100 L

I get my students to count on their fingers to ensure they get the correct term or day.

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October 22, 2024 · 6:55 am

Year 12 Mathematics Applications – Finance Question

What happens with an investment or a loan when the compounding periods are not the same as the payment periods?

For example,

Gerald invests $450 000 at a rate of 6.4% p.a. compounding monthly. At the end of every quarter he receives $35 000 from his investment.

(a) Write a recursive rule that will enable you to find the balance of the account, T_n after n payments.

(b) Find the balance of the account after 3 years.

(c) How long will it take for the balance in the account to reach zero?

(d) What will be the amount of Gerald’s last payment?

The interest is compounding at a different rate to the payments.
T_{n+1}=T_n(1+\frac{6.4}{100\times12})^{12\div4}-35 000, T_0=450 000

We need to raise the multiplier (1+\frac{6.4}{100\times12}) by the number of compound periods per payment, i.e. 12\div4

(a) T_{n+1}=T_n(1.0161)^3-35 000, T_0=450 000

(b) 3 years is 12 payments, \therefore n=12

T_12=245548.28

(c)

21 years

(d) The final repayment is 35 000-6535.03=28 464.97

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August 28, 2024 · 6:42 am

Using the Hungarian Algorithm for an Assignment Problem

The following question is from the 2019 ATAR Mathematics Applications Paper – Calculator Free Question 3.

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June 12, 2024 · 7:37 am

Year 12 Mathematics Applications Finance

Ming, a former high school student and now a successful business owner, wishes to set up a perpetuity of $6000 per year to be paid to a deserving student from her school. The perpetuity is to be paid at the start of the year in one single payment.

(a) A financial institution has agreed to maintain an account for the perpetuity paying a fixed rate of 5.9% p.a. compounded monthly. Show that an amount of $98 974, to the nearest dollar, is required to maintain this perpetuity.

(b) Ming allows herself five years to accumulate the required $98 974 by making regular quarterly payments into an account paying 5.4% p.a. compounded monthly. Determine the quarterly payment needed to reach the required amount after five years if Ming starts the account with an initial deposit of $1000.

SCSA 2017 CA 8

(a) For a perpetuity, we want the interest to equal the payment.

Remember the compound interest formula is

A=P(1+\frac{r}{100n})^{nt}

Where P is the principal, r is the interest rate (as a percent), n is the number of compounding periods in a year, and t is the time.

\therefore I=A-P

I=P(1+\frac{r}{100n})^{nt}-P

6000=P(1.00492)^12-P

P=\frac{6000}{(1.00492^12-1)}

P=98 974.14

Therefore an amount of $98 974 is required to maintain this perpetuity

For part (b) I will use the Finance Solver on a Classpad (Casio).


N is the number of payments, 4\times 5=20

PV is the principal value. To get the signs correct it is
helpful to think about the direction of the flow of the
money. The $1000 is going away from Ming so it is negative.

FV is the future value.

P/Y is the number of payments per year (quarterly so 4)

C/Y is the number of compounding periods per year (monthly so 12).

PMT is the payment

The quarterly payments are $4283.77.

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