Category Archives: Finance

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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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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