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Financial Maths, STD2 EQ-Bank 29

Mei sets up a superannuation account to save for retirement. She contributes $5000 at the end of each year into the account which earns interest at 6% per annum, compounded annually.

The spreadsheet shown models the first 4 years of the account.
  

  1. Calculate the value in cell C9.   (1 mark)
  2. From the end of year 4, Mei increases her annual contribution from $5000 to $8000. Calculate the balance in her superannuation account at the end of year 7, and determine how much more this is than if she had continued contributing $5000.   (3 marks)
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a.    \(\text{C9} = \$618.00\)

b.    \(\text{Balance at end of year 7} = \$51\,519.99\)

\(\text{Mei has}\ \$9550.80\ \text{more than at the standard contribution.}\)

Show Worked Solution

a.    \(\text{C9 (Year 3 interest)} = \text{balance at start} \times \text{rate}\)

\(\text{C9} = 10\,300 \times 0.06 = \$618.00\)
  

b.    \(\text{Using}\ \ P+I+C\ \ \text{from end of year 4 balance}\ \$21\,873.08:\)

\(\text{Increased contributions of}\ \$8000\ \text{from year 5:}\)

\(\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+8000=\$31\,185.46\)

\(\text{Year 6:}\ 31\,185.46+31\,185.46 \times 0.06+8000=\$41\,056.59\)

\(\text{Year 7:}\ 41\,056.59+41\,056.59 \times 0.06+8000=\$51\,519.99\)
  

\(\text{Standard contributions of}\ \$5000\ \text{from year 5:}\)

\(\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+5000=\$28\,185.46\)

\(\text{Year 6:}\ 28\,185.46+28\,185.46 \times 0.06+5000=\$34\,876.59\)

\(\text{Year 7:}\ 34\,876.59+34\,876.59 \times 0.06+5000=\$41\,969.19\)
  

\(\text{Difference}=51\,519.99-41\,969.19=\$9550.80\)

\(\therefore\ \text{Mei has}\ \$9550.80\ \text{more by increasing her contributions.}\)

Financial Maths, STD2 EQ-Bank 27

Haruki opens a savings account. He deposits $2000 at the end of each year into an account earning 3% per annum, compounded annually.

  1. Complete the table below to show the growth of Haruki's savings over the first 4 years.   (3 marks)

  2. Hence, calculate the total interest Haruki earns over the 4 years.   (1 mark)

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a.    

b.     \(\$367.25\)

Show Worked Solution

a.    \(\text{Year 2:}\)

\(\text{Interest} = 2000 \times 0.03 = \$60.00\)

\(\text{Closing balance} = 2000+60+2000 = \$4060.00\)

\(\text{Year 3:}\)

\(\text{Interest} = 4060 \times 0.03 = \$121.80\)

\(\text{Closing balance} = 4060+121.80+2000 = \$6181.80\)

\(\text{Year 4:}\)

\(\text{Interest} = 6181.80 \times 0.03 = 185.454\ldots = \$185.45\)

\(\text{Closing balance} = 6181.80+185.45+2000 = \$8367.25\)
 

 
b.    \(\text{Total deposits} = 4 \times 2000 = \$8000.00\)

\(\text{Total interest} = 8367.25-8000 = \$367.25\)

Financial Maths, STD2 F5 2020 HSC 14 MC

An annuity consists of ten payments, each equal to $1000. Each payment is made on 30 June each year from 2021 through to 2030 inclusive.

The rate of compound interest is 5% per annum.

The present value of the annuity is calculated at 30 June 2020.

The future value of the annuity is calculated at 30 June 2030.

Without performing any calculations, which of the following statements is true?

  1. Present value of the annuity  <  $10 000  <  future value of the annuity
  2. $10 000  <  present value of the annuity  <  future value of the annuity
  3. Future value of the annuity  <  $10 000  <  present value of the annuity
  4. $10 000  <  future value of the annuity  <  present value of the annuity
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`A`

Show Worked Solution

Mean mark 53%.

`PV\ text{(30 June 2020)}  < $10\ 000\ \ text{(each payment discounted to 30-Jun-20 value)}`

`FV\ text{(30 June 2030)}  => text{annuity has received}\ \  10 xx $1000`

`text{payments plus interest}`

`therefore \ FV\ text{(30 June 2030)} \ > \ $10\ 000 `
 

`=> \ A`

Financial Maths, STD2 F5 2013 23 MC

Zina opened an account to save for a new car. Six months after opening the account, she made first deposit of $1200 and continued depositing $1200 at the end of each six month period. Interest was paid at 3% per annum, compounded half-yearly.

How much was in Zina's account two years after first opening it?

  1. $4909.08
  2. $4982.72
  3. $5018.16
  4. $5094.55
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`A`

Show Worked Solution

`text(Interest: 3% p.a ⇒ 1.5% per 6 months)`

♦ Mean mark 41%.

`text(After 2 years,)`

`text(Value of 1st deposit) = 1200(1.015)^3 = 1254.81`

`text(Value of 2nd deposit) = 1200(1.015)^2 = 1236.27`

`text(Value of 3rd deposit) = 1200(1.015) = 1218`

`text(Value of 4th deposit) = 1200`
 

`:.\ text(Amount in account after 2 years)`

`= 1254.81 + 1236.27 + 1218 + 1200`

`=$4909.08`

`=> A`