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Financial Maths, GEN1 2022 VCAA 24 MC

On 1 January 2020, Dion invested $10 500 into an investment account paying compound interest of 0.52% quarterly.

At the end of each quarter, after the interest was credited, Dion added an additional amount of money.

Let \(D_n\) represent the additional amount, in dollars, added at the end of quarter \(n\).

This additional amount per quarter is modelled by the recurrence relation

\(D_1=C,\ \ \ D_{n+1}=D_n\)

The balance of Dion's investment account on 1 January 2022 was $12 700.95

The value of \(C\) is

  1. $71.69
  2. $215.55
  3. $260.22
  4. $270.15
  5. $275.12
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Annual interest rate}\ = 0.52  \times 4 = 2.08\%\)

\(D_n = D_{n+1}\ \text{indicates the additional payment is constant}\) 

\(\text{By TVM Solver:}\)

\(N\) \(=4 \times 2 = 8\)  
\(I\%\) \(=2.08\)  
\(PV\) \(=-10\ 500\)  
\(PMT\) \(=?\)  
\(FV\) \(=12\ 700.95\)
  
\(\text{P/Y}\) \(=4\)  
\(\text{C/Y}\) \(=4\)  

 
\(PMT = -215.55\)

\(\Rightarrow B\)


♦♦ Mean mark 32%.

CORE, FUR1 2021 VCAA 21 MC

Enrico invests $3000 in an account that pays interest compounding monthly.

After four years, the balance of the account is $3728.92

The effective annual interest rate for this investment, rounded to two decimal places, is

  1. 5.45%
  2. 5.52%
  3. 5.56%
  4. 5.59%
  5. 5.60%
Show Answers Only

`D`

Show Worked Solution

`text{By TVM Solver:}`

♦ Mean mark 40%.
`N` `= 48`
`I(%)` `= ?`
`PV` `= -3000`
`PMT` `= 0`
`FV` `= 3728.92`
`text(P/Y)` `= text(C/Y) = 12`

 
`=> I(%) = 5.45text(%)`

`r_text{effective}` `= [(1 + {5.45}/{100 xx 12})^12 -1] xx 100text(%)`
  `= 5.588 …%`

`=> D`

CORE, FUR2 2020 VCAA 9

Samuel opens a savings account.

Let `B_n` be the balance of this savings account, in dollars, `n` months after it was opened.

The month-to-month value of `B_n` can be determined using the recurrence relation shown below.

`B_0 = 5000, qquad B_(n+1) = 1.003B_n`

  1. Write down the value of `B_4`, the balance of the savings account after four months.
  2. Round your answer to the nearest cent.   (1 mark)

  3. Calculate the monthly interest rate percentage for Samuel’s savings account.   (1 mark)

  4. After one year, the balance of Samuel’s savings account, to the nearest dollar, is $5183.

     

    If Samuel had deposited an additional $50 at the end of each month immediately after the interest was added, how much extra money would be in the savings account after one year?

     

    Round your answer to the nearest dollar.   (1 mark)

Show Answers Only
  1. `$5060.27`
  2. `0.3 text(%)`
  3. `$610`
Show Worked Solution
a.   `B_1` `= 1.003 (5000)`
  `B_2` `= 1.003^2 (5000)`

`vdots`

`:. B_4` `= 1.003^4 (5000)`
  `= $5060.27`

 

b.  `text(Monthly interest rate)`

`= (1.003-1) xx 100`

`= 0.3%`

 

c.   `text(Extra)\ =\ text(value of annuity after 12 months)`

`text(By TVM solver:)`

`N` `= 12`
`I(%)` `= 3.6`
`PV` `= 0`
`PMT` `= 50`
`FV` `= ?`
`text(PY)` `= text(CY) = 12`

 
`FV = 609.84`

`:.\ text(Extra money) = $610`

CORE, FUR1 2020 VCAA 30 MC

Twenty years ago, Hector invested a sum of money in an account earning interest at the rate of 3.2% per annum, compounding monthly.

After 10 years, he made a one-off extra payment of $10 000 to the account.

For the next 10 years, the account earned interest at the rate of 2.8% per annum, compounding monthly.

The balance of his account today is $686 904.09

The sum of money Hector originally invested is closest to

  1. $355 000
  2. $370 000
  3. $377 000
  4. $384 000
  5. $385 000
Show Answers Only

`B`

Show Worked Solution

`text(Let)\ I = text(original investment)`

♦ Mean mark 41%.

`text(Strategy 1:)`

`text(Balance)` `= [I(1 + 3.2/(12 xx 100))^120 + 10\ 000](1 + 2.8/(12 xx 100))^120`
  `= $686\ 904.09`

 
`text(Test each option in the equation)`

`text(Option)\ B:\ \ I = $370\ 000\ \ text(is correct)`
 

`text{Strategy 2 (By TVM Solver):}`

`N` `=120`  
`Itext{(%)}` `= 2.8`  
`PMT` `=0`  
`PV` `= ?`  
`FV` `= 686\ 904.09`  
`text(PY)` `= text(CY) =12`  

 
`:. PV = 519\ 320.3`
 

`N` `=120`  
`Itext{(%)}` `= 3.2`  
`PMT` `=0`  
`PV` `= ?`  
`FV` `= 509\ 320.3`  
`text(PY)` `= text(CY) =12`  

 
`:. PV = 370\ 000`

`=>  B`