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Recursion and Finance, GEN1 2024 VCAA 19 MC

Liv bought a new car for $35 000. The value of the car will be depreciated by 18% per annum using the reducing balance method.

A recurrence relation that models the year-to-year value of her car is of the form

\(L_0=35\,000, \quad L_{n+1}=k \times L_n\)

The value of \(k\) is

  1. 0.0082
  2. 0.18
  3. 0.82
  4. 1.18
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Reducing balance interest rate}\ =1-0.18=0.82\)

\(\Rightarrow C\)

Recursion and Finance, GEN1 2024 VCAA 18 MC

Trevor took out a reducing balance loan of $400 000, with interest calculated weekly. The balance of the loan, in dollars, after \(n\) weeks,  \(T_n\), can be modelled by the recurrence relation

\(T_0=400\,000, \quad T_{n+1}=1.00075 T_n-677.55\)

Assume that there are exactly 52 weeks in a year.

The interest rate, per annum, for this loan is

  1. 0.75%
  2. 3.9%
  3. 4.5%
  4. 7.5%
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Interest rate}\ = 0.00075\times 52 =0.039=3.9\%\)

\(\Rightarrow B\)

Recursion and Finance, GEN2 2023 VCAA 7

Arthur takes out a new loan of $60 000 to pay for an overseas holiday.

Interest on this loan compounds weekly.

The balance of the loan, in dollars, after \(n\) weeks, \(V_n\), can be determined using a recurrence relation of the form

\(V_0=60\ 000, \quad V_{n+1}=1.0015\,V_n-d\)

  1. Show that the interest rate for this loan is 7.8% per annum.  (1 mark)

  2. Determine the value of \(d\) in the recurrence relation if
  3.   i. Arthur makes interest-only repayments  (1 mark)

  4.  ii. Arthur fully repays the loan in five years. Round your answer to the nearest cent.  (1 mark)

  5. Arthur decides that the value of \(d\) will be 300 for the first year of repayments.  
  6. If Arthur fully repays the loan with exactly three more years of repayments, what new value of \(d\) will apply for these three years? Round your answer to the nearest cent.  (1 mark)

  7. For what value of \(d\) does the recurrence relation generate a geometric sequence?  (1 mark)

Show Answers Only

 

Show Worked Solution

a.   \(\text{Weekly interest rate factor}\ = 1.0015-1 = 0.0015 = 0.15\% \)

\(I\%(\text{annual}) = 52 \times 0.15 = 7.8\%\)

♦♦ Mean mark (a) 32%.

 
b.i.  \(d= 0.15\% \times 60\ 000 = \dfrac{0.15}{100} \times 60\ 000 = 90\)
 

b.ii. \(\text{By TVM solver:} \)

\(N\) \(=5 \times 52 = 260\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -60\ 000\)  
\(PMT\) \(= ?\)  
\(FV\) \(=0\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow PMT = d= $278.86\)
 

♦ Mean mark (b)(i) 47%.
♦ Mean mark (b)(ii) 40%.

c.    \(d=300\ \text{for the 1st 52 weeks.}\)

\(\text{Find}\ FV\ \text{after 52 weeks:}\)

\(N\) \(=52\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -60\ 000\)  
\(PMT\) \(= 300\)  
\(FV\) \(=?\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow FV = $48\ 651.67\)

 
\(\text{Find}\ PMT\ \text{given}\ FV=0\ \text{after 156 more weeks:}\)

\(N\) \(=156\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -48\ 651.67\)  
\(PMT\) \(= ?\)  
\(FV\) \(=0\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow PMT = d= $350.01\)
 

♦♦ Mean mark (c) 32%.

d.    \(\text{Geometric sequence when}\ \ d=0\)

\(V_0=60\ 000, V_1=60\ 000(1.0015), V_2=60\ 000(1.0015)^2, …\)

♦♦♦ Mean mark (d) 11%.

Recursion and Finance, GEN1 2022 VCAA 23 MC

Li invests $4000 for five years at 3.88% per annum, compounding annually.

Joseph invests a sum of money for five years, which earns simple interest paid annually.

Let \(J_n\) be the value, in dollars, of Joseph's investment after \(n\) years.

The two investments will finish at the same value, rounded to the nearest cent, if Joseph's investment is modelled by which one of the following recurrence relations?

  1. \(J_0=2000,\ \ \ J_{n+1}=J_n+467.72\)
  2. \(J_0=2500,\ \ \  J_{n+1}=J_n+367.72\)
  3. \(J_0=3000,\ \ \  J_{n+1}=J_n+317.72\)
  4. \(J_0=3500,\ \ \ J_{n+1}=J_n+267.72\)
  5. \(J_0=4000,\ \ \ J_{n+1}=J_n+67.72\)
Show Answers Only

\(D\)

Show Worked Solution
\(L_n\) \( = L_0 \times 1.0388^n\)  
\(L_5\) \( = 4000 \times 1.0388^5 \approx 4838.60\ \ \text{(2 d.p.)} \)  

 
\(\text{Check Li’s total versus each option:}\)

\(J_{n+1}\) \(= J_n + 267.72\)  
\(J_5\) \(= 3500 + 267.72 \times 5\)  
  \(= 4838.60\)  

 
\(\Rightarrow D\)


♦♦ Mean mark 49%.

CORE, FUR2 2021 VCAA 8

For renovations to the coffee shop, Sienna took out a reducing balance loan of $570 000 with interest calculated fortnightly.

The balance of the loan, in dollars, after `n` fortnights, `S_n` can be modelled by the recurrence relation
 

`S_0 = 570 \ 000,`                  `S_{n+1} = 1.001 S_n - 1193`
 

  1. Calculate the balance of this loan after the first fortnightly repayment is made.  (1 mark)
  2. Show that the compound interest rate for this loan is 2.6% per annum.  (1 mark)
  3. For the loan to be fully repaid, to the nearest cent, Sienna's final repayment will be a larger amount.
  4. Determine this final repayment amount.
  5. Round your answer to the nearest cent.  (1 mark)
Show Answers Only
  1. `$ 569 \ 377`
  2. `2.6text(%)`
  3. `$ 1198.56`
Show Worked Solution
a.   `S_1` `= 1.001 xx 570 \ 000 – 1193`
    `= $ 569 \ 377`


b.  `text{Fortnights in 1 year} = 26`

`text{Rate per fortnight} = (1.001 – 1) xx 100text(%) = 0.1text(%)`
 

`:.\ text(Annual compound rate)` `= 26 xx 0.1text(%)`
  `= 2.6text(%)`

 

c.  `text{Find}\ N \ text{by TVM Solver:}`

`N` `= ?`
`I text{(%)}` `= 2.6`
`PV` `= 570 \ 000`
`PMT` `= -1193`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 26`

 
`=> N = 650.0046 …`
 

`text{Find} \ FV \ text{after exactly 650 payments:}`

`N` `= 650`
`Itext{(%)}` `= 2.6`
`PV` `= 570 \ 000`
`PMT` `= -1193`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 26`

 
`=> FV = -5.59`
 

`:. \ text{Final repayment}` `= 1193 + 5.59`
  `= $ 1198.59`

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 23 MC

Consider the following four recurrence relations representing the value of an asset after `n` years, `V_n`.
 

  • `V_0 = 20\ 000, qquad V_(n+1) = V_n + 2500`
  • `V_0 = 20\ 000, qquad V_(n+1) = V_n - 2500`
  • `V_0 = 20\ 000, qquad V_(n+1) = 0.875 V_n`
  • `V_0 = 20\ 000, qquad V_(n+1) = 1.125V_n - 2500`

How many of these recurrence relations indicate that the value of an asset is depreciating?

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

`C`

Show Worked Solution
`V_(n+1)` `= V_n + 2500\ => \ text(appreciating)`
`V_(n-1)` `= V_n – 2500\ => \ text(depreciating)`
`V_(n+1)` `= 0.875 V_n\ => \ text(depreciating)`

 
`text(When)\ \ V_(n+1) = 1.125 V_n – 2500`

`V_1` `= 1.125 xx 20\ 000 – 2500`
  `= 20\ 000\ text{(same price)}`

 
`=>  C`

CORE, FUR1-NHT 2019 VCAA 19 MC

Consider the recurrence relation shown below.

`V_0 = 125\ 000,quadqquadV_(n + 1) = 1.013V_n - 2000`

This recurrence relation could be used to determine the value of

  1. a perpetuity with a payment of $2000 per quarter.
  2. an annuity with withdrawals of $2000 per quarter.
  3. an annuity investment with additional payments of $2000 per quarter.
  4. an item depreciating at a flat rate of 1.3% of the purchase price per quarter.
  5. a compound interest investment earning interest at the rate of 1.3% per annum.
Show Answers Only

`B`

Show Worked Solution

`text(The investment earns interest of 13% each period)`

`text(and a withdrawal of $2000 is made at the end of)`

`text(each period.)`

`=>\ B`

CORE, FUR2 2019 VCAA 8

Phil invests $200 000 in an annuity from which he receives a regular monthly payment.

The balance of the annuity, in dollars, after `n` months, `A_n`, can be modelled by the recurrence relation

`A_0 = 200\ 000, qquad A_(n + 1) = 1.0035\ A_n - 3700`

  1. What monthly payment does Phil receive?   (1 mark)
  2. Show that the annual percentage compound interest rate for this annuity is 4.2%.   (1 mark)

At some point in the future, the annuity will have a balance that is lower than the monthly payment amount.

  1. What is the balance of the annuity when it first falls below the monthly payment amount?

     

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

  2. If the payment received each month by Phil had been a different amount, the investment would act as a simple perpetuity.

     

    What monthly payment could Phil have received from this perpetuity?  (1 mark)

Show Answers Only
  1. `$3700`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `$92.15`
  4. `$700`
Show Worked Solution

a.  `$3700`

 

b.   `text(Monthly rate)` `= 0.0035 = 0.35%`
  `text(Annual rate)` `= 12 xx 0.35 = 4.2%`

 

c.  `text(Find)\ N\ text(when)\ FV = 0\ \ text{(by TVM solver)}:`

`N` `= ?`
`I(%)` `= 4.2`
`PV` `= 200\ 000`
`PMT` `= 3700`
`FV` `= 0`
`text(P/Y)` `= 12`
`text(C/Y)` `= 12`

 
`=> N = 60.024951`

 
`text(Find)\ \ FV\ \ text(when)\ \ N = 60.024951\ \ text{(by TVM solver):}`

`=>FV = $92.15`

 

d.  `text(Perpetuity) => text(monthly payment) = text(monthly interest)`

`:.\ text(Perpetuity payment)` `= 200\ 000 xx 4.2/(12 xx 100)`
  `= $700`

CORE, FUR1 2019 VCAA 18 MC

The value of a compound interest investment, in dollars, after `n` years, `V_n`, can be modelled by the recurrence relation shown below.

`V_0 = 100\ 000, qquad V_(n + 1) = 1.01 V_n`

The interest rate, per annum, for this investment is

  1.     `0.01 text(%)`
  2.     `0.101 text(%)`
  3.     `1 text(%)`
  4.     `1.01 text(%)`
  5. `101 text(%)`
Show Answers Only

`C`

Show Worked Solution

`V_(n + 1) = 1.01 V_n`

`text(Interest) = 1%`

`=>  C`

CORE, FUR2 2018 VCAA 5

After three years, Julie withdraws $14 000 from her account to purchase a car for her business.

For tax purposes, she plans to depreciate the value of her car using the reducing balance method.

The value of Julie’s car, in dollars, after `n` years, `C_n`, can be modelled by the recurrence relation shown below

`C_0 = 14\ 000, qquad C_(n + 1) = R xx C_n`

  1. For each of the first three years of reducing balance depreciation, the value of `R` is 0.85

     

    What is the annual rate of depreciation in the value of the car during these three years? (1 mark)

  2. For the next five years of reducing balance depreciation, the annual rate of depreciation in the value of the car is changed to 8.6%.

     

    What is the value of the car eight years after it was purchased?
    Round your answer to the nearest cent. (2 marks)

Show Answers Only
  1. `15 text(%)`
  2. `$5484.23\ text{(nearest cent)}`
Show Worked Solution

a.  `text(S)text(ince)\ \ R = 0.85,`

COMMENT: Note almost half of students answered incorrectly here!

`=> 85 text(% of the car’s value remains at the end of each)`

      `text{year (vs the value at the start of the same year.)}`

`:.\ text(Annual rate of depreciation) = 15 text(%)`

 

b.   `text(Value after 3 years)`

♦ Mean mark 42%.

`C_3` `= (0.85)^3 xx 14\ 000`
  `= $8597.75`

 
`:.\ text(Value after 8 years)`

`C_8` `= (0.914)^5 xx C_3`
  `= (0.914)^5 xx 8597.75`
  `= $5484.23\ text{(nearest cent)}`

CORE, FUR1 2018 VCAA 17-18 MC

The value of an annuity investment, in dollars, after `n` years, `V_n` , can be modelled by the recurrence relation shown below.

`V_0 = 46\ 000, quadqquadV_(n + 1) = 1.0034V_n + 500`

 
Part 1

What is the value of the regular payment added to the principal of this annuity investment?

  1.   $34.00
  2. $156.40
  3. $466.00
  4. $500.00
  5. $656.40

 
Part 2

Between the second and third years, the increase in the value of this investment is closest to

  1.   $656
  2.   $658
  3.   $661
  4. $1315
  5. $1975
Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Regular payment = $500)`

`=> D`
 

`text(Part 2)`

`V_1` `= 1.0034 xx 46\ 000 + 500 = $46\ 656.40`
`V_2` `= 1.0034 xx 46\ 656.40 + 500 = $47\ 315.03`
`V_3` `= 1.0034 xx 47\ 315.03 + 500 = $47\ 975.90`

 

`:.\ text(Increase)` `= 47\ 975.90 – 47\ 315.03`
  `= $660.87`

`=> C`