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Financial Maths, 2ADV M1 2023 HSC 25

On the first day of November, Jia deposits $10 000 into a new account which earns 0.4% interest per month, compounded monthly. At the end of each month, after the interest is added to the account, Jia intends to withdraw \($M\) from the account.

Let \(A_n\) be the amount (in dollars) in Jia's account at the end of \(n\) months.

  1. Show that  \(A_2 = 10\ 000(1.004)^2-M(1.004)-M\).  (1 mark)

  2. Show that  \(A_n = (10\ 000-250M)(1.004)^n + 250M\).  (3 marks)

  3. Jia wants to be able to make at least 100 withdrawals.
  4. What is the largest value of \(M\) that will enable Jia to do this?  (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \($121.52\)
Show Worked Solution

a.    \(A_1=10\ 000(1.004)-M\)

\(A_2\) \(=A_1(1.004)-M\)  
  \(=[(10\ 000(1.004)-M)](1.004)-M\)  
  \(=10\ 000(1.004)^2-M(1.004)-M\)  

 
b.    \(A_3=10\ 000(1.004)^3-M(1.004)^2-M(1.004)-M\)

\( \vdots\)

♦ Mean mark (b) 44%.
\(A_n\) \(=10\ 000(1.004)^n-M(1.004)^{n-1}-…-M(1.004)-M\)  
  \(=10\ 000(1.004)^n-M \underbrace{(1+1.004+1.004^2+…+1.004^{n-1})}_{\text{GP where}\ a=1, r=1.004, n=n} \)  
\(A_n\) \(=10\ 000(1.004)^n-M(\dfrac{1.004^n-1}{1.004-1}) \)  
  \(=10\ 000(1.004)^n-M(\dfrac{1.004^n-1}{0.004}) \)  
  \(=10\ 000(1.004)^n-250M(1.004^n-1) \)  
  \(=10\ 000(1.004)^n-250M(1.004)^n+250M \)  
  \(=(10\ 000-250M)(1.004)^n + 250M\)  

 
c.    \(\text{Find}\ M\ \text{when}\ A_{100}=0: \)

\(A_{100}\) \(= (10\ 000-250M)(1.004)^{100}+250M\)  
\(0\) \(=10\ 000(1.004)^{100}-250M(1.004)^{100}+250M \)  
\(250M(1.004^{100}-1)\) \(=10\ 000(1.004)^{100} \)  
\(M\) \(=\dfrac{10\ 000(1.004)^{100}}{250(1.004^{100}-1)} \)  
  \(=121.527…\)  

 
\( \therefore M_\text{max} = $121.52\)

Mean mark (c) 53%.

Financial Maths, 2ADV M1 2021 HSC 29

  1. On the day that Megan was born, her grandfather deposited $5000 into an account earning 3% per annum compounded annually. On each birthday after this, her grandfather deposited $1000 into the same account, making his final deposit on Megan’s 17th birthday. That is, a total of 18 deposits were made.
  2. Let `A_n` be the amount in the account on Megan’s `n`th birthday, after the deposit is made.
  3. Show that  `A_3 = $8554.54`.  (2 marks)

  4. On her 17th birthday, just after the final deposit is made, Megan has $30 025.83 in her account. You are NOT required to show this.
  5. Megan then decides to leave all the money in the same account continuing to earn interest at 3% per annum compounded annually. On her 18th birthday, and on each birthday after this, Megan withdraws $2000 from the account.
  6. How many withdrawals of $2000 will Megan be able to make?  (3 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `20`
Show Worked Solution

a.   `A_1 = 5000(1.03) + 1000`

`A_2` `= [5000(1.03) + 1000](1.03) + 1000`
  `= 5000(1.03)^2 + 1000(1.03) + 1000`
`A_3` `= 5000(1.03)^3 + 1000(1.03)^2 + 1000(1.03) + 1000`
  `= 8554.535`
  `= $8554.54`

 

b.   `text(Megan’s 18th birthday onwards:)`

`A_1 = 30\ 025.83(1.03) – 2000`

`A_2` `= [30\ 025.83(1.03) – 2000](1.03) – 2000`
  `= 30\ 025.83(1.03)^2 – 2000(1.03) – 2000`
  `vdots`
`A_n` `= 30\ 025.83(1.03)^n – 2000(1 + 1.03 + 1.03^2 + … + 1.03^(n – 1))`

 

`text(Find)\ n\ text(when)\ \ A_n = 0:`

`30\ 025.83(1.03)^n` `= 2000[(a(r^n – 1))/(r – 1)]`
  `= 2000((1.03^n – 1)/(1.03 – 1))`
`30\ 025.83(1.03)^n` `= 66\ 666.67(1.03)^n – 66\ 666.67`
`36\ 640.84(1.03)^n` `= 66\ 666.67`
`1.03^n` `= (66\ 666.67)/(36\ 640.84)`
`n ln 1.03` `= ln ((66\ 666.67)/(36\ 640.84))`
`n` `= (ln ((66\ 666.67)/(36\ 640.84)))/(ln 1.03)`
  `= 20.24…`

 
`:.\ text(Megan can make 20 withdrawals of $2000.)`

Financial Maths, 2ADV M1 2020 HSC 26

Tina inherits $60 000 and invests it in an account earning interest at a rate of 0.5% per month. Each month, immediately after the interest has been paid, Tina withdraws $800.

The amount in the account immediately after the `n`th withdrawal can be determined using the recurrence relation

`A_n = A_(n - 1)(1.005) - 800`,

where `n = 1, 2, 3, …`  and  `A_0 = 60\ 000`

  1. Use the recurrence relation to find the amount of money in the account immediately after the third withdrawal.  (2 marks)

  2. Calculate the amount of interest earned in the first three months.  (2 marks)

  3. Calculate the amount of money in the account immediately after the 94th withdrawal.  (3 marks) 

Show Answers Only
  1. `$58\ 492.49`
  2. `$892.49`
  3. `$187.85`
Show Worked Solution
a.    `A_1` `= 60\ 000(1.005) – 800 = $59\ 500`
  `A_2` `= 59\ 500(1.005) – 800 = $58\ 997.50`
  `A_3` `= 58\ 997.50(1.005) – 800 = $58\ 492.49`

 

b.   `text{Amount (not interest)}`

`= 60\ 000 – (3 xx 800)`

`= $57\ 600`
 

`:.\ text(Interest earned in 3 months)`

`= A_3 – 57\ 600`

`= 58\ 492.49 – 57\ 600`

`= $892.49`
 

c.   `A_1 = 60\ 000(1.005) – 800`

`A_2` `= [60\ 000(1.005) – 800](1.005) – 800`
  `= 60\ 000(1.005)^n – 800(1.005 + 1)`
`vdots`  
`A_n` `= 60\ 000(1.005)^n – 800(1 + 1.005 + … + 1.005^(n – 1))`
`A_94` `= 60\ 000(1.005)^94 – 800\ underbrace((1 + 1.005 + … + 1.005^93))_(text(GP where)\ a = 1,\ r = 1.005,\ n = 94)`
  `= 60\ 000(1.005)^94 – 800 ((1(1.005^94 – 1))/(1.005 – 1))`
  `= 60\ 000(1.005)^94 – 160\ 000(1.005^94 – 1)`
  `= $187.85`

Financial Maths, 2ADV M1 2019 HSC 16a

A person wins $1 000 000 in a competition and decides to invest this money in an account that earns interest at 6% per annum compounded quarterly. The person decides to withdraw $80 000 from this account at the end of every fourth quarter. Let  `A_n`  be the amount remaining in the account after the `n`th withdrawal.

  1.  Show that the amount remaining in the account after the withdrawal at the end of the eighth quarter is

     

    `qquad A_2 = 1\ 000\ 000 xx 1.015^8 - 80\ 000(1 + 1.015^4)`.  (2 marks)

  2.  For how many years can the full amount of $80 000 be withdrawn?  (3 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `23\ text(years)`
Show Worked Solution

i.  `6 text(% p.a.)= 1.5 text(%)\ text(per quarter)`

  `A_1` `= 1\ 000\ 000 xx 1.015^4 – 80\ 000`
  `A_2` `= A_1 xx 1.015^4 – 80\ 000`
    `= (1\ 000\ 000 xx 1.015^4 – 80\ 000) xx 1.015^4 – 80\ 000`
    `= 1\ 000\ 000 xx 1.015^8 – 80\ 000 xx 1.015^4 – 80\ 000`
    `= 1\ 000\ 000 xx 1.015^8 – 80\ 000 (1 + 1.015^4)`

 

ii.    `A_3` `= 1\ 000\ 000 xx 1.015^12 – 80\ 000 (1 + 1.015^4 + 1.015^8)`
  `vdots`  
  `A_n` `= 1\ 000\ 000 xx 1.015^(4n) – 80\ 000\ underbrace{(1 + 1.015^4 + … + 1.015^(4n – 4))}_{text(GP where)\ a = 1,\ r = 1.015^4}`
    `= 1\ 000\ 000 xx 1.015^(4n) – 80\ 000 [(1(1.015^(4n) – 1))/(1.015^4 – 1)]`
    `= 1\ 000\ 000 xx 1.015^(4n) – 1\ 303\ 706 (1.015^(4n) – 1)`

♦♦♦ Mean mark part (ii) 19%.

`text(Find)\ \ n\ \ text(when)\ \ A_n =0:`

`0` `= 1\ 000\ 000 xx 1.015^(4n) – 1\ 303\ 706 xx 1.015^(4n) + 1\ 303\ 706`
`-1\ 303\ 706` `> -303\ 706 xx 1.015^(4n)`
`1.015^(4n)` `> (1\ 303\ 706)/(303\ 706)`
`4n` `> (ln((1\ 303\ 706)/(303\ 706)))/(ln 1.015)`
  `> 97.853…`
`n` `> 24.46…`

 
`:.\ text(Full amount can be drawn for 24 years.)`

Financial Maths, 2ADV M1 2018 HSC 16c

Kara deposits an amount of $300 000 into an account which pays compound interest of 4% per annum, added to the account at the end of each year. Immediately after the interest is added, Kara makes a withdrawal for expenses for the coming year. The first withdrawal is `$P`. Each subsequent withdrawal is 5% greater than the previous one.

Let  `$A_n`  be the amount in the account after the `n`th withdrawal.

  1. Show that  `A_2 = 300\ 000(1.04)^2 - P[(1.04) + (1.05)]`.  (1 mark)

  2. Show that  `A_3 = 300\ 000 (1.04)^3 - P[(1.04)^2 + (1.04)(1.05) + (1.05)^2]`.  (1 mark)

  3. Show that there will be money in the account when
     
    `qquad (105/104)^n < 1 + 3000/P`  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

♦ Mean mark 47%.

i.   `A_1` `= 300\ 000 (1.04) – P`
  `A_2` `= [300\ 000 (1.04) – P](1.04) – P(1.05)`
    `= 300\ 000 (1.04)^2 – P(1.04) – P(1.05)`
    `= 300\ 000 (1.04)^2 – P[1.04 + 1.05]`

 

♦ Mean mark part (ii) 30%.

ii.   `A_3` `= [300\ 000 (1.04)^2 – P(1.04 + 1.05)](1.04) – P(1.05)^2`
    `= 300\ 000 (1.04)^3 – P(1.04)^2 – P(1.04)(1.05) – P(1.05)^3`
    `= 300\ 000 (1.04)^3 – P[(1.04)^2 + P(1.04)(1.05) + (1.05)^2]`

 

iii.   `A_4` `= 300\ 000 (1.04)^4 – P[(1.04)^3 + (1.04)^2(1.05) + … + (1.05)^3]`
    `vdots`
  `A_n` `= 300\ 000 (1.04)^n`
    `- P underbrace{[(1.04)^(n-1) + (1.04)^(n-2) (1.05) + … + (1.04)(1.05)^(n-2) + (1.05)^(n-1)]}_{text(GP),\ a = (1.04)^(n-1),\ r = 1.05/1.04}`

 
`text(Money in account when)\ \ A_n > 0:`

♦♦♦ Mean mark part (iii) 8%.

`(300\ 000(1.04)^n)/P` `> (1.04)^(n-1)[((1.05/1.04)^n-1)/((1.05/1.04)-1)]`
`(300\ 000)/P` `> 1/1.04 [((1.05/1.04)^n-1)/((1.05/1.04)-1)]`
`(300\ 000)/P` `> ((1.05/1.04)^n – 1)/(1.05 – 1.04)`
`3000/P` `> (1.05/1.04)^n – 1`
`3000/P + 1` `> (105/104)^n qquad (text{s}text{ince}\ \ (1.05/1.04)^n = (105/104)^n)`

Financial Maths, 2ADV M1 2008 HSC 9b

Peter retires with a lump sum of $100 000. The money is invested in a fund which pays interest each month at a rate of 6% per annum, and Peter receives a fixed monthly payment `$M` from the fund. Thus the amount left in the fund after the first monthly payment is   `$(100\ 500-M)`.

  1. Find a formula for the amount, `$A_n`, left in the fund after `n\ ` monthly payments.   (2 marks)

  2. Peter chooses the value of `M` so that there will be nothing left in the fund at the end of the 12th year (after 144 payments). Find the value of `M`.   (3 marks)

Show Answers Only
  1.  `100\ 000(1.005^n)-M((1.005^n-1)/0.005)`
  2. `$975.85`
Show Worked Solutions

i.    `r=(1+0.06/12)=1.005`

MARKER’S COMMENT: Students who developed this answer from writing the calculations of `A_1`, `A_2`, `A_3` to then generalising for `A_n` were the most successful.
`A_1` `=(100\ 500-M)`
  `=100\ 000(1.005)^1-M`
`A_2` `=A_1 (1.005)-M`
  `=[100\ 000(1.005^1)-M](1.005)-M`
  `=100\ 000(1.005^2)-M(1.005)-M`
  `=100\ 000(1.005^2)-M(1+1.005)`
`A_3` `=100\ 000(1.005^3)-M(1+1.005^1+1.005^2)`

`\ \ \ \ \ vdots`

`A_n` `=100\ 000(1.005^n)-M(1+1.005+\ …\ +1.005^(n-1))`
  `=100\ 000(1.005^n)-M((a(r^n-1))/(r-1))`
  `=100\ 000(1.005^n)-M((1(1.005^n-1))/(1.005-1))`
  `=100\ 000(1.005^n)-M((1.005^n-1)/0.005)`

 

ii.  `text(Find)\ M\  text(such that)\ \ $A_n=0\ \ text(when)\ \ n=144`

`A_144=100\ 000(1.005)^144-M((1.005^144-1)/0.005)=0`

`M((1.005^144-1)/0.005)` `=100\ 000(1.005^144)`
`M(1.005^144-1)` `=500(1.005^144)`
`M` `=(500(1.005^144))/(1.005^144-1)`
  `=975.85`

 

`:. M=$975.85\ \ text{(nearest cent).}`

Financial Maths, 2ADV M1 2010 HSC 9a

  1. When Chris started a new job, $500 was deposited into his superannuation fund at the beginning of each month. The money was invested at 0.5% per month, compounded monthly. 

     

    Let  `$P`  be the value of the investment after 240 months, when Chris retires.

     

    Show that  `P=232\ 175.55`     (2 marks)

  2. After retirement, Chris withdraws $2000 from the account at the end of each month, without making any further deposits. The account continues to earn interest at 0.5% per month.

     

    Let  `$A_n`  be the amount left in the account  `n`  months after Chris's retirement.

     

      (1)  Show that  `A_n=(P-400\ 000)xx1.005^n+400\ 000`.     (3 marks)

     

      (2)  For how many months after retirement will there be money left in the account?     (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. (1)  `text{Proof (See Worked Solutions)}`
  3. (2)  `text(175 months)`
Show Worked Solutions
i.     `P_1` `=500(1.005)`
`P_2` `=500(1.005^2)+500(1.005^1)`
`P_3` `=500(1.005^3+1.005^2+1.005)`
  `vdots`
`P_240` `=500(1.005+1.005^2+1.005^3 …+1.005^240)`

 

`=>\ text(GP where)\ \ a=1.005,\ text(and)\ \ r=1.005`

MARKER’S COMMENT: Common errors included using `r=1.05`, and taking the first term of the GP as 1 instead of 1.005 (note that the $500 goes in at the start of the month and earns interest before it is included in `$P_n`).
`P_240` `=500((a(r^n-1))/(r-1))`
  `=500((1.005(1.005^240-1))/(1.005-1))`
  `=100\ 000[1.005(1.005^240-1)]`
  `=232\ 175.55`

 

`:.\ text(The value of Chris’ investment after 240 months)`

`text(is) \ $232\ 175.55 text(  … as required)`

 

ii. (1)  `text(After 1 month,)\  A_1=P(1.005)-2000`

IMPORTANT: At the end of the month, `$P` earns interest for the month BEFORE any withdrawal is made. Many students mistakenly had `$A_1=(P-2000)(1.005)`.
`A_2` `=A_1(1.005)-2000`
  `=[P(1.005)-2000](1.005)-2000`
  `=P(1.005^2)-2000(1.005)-2000`
  `=P(1.005^2)-2000(1+1.005)`
  ` vdots`
`A_n` `=P(1.005^n)-2000(1+1.005+…+1.005^(n-1))`

`=>\ text(GP where)\ \ a=1\ \ text(and)\ \ r=1.005`

`A_n` `=P(1.005^n)-2000((1(1.005^n-1))/(1.005-1))`
  `=P(1.005^n)-400\ 000(1.005^n-1)`
  `=P(1.005^n)-400\ 000(1.005^n)+400\ 000`
  `=(P-400\ 000)xx1.005^n+400\ 000\ \ text(… as required)`

 

ii. (2)  `text(Find)\ n\ text(such that)\ A_n<=0`

♦ Mean mark 38%

`text(S)text(ince)\  P=232\ 175.55`,

`(232\ 175.55-400\ 000)(1.005^n)+400\ 000<=0`

`167\ 824.45(1.005^n)` `>=400\ 000`
`1.005^n` `>=(400\ 000)/(167\ 824.45)`
`n ln1.005` `>=ln2.383443`
`n` `>=ln2.383443/ln1.005`
`n` `>=174.14\ \ text{(to 2 d.p.)}`

 

`:.\ text(There will be money left in the account for 175 months.)`