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Financial Maths, 2ADV M1 2012 HSC 15c

Ari takes out a loan of $360 000. The loan is to be repaid in equal monthly repayments, `$M`, at the end of each month, over 25 years (300 months). Reducible interest is charged at 6% per annum, calculated monthly.

Let  `$A_n`  be the amount owing after the `n`th repayment.

  1. Write down an expression for the amount owing after two months, `$A_2`.   (1 mark)

  2. Show that the monthly repayment is approximately $2319.50.   (2 marks)

  3. After how many months will the amount owing, `$A_n`, become less than $180 000.   (3 marks)

Show Answers Only
  1.  `$A_2=(360\ 000)(1.005^2)-M(1+1.005)`
  2. `text{Proof (See Worked Solutions)}`
  3. `202\ text(months)`
Show Worked Solutions
i.    `A_1` `=360\ 000(1+text(6%)/12)-M`
  `=360\ 000(1.005)-M`
`A_2` `=[360\ 000(1.005)-M](1.005)-M`
  `=360\ 000(1.005^2)-M(1.005)-M`
  `=360\ 000(1.005^2)-M(1+1.005)`

 

ii.  `A_n=360\ 000(1.005^n)-M(1+1.005^1+ … +1.005^(n-1))`

`text(When)\  n=300,\ A_n=0`

`0=360\ 000(1.005^300)-M(1+1.005^1+….+1.005^299)`

`360\ 000(1.005^300)` `=M((a(r^n-1))/(r-1))`
`M((1(1.005^300-1))/(1.005-1))` `=360\ 000(1.005^300)`
`:.M` `=((1\ 607\ 389.13)/692.994)`
  `~~2319.50\ \ \ text(… as required)`

 

iii.   `text(Find)\ n\ text(such that)\  $A_n<$180\ 000`

`360\ 000(1.005^n)-2319.50((1.005^n-1)/(1.005-1))` `<180\ 000`
`360\ 000(1.005^n)-463\ 900(1.005^n-1)` `<180\ 000`
`-103\ 900(1.005^n)+463,900` `<180\ 000`
♦ Mean mark 38%
MARKER’S COMMENT: Challenging calculations using logarithms are common in this topic. A high percentage of students consistently struggle in this area.
`103\ 900(1.005^n)` `>283\ 900`
`1.005^n` `>(283\ 900)/(103\ 900)`
`n(ln1.005)` `>ln((283\ 900)/(103\ 900))`
`n` `>1.005193/0.0049875`
`n` `>201.54`

 

`:.\ text(After 202 months,)\  $A_n< $180\ 000.`

Probability, 2ADV S1 2013 HSC 15d

Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

  1. Find the probability that Pat wins the game on the first throw.     (1 mark)

  2. What is the probability that Pat wins the game on the first or on the second throw?     (2 marks)

  3. Find the probability that Pat eventually wins the game.     (2 marks)

Show Answers Only
  1. `1/36`
  2. `(2521)/(46\ 656)\ \ text(or)\ \ 0.054`
  3. `36/71`
Show Worked Solutions

i.   `P\ text{(Pat wins on 1st throw)}=P(W)`

`P(W)` `=P\ text{(Pat throws 2 sixes)}`
  `=1/6 xx 1/6`
  `=1/36`

 

ii.  `text(Let)\ P(L)=P text{(loss for either player on a throw)}=35/36`

`P text{(Pat wins on 1st or 2nd throw)}` 

♦♦ Mean mark 33%
MARKER’S COMMENT: Many students did not account for Chandra having to lose when Pat wins on the 2nd attempt.

`=P(W) + P(LL W)`

`=1/36\ + \ (35/36)xx(35/36)xx(1/36)`

`=(2521)/(46\ 656)`

`=0.054\ \ \ text{(to 3 d.p.)}`

 

iii.  `P\ text{(Pat wins eventually)}`

`=P(W) + P(LL\ W)+P(LL\ LL\ W)+ … `

`=1/36\ +\ (35/36)^2 (1/36)\ +\ (35/36)^2 (35/36)^2 (1/36)\ +…`

 
`=>\ text(GP where)\ \ a=1/36,\ \ r=(35/36)^2=(1225)/(1296)`

♦♦♦ Mean mark 8%!
 COMMENT: Be aware that diminishing probabilities and `S_oo` within the Series and Applications are a natural cross-topic combination.

 
`text(S)text(ince)\ |\ r\ |<\ 1:`

`S_oo` `=a/(1-r)`
  `=(1/36)/(1-(1225/1296))`
  `=1/36 xx 1296/71`
  `=36/71`

 

`:.\ text(Pat’s chances to win eventually are)\  36/71`.

Financial Maths, 2ADV M1 2013 HSC 13d

A family borrows $500 000 to buy a house. The loan is to be repaid in equal monthly instalments. The interest, which is charged at 6% per annum, is reducible and calculated monthly. The amount owing after  `n`  months, `$A_n`, is given by

`qquad qquadA_n=Pr^n-M(1+r+r^2+ \ .... +r^(n-1))\ \ \ \ \ \ \ \ \ ` (DO NOT prove this)

where  `$P`  is the amount borrowed, `r=1.005`  and  `$M`  is the monthly repayment.

  1. The loan is to be repaid over 30 years. Show that the monthly repayment is $2998 to the nearest dollar.     (2 marks)

  2. Show that the balance owing after 20 years is $270 000 to the nearest thousand dollars.             (1 mark)

After 20 years the family borrows an extra amount, so that the family then owes a total of $370 000. The monthly repayment remains $2998, and the interest rate remains the same.

  1. How long will it take to repay the $370 000?     (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text(193 months)`
Show Worked Solutions

i.    `text(Find)\  $M\  text(such that the loan is repaid over 30 years.)`

`n=30xx12=360\ text(periods)\ \ \   r=1+6/12%=1.005`

`A_360` `=500\ 000 (1.005^360)-M(1+1.005+..+1.005^359)=0`

`=>GP\ text(where)\ a=1,\ \ r=1.005,\ \ \ n=360`

`M((1(1.005^360-1))/(1.005-1))` `=500\ 000(1.005^360)`
`M(1004.515)` `=3\ 011\ 287.61`
`M` `=2997.75`

 

`:.$M=$2998\ \ text{(nearest dollar) … as required}`

 

 ii.    `text(Find)\  $A_n\ text(after 20 years)\ \ \ =>n=20xx12=240` 

`A_240` `=500\ 000(1.005^240)-2998(1+1.005+..+1.005^239)`
  `=1\ 655\ 102.24-2998((1(1.005^240-1))/(1.005-1))`
  `=269\ 903.63`
  `=270\ 000\ \ text{(nearest thousand) … as required}`
MARKER’S COMMENT: Within the GP formula, many students incorrectly wrote the last term as `1.005^240` rather than `1.005^239`. Note `T_n=ar^(n-1)`.

 

 

iii.  `text(Loan)=$370\ 000`

`text(Find)\  n\  text(such that)\  $A_n=0,\ \ \ M=$2998`

`A_n=370\ 000(1.005^n)-2998(1+1.005+..+1.005^(n-1))=0`

♦♦ Mean mark 33%
COMMENT: Another good examination of working with logarithms. Students should understand why they must ’round up’ their answer in this question.
`370\ 000(1.005^n)` `=2998((1(1.005^n-1))/(1.005-1))` 
`370\ 000(1.005^n)` `=599\ 600(1.005^n-1)`
`229\ 600(1.005^n)` `=599\ 600`
`ln1.005^n` `=ln((599\ 600)/(229\ 600))`
`n` `=ln2.6115/ln1.005`
`n` `=192.4`

 
`:.\ text(The loan will be repaid after 193 months.)`

Financial Maths, 2ADV M1 2013 HSC 12c

Kim and Alex start jobs at the beginning of the same year. Kim's annual salary in the first year is $30,000 and increases by 5% at the beginning of each subsequent year. Alex's annual salary in the first year is $33,000, and increases by $1,500 at the beginning of each subsequent year.

  1. Show that in the 10th year, Kim's annual salary is higher than Alex's annual salary.     (2 marks)

  2. In the first 10 years how much, in total, does Kim earn?     (2 marks)

  3. Every year, Alex saves `1/3` of her annual salary. How many years does it take her to save $87,500?     (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `$377\ 336.78`
  3. `text(7  years)`
Show Worked Solutions

i.    `text(Let)\ \ K_n=text(Kim’s salary in Year)\  n`

`{:{:(K_1=a=30\ 000),(K_2=ar^1=30\ 000(1.05^1)):}}{:(\ =>\ GP),(\ \ \ \ \ \ a=30\ 000),(\ \ \ \ \ \ r=1.05):}`

`vdots`

`:.K_10=ar^9=30\ 000(1.05)^9=$46\ 539.85`

 

`text(Let)\ \ A_n=text(Alex’s salary in Year)\ n`

`{:{:(A_1=a=33\ 000),(A_2=33\ 000+1500=34\ 500):}}{:(\ =>\ AP),(\ \ \ \ \ \ a=33\ 000),(\ \ \ \ \ \ d=1500):}`

`vdots`

`A_10=a+9d=33\ 000+1500(9)=$46\ 500`

`=>K_10>A_10`
 

`:.\ text(Kim earns more than Alex in the 10th year)`

 

ii.    `text(In the first 10 years, Kim earns)`

`K_1+K_2+\  ….+ K_10`

`S_10` `=a((r^n-1)/(r-1))`
  `=30\ 000((1.05^10-1)/(1.05-1))`
  `=377\ 336.78`

 

`:.\ text(In the first 10 years, Kim earns $377 336.78)`

 

iii.   `text(Let)\ T_n=text(Alex’s savings in Year)\ n`

`{:{:(T_1=a=1/3(33\ 000)=11\ 000),(T_2=a+d=1/3(34\ 500)=11\ 500),(T_3=a+2d=1/3(36\ 000)=12\ 000):}}{:(\ =>\ AP),(\ \ \ \ a=11\ 000),(\ \ \ \ d=500):}`
 

`text(Find)\ n\ text(such that)\ S_n=87\ 500`

♦ Mean mark 45%.
IMPORTANT: Using the AP sum formula to create and then solve a quadratic in `n` is challenging and often examined. Students need to solve and interpret the solutions.
`S_n` `=n/2[2a+(n-1)d]`
`87\ 500` `=n/2[22\ 000+(n-1)500]`
`87\ 500` `=n/2[21\ 500+500n]`
`250n^2+10\ 750n-87\ 500` `=0`
`n^2+43n-350` `=0`
`(n-7)(n+50)` `=0`

 
`:.n=7,\ \ \ \ n>0`

`:.\ text(Alex’s savings will be $87,500 after 7 years).`