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Financial Maths, 2ADV M1 2009 HSC 8b

One year ago Daniel borrowed $350 000 to buy a house. The interest rate was 9% per annum, compounded monthly. He agreed to repay the loan in 25 years with equal monthly repayments of $2937.

  1. Calculate how much Daniel owed after his first monthly repayment.    (1 mark)

Daniel has just made his 12th monthly repayment. He now owes $346 095. The interest rate now decreases to 6% per annum, compounded monthly.

 

The amount  `$A_n`, owing on the loan after the `n`th monthly repayment is now calculated using the formula  
 
`qquad qquad A_n=346,095xx1.005^n-1.005^(n-1)M-\ ... -1.005M-M`
  
where `$M` is the monthly repayment, and `n=1,2,\ ...,288`.   (DO NOT prove this formula.)

  1. Calculate the monthly repayment if the loan is to be repaid over the remaining 24 years (288 months).    (3 marks)

  2. Daniel chooses to keep his monthly repayments at $2937. Use the formula in part (ii) to calculate how long it will take him to repay the $346 095.   (3 marks)

  3. How much will Daniel save over the term of the loan by keeping his monthly repayments at $2937, rather than reducing his repayments to the amount calculated in part (ii)?   (1 mark)

Show Answers Only
  1. `$349\ 688`
  2. `$2270.31\ \ text{(nearest cent)}`
  3. `178.37\ text(months)`
  4. `$129\ 976.59\ \ text{(nearest cent)}`
Show Worked Solutions

i.   `text(Let)\ L_n= text(the amount owing after)\ n\ text(months)`

`text(Repayment)\ =M=$2937\ \ text(and)\ \ r=text(9%)/12=0.0075\ text(/month)`

`:.\ L_1` `=350\ 000(1+r)-M`
  `=350\ 000(1.0075)-2937`
  `=349\ 688`

 

`:.\ text(After 1 month, the amount owing is)\  $349\ 688`

 

ii.   `text(After 12 repayments, Daniel owes)\ $346\ 095,\  text(and)\ r darr 6%`

`:.\ r=(6%)/12=0.005`

`text(Loan is repaid over the next 24 years. i.e.)\ $A_n=0\ text(when)\  n=288`

`A_n` `=346\ 095(1.005^n)-1.005^(n-1)M-\ ..\ -1.005M-M`
  `=346\ 095(1.005^n)-M(1+1.005+..+1.005^(n-1))`
`A_288` `=346\ 095(1.005^288)-M(1+1.005+..+1.005^287)=0`

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

MARKER’S COMMENT: Careless setting out and poor handwriting, especially where indexes were involved, was a major contributor to errors in this question.

`M((1(1.005^288-1))/(1.005-1))=346\ 095(1.005^288)`

`M` `=(1\ 455\ 529.832)/641.1158`
  `=2270.31`

 

`:.\ text{Monthly repayment is $2270.31  (nearest cent)}` 

 

iii.  `text(Given)\ $M\ text(remains at $2937, find)\  n\ text(such that)`

`$A_n=0\ text{(i.e. loan fully paid off)}`

`:. 346\ 095(1.005^n)-2937((1(1.005^n-1))/(1.005-1))` `=0`
`346\ 095(1.005^n)-587\ 400(1.005^n-1)` `=0`
`(346\ 095-587\ 400)(1.005^n)+587\ 400` `=0`
♦♦ A poorly answered question.
MARKER’S COMMENT: Many students struggled to handle the exponential and logarithm calculations in this question.
ALGEBRA TIP: Dividing by `(1.005-1)` in part (iii) is equivalent to multiplying by 200, and cleans up working calculations (see Worked Solutions).
 

`241\ 305(1.005^n)` `=587\ 400`
`ln1.005^n` `=ln((587\ 400)/(241\ 305))`
`n` `=ln2.43426/ln1.005`
  `=178.37..`

 

`:.\ text{He will pay off the loan in 179 months (note the}`

`text{last payment will be a partial payment).}`

 

iv.  `text(Total paid at $2937 per month)`

`= 2937xx178.37=$523\ 872.69`

`text(Total paid at $2270.31 per month)`

`=2270.31xx288=$653,849.28`

 

`:.\ text(The amount saved)`

`=653\ 849.28-523\ 872.69`

`=$129\ 976.59\ \ 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.)`

Financial Maths, 2ADV M1 2010 HSC 4a

Susanna is training for a fun run by running every week for 26 weeks. She runs 1 km  in the first week and each week after that she runs 750 m more than the previous week, until she reaches 10 km in a week. She then continues to run 10 km each week.

  1. How far does Susannah run in the 9th week?     (1 mark)

  2. In which week does she first run 10 km?     (1 mark)

  3. What is the total distance that Susannah runs in 26 weeks?     (2 marks)

Show Answers Only
  1. `7\ text(km)`
  2. `13 text(th week)`
  3. `201.5\ text(km)`
Show Worked Solutions

i.    `T_1=a=1`

`T_2=a+d=1.75`

`T_3=a+2d=2.50`

`=>\ text(AP where)\  a=1  \ \ d=0.75`

`\ \ vdots`

`T_9` `=a+8d`
  `=1+8(0.75)`
  `=7`

 

`:.\ text(Susannah runs 7 km in the 9th week.)`

 

ii.  `text(Find)\ n\ text(such that)\ T_n=10\ text(km)`

`text(Using)\ T_n=a+(n-1)d`

MARKER’S COMMENT: Better responses wrote the formula for the `nth` term before clearly substituting in known values `a` and `d`.
`1+(n-1)(0.75)` `=10`
`0.75n-0.75` `=9`
`n` `=9.75/0.75`
  `=13`

 

`:.\ text(Susannah runs 10 km for the first time in the 13th Week.)`

 

iii.  `text{Let D = the total distance Susannah runs in 26 weeks}`

MARKER’S COMMENT: Many students incorrectly calculated `S_26`, not taking into account the AP stopped at the 13th term.
`text(D)` `=S_13+13(10)`
  `=n/2[2a+(n-1)d]+13(10)`
  `=13/2[2(1)+(13-1)(0.75)]+130`
  `=13/2(2+9)+130`
  `=201.5`

 

`:.\ text(Susannah runs a total of 201.5 km in 26 weeks.)`

Financial Maths, 2ADV M1 2011 HSC 8c

When Jules started working she began paying $100 at the beginning of each month into a superannuation fund.

The contributions are compounded monthly at an interest rate of 6% per annum.

She intends to retire after having worked for 35 years.

  1. Let  `$P`  be the final value of Jules's superannuation when she retires after 35 years (420 months). Show that  `$P=$143\ 183`  to the nearest dollar.     (2 marks)

  2. Fifteen years after she started working Jules read a magazine article about retirement, and realised that she would need `$800\ 000` in her fund when she retires. At the time of reading the magazine article she had `$29\ 227` in her fund. For the remaining 20 years she intends to work, she decides to pay  `$M`  into her fund at the beginning of each month. The contributions continue to attract the same interest rate of 6% per annum, compounded monthly.

  3.  

    At the end of  `n`  months after starting the new contributions, the amount in the fund is  `$A_n`.

  4.  

      (1)  Show that  `A_2=29\ 227xx1.005^2+M(1.005+1.005^2)`.     (1 mark)

  5.  

      (2)  Find the value of  `M`  so that Jules will have $800 000 in her fund after the remaining 20 years (240 months).     (3 marks)

Show Answers Only
  1.  `text{Proof (See Worked Solutions)}`
  2. (1) `text{Proof (See Worked Solutions)}`
    (2) `$1514.48\ text{(nearest cent)}`
Show Worked Solutions
i.    `P_1` `=Pxxr=Pxx(1+(6%)/12)=100(1.005)`
  `P_2` `=P_1(1.005)+100(1.005^1)`
    `=100(1.005^2)+100(1.005^1)`
    `=100(1.005^2+1.005)`
  `P_3` `=(1.005)[100(1.005^2+1.005)]+100(1.005)`
    `=100(1.005+1005^2+1.005^3)`
    `\ \ \ \ vdots`
  `P_420` `=100(1.005+1.005^2+1.005^3 …+1.005^420)`

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

MARKER’S COMMENT: Common errors in this part included having the first term of the GP as 1 instead of 1.005 (note that the $100 goes in at the start of the month and earns interest before it is included in `$P_n)`.
`P_420` `=100((a(r^n-1))/(r-1))`
  `=100((1.005(1.005^420-1))/(1.005-1))`
  `=20\ 000(1.005(1.005^420-1))`
  `=$143\ 183.39`

 

`:.\ text{The final value of Jules’s superannuation is}`

`$143\ 183\ \ text{(to the nearest dollar)   … as required}`

 

♦ Mean mark 34% for part (ii)(1)

ii. (1)  `text(After 1 month,)\  A_1=29\ 227(1.005)+M(1.005)`

`A_2` `=A_1 (1.005)+M(1.005)`
  `=[29\ 227(1.005)+M(1.005)](1.005)+M(1.005)`
  `=29\ 227(1.005^2)+M(1.005^2)+M(1.005)`
  `=29\ 227(1.005^2)+M(1.005+1.005^2)\ \ text(… as required)`

 

ii. (2)  `text(Find)\ $M\ text(such that)\  A_n=$800\ 000\ text(after 240 months.)`

♦ Mean mark 49%

`A_240=29\ 227(1.005^240)+M(1.005+1.005^2+..+1.005^240)`

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

`800\ 000=29\ 227(1.005^240)+M((1.005(1.005^240-1))/(1.005-1))`

`M((1.005(1.005^240-1))/(1.005-1))=800\ 000-29\ 227(1.005^240)`

`M` `=(703\ 252.65)/(464.3511)`
  `=1514.484`..
`:.M` `=$1514.48\ \ text{(to the nearest cent)}`

Financial Maths, 2ADV M1 2011 HSC 5a

The number of members of a new social networking site doubles every day. On Day 1 there were 27 members and on Day 2 there were 54 members.

  1. How many members were there on Day 12?     (1 mark)

  2. On which day was the number of members first greater than 10 million?     (2 marks)

  3. The site earns 0.5 cents per member per day. How much money did the site earn in the first 12 days? Give your answer to the nearest dollar.     (2 marks)

Show Answers Only
  1. `55\ 296`
  2. `text(20th)`
  3. `$553`
Show Worked Solutions
MARKER’S COMMENT: Better responses stated the general term, `T_n` before any substitution was made, as shown in the worked solutions.

i.   `T_1=a=27`

`T_2=27xx2^1=54`

`T_3=27xx2^2=108`

`=>\ text(GP where)\ \ a=27,\ \ r=2`

`\ \ \ vdots`

`T_n` `=ar^(n-1)`
`T_12` `=27 xx 2^11=55\ 296`

 

`:.\ text(On Day 12, there are 55 296 members.)`

 

ii.   `text(Find)\ n\ text(such that)\  T_n>10\ 000\ 000`

MARKER’S COMMENT: Many elementary errors were made by students in dealing with logarithms. BE VIGILANT.
`T_n` `=27(2^(n-1))`
`27xx2^(n-1)` `>10\ 000\ 000`
`2^(n-1)` `>(10\ 000\ 000)/27`
`ln 2^(n-1)` `>ln((10\ 000\ 000)/27)`
`(n-1)ln2` `>ln(370\ 370.370)`
`n-1` `>ln(370\ 370.370)/ln 2`
`n-1` `>18.499…`
`n` `>19.499…`

 

`:.\ text(On the 20th day, the number of members >10 000 000.)`

 

iii.  `text(If the site earns 0.5 cents per day per member,)`

`text(On Day 1, it earns)\  27 xx 0.5 = 13.5\ text(cents)`

`text(On Day 2, it earns)\  27 xx 2 xx 0.5 = 27\ text(cents)`

`T_1=a=13.5`

`T_2=27`

`T_3=54`

`=>\ text(GP where)\ \ a=13.5,\ \ r=2`
 

`S_12=text(the total amount of money earned in the first 12 Days)`

♦ Mean mark 44%.
NOTE: This question can also be easily solved by making `S_12` the total sum of members (each day) and then multiplying by 0.5 cents.
`S_12` `=(a(r^n-1))/(r-1)`
  `=(13.5(2^12-1))/(2-1)`
  `=55\ 282.5\ \ text(cents)`
  `=552.825\ \ text(dollars)`

 

`:.\ text{The site earned $553 in the first 12 Days (nearest $).}`

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.`

Financial Maths, 2ADV M1 2012 HSC 15a

Rectangles of the same height are cut from a strip and arranged in a row. The first rectangle has width 10cm. The width of each subsequent rectangle is 96% of the width of the previous rectangle.
 

2012 15a
 

  1. Find the length of the strip required to make the first ten rectangles.     (2 marks)

  2. Explain why a strip of 3m is sufficient to make any number of rectangles.     (1 mark)

Show Answers Only
  1. `83.8\ text{cm  (1 d.p.)}`
  2. `S_oo=2.5\ text(m)\ \ =>\ \ text(sufficient.)`
Show Worked Solutions
i.    `T_1` `=a=10`
`T_2` `=ar=10xx0.96=9.6`
`T_3` `=ar^2=10xx0.96^2=9.216`

 
`=>\ text(GP where)\ \ a=10\ \ text(and)\ \ r=0.96`

MARKER’S COMMENT: A common error was to find `T_10` instead of `S_10`
`S_10` `=\ text(Length of strip for 10 rectangles)`
  `=(a(1-r^n))/(1-r)`
  `=10((1-0.96^10)/(1-0.96))`
  `=83.8\ text{cm   (to 1 d.p.)}`

 

ii.   `text(S)text(ince)\ |\ r\ |<\ 1`

`S_oo` `=a/(1-r)`
  `=10/(1-0.96)`
  `=250\ text(cm)`

 

`:.\ text(S)text(ince  3 m > 2.5 m, it is sufficient.)`

Financial Maths, 2ADV M1 2012 HSC 12c

Jay is making a pattern using triangular tiles. The pattern has 3 tiles in the first row, 5 tiles in the second row, and each successive row has 2 more tiles than the previous row.

2012 12c

  1. How many tiles would Jay use in row 20?     (2 marks)

  2. How many tiles would Jay use altogether to make the first 20 rows?     (1 mark)

  3. Jay has only 200 tiles. How many complete rows of the pattern can Jay make?     (2 marks)

Show Answers Only
  1. `\ 41`
  2. `440`
  3. `13\ text(rows)`
Show Worked Solutions
i.    `T_1` `=a=3`
  `T_2` `=a+d=5`
  `T_3` `=a+2d=7`

 
`=>\ text(AP where)\ \ a=3,\ \ d=2`

`\ \ \ \ \ vdots`

`T_20` `=a+19d`
  `=3+19(2)`
  `=41`

 

`:.\ text(Row 20 has 41 tiles.)`

 

MARKER’S COMMENT: Better responses stated the formula BEFORE any calculations were performed. This enabled students to get some marks if they made an error in their working.
ii.    `S_20` `=\ text(the total number of tiles in first 20 rows)`
`S_20` `=n/2(a+l)`
  `=20/2(3+41)`
  `=440`

 

`:.\ text(There are 440 tiles in the first 20 rows.)`

 

 iii.   `text(If Jay only has 200 tiles, then)\ \ S_n<=200`

NOTE: Examiners often ask questions requiring `n` to be found using the formula `S_n=n/2[2a+(n-1)d]` as this requires the solving of a quadratic, and interpretation of the answer.
`n/2(2a+(n-1)d)` `<=200`
`n/2(6+2n-2)` `<=200`
`n(n+2)` `<=200`
`n^2+2n-200` `<=0`
`n` `=(-2+-sqrt(4+4*1*200))/(2*1)`
  `=(-2+-sqrt804)/2`
  `=-1+-sqrt201`
  `=13.16\ \ text{(answer must be positive)}`

 

`:.\ text(Jay can complete 13 rows.)`

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).`