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Financial Maths, 2ADV M1 2018 HSC 14d

An artist posted a song online. Each day there were  `2^n + n`  downloads, where `n` is the number of days after the song was posted.

  1. Find the number of downloads on each of the first 3 days after the song was posted.  (1 mark)

  2. What is the total number of times the song was downloaded in the first 20 days after it was posted?  (2 marks)

Show Answers Only
  1. `text(Day 1) : 3`
    `text(Day 2) : 6`
    `text(Day 3) : 11`

  2. `2\ 097\ 360`
Show Worked Solution

i.   `text(Day 1:)\ \ 2^1 + 1 = 3`

`text(Day 2:)\ \ 2^2 + 2 = 6`

`text(Day 3:)\ \ 2^3 + 3 = 11`

 

ii.  `text{Total downloads (20 days)}`

♦ Mean mark 38%.

`= 2^1 + 1 + 2^2 + 2 + … + 2^20 + 20`

`= underbrace(2^1 + 2^2 + … + 2^20)_{text(GP),\ a = 2,\ r=2} + underbrace(1 + 2 + … + 20)_{text(AP),\ a = 1,\ d = 1}`

`= (2(2^20 – 1))/(2 – 1) + 20/2(1 + 20)`

`= 2\ 097\ 150 + 210`

`= 2\ 097\ 360`

Financial Maths, 2ADV M1 SM-Bank 4 MC

The first four terms of a geometric sequence are 

`4, – 8, 16, – 32`

The sum of the first ten terms of this sequence is

  1. `– 2048`
  2. `– 1364`
  3. `684`
  4. `1367`
Show Answers Only

`B`

Show Worked Solution

`4, –8, 16, –32,\ …`

`text(GP where)\ \ \ a` `= 4`
`r` `=t_(2)/t_(1)= (–8)/4=–2`
`S_n` `=(a(r^n – 1))/(r – 1)`
`:.S_10`  `=[4[(–2)^10 – 1]]/(–2 – 1)`
  `= – 1364`

`=> B`

Financial Maths, 2ADV M1 2005 HSC 7a

Anne and Kay are employed by an accounting firm.

Anne accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by $2500.

Kay accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by 4%.

  1. What is Anne’s annual salary in her thirteenth year?  (2 marks)

  2. What is Kay’s annual salary in her thirteenth year?  (2 marks)

  3. By what amount does the total amount paid to Kay in her first twenty years exceed that paid to Anne in her first twenty years?  (3 marks)

Show Answers Only
  1. `$80\ 000`
  2. `text{$80 052 (nearest dollar)}`
  3. `text{$13 904 (nearest $)}`
Show Worked Solution

i.   `text(Let)\ T_n = text(Anne’s salary in year)\ n`

`T_1 = a = $50\ 000`

`T_2 = a + d = $52\ 500`

`⇒\ text(AP where)\ a = $50\ 000,\ \ d = $2500`

`T_n = a + (n − 1)d`

`T_13` `= 50\ 000 + (13 − 1) xx 2500`
  `=80\ 000`

 

`:.\ text(Anne’s salary in her 13th year is $80 000.)`

 

ii.  `text(Let)\ K_1 =text(Kay’s salary in year)\ n`

`K_1` `= a` `= 50\ 000`
`K_2` `= ar` `= 50\ 000 xx 1.04 = 52\ 000`
`⇒\ text(GP where)\ \ a = 50\ 000, \ \ r = 1.04`
`K_n` `= ar^(n − 1)`
 `K_13` `= 50\ 000 xx (1.04)^12`
  `= $80\ 051.61…`
  `= $80\ 052\ \ \ text{(nearest dollar)}`

 

iii.  `text(Anne)`

`S_n` `= n/2[2a + (n − 1)d]`
 `S_20` `= 20/2[2 xx 50\ 000 + (20 − 1)2500]`
  `= 10[100\ 000 + 47\ 500]`
  `= $1\ 475\ 000`

 

`text(Kay)`

`S_n` `= (a(r^n − 1))/(r − 1)`
`S_20`  `= (50\ 000(1.04^20 -1))/(1.04 − 1)`
  `= $1\ 488\ 903.929…`

 

`text(Difference)`

`= 1\ 488\ 903.929… − 1\ 475\ 000`

`= $13\ 903.928…`

`= $13\ 904\ \ \  text{(nearest $)}`

 

`:.\ text(Kay’s total salary exceeds Anne’s by)\ $13\ 904`

Financial Maths, 2ADV M1 2011 HSC 9d

 

  1. Rationalise the denominator in the expression  `1/(sqrtn + sqrt(n+1))`  where  `n`  is an integer and  `n >= 1`.   (1 mark)

  2. Using your result from part (i), or otherwise, find the value of the sum
     
         `1/(sqrt1+ sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + ... + 1/(sqrt99 + sqrt100)`   (2 marks)

Show Answers Only
  1. `sqrt(n+1)\ – sqrtn`
  2. `9`
Show Worked Solution

i.  `1/(sqrtn + sqrt(n+1)) xx (sqrtn\ – sqrt(n+1))/(sqrtn\ – sqrt(n+1))`

MARKER’S COMMENT: Students connecting the information between parts (i) and (ii) were easily the most successful. AGAIN, the information from earlier parts of multi-part questions is gold plated information for directing your answer.

`= (sqrtn\ – sqrt(n+1))/((sqrtn)^2\ – (sqrt(n+1))^2)`

`= (sqrtn\ – sqrt(n+1))/(n\ – (n + 1))`

`= (sqrtn\ – sqrt(n+1))/-1`

`= sqrt(n+1)\ – sqrtn`

 

ii.  `1/(sqrt1 + sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + … + 1/(sqrt99 + sqrt100)`

`= (sqrt2\ – sqrt1) + (sqrt3\ – sqrt2) + (sqrt4\ – sqrt3) + … + (sqrt100\ – sqrt99)`

`= – sqrt1 + sqrt 100`

♦♦♦ Mean mark 15%.

`= -1 + 10`

`= 9`

 

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 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 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).`