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Financial Maths, 2ADV M1 EQ-Bank 2

An army training fitness station requires soldiers to climb up steel cylinders steps that gradually ascend in height, as shown below.
 

The shortest cylinder is 0.7 metres high and the highest is 4.0 metres. The height of each cylinder increases by the same amount and the fitness station is made up of a total of 23 steel cylinders.

  1. Determine the height of the 13th steel cylinder.   (2 marks)

  2. Show that the total height of the steel cylinder steps in the fitness station is 54.05 metres.   (1 mark)

  3. The army wants to build another fitness station exactly the same but only has a 41 metre steel cylinder it can use for construction. Determine how many steel cylinder steps can be made.   (2 marks)

Show Answers Only

a.   \(2.5\ \text{metres}\)

b.   \(\text{See worked solutions} \)

c.   \(\text{19 cylinders} \)

Show Worked Solution

a.     \(a=0.7\)

\(T_{23}\) \(=0.7+22 d\)
\(22 d\) \(=4.0-0.7\)
\(d\) \(=0.15\)

 
\(\therefore T_{13}=0.7+(13-1) 0.15=2.5\ \text{metres}\)
 

b.     \(S_{23}\) \(=\dfrac{n}{2}(a+l)\)
    \(=\dfrac{23}{2}(0.7+4.0)\)
    \(=54.05 \ \text{metres … as required}\)

 

c.    \(\text{Find} \ n \ \text{such that}\ \  S_n=41:\)

\(S_n\) \(=\dfrac{n}{2}(2a+(n-1)d) \)
\(41\) \(=\dfrac{n}{2}[2 \times 0.7+(n-1) 0.15]\)
\(82\) \(=1.4 n+0.15 n^2-0.15 n\)
\(0\) \(=0.15n^2+1.25n-82\)
\(n\) \(=\dfrac{-1.25 \pm \sqrt{1.25^2+4 \times 0.15 \times 82}}{2 \times 0.15}\)
  \(=19.58…\ \ (n>0) \)

 
\(\therefore\ \text{19 cylinders can be made.}\)

Financial Maths, 2ADV M1 2022 HSC 17

Cards are stacked to build a 'house of cards'. A house of cards with 3 rows is shown.

A house of cards requires 3 cards in the top row, 6 cards in the next row, and each successive row has 3 more cards than the previous row.

  1. Show that a house of cards with 12 rows has a total of 234 cards.  (2 marks)

  2. Another house of cards has a total of 828 cards.
  3. How many rows are in this house of cards?  (3 marks)

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

a.   `a=3, \ d=3`

`S_n` `=n/2[2a+(n-1)d]`  
`S_12` `=12/2(2xx3 + 11xx3)`  
  `=6(6+33)`  
  `=234\ \ text{… as required}`  

 
b.   `text{Find}\ \ n\ \ text{given}\ \ S_n=828:`

`828` `=n/2[6+(n-1)3]`  
`1656` `=n(3+3n)`  
  `=3n^2+3n`  

 

`3n^2+3n-1656` `=0`  
`n^2+n-552` `=0`  
`(n+24)(n-23)` `=0`  

 
`:. n=23\ text{rows}\ \ (n>0)`

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 2018 HSC 11d

In an arithmetic series, the third term is 8 and the twentieth term is 59.

  1. Find the common difference.  (1 mark)

  2. Find the 50th term.  (2 marks)

Show Answers Only
  1. `d = 3`
  2. `149`
Show Worked Solution
i.   `a + 2d` `= 8 qquad text{… (1)}`
  `a + 19d` `= 59 qquad text{… (2)}`

 
`text(Substract)\ \ (2) – (1)`

`17d` `= 51`
`:. d` `= 3`

 

ii.  `text(Find)\ \ T_50`

`text(Substitute)\ \ d = 3\ \ text{into (1)}`

`=> a = 12`

`T_n` `=a+(n-1)d`
`:. T_50` `= 2 + 49 xx 3`
  `= 149`

Financial Maths, 2ADV M1 2017 HSC 12c

In an arithmetic series, the fifth term is 200 and the sum of the first four terms is 1200.

Find the value of the tenth term.  (3 marks)

Show Answers Only

`0`

Show Worked Solution

`T_n = a + (n – 1) d`

`=>a + 4d = 200 …\ (1)`

 

`S_n = n/2 [2a + (n – 1) d]`

`=>4a + 6d = 1200 …\ (2)`

 

`text(Multiply)\ (1) xx 4`

`=>4a + 16d = 800 …\ (1 prime)`

 

`text(Subtract)\ \ (1 prime) – (2)`

`10d` `= -400`
`d` `= -40`

 

`text(Substitute)\ \ d = -40\ \ text(into)\ (1)`

`a – 160` `= 200`
`a` `= 360`
`:. T_10` `= 360 – 9(-40)`
  `= 0`

Financial Maths, 2ADV M1 SM-Bank 6

A toy train track consists of a number of pieces of track which join together.

The shortest piece of the track is 15 centimetres long and each piece of track after the shortest is 2 centimetres longer than the previous piece.

The total length of the complete track is 7.35 metres.

Find the length of the longest piece of track. (3 marks)

Show Answers Only

`55\ text(cm)`

Show Worked Solution

`text(Sequence is 15, 17, 19 , . . .)`

`text(AP where)\ \ \ a` `= 15, and`
`d` `= 2`
`S_n` `= n / 2 [ 2a + (n – 1) d ]`
`:. 735` `= n / 2 [ 2 xx 15 + (n – 1) 2 ]`
  `= n / 2 [ 30 + 2n – 2 ]`
  `= n^2 + 14n`
`0`  `= n^2 + 14n – 735`

 

`text(Using the quadratic formula)`

`n` `= {–14 +- sqrt (14^2 – 4. 1. (–735))} / (2 xx 1)`
  `= (–14 +-56) / 2`
  `= 21 \ \ \ \  (n > 0)`

 

`:.\ text(Longest piece of track)` `= a + (n – 1) d`
  `= 15 + (21 – 1) 2`
  `= 55\ text(cm)`

Financial Maths, 2ADV M1 2004 HSC 5a

Clare is learning to drive. Her first lesson is 30 minutes long. Her second lesson is 35 minutes long. Each subsequent lesson is 5 minutes longer than the lesson before.

  1. How long will Clare’s twenty-first lesson be?  (1 mark)

  2. How many hours of lessons will Clare have completed after her twenty-first lesson?  (2 marks)

  3. During which lesson will Clare have completed a total of 50 hours of driving lessons?  (2 marks)

Show Answers Only
  1. `text(130 minutes)`
  2. `28\ text(hours)`
  3. `text(30th lesson)`
Show Worked Solution

i.   `30, 35, 40, …`

`=>\ text(AP where)\ \ a = 30,\ \ d = 5`

`T_n` `= a + (n − 1) d`
`T_(21)` `= 30 + 20(5)`
  `= 30 + 100`
  `= 130`

 

`:.\ text(Clare’s twenty-first lesson will be 130)`

`text(minutes long.)`

 

ii.  `text(Find)\ \ S_21\ \ text(given)\ \ a = 30, \ d = 5`

`S_n` `= n/2[2a + (n − 1)d]`
`:.S_21` `= 21/2[2(30) + 20 × 5]`
  `= 21/2[60 + 100]`
  `= 21/2 × 160`
  `= 1680\ text(minutes)`
  `= 28\ text(hours.)`

 

(iii)  `text(50 hours)` `= 50 × 60`
  `= 3000\ text(minutes)`

 

`text(Find)\ \ n,\ \ text(given)\ \ a = 30, \ d = 5, \ S_n = 3000`

`S_n` `= n/2[2a + (n − 1)d]`
`:. 3000` `= n/2[2(30) + (n − 1)5]`
`6000` `= n[60 + 5n − 5]`
`6000` `= n[55 + 5n]`
`6000` `= 55n + 5n^2`
`5n^2 + 55n − 6000` `= 0`
`n^2 + 11n − 1200` `= 0`
`:.n` `= (−11 ± sqrt((11)^2 − 4(1)(−1200)))/(2(1))`
  `= (−11 ± sqrt(121 + 4800))/(2)`
  `= (−11 ± sqrt4921)/2`
  `= 29.5749…\ \ \ (n> 0)`

 

`:.\ text(Clare completes 50 hours of lessons during)`

`text(her 30th lesson.)`

Financial Maths, 2ADV M1 2007 HSC 3b

Heather decides to swim every day to improve her fitness level.

On the first day she swims 750 metres, and on each day after that she swims `100` metres more than the previous day. That is, she swims 850 metres on the second day, 950 metres on the third day and so on.

  1. Write down a formula for the distance she swims on the `n`th day.  (1 mark)

  2. How far does she swim on the 10th day?  (1 mark)

  3. What is the total distance she swims in the first 10 days?  (1 mark)

  4. After how many days does the total distance she has swum equal the width of the English Channel, a distance of 34 kilometres?  (2 marks)

Show Answers Only
  1. `T_n = 650 + 100n`
  2. `1650\ text(metres)`
  3. `12\ text(km)`
  4. `34\ text(km)`
Show Worked Solution
i.  `T_1` `= 750`
`T_2` `= 850`

`=>  text(AP)\ text(where)\ a = 750\ ,\ d = 100`

`:. T_n` `= a + (n-1) d`
  `= 750 + (n – 1) 100`
  `= 750 + 100n – 100`
  `= 650 + 100n`

 

ii.  `T_10` `= 650 + 100 xx 10`
  `= 1650\ text(metres)`

 

`:.\ text(She swims 1650 metres on the 10th day.)`

 

iii.  `S_n = n/2 [2a + (n – 1) d]`

`:. S_10` `= 10/2 [2 xx 750 + (10 -1) 100]`
  `= 5 [1500 + 900]`
  `= 12\ 000`

 

`:.\ text(She swims 12 km in the first 10 days.)`

 

iv.  `text(Find)\ n\ text(such that)\ S_n = 34\ text(km)`

`n/2 [2 xx 750 + (n – 1) 100]` `= 34\ 000`
`n/2 [1500 + 100n – 100]` `= 34\ 000`
`n/2 [1400 + 100n]` `= 34\ 000`
`700n + 50n^2` `= 34\ 000`
`50 n^2 + 700n – 34\ 000` `= 0`
`50 (n^2 + 14 n – 680)` `= 0`

 

`text(Using the quadratic formula)`

`n` `= {-b +- sqrt(b^2 – 4ac)}/(2a)`
  `= {-14 +- sqrt(14^2 – 4 xx 1 xx (-680))}/(2 xx 1)`
  `= (-14 +- sqrt 2916)/2`
  `= (-14 +- 54)/2`
  `= 20 or -34`
  `= 20\ ,\ n > 0`

 

`:.\ text(Her total distance equals 34 km after 20 days.)`

Financial Maths, 2ADV M1 2006 HSC 3c

On the first day of the harvest, an orchard produces 560 kg of fruit. On the next day, the orchard produces 543 kg, and the amount produced continues to decrease by the same amount each day.

  1. How much fruit is produced on the fourteenth day of the harvest?  (2 marks)

  2. What is the total amount of fruit that is produced in the first 14 days of the harvest?  (1 mark)

  3. On what day does the daily production first fall below 60 kg?  (2 marks)

Show Answers Only
  1. `text(339 kg of fruit is produced on 14th day.)`
  2. `text(6293 kg of fruit is produced in the 1st 14 days.)`
  3. `text(On the 31st day, production will first drop below 60 kg.)`
Show Worked Solution

i.   `T_1 = a = 560`

`T_2 = a + d = 543`

`=>\ text(AP where)\ a = 560, d = -17`

`vdots`

`T_14` `= a + 13d`
  `= 560 – (13 xx 17)`
  `= 339`

 

`:.\ text(339 kg of fruit is produced on 14th day.)`

 

ii.  `S_14 = text(total fruit produced in 1st 14 days)`

`S_14` `= n/2 [2a + (n – 1)d]`
  `= 14/2 [2 xx 560 – (14 – 1) xx 17]`
  `= 7 [1120 – 221]`
  `= 6293`

 

`:.\ text(6293 kg of fruit is produced in the 1st 14th days.)`

 

iii.  `text(Find)\ n\ text(such that)\ T_n < 60`

`T_n = a + (n – 1)d` `< 60`
`560 – 17(n – 1)` `< 60`
`560 – 17n + 17` `< 60`
`17n` `> 517`
`n` `> 30.41…`

 
`:.\ text(On the 31st day, production will first drop)`

`text(below 60 kg.)`

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 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 3a

A skyscraper of 110 floors is to be built. The first floor to be built will cost $3 million. The cost of building each subsequent floor will be $0.5 million more than the floor immediately below.

  1. What will be the cost of building the 25th floor?     (2 marks)

  2. What will be the cost of building all 110 floors of the skyscraper?     (2 marks)

Show Answers Only
  1. `$15\ text(million)`
  2. `$3,327.5\  text(million)`
Show Worked Solutions
i.     `T_1` `=a=3`
`T_2` `=a+d=3.5`
`T_3` `=a+2d=4`

 
`=>\ text(AP where)\ \ a=3\ \ d=0.5` 

MARKER’S COMMENT: Better responses listed the sequence of terms in the series as illustrated.
`T_25` `=a+24d`
  `=3+24(0.5)`
  `=15`

 
 `:.\ text(The 25th floor costs  $15,000,000.)`

MARKER’S COMMENT: Better responses stated the formula BEFORE any calculations were performed. This allows markers to allocate part marks to students who had errors in calculation.

 

ii.     `S_110` `=\ text(Total cost of 110 floors)`
  `=n/2(2a+(n-1)d)` 
  `=110/2(2xx3\ 000\ 000+(110-1)500\ 000)`
  `=55(6\ 000\ 000+49\ 500\ 000)`
  `=$3327.5\  text(million)`

 

`:.\ text{The total cost of 110 floors is $3327.5 million}`

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 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).`