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Probability, 2ADV S1 2005 HSC 5d

A total of 300 tickets are sold in a raffle which has three prizes. There are 100 red, 100 green and 100 blue tickets.

At the drawing of the raffle, winning tickets are NOT replaced before the next draw.

  1. What is the probability that each of the three winning tickets is red?  (2 marks)

  2. What is the probability that at least one of the winning tickets is not red?  (1 mark)

  3. What is the probability that there is one winning ticket of each colour?  (2 marks)

Show Answers Only
  1. `1617/(44\ 551)`
  2. `(42\ 934)/(44\ 551)`
  3. `0.224\ \ text{(to 3 d.p.)}`
Show Worked Solution
i.   `P(R R R)` `= 100/300 xx 99/299 xx 98/298`
    `= 1617/(44\ 551)`

 

ii. `Ptext{(at least 1 winner NOT red)}`

`= 1 − P(R R R)`

`= 1− 1617/(44\ 551)`

`= (42\ 934)/(44\ 551)`

 

iii. `text(# Combinations of winning tickets)`

`= 3 xx 2 xx 1`

`= 6`
 

`:.P text{(one winner from each colour)}`

`= 6 xx 100/300 xx 100/299 xx 100/298`

`= 0.22446…`

`= 0.224\ \ text{(to 3 d.p.)}`

Trigonometry, 2ADV T1 2005 HSC 4a

 Trig Calculus, 2UA 2005 HSC 4a
 

A pendulum is 90 cm long and swings through an angle of 0.6 radians. The extreme positions of the pendulum are indicated by the points `A` and `B` in the diagram.

  1. Find the length of the arc  `AB`.  (1 mark)

  2. Find the straight-line distance between the extreme positions of the pendulum.  (2 marks)

  3. Find the area of the sector swept out by the pendulum.  (1 mark)

Show Answers Only
  1. `text(54 cm)`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(2430 cm²)`
Show Worked Solution
i.    `text(Arc)\ AB` `= theta/(2 pi) xx 2 pi r`
    `= rtheta`
    `= 90 xx 0.6`
    `= 54\ text(cm)`

 

ii. 

 Trig Calculus, 2UA 2005 HSC 4a Answer

`text(Using the cosine rule)`

`text(Distance)\ AB\ text(in straight line)`

`AB^2` `= 90^2 xx 90^2 − 2 xx 90 xx 90 xx cos\ 0.6`
  `= 2829.563…`
`:.AB` `= 53.193…`
  `= 53.2\ text{cm  (to 1 d.p.)}`

 

iii. `text(Area of Sector)`

`= 0.6/(2pi) xx pir^2`

`= 0.3 xx 90^2`

`= 2430\ text(cm²)`

Plane Geometry, 2UA 2005 HSC 3c

2005 3c

In the diagram, `A`, `B`  and  `C`  are the points  `(6, 0), (9, 0)`  and  `(12, 6)` respectively. The equation of the line  `OC`  is  `x - 2y = 0`. The point  `D`  on  `OC`  is chosen so that  `AD`  is parallel to  `BC`. The point  `E`  on  `BC`  is chosen so that  `DE`  is parallel to the `x`-axis.

  1.  Show that the equation of the line `AD` is `y = 2x - 12`.  (2 marks)
  2. Find the coordinates of the point `D`.  (2 marks)
  3. Find the coordinates of the point `E`.  (1 marks)
  4. Prove that  `ΔOAD\ text(|||)\ ΔDEC`.  (2 marks)
  5. Hence, or otherwise, find the ratio of the lengths `AD` and `EC`.  (1 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(8, 4)`
  3. `(11, 4)`
  4. `text(See Worked Solutions)`
  5. `2:1`
Show Worked Solution

(i)   `text(S)text(ince)\ DA\ text(||)\ CB`

`m_(DA)` `= (y_2 − y_1)/(x_2 − x_1)`
  `= (6 − 0)/(12 − 9)`
  `= 2`

 

`:.\ text(Equation of)\ AD, m = 2,\ text(through)\ A(6, 0)`

`y − y_1` `= m(x − x_1)`
 `y − 0` `= 2(x − 6)`
 `y` `= 2x − 12\ \ \ \ …\ text(as required)`

 

(ii)  `D\ text(is at the intersection of)`

`x − 2y` `= 0` `\ \ …\ (1)`
 `y` `= 2x − 12\ ` `\ \ …\ (2)` 

`text(Substitute)\ y = 2x − 12\ text{into (1)}`

`x − 2(2x − 12)` `= 0`
`x − 4x + 24` `= 0`
`text(−3)x + 24` `= 0`
`3x` `= 24`
`x` `= 8`

`text(Substitute)\ x = 8\ text{into (2)}`

`y` `= 2 xx 8 − 12 = 4`
`:.D` `= (8, 4)`

 

(iii)  `text(Distance)\ \ AB=3`

`:. E\ \ text(has coordinates)\ \ (8+3,4)-=(11,4)`

 

(iv)  `text(Prove)\ ΔOAD\ text(|||)\ ΔDEC`

`∠ODA = ∠DCE`

`text{(corresponding angles,}\ AD\ text(||)\ BC)`

`∠OAD = ∠ABE`

`text{(corresponding angles,}\ AD\ text(||)\ BC)`

`∠ABE = ∠DEC`

`text{(corresponding angles,}\ AB\ text(||)\ DE)`

`:.∠OAD = ∠DEC`

`:.ΔOAD\ text(|||)\ ΔDEC\ text{(equiangular)}`

 

(v)     `(AD)/(EC)` `= (OA)/(DE)` 
    `= 6/3` 
    `=2` 

 `text{(corresponding sides of similar triangles)}`

 

`:.\ text(Ratio of lengths)\ AD:EC = 2:1` 

Financial Maths, 2ADV M1 2006 HSC 8b

Joe borrows $200 000 which is to be repaid in equal monthly instalments. The interest rate is 7.2% per annum reducible, calculated monthly.

It can be shown that the amount, `$A_n`, owing after the `n`th repayment is given by the formula:

`A_n = 200\ 000r^n - M(1 + r + r^2 + … + r^(n-1))`,

where  `r = 1.006`  and  `$M`  is the monthly repayment.  (Do NOT show this.)

  1. The minimum monthly repayment is the amount required to repay the loan in 300 instalments.

     

    Find the minimum monthly repayment.  (3 marks)

  2. Joe decides to make repayments of $2800 each month from the start of the loan.

     

    How many months will it take for Joe to repay the loan?  (2 marks)

Show Answers Only
  1. `$1439`
  2. `94\ \  text(months)`
Show Worked Solution

i.  `A_n = 200\ 000r^n – M(1 + r + r^2 + … + r^(n-1))`

`text(Find)\ M\ text(such that)\ A_n = 0\ text(when)\ n = 300`

`0 = 200\ 000(1.006^300) – M(1 + 1.006+ … + 1.006^299)`

`=>  text(Note that ) (1 + 1.006+ … + 1.006^299)\ \ text(is a)`

`text(GP)\ text(where)\ a = 1, r = 1.006, n = 300`

`0 = 200\ 000(1.006^300) – M ({1(1.006^300-1)}/(1.006 – 1))`

`M ({(1.006^300-1)}/(0.006))` `= 200\ 000(1.006^300)`
`M` `= (1\ 203\ 439.36…)/(836.199…)`
  `= 1439.177…`
  `= $1439\ \ text{(nearest dollar)}`

 
`:.\ text(The minimum monthly repayment is $1439.)`

 

ii.  `text(If)\ M = 2800,\ text(find)\ n\ text(when)\ A_n = 0`

`0 = 200\ 000(1.006^n) – 2800({(1.006^n – 1)}/0.006)`

`0 = 1200(1.006^n) – 2800(1.006^n) + 2800`

`0 = -1600(1.006^n) + 2800`

`1600(1.006^n)` `= 2800`
`1.006^n` `= 2800/1600`
`n xx ln 1.006` `= ln­ \ 1.75`
`n` `= (ln­ \ 1.75)/ln 1.006`
  `= 93.54…`
  `=94\ \ text{months  (nearest month)}`

 

`:.\ text(It will take Joe 94 months to repay the loan.)`

Calculus, 2ADV C1 2006 HSC 8a

A particle is moving in a straight line. Its displacement, `x` metres, from the origin, `O`, at time `t` seconds, where  `t ≥ 0`, is given by  `x = 1 - 7/(t + 4)`.

  1. Find the initial displacement of the particle.  (1 mark)

  2. Find the velocity of the particle as it passes through the origin.  (3 marks)

  3. Show that the acceleration of the particle is always negative.  (1 mark)

  4. Sketch the graph of the displacement of the particle as a function of time.  (2 marks)

Show Answers Only
  1. `text(–3/4 m)`
  2. `1/7\ text(ms)^-1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4.  
Show Worked Solution

i.   `x = 1 – 7/(t + 4)`
 

`text(When)\ \ t = 0,`

`x` `= 1 – 7/4`
  `= -3/4 \ \ text(m)`

 
`:.\ text(Initial displacement is)\ 3/4\ text(metres to)`

`text(the left of the origin.)`

 

ii.  `x = 1 – 7/(t+4) = 1 – 7(t + 4)^-1`

`dot x` `= (-1)  -7(t + 4)^-2 xx d/(dt)(t + 4)`
  `= 7 (t + 4)^-2 xx 1`
  `= 7/(t + 4)^2`

 

`text(Find)\ t\ text(when)\ x = 0`

`0` `= 1 – 7/(t + 4)`
`7/(t + 4)` `= 1`
`7` `= (t + 4)`
`t` `= 3`

 

`text(When)\ t = 3`

`dot x` `= 7/(3 + 4)^2`
  `= 1/7\ text(ms)^-1`

 

`:.\ text(The velocity of the particle as it passes)`

`text(through the origin is)\ 1/7\ text(ms)^-1.`

 

iii.  `dot x` `= 7(t + 4)^-2`
`ddot x` `= (d dot x)/(dt) = -14 (t +4)^-3`

 
`text(Given)\ t >= 0`

`=>  (t + 4)^-3 >= 0`

`=> -14 (t + 4)^-3 <= 0`

`:. ddot x\ text(is always negative.)`
 

(iv)  2UA HSC 2006 8a

Probability, STD2 S2 2006 HSC 28a

On a bridge, the toll of $2.50 is paid in coins collected by a machine. The machine only accepts two-dollar coins, one-dollar coins and fifty-cent coins.

  1. List the different combinations of coins that could be used to pay the $2.50 toll.  (1 mark)

  2. Jill has three two-dollar coins, six one-dollar coins and two fifty-cent coins. She selects two coins at random.

     

    What is the probability that she selects exactly $2.50?  (3 marks)

  3. At the end of a day, the machine contains `x` two-dollar coins, `y` one-dollar coins and `w` fifty-cent coins.

     

    Write an expression for the total value of coins in dollars in the machine.  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `6/55`
  3. `$(2x + y + 0.5w)`
Show Worked Solution

i.  `text(Combinations for $2.50)`

`$2, 50c`

`$1, $1, 50c`

`$1, 50c, 50c, 50c`

`50c, 50c, 50c, 50c, 50c`

 

ii.  `3 xx $2, 6 xx $1, 2 xx 50c`

`P($2.50)` `= (3/11 xx 2/10) + (2/11 xx 3/10)`
  `= 6/110 + 6/110`
  `= 6/55`

 

iii.  `text(Total value) = $ (2x + y + 0.5w)`

Algebra, STD2 A4 2005 HSC 28b

Sue and Mikey are planning a fund-raising dance. They can hire a hall for $400 and a band for $300. Refreshments will cost them $12 per person.

  1. Write a formula for the cost ($C) of running the dance for `x` people. (1 mark)

The graph shows planned income and costs when the ticket price is $20 

2005 28b

  1. Estimate the minimum number of people needed at the dance to cover the costs.  (1 mark)

  2. How much profit will be made if 150 people attend the dance? (1 mark)

Sue and Mikey plan to sell 200 tickets. They want to make a profit of $1500.

  1. What should be the price of a ticket, assuming all 200 tickets will be sold?  (3 marks)

Show Answers Only
  1. `700 + 12x`
  2. `text(Approximately 90)`
  3. `$500`
  4. `$23`
Show Worked Solution
i.    `$C` `= 400 + 300 + (12 xx x)`
    `= 700 + 12x`

 

ii.  `text(Using the graph intersection)`

`text(Approximately 90 people are needed)`

`text(to cover the costs.)`

 

iii.  `text(If 150 people attend)`

`text(Income)` `= 150 xx $20`
  `= $3000`
`text(C)text(osts)` `= 700 + (12 xx 150)`
  `= $2500`

 

`:.\ text(Profit)` `= 3000 − 2500`
  `= $500`

 

iv.  `text(C)text(osts when)\ x = 200:`

`C` `= 700 + (12 xx 200)`
  `= $3100`

 

`text(Income required to make $1500 profit)`

`= 3100 + 1500`

`= $4600`
 

`:.\ text(Price per ticket)` `= 4600/200`
  `= $23`

Measurement, STD2 M1 2005 HSC 28a

The Mitchell family has moved to a new house which has an empty swimming pool. The base of the pool is in the shape of a rectangle, with a semicircle on each end.

2005 28a1

  1. Explain why the expression for the area of the base of the pool is  `2xy + πy^2`.   (1 mark) 

      
            2005 28a2
      

The pool is 1.1 metres deep.

  1. The sides and base of the pool are covered in tiles. If  `x =6`  and  `y = 2.5`, find the total area covered by tiles. (Give your answer correct to the nearest square metre.)   (4 marks)

Before filling the pool, the Mitchells need to install a new shower head, which saves 6 litres of water per minute.

The shower is used 5 times every day, for 3 minutes each time.

  1. If the charge for water is $1.013 per kilolitre, how much money would be saved in one year by using this shower head? (Assume there are 365 days in a year.)   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(80 m)^2\ text{(nearest m}^2text{)}`
  3. `$33.28\ \ text{(nearest cent)}`
Show Worked Solution
i.    `text(Area of base)` `=\ text(Area of rectangle +)`
    `\ \ \ \ \ \2 xx text(Area of semi-circle)`
    `= (x xx 2y) + 2 xx (1/2 xx pi xx y^2)`
    `= 2xy + piy^2`

 

ii.   `text(Area of base)` `= (2 xx 6 xx 2.5) + (pi xx 2.5^2)`
    `= 49.634…\ text(m²)`

 

`text(Area of walls) = text(Length) xx text(Height)`

`text(Length)` `= 2x + 2 xx text(semi-circle perimeter)`
  `= (2 xx 6) + 2 xx (1/2 xx 2 xx pi xx 2.5)`
  `= 12 + 15.707…`
  `= 27.707…\ text(m)`

 

`:.\ text(Area of walls)` `= 27.707 xx 1.1`
  `= 30.478…\ text(m)^2`

 

`:.\ text(Total Area covered by tiles)`

`= 49.634… + 30.478…`

`= 80.11…`

`= 80\ text{m²  (nearest m²)}`

 

iii.  `text(Water saved)` `= 5 xx 3 xx 6`
    `= 90\ text(L per day.)`

 

`text(Water saved per year)`

`= 90 xx 365`

`= 32\ 850\ text(L)`

`= 32.85\ text(kL)`

`:.\ text(Money saved)` `= 32.85 xx $1.013`
  `= $33.277…`
  `= $33.28\ text{(nearest cent)}`

Statistics, STD2 S5 2005 HSC 26c

The weights of boxes of Brekky Bicks are normally distributed. The mean is 754 grams and the standard deviation is 2 grams.

  1. What is the `z`-score of a box of Brekky Bicks with a weight of 754 g?   (1 mark)

  2. What is the weight of a box that has a `z`-score of  –1?   (1 mark)

  3. Brekky Bicks boxes are labelled as having a weight of 750 g. What percentage of boxes will have a weight less than 750 g?   (2 marks)

Show Answers Only
  1. `0\ text{(mean)}`
  2.  `752\ text(grams)`
  3. `text(2.5%)`
Show Worked Solution

i.    `text{z-score (754 g) = 0}`

`text{(Noting that 754 g is the mean)}`

 

ii.   `ztext(-score)` `= (x − mu)/sigma`
`-1` `= (x − 754)/2`
 `x – 754` `= -2`
`x`  `= 752\ text(grams)`

 

iii. `text{z-score (750)}` `= (750 − 754)/2`
    `= -2`

 

 
  `:.\ text(Graph shows that 2.5% of boxes will weigh)`

`text(less than 750 g.)`

Financial Maths, STD2 F5 2005 HSC 26b

Rod is saving for a holiday. He deposits $3600 into an account at the end of every year for four years. The account pays 5% per annum interest, compounding annually.

The table shows future values of an annuity of $1.
 

2UG-2005-26b
 

  1. Use the table to find the value of Rod’s investment at the end of four years.   (2 marks)

  2. How much interest does Rod earn on his investment over the four years?   (2 marks)

Show Answers Only
  1. `$15\ 516.36`
  2. `$1116.36`
Show Worked Solution

i.   `text(Using the table),\ r =\ text(5% and)\ n = 4`

`text(Annuity factor = 4.3101)`

`:.\ text(Value of investment)`

`= 3600 xx 4.3101`

`= $15\ 516.36`

 

ii.  `text(Interest)` `= text(Value) − text(Contributions)`
  `= 15\ 516.36 − (4 xx 3600)`
  `= $1116.36`

Financial Maths, STD2 F4 2006* HSC 25b

In June, Ms Bigspender received a statement for her credit card account.

The account has no interest-free period. Compound interest is calculated daily and charged to her account on the statement date.
 

  1. For how many days is she charged interest on her purchase?  (1 mark)

  2. Calculate the interest charged to her account.  (2 marks)

Show Answers Only
  1.  `29`
  2.  `$8.98`
Show Worked Solution

i.  `text(# Days interest charged)`

`=\ text{9 (in May) + 20 (in June)}`

`= 29`

 

ii.   `text(Daily interest rate)\ (r) = 0.0498/100=0.000498`

`text(Closing Balance)\ (FV)` `= PV(1+r)^n`
  `= 617.72(1.000498)^29`
  `= $626.703`

 

`:.\ text(Interest charged)` `=626.70 – 617.72`  
  `=$8.98`  

Probability, STD2 S2 2006 HSC 25a

Three cards labelled `C`, `J` and `M` can be arranged in any order.

  1. In how many different ways can the cards be arranged?  (1 mark)

  2. What is the probability that the second card in an arrangement is a `J`?  (1 mark)

  3. What is the probability that the last card in an arrangement is not a `C`?  (1 mark)

Show Answers Only
  1. `6`
  2. `1/3`
  3. `2/3`
Show Worked Solution
i.  `text(# Arrangements)` `= 3 xx 2 xx 1`
  `= 6`

 

ii.  `P\ text{(second card is J)}`

`= 1/3`

 

iii.  `P\ text{(last card is not a C)}`

`= 1 – P\ text{(last card is a C)}`

`= 1 – 1/3`

`= 2/3`

Measurement, STD2 M6 2006 HSC 24b

A 130 cm long garden rake leans against a fence. The end of the rake is 44 cm from the base of the fence.

  1. If the fence is vertical, find the value of `theta` to the nearest degree.  (2 marks)
      
          2UG-2006-24b-i

     

  2. The fence develops a lean and the rake is now at an angle of 53° to the ground. Calculate the new distance (`x` cm) from the base of the fence to the head of the rake. Give your answer to the nearest centimetre.  (2 marks)
     
          2UG-2006-24b-ii

Show Answers Only
  1. `text{70°  (nearest degree)}`
  2. `text{109 cm  (nearest cm)}`
Show Worked Solution
i.   

2UG-2006-24b1 Answer
`cos theta` `= 44/130`
  `= 70.216… ^@`
  `= 70^@\ \ \ text{(nearest degree)}`

 

ii.   

2UG-2006-24b2 Answer

`text(Using cosine rule)`

`x^2` `= 130^2 + 44^2-2 xx 130 xx 44 xx cos 53^@`
  `= 11\ 951.23…`
`x` `= 109.32…`
  `= 109\ text{cm  (nearest cm)}`

Statistics, STD2 S1 2006 HSC 23c

Vicki wants to investigate the number of hours spent on homework by students at her high school.

  1. Briefly describe a valid method of randomly selecting 200 students for a sample.  (1 mark)

  2. Vicki chooses her sample and asks each student how many hours (to the nearest hour) they usually spend on homework during one week.

     

    The responses are shown in the frequency table.
     
         2UG-2006-23c

    What is the mean amount of time spent on homework?  (2 marks)

Show Answers Only
  1. `text(A valid method would be using a stratified sample.)`

     

    `text(The number of students sampled in each year is)`

     

    `text(proportional to the size of each year.)`

  2. `text(7.275 hours)`
Show Worked Solution

i.   `text(A valid method would be using a stratified sample.)`

`text(The number of students sampled in each year is)`

`text(proportional to the size of each year.)`

MARKER’S COMMENT: This “routine” exercise of finding a mean from grouped data was incorrectly answered by most students! The best responses copied the table and inserted a class-centre column (see solution).

 

ii.    2UG-2006-23c Answer

 

`text(Mean)` `= text(Sum of Scores) / text(Total scores)`
  `= 1455/200`
  `= 7.275\ text(hours)`

Statistics, STD2 S1 2006 HSC 24c

The heights of the 60 members of a choir were recorded. These results were grouped and then displayed as a cumulative frequency histogram and polygon.

The shortest person in the choir is 140 cm and the tallest is 190 cm.
 

2UG-2006-24c1

Draw an accurate box-and-whisker plot to represent the data.  (3 marks)

Show Answers Only

`text{See Worked Solutions}`

Show Worked Solution

`text(Low) = 140`

`text(High) = 190`

`text(Median) = 150\ \ \ \ text{(# People = 30)}`

`Q_1 = 145\ \ \ \ text{(# People = 15)}`

`Q_3 = 170\ \ \ \ text{(# People = 45)}`

`text(Box and Whisker)`

HSC Data 13

Algebra, 2UG 2004 HSC 28b

2004 28b

A set of garden gnomes is made so that the cost (`$C`) varies directly with the cube of the base length (`b` centimetres). A gnome with a base length of `text(10 cm)` has a cost of `$50`.

  1. Write an equation relating the variables `C` and `b`, and a constant `k`.  (1 mark)
  2. Find the value of `k`.  (1 mark)
  3. Felicity says, ‘If you double the base length, you double the cost.’ Is she correct? Justify your answer with mathematical calculations.  (2 marks)

 

Show Answers Only
  1. `C = kb^3`
  2. `0.05`
  3. `text(Felicity is incorrect, because doubling)`
    `text(the base from 10 cm to 20 cm causes the)`
    `text(cost to increase 8 times.)`

 

Show Worked Solution
(i)  `C` `∝ b^3`
 `:.C` `= kb^3`

 

(ii)  `C = 50\ \ text(when)\ \ b = 10`

`50` `= k xx 10^3`
`:.k` `= 50/10^3`
  `= 0.05`

 

(iii)  `text(If)\ b = 20`

`C` `= 0.05 xx 20^3`
  `= $400`

 

`:.\ text(Felicity is incorrect, because doubling)`

`text(the base from 10 cm to 20 cm causes the)`

`text(cost to increase 8 times.)`

Probability, 2UG 2005 HSC 25c

Robyn plays a game in which she randomly chooses one of these five cards. She plays the game `60` times, replacing the card after each game.

2UG-2005-25c

  1. How many times would she expect to win `$4`?   (1 mark)
  3. What is the financial expectation of the game?   (2 marks)
  4. Another card is added to the game with ‘Win nothing $0’ written on it.
    Robyn claims that the financial expectation will not change.
  6. Do you agree? Justify your answer with suitable calculations.   (2 marks)
Show Answers Only
  1. `36`
  2. `text(Financial Expectation is to win)\ $0.80.`
  3. `text(The financial expectation will reduce)`
  4. `text(from win $0.80 to win $0.677…)`

 

Show Worked Solution

(i)   `Ptext{(win $4)}\ = 3/5`

`:.\ text(Expected times to win $4)`

`= 3/5 xx 60`

`= 36`

 

(ii)  `P($0) = 1/5`

`Ptext{(lose $8)}\ = 1/5`

`text(Financial Expectation)`

`= (3/5 xx 4) + (1/5 xx 0) − (1/5 xx 8)`

`= 12/5 + 0 − 8/5`

`= 4/5`

`:.\ text(Financial Expectation is to win $0.80)`

 

(iii)  `text(If an extra card to win $0 was added)`

`text(Financial expectation)`

`= (3/6 xx 4) + (2/6 xx 0) − (1/6 xx 8)`

`= 2 + 0 − 4/3`

`= 2/3`

`= $0.677…`

 

`:.\ text(The financial expectation will reduce from)`

`text(win $0.80 to win $0.677…)`

Algebra, STD2 A1 2005 HSC 24c

Make  `L`  the subject of the equation  `T = 2piL^2`.   (2 marks)

Show Answers Only

`± sqrt(T/(2pi))`

Show Worked Solution
`T` `= 2piL^2`
`L^2` `= T/(2pi)`
`:.L` `= ±sqrt(T/(2pi))`

Statistics, STD2 S1 2005 HSC 24a

  1. Draw a stem-and-leaf plot for the following set of scores.

  2.  

     

    `21\ \ \ 45\ \ \ 29\ \ \ 27\ \ \ 19\ \ \ 35\ \ \ 23\ \ \ 58\ \ \ 34\ \ \ 27`  (2 marks)

  3. What is the median of the set of scores?   (1 mark)

  4. Comment on the skewness of the set of scores.   (1 mark)

Show Answers Only
  1.  
  2. `28`
  3. `text(The data has a tail that stretches to the right)`

  4.  

    `:.\ text(Data is positively Skewed.)`

Show Worked Solution
i.    HSC 2005 24a

 

ii.  `text(10 scores)`

`:.\ text(Median)` `= text{(5th + 6th)}/2`
  `= (27 + 29)/2`
  `= 28`

 

iii.  `text(The data has a tail that stretches to the right)`

`:.\ text(Data is positively skewed.)`

Probability, STD2 S2 2005 HSC 23c

Moheb owns five red and seven blue ties. He chooses a tie at random for himself and puts it on. He then chooses another tie at random, from the remaining ties, and gives it to his brother.

  1. What is the probability that Moheb chooses a red tie for himself?  (1 mark)

Copy the tree diagram into your writing booklet.
 

2UG-2005-23c
 

  1. Complete your tree diagram by writing the correct probability on each branch.  (2 marks)
  2. Calculate the probability that both of the ties are the same colour.  (2 marks)

Show Answers Only
  1. `5/12`
  2.  
  3. `31/66`
Show Worked Solution
i. `P(R)` `= (#\ text(red ties))/(#\ text(total ties))`
    `= 5/12`

 

ii.  

 

iii. `Ptext((same colour))`

`= P(text(RR)) + P(text(BB))`

`= 5/12 × 4/11\ \ +\ \ 7/12 × 6/11`

`= 20/132 + 42/132`

`= 31/66`

Algebra, STD2 A4 2004 HSC 28a

A health rating, `R`, is calculated by dividing a person’s weight, `w`, in kilograms by the square of the person’s height, `h`, in metres.

  1. Fred is 150 cm and weighs 72 kg. Calculate Fred’s health rating.  (1 mark)

  2. Over several years, Fred expects to grow 10 cm taller. By this time he wants his health rating to be 25. How much weight should he gain or lose to achieve his aim? Justify your answer with mathematical calculations.  (2 marks)

Show Answers Only
  1. `32`
  2. `text(8 kg)`
Show Worked Solution

i.  `R = w/h^2`

`text(When)\ \ w = 72\ \ and\ \ h = 1.5\ text(m)`

`R` `= 72/1.5^2`
  `= 32`

 

ii.  `text(Find)\ \ w\ \ text(if)\ \ R = 25\ \ and\ \ h = 1.6`

`25` `= w/1.6^2`
`w` `= 25 xx 1.6^2`
  `= 64\ text(kg)`

 

`:.\ text(Weight Fred should lose)`

`= 72 – 64`

`= 8\ text(kg)`

Financial Maths, STD2 F1 2004 HSC 27b

David is paid at these rates:
  

 

His time sheet for last week is:
  

  1. Calculate David’s gross pay for last week.  (3 marks)

  2. David decides not to work on Saturdays. He wants to keep his weekly gross pay the same. How many extra hours at the weekday rate must he work?  (1 mark)

Show Answers Only
  1. `$414.00`
  2. `text(9 extra hours)`
Show Worked Solution
i.   `text{Pay (Fri)}` `= text(4 hours) xx 18.00`
  `= $72.00`
`text{Pay (Sat)}` `= 6\ text(hours) xx 1.5 xx 18.00`
  `= $162.00`
`text{Pay (Sun)}` `= 5\ text(hours) xx 2 xx 18.00`
  `= $180.00`

 

`:.\ text(Gross pay)` `= 72 + 162 + 180`
  `= $414.00`

 

 ii.  `text(Pay on Sat) = $162.00`

`text(Weekly equivalent hours)`

`= 162/18`

`= 9\ text(hours)`

`:.\ text(He will have to work 9 extra hours on)`

`text(a weekday for the same gross pay)`

 

Financial Maths, STD2 F4 2004 HSC 27a

Aaron decides to borrow  $150 000  over a period of 20 years at a rate of 7.0% per annum.

2004 27a

  1. Using the Monthly Repayment Table, calculate Aaron’s monthly repayment.  (2 marks)

  2. How much interest does he pay over the 20 years?  (2 marks)

  3. Aaron calculates that if he repays the loan over 15 years, his total repayments would be `$242\ 730`.

     

    How much interest would he save by repaying the loan over 15 years instead of 20 years?  (2 marks)

Show Answers Only
  1. `$1162.50`
  2. `$129\ 000`
  3. `$36\ 270`
Show Worked Solution

i.   `text(Using the table:)`

`text(Monthly repayment on $1000 at 7.0% over 20 years = $7.75)`
 

`:.\ text(Monthly repayment on $150 000 loan)`

`= 150 xx 7.75`

`= $1162.50`

 

ii.  `text(Total repayments over 20 years)`

`= 20 xx 12 xx 1162.50`

`= $279\ 000`
 

`:.\ text(Interest paid over 20 years)`

`= 279\ 000 – 150\ 000`

`= $129\ 000`

 

iii.  `text(Savings)` `=\ text{Total paid (20 years) – Total paid (15 years)`
  `= 279\ 000 – 242\ 730`
  `= $36\ 270`

Algebra, STD2 A4 2004 HSC 26a

  1. The number of bacteria in a culture grows from 100 to 114 in one hour.

     

    What is the percentage increase in the number of bacteria?   (1 mark)

  2. The bacteria continue to grow according to the formula  `n = 100(1.14)^t`, where  `n`  is the number of bacteria after  `t`  hours.

     

    What is the number of bacteria after 15 hours?   (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\;  &  \;\; 5 \;\;  & \;\; 10 \;\;  & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\;  &  \;\; 193 \;\;  & \;\; 371 \;\;  & \;\; ? \;\; \\
\hline
\end{array}

  1. Use the values of `n` from  `t = 0`  to  `t = 15`  to draw a graph of  `n = 100(1.14)^t`.

     

    Use about half a page for your graph and mark a scale on each axis.   (4 marks)

  2. Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300.   (1 mark)

Show Answers Only
  1. `text(14%)`
  2. `714`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(8.4 hours)`
Show Worked Solution

i.   `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

 

ii.  `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n` `= 100(1.14)^15`
  `= 713.793\ …`
  `= 714\ \ \ text{(nearest whole)}`

 

iii. 

 

iv.  `text(Using the graph)`

`text(The number of bacteria reaches 300 after)`

`text(approximately 8.4 hours.)`

Probability, 2UG 2004 HSC 25b

Joe sells three different flavours of ice-cream from three different tubs in a cabinet. The flavours are chocolate, strawberry and vanilla.

2004 25b

  1. In how many different ways can he arrange the tubs in a row? Show working to justify your answer.  (2 marks)
  3. Paul buys an ice-cream from Joe on two different days. He chooses the flavour at random. What is the probability that he chooses chocolate on both days?  (1 mark)
  5. Mei-Ling buys an ice-cream from Joe and chooses any two different flavours at random. What is the probability that she chooses chocolate first and then strawberry?  (2 marks)

 

Show Answers Only
  1. `6`
  3. `1/9`
  5. `1/6`
Show Worked Solution

(i)  `text{# Arrangements (order matters)}`

`= 3 xx 2 xx 1`

`= 6`

 

(ii)  `P(C, C)` `= 1/3 xx 1/3`
  `= 1/9`

 

(iii)  `P(C, S)` `= 1/3 xx 1/2`
  `= 1/6`

Measurement, STD2 M1 2005 HSC 23b

A clay brick is made in the shape of a rectangular prism with dimensions as shown.
 

  1. Calculate the volume of the clay brick.  (1 mark)

Three identical cylindrical holes are made through the brick as shown. Each hole has a radius of 1.4 cm.  
 

  1. What is the volume of clay remaining in the brick after the holes have been made? (Give your answer to the nearest cubic centimetre.)  (3 marks)

  2. What percentage of clay is removed by making the holes through the brick? (Give your answer correct to one decimal place.)  (1 mark)

Show Answers Only
  1. `text(1512 cm)^3`
  2. `text{1364 cm}^3`
  3. `text{9.8%}`
Show Worked Solution
i.    `V` `= l × b × h`
    `= 21 × 8 × 9`
    `= 1512\ text(cm)^3`

 

ii.  `text(Volume of each hole)`

`= pir^2h`

`= pi × 1.4^2 × 8`

`= 49.260…\ text(cm)^3`

 

`:.\ text(Volume of clay still in brick)`

`= 1512 − (3 × 49.260…)`

`= 1364.219…`

`= 1364\ text{cm}^3\ text{(nearest whole)}`

 

iii. `text(Percentage of clay removed)`

`= ((3 × 49.260…))/1512 × 100`

`= 9.773…`

`= 9.8 text{%   (1 d.p.)}`

Data, 2UG 2005 HSC 21 MC

Yousef used the ‘capture-recapture’ technique to estimate the number of kangaroos living
in a particular area.

• He caught, tagged and released `50` kangaroos.
• Later, he caught `200` kangaroos at random from the same area.
• He found that `5` of these `200` kangaroos had been tagged.

What is the correct estimate for the total number of kangaroos living in this area, using
the ‘capture-recapture’ technique?

(A)   `245`

(B)   `250`

(C)   `2000`

(D)   `10\ 000`

Show Answers Only

`C`

Show Worked Solution

`text(Let P = population of kangaroos)`

`text(Capture)= 50/text(P)`

`text(Recapture)= 5/200`

`:. 50/text(P)` `= 5/200`
`text(5P)` `= 10\ 000`
`text(P)` `= 2000`

 

`=>  C`

Measurement, 2UG 2005 HSC 19 MC

The location of Town A is `text(25°N 45°E)`. The location of Town B is `10°N \ 105°E`.

Which of the following is true? (Ignore time zones.)

(A)   Town A is four hours behind Town B.

(B)   Town A is four hours ahead of Town B.

(C)   Town A is one hour behind Town B.

(D)   Town A is one hour ahead of Town B.

Show Answers Only

`A`

Show Worked Solution

`text(Town A is 25)^@N \ 45^@E`

`text(Town B is 10)^@N \ 105^@E`

`text(Angular difference (longitude))`

`= 105^@ − 45^@`

`= 60^@`

`text(Using 15)^@ = text(1 hour)`

`text(Time Difference)` `= 60/15`
  `= 4\ text(hours)`

 

`text(S)text(ince Town A is further West, Town A)`

`text(is 4 hours behind Town B.)`

`=>  A`

Measurement, 2UG 2005 HSC 18 MC

A model yacht has two triangular sails. These triangles are similar to each other. Some dimensions of the sails, in centimetres, are shown on the diagram.

2UG-2005-18MC

What is the total area of both sails?

(A)   `24\ text(cm²)`

(B)   `27\ text(cm²)`

(C)   `60\ text(cm²)`

(D)   `97\ text(cm²)`

Show Answers Only

`C`

Show Worked Solution
`text(Ratio of hypotenuses)` `= 15:5`
  `= 3:1`

`:.\ text(Sails have the dimensions)`

2UG-2005-18MC Answer

`:.\ text(Area of both sails)`

`= (1/2 × 3 × 4) +(1/2 × 12 × 9)`

`= 6 + 54`

`= 60\ text(cm²)`

`=>  C`

Algebra, STD2 A2 2005 HSC 17 MC

The total cost, `$C`, of a school excursion is given by  `C = 2n + 5`, where `n` is the number of students.

If three extra students go on the excursion, by how much does the total cost increase?

  1. `$6`
  2. `$11`
  3. `$15`
  4. `$16`
Show Answers Only

`A`

Show Worked Solution

`C = 2n + 5`

`text(If)\ n\ text(increases to)\ n + 3`

`C` `= 2(n + 3) + 5`
  `= 2n + 6 + 5`
  `= 2n + 11`

 

`:.\ text(Total cost increases by $6)`

`=>  A`

Financial Maths, STD2 F4 2005 HSC 13 MC

Last year, Helen bought 150 shares at $2.00 per share. They are now worth $2.50 per share. Helen receives a dividend of $0.10 per share.

What is the dividend yield?

  1.    `text(4%)`
  2.    `text(20%)`
  3.    `$15`
  4.    `$75`
Show Answers Only

`A`

Show Worked Solution
`text(Dividend yield)` `= text(Dividend)/text(Share Value)`
  `= 0.10/2.50`
  `= 0.04`
  `=\ text(4%)`

`=>  A`

Statistics, STD2 S1 2005 HSC 9 MC

A set of data is represented by the cumulative frequency histogram and ogive.
 

2UG-2005-9MC
 

What is the best approximation for the interquartile range for this set of data?

  1.    25
  2.    30
  3.    35
  4.    40
Show Answers Only

`C`

Show Worked Solution

2UG-2005-9MC Answer

`IQR` `= Q3 − Q1`
  `= 80 − 45`
  `= 35`

`=>  C`

Probability, STD2 S2 2004 HSC 25c

Lie detector tests are not always accurate. A lie detector test was administered to 200 people.

The results were:

• 50 people lied. Of these, the test indicated that 40 had lied;
• 150 people did NOT lie. Of these, the test indicated that 20 had lied.

  1. Complete the table using the information above   (2 marks)
      
        

  2. For how many of the people tested was the lie detector test accurate?   (1 mark)

  3. For what percentage of the people tested was the test accurate?   (1 mark)

  4. What is the probability that the test indicated a lie for a person who did NOT lie?   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `170`
  3. `text(85%)`
  4. `2/15`
Show Worked Solution

i.

ii.  `text(# Accurate readings)`

`= 40 + 130`

`= 170`
 

iii.  `text(Percentage of people with accurate readings)`

`= text(# Accurate readings)/text(Total readings) xx 100`

`= 170/200`

`= 85 text(%)`
 

iv.  `text{P(lie detected when NOT a lie)}`

`= 20/150`

`= 2/15`

Statistics, STD2 S5 2004 HSC 24c

The normal distribution shown has a mean of 170 and a standard deviation of 10.
 


 

  1. Roberto has a raw score in the shaded region. What could his `z`-score be?  (1 mark)

  2. What percentage of the data lies in the shaded region?  (2 marks)

Show Answers Only
  1. `1 <= text(z-score) <= 2`
  2. `text(13.5%)`
Show Worked Solution
i.  `ztext{-score(180)}` `= (x – mu)/sigma`
  `= (180 – 170) / 10`
  `= 1`

 

`ztext{-score(190)}` `= (190 -170)/10`
  `= 2`

 
`:. 1 <= ztext(-score) <= 2`

 

ii.   

`text(From the graph above,)`

`text(13.5% lies in the shaded area.)`

Measurement, STD2 M6 2004 HSC 24b

The diagram shows a radial survey of a piece of land.
 


 

  1. `Q` is south-east of `A`. What is the size of angle `PAQ`?  (1 mark)

  2. What is the bearing of `R` from `A`?  (1 mark)

  3. Find the size of angle `PAB` to the nearest degree.  (3 marks)

Show Answers Only
  1. `135°`
  2. `185°`
  3. `text{73° (nearest degree)}`
Show Worked Solution
i.   

`text(Let)\ E\ text(be directly east of)\ A`

`/_ EAQ = 45^@\ \ \ \ text{(given)}`

`:. /_ PAQ` `= 90^@ + 45^@`
  `= 135^@`

 

ii.   `text(Bearing of)\ R\ text(from)\ A`

`= 90^@ + 45^@ + 50^@`

`= 185^@\ \ \ text{(or  S 5°W)}`

 

iii.  `text(Using cosine rule in)\ Delta PAB`

` cos\ /_ PAB` `= (31^2 + 28^2 – 35^2) / (2 xx 31 xx 28)`
  `= 0.2995…`
`:. /_ PAB` `= 72.570…^@`
  `= 73^@\ \ \ \ text{(nearest degree)}`

Measurement, 2UG 2004 HSC 23c

Calculate the height  `(h\ text{metres})`  of the tree in the diagram. All measurements are
in metres.  (2 marks)

2004 23c

Show Answers Only

`text(6.75 m)`

Show Worked Solution

`text(Using similar triangles,)`

`h/5` `= 2.7/2`
`:.h` `= (5 xx 2.7)/2`
  `= 6.75\ text(m)`

Algebra, STD2 A2 2004 HSC 22 MC

John knows that

• one Australian dollar is worth 0.62 euros
• one Vistabella dollar  `text{($V)}`  is worth 1.44 euros.

John changes 25 Australian dollars to Vistabella dollars.

How many Vistabella dollars will he get?

  1. `text($V10.76)`
  2. `text($V22.32)`
  3. `text($V28.00)`
  4. `text($V58.06)`
Show Answers Only

`A`

Show Worked Solution

`text(John has 25 Aust dollars.)`

`text(Converting to Euros)`

`text(25 Aust)` `= 25 xx 0.62`
  `= 15.5\ text(Euros)`

 

`text(Converting to Vistabella dollars)`

`text(15.5 Euros)` `= 15.5/1.44`
  `=\ text($V10.76)`

`=>  A`

Algebra, STD2 A4 2004 HSC 21 MC

The time `(t)` taken to clean a house varies inversely with the number `(n)` of people
cleaning the house.

Which graph represents this relationship?

Show Answers Only

`D`

Show Worked Solution

`t ∝ 1/n \ => \ t = k/n`

`text(The graph is a hyperbola that sees)\ t\ text(decrease as)\ n\ text{increases (eliminate A and B).}`

`text(Also,)\ t\ text{cannot be zero (eliminate C).}`

`=>  D`

Financial Maths, STD2 F4 2006 HSC 21 MC

Bill borrows  $420 000  to buy a house. Interest is charged at 7.2% per annum, compounded monthly.

How much does he owe at the end of the first month, after he has made a $4000 repayment?

  1.   $418 496
  2.   $418 520
  3.   $445 952
  4.   $446 240
Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ L\ =\ text(Amount of the loan after 1 month)`

`r= (7.2 text(%))/12=0.6 text(%)\ =0.006`
 

`text(Using)\ \ FV=PV(1+r)^n`

`L` `= 420\ 000 (1 + 0.006)^1 – text(repayment)`
  `= 420\ 000 (1.006) – 4000`
  `= $418\ 520`

 
`=>  B`

Measurement, 2UG 2006 HSC 19 MC

Makoua and Macapá are two towns on the equator.

The longitude of Makoua is `text(16°E)` and the longitude of Macapá is `text(52°W)`.

How far apart are these two towns if the radius of Earth is approximately `6400` km?

(A)  `4000\ text(km)`

(B)  `7600\ text(km)`

(C)  `1\ 447\ 600\ text(km)`

(D)  `2\ 734\ 400\ text(km)`

Show Answers Only

`B`

Show Worked Solution

2UG-2006-19MC Answer

`text(Angular Difference)` `= 52 + 16`
  `= 68°`
`text(Arc)\ AB` `=\ text(Distance between two towns)`
  `= 68/360 xx 2 xx pi xx r`
  `= 68/360 xx 2 xx pi xx 6400`
  `= 7595.67…\ text(km)`

`=>  B`

Algebra, STD2 A1 2006 HSC 18 MC

What is the formula for `q` as the subject of `4p =5t + 2q^2`?

  1. `q = +- sqrt(5t - 4p)/2`
  2. `q = +- sqrt(4p + 5t)/2`
  3. `q = +- sqrt{(5t - 4p)/2}`
  4. `q = +- sqrt{(4p - 5t)/2}`
Show Answers Only

`D`

Show Worked Solution
`4p` `= 5t + 2q^2`
`2q^2` `= 4p – 5t`
`q^2` `= (4p – 5t)/2`
`q` `= +- sqrt{(4p – 5t)/2}`

 
`=>  D`

Statistics, STD2 S5 2006 HSC 17 MC

In a normally distributed set of scores, the mean is 23 and the standard deviation is 5.

Approximately what percentage of the scores will lie between 18 and 33?

  1. `text(34%)`
  2. `text(47.5%)`
  3. `text(68%)`
  4. `text(81.5%)`
Show Answers Only

`D`

Show Worked Solution

`mu = 23`

`sigma = 5`

`z text{-score (18)}` `= (x – mu)/sigma`
  `= (18 – 23)/5`
  `= -1`
`z text{-score (33)}` `= (33 – 23)/5`
  `= 2`

z-score

`:.\ text(Percentage between –1 and 2)`

`= 34 + 34 + 13.5`

`=\ text(81.5%)`

`=>  D`

Probability, 2UG 2006 HSC 15 MC

Two people are to be selected from a group of six people to form a team.

How many different teams can be formed?

(A)  15

(B)  18

(C)  30

(D)  36

Show Answers Only

`A`

Show Worked Solution
`text(# Different teams)` `= (6 xx 5)/(1 xx 2)`
  `= 15`

`=>  A`

Statistics, STD2 S1 2006 HSC 8 MC

Which of these graphs best represents positively skewed data with the smaller standard deviation?
 

2UG-2006-8abMC

2UG-2006-8cdMC

Show Answers Only

`C`

Show Worked Solution

`text(By elimination)`

`text(Positive skew when the tail on the`

`text(right side is longer.)`

`:.\ text(NOT)\ B\ text(or)\ D`

`text(A smaller standard deviation occurs)`

`text(when data is clustered more closely.)`

`:.\ text(NOT)\ A\ text(where data is more widely spread.)`

`=>  C`

Financial Maths, STD2 F4 2004 HSC 19 MC

Kerry has a credit card. She is charged 0.05% compound interest per day on outstanding
balances.

How much interest is Kerry charged on an amount of $250, which is outstanding on her
credit card for 30 days?

  1.   $3.75
  2.   $3.78
  3.   $253.75
  4.   $253.78
Show Answers Only

`B`

Show Worked Solution

`text(Using)\ A = P(1 + r)^n`

`A` `= 250 (1 + 0.0005)^30`
  `= 250 (1.0151…)`
  `= 253.777…`

 

`:.\ text(Amount of interest charged)`

`= 253.777… – 250`

`= $3.777…`

`=>  B`

Algebra, STD2 A4 2004 HSC 16 MC

George drew a correct diagram that gave the solution to the simultaneous equations

`y = 2x − 5`  and  `y = x + 6`.

Which diagram did he draw?

Show Answers Only

`D`

Show Worked Solution

`text(By elimination)`

`y = 2x – 5\ \ text(cuts the y-axis at)\ -5`

`:.\ text(Cannot be A or B)`

 

`y = x + 6\ \ text(cuts the y-axis at)\ 6`

`text(AND has a positive gradient)`

`:.\ text(Cannot be C)`

`=>  D`

Probability, STD2 S2 2004 HSC 18 MC

Two dice are rolled. What is the probability that only one of the dice shows a six?

  1. `5/36`
  2. `1/6`
  3. `5/18`
  4. `11/36`
Show Answers Only

`C`

Show Worked Solution

`text{P (Only 1 six)}`

`= P text{(6, not 6)} + P text{(not 6, 6)}`

`= 1/6 xx 5/6 + 5/6 xx 1/6`

`= 10/36`

`= 5/18`

`=>  C`

Probability, 2UG 2004 HSC 14 MC

Mike plays a game in which he has:
• `1/10` chance of winning `$20`

• `1/2` chance of winning `$1`

• `2/5` chance of losing `$2`.

 

What is Mike’s financial expectation when playing this game?

(A)  `$1.70`

(B)  `$3.30`

(C)  `$17.00`

(D)  `$19.00`

Show Answers Only

`A`

Show Worked Solution

`text(Financial Expectation)`

`= (1/10 xx 20) + (1/2 xx 1) – (2/5 xx 2)`

`= 2 + 0.5 – 0.8`

`= 1.7`

`:.\ text(Mike should win)\ $1.70`

`=>  A`

CORE*, FUR1 2009 VCAA 3 MC

Jamie bought a $500 games console on a hire-purchase plan.

He paid $50 deposit and monthly instalments of $25 for two years.

The flat interest rate charged per annum is closest to

A.   15.0%

B.   16.7%

C.   30.0%

D.   33.3%

E.   66.7%

Show Answers Only

`B`

Show Worked Solution
`text(Total owing)` `= 500 − text(deposit)`
  `= 500 − 50`
  `= 450`

 
`text(Total paid in instalments)`

♦ Mean mark 44%.

`= $25 xx 12\ text(months) xx 2\ text(years)`

`= 600`
 

`:.\ text(Interest paid)` `= 600 − 450`
  `= 150`

 

`I` `= (PrT)/100`
`150` `= ((450)(r)(2))/100`
`:. r`  `= (150 xx 100)/(450 xx 2)`
  `= 16.66… %`

 
`=>  B`

CORE*, FUR1 2008 VCAA 3 MC

A computer originally purchased for $6000 is depreciated each year using the reducing balance method.

If the computer is valued at $2000 after four years, then the annual rate of depreciation is closest to

A.   17%

B.   24%

C.   25%

D.   28%

E.   33%

Show Answers Only

`B`

Show Worked Solution

`A = 2000, \ P = 6000, \ n = 4`

♦ Mean mark 47%.
`text(Using)\ \ \ A` `= PR^n`
`2000` `= 6000 xx R^4`
 `R^4` `= 1/3`
`R`  `= 0.7598…`

 

`text(S)text(ince)\ \ \ R` `= 1 – (r/100)`
`:.\ 0.7598` `= 1 – r/100`
`r/100` `= 0.2402`
 `r` `= 24.02 text(%)`

 
`=>  B`

CORE*, FUR1 2007 VCAA 9 MC

Petra borrowed $250 000 to buy a home. The interest rate is 7% per annum, calculated monthly on the reducing balance over the life of the loan. She will fully repay the loan over 20 years with equal monthly instalments.

The total amount of interest she will pay on the loan is closest to

A.   $215 000

B.   $266 000

C.   $281 000

D.   $350 000

E.   $465 000

Show Answers Only

`A`

Show Worked Solution

`text(By TVM Solver:)`

♦ Mean mark 37%.
`N` `= 20 xx 12 = 240`
`I(%)` `= 7text(%)`
`PV` `= − 250\ 000`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 
`:. \ PMT = $1938.25`

 

`:.\ text(Total interest paid)`

`= 240 xx PMT – 250\ 000`

`= 240 xx $1938.25 – 250\ 000`

`= $215\ 180`

`=>  A`

CORE*, FUR1 2007 VCAA 8 MC

Brad investigated the cost of buying a $720 washing machine under a hire purchase agreement.

A deposit of $180 is required and the balance will be paid in 24 equal monthly repayments.

A flat interest rate of 12% per annum applies to the balance.

Brad correctly calculated the monthly repayment to be

A.   $22.50

B.   $25.20

C.   $26.10

D.   $27.90

E.   $29.70

Show Answers Only

`D`

Show Worked Solution
♦ Mean mark 36%.
`text(Amount owing` `= 720 − 180`
  `= $540`
`I` `= (PrT)/100`
  `=(540 xx 12 xx2)/100`
  `=$129.60`

 

`:. \ text(Total Amount Owed)` `= 129.60 + 540`
  `= $669.60`

 
`:. \ text(Monthly repayments)\ =669.60/24=$27.90` 

`=>  D`

CORE*, FUR1 2007 VCAA 6 MC

$10 000 is invested at a rate of 10% per annum compounding half yearly.

The value, in dollars, of this investment after five years, is given by

A.  `10\ 000 xx 0.10 xx 5`

B.  `10\ 000 xx 0.05 xx 10`

C.  `10\ 000 xx 0.05^10`

D.  `10\ 000 xx 1.05^10`

E.  `10\ 000 xx 1.10^5`

Show Answers Only

`D`

Show Worked Solution

`text(Interest rate)\ = \frac{10%}{2} = 5%\ \ \text{(per 6 months)}`

♦ Mean mark 42%.

`text{Compounding periods}\ (n) =5 xx 2=10`

`:.A` ` = PR^n`
  `= 10\ 000 xx 1.05^(10)`

 
`=>  D`

CORE*, FUR1 2006 VCAA 8 MC

The points on the graph below show the balance of an investment at the start of each quarter for a period of six years.

The same rate of interest applied for these six years.
 

In relation to this investment, which one of the following statements is true?

A.   interest is compounding annually and is credited annually

B.   interest is compounding annually and is credited quarterly

C.   interest is compounding quarterly and is credited quarterly

D.   simple interest is paid on the opening balance and is credited annually

E.   simple interest is paid on the opening balance and is credited quarterly

Show Answers Only

`A`

Show Worked Solution

`text(From the graph, as balance increases after)`

♦ Mean mark 37%.
MARKERS’ COMMENT: Graphical analysis of financial situations is a requirement of the study design and deserves attention.

`text(each year, interest is credited annually.)`

`:.\ text(Eliminate)\ B, C\ text(and)\ E.`

`text(The difference of the balances between successive)`

`text(years is increasing which indicates that interest is)`

`text(compounding.)`

`:.\ text(Eliminate)\ D.`

`=>  A`

CORE*, FUR1 2006 VCAA 7 MC

Mervyn bought a new lawn mower at a sale.

First, there was a 20% discount from the original price.

Then, an $80 trade-in for his old mower was subtracted from this reduced price.

This left Mervyn with $368 to pay for the new lawn mower.

The original price of the new lawn mower was

A.   $468.00

B.   $537.50

C.   $540.00

D.   $560.00

E.   $580.00

Show Answers Only

`D`

Show Worked Solution

 `text(Let original price) = P`

`:.\ text(Original price less 20% discount) = 0.8P`

 

`text($80 trade-in for old mower subtracted)`

`text{from discounted price = $368 (given)}`

`:.\ 0.8P − 80` `= $368`
`0.8P` `= 448`
`P` `= $560`

`=>  D`

CORE*, FUR1 2006 VCAA 6 MC

A $2000 lounge suite was sold under a hire-purchase agreement.

A deposit of $200 was paid.

The balance was to be paid in 36 equal monthly instalments of $68.

The annual flat rate of interest applied to this agreement is

A.   10.0%

B.   11.4%

C.   12.0%

D.   22.4%

E.   36.0%

Show Answers Only

`C`

Show Worked Solution
`text(Amount owing)` `= 2000 − 200`
  `= $1800`

 
`text(Total amount of instalments)`

`= 68 xx 36`

`= $2448`
 

`:.\ text(Interest paid)` `= 2448 − 1800`
  `= $648`
`I` `= (PrT)/100`
`648` `= (1800 xx r xx 3)/100`
`:.r` `= (648 xx 100)/(1800 xx 3)`
  `= 12text(% p.a.)`

`=>  C`

CORE*, FUR1 2006 VCAA 4 MC

An item was purchased for a price of $825.

The price included 10% GST (Goods and Services Tax).

The amount of GST included in the price is

A.       $8.25

B.     $75.00

C.     $82.50

D.     $90.75

E.   $125.00

Show Answers Only

`B`

Show Worked Solution

`text(Let price without GST) = x`

♦ Mean mark 42%.
MARKERS’ COMMENT: Almost half of all students incorrectly calculated the GST as 10% of the purchase price and answered C.
`text(Price with GST)` `= x + 10text(%) xx x`
  `= 1.1x`
`825`  `= 1.1x`
`x`  `= 750`

 

`:.\ text(GST)` `= 750 xx 0.1`
  `= $75`

 `=>  B`

CORE*, FUR1 2005 VCAA 8 MC

On Monday, a clock has a certain price.

In a crazy week of sales the price goes up and down as follows.

Tuesday – the price of the clock is increased by 10%.
Wednesday – Tuesday’s price is reduced by 10%.
Thursday – Wednesday’s price is increased by 20%.
Friday – Thursday’s price is reduced by 20%.

When compared with Monday’s price, the price on Friday is closest to

A.   the same as Monday’s price.

B.   4% lower than Monday’s price.

C.   4% higher than Monday’s price.

D.   5% lower than Monday’s price.

E.   5% higher than Monday’s price.

Show Answers Only

`D`

Show Worked Solution

`text(Let Monday price be)\ $x.`

`text(Tuesday price)` `= x xx text(110%)`
  `= 1.1x`
`text(Wednesday price)` `= 1.1x × text(90%)`
  `= 0.99x`
`text(Thursday price)`  `= 0.99x xx text(120%)`
  `= 1.188x`
`text(Friday price)` `= 1.188x xx text(80%)`
  `= 0.9504x`

 

`:.\ text(Friday’s price is around 95% of Monday’s)`

`text(price, or 5% lower compared to Monday.)`

`=>  D`

CORE*, FUR1 2005 VCAA 7 MC

Gregor invests $10 000 and earns interest at a rate of 6% per annum compounding quarterly.

Every quarter, after interest has been added, he withdraws $500.

At the end of four years, after interest has been added and he has made the $500 withdrawal, the value of the remaining investment will be closest to

A.     $3720

B.     $4220

C.     $5440

D.   $21 660

E.   $22 160

Show Answers Only

`A`

Show Worked Solution

`text(By TVM Solver:)`

`N` `= 16`  
`I text(%)` `=6`  
`PV` `=-10\ 000`  
`PMT` `= 500`  
`FV` `=?`  
`text(PY)` `= text(C/Y) = 4`  

 
`:. FV = 3723.67`

`=>  A`

CORE*, FUR1 2005 VCAA 6 MC

Tim invests $3000 in a term deposit account that adds 6.5% interest annually, calculated on the account balance at the end of each year.

The interest paid in the fourth year is

A.     `$195.00`

B.     `$221.16`

C.     `$235.55`

D.   `$3623.85`

E.   `$3859.40`

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ A = PR^n`

`text(Total after 3rd year)` `= 3000(1.065)^3 \ \ \ \ \ (n =3)`
  `= $3623.848875…`
  `= $3623.85 \ \ (text(2 d.p.))`
 `text(Total after 4th year)` `= 3000(1.065)^4 \ \ \ \ \ (n = 4)`
  `= $3859.399…`
  `= $3859.40 \ \ (text(2 d.p.))`

 

 `:.\ text(Interest paid in 4th year)` `= 3859.40 − 3623.85`
  `= $235.55`

`=>  C`