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Financial Maths, STD2 F4 2006 HSC 27a

Liliana wants to borrow money to buy a house. The bank sent her an email with the following table attached.

2006 27a

  1. Liliana decides that she can afford $1000 per month on repayments.

     

    What is the maximum amount she can borrow, and how many years will she have to repay the loan?  (1 mark)

  2. Zali intends to borrow  $160 000  over 15 years from the same bank.

     

    If she chooses to borrow  $160 000  over 20 years instead, how much more interest will she pay?  (2 marks)

Show Answers Only
  1. `text{$130 000 (over 30 years)}`
  2. `$45\ 964.80`
Show Worked Solution

i.  `text(From table)`

`text{$130 000 (over 30 years)}`

 

ii.  `text(Total repayments over 15 years)`

`= $1529.04 xx 180`

`= $275\ 227.20`

`text(Total repayments over 20 years)`

`= $1338.30 xx 240`

`= $321\ 192.00`

 

`:.\ text(Extra interest over 20 years)`

`= 321\ 192.00 – 275\ 227.20`

`= $45\ 964.80`

Measurement, STD2 M2 2006 HSC 26d

Cassie flew from London to Manila. Manila is 8 hours ahead of London.

Her plane left London at 9.30 am Monday (London time), stopped for 5 hours in Singapore and arrived in Manila at 4.00 pm Tuesday (Manila time).

What was the total flying time? (Ignore time zones.)  (2 marks)

Show Answers Only

`text(17.5 hours)`

Show Worked Solution
`:.\ text(Flight time)` `=\ text{9:30 (Mon) to 4:00 pm (Tues)}`
         `- \ text{(stopover + time difference)}`
  `= 14.5 + 16-(5 + 8)`
  `= 17.5\ text(hours)`

Probability, STD2 S2 2006 HSC 26c

A new test has been developed for determining whether or not people are carriers of the Gaussian virus.

Two hundred people are tested. A two-way table is being used to record the results.
 

  1.  What is the value of  `A`?  (1 mark)

  2.  A person selected from the tested group is a carrier of the virus.

     

     What is the probability that the test results would show this?  (2 marks) 

  3.  For how many of the people tested were their test results inaccurate?  (1 mark)

Show Answers Only
  1. `98`
  2. `37/43`
  3. `28`
Show Worked Solution
i.  `A` `= 200-(74 + 12 + 16)`
  `= 98`

 

ii.  `P` `= text(# Positive carriers)/text(Total carriers)`
  `= 74/86`
  `= 37/43`

 

iii.  `text(# People with inaccurate results)`

`= 12 + 16`

`= 28`

Measurement, STD2 M6 2006 HSC 26a

Daniel conducts an offset survey to sketch a diagram, `ABCD`, of a block of land.

Daniel walks from `A` to `C`, a distance of 62 m.

When he is 15 m from `A`, he notes that point `D` is 25 m to his right.

When he is 57 m from `A`, he notes that point `B` is 20 m to his left.

This is his notebook entry.

  1. Draw a neat sketch of the block of land. Label `A, B, C` and `D` on your diagram.  (1 mark)

  2. Calculate the distance from `C` to `D`. (Give your answer to the nearest metre.)  (2 marks)

Show Answers Only
  1.  
  2. `text{53 m (nearest m)}`
Show Worked Solution
i.   

 

ii.  `text(Using Pythagoras:)`

`CD^2` `= 47^2 + 25^2`
  `= 2834`
`:. CD` `= 53.235…`
  `= 53\ text{m  (nearest m)}`

Probability, STD2 S2 2006 HSC 25c

Sonia buys three raffle tickets.

HSC 2006 25c

  1. What is the probability that Sonia wins first prize?  (1 mark)

  2. What is the probability that she wins both prizes?  (2 marks)

Show Answers Only
  1. `1/60`
  2. `1/5370`
Show Worked Solution
i.  `text{P (wins 1st prize)}` `= text(# tickets bought) / text(total tickets)`
  `= 3/180`
  `= 1/60`

 

ii.  `text{P (wins both)}` `= text{P (wins 1st)} xx text{P (wins 2nd)}`
  `= 1/60 xx 2/179`
  `= 1/5370`

Calculus, EXT1* C3 2005 HSC 6c

2005 6c
 

The graphs of the curves  `y = x^2`  and  `y = 12 - 2x^2`  are shown in the diagram.

  1. Find the points of intersection of the two curves.  (1 mark)

  2. The shaded region between the curves and the `y`-axis is rotated about the `y`-axis. By splitting the shaded region into two parts, or otherwise, find the volume of the solid formed.  (3 marks)

Show Answers Only
  1. `text{(2, 4), (–2, 4)`
  2. `24pi\ \ text(u³)`
Show Worked Solution
i.    `y` `= x^2` `\ …\ (1)`
  `y` `= 12 − 2x^2` `\ …\ (2)`

 

`text(Substitute)\ \ y = x^2\ \ text(into)\ (2)`

`x^2` `= 12 − 2x^2`
`3x^2 − 12` `= 0`
`3(x^2 − 4)` `= 0`
`x` `= ±2`
`text(When)` `\ x = 2,` `\ y = 4`
`text(When)` `\ x = text(−2),` `\ y = 4`

 
`:.\ text{Intersection at (2, 4), (−2, 4)}`
 

ii.  `text{In (1),}\ \ x^2=y`

`text{In (2),}\ \ \ y` `= 12 − 2x^2`
`2x^2` `= 12 − y`
`x^2` `= (12 − y)/2`
  `= 6 − 1/2y`

 
`:.\ text(Volume)`

`= pi int_0^4 y\ dy + pi int_4^12 6 − 1/2y\ dy`
`= pi[y^2/2]_0^4 + pi[6y − y^2/4]_4^12`
`= pi[16/2 − 0] + pi[(6 xx 12 − 12^2/4) − (6 xx 4 − 4^2/4)]`
`= 8pi + pi[36 − 20]`
`= 8pi + 16pi`
`= 24pi\ \ text(u³)`

Calculus, 2ADV C3 2005 HSC 6b

A tank initially holds 3600 litres of water. The water drains from the bottom of the tank. The tank takes 60 minutes to empty.

A mathematical model predicts that the volume, `V`  litres, of water that will remain in the tank after  `t`  minutes is given by
  

`V = 3600(1 − t/60)^2,\ \ text(where)\ \ 0 ≤ t ≤ 60`.
 

  1. What volume does the model predict will remain after ten minutes?  (1 mark)

  2. At what rate does the model predict that the water will drain from the tank after twenty minutes?  (2 marks)

  3. At what time does the model predict that the water will drain from the tank at its fastest rate?  (2 marks)

Show Answers Only
  1. `text(2500 L)`
  2. `80\ text(liters per minute)`
  3. `0`
Show Worked Solution

i.    `V = 3600(1 − t/60)^2`

`text(When)\ t = 10,`

`V` `= 3600(1 − 10/60)^2`
  `= 3600 xx (5/6)^2`
  `= 2500\ text(L)`

 

ii.   `V = 3600(1 -t/60)^2`

`text(Using chain rule:)`

`(dV)/dt` `= 3600 xx 2 xx (1 – t/60) xx d/dt(1 – t/60)`
  `= 7200(1 – t/60) xx -1/60`
  `= −120(1 – t/60)`

 

`text(When)\ \ t =20`

`(dV)/dt` `= −120(1 – 20/60)`
  `= −80`

 

`:.\ text(After 20 minutes, the water will drain)`

`text(at 80 litres per minute.)`

 

iii. `(dV)/dt` `= −120(1 − t/60)`
    `= −120 + 2t`
  `(d^2V)/dt^2` `= 2`

 
`text(S)text(ince)\ (d^2V)/dt^2\ text(is a constant, no S.P.’s)`

 

`text(Checking limits of)\ \ 0 ≤ t ≤ 60`

`text(At)\ t = 0,`

`(dV)/dt = −120(1-0) = −120\ text(L/min)`

`text(At)\ t = 60,`

`(dV)/dt = −120(1 − 60/60) = 0\ text(L/min)`

 

`:.\ text(The model predicts water will drain)`

`text(out the fastest when)\ \ t = 0.`

Integration, 2UA 2005 HSC 6a

Five values of the function `f(x)` are shown in the table.

Integration, 2UA 2005 HSC 6a

Use Simpson’s rule with the five values given in the table to estimate

`int_0^20 f(x)\ dx`.  (3 marks)

Show Answers Only

`401 2/3`

Show Worked Solution

Integration, 2UA 2005 HSC 6a Answer

`int_0^20 f(x)\ dx` `= h/3[y_0 + y_4 + 4text{(odds)} + 2text{(evens)}]`
  `= 5/3[y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]`
  `= 5/3[15 + 10 + 4(25 + 18)+ 2(22)]`
  `= 5/3[25 + 172 + 44]`
  `= 401 2/3`

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.)}`

Calculus, 2ADV C3 2005 HSC 5c

Find the coordinates of the point  `P`  on the curve  `y = 2e^x + 3x`  at which the tangent to the curve is parallel to the line  `y = 5x - 3`.  (3 marks)

Show Answers Only

`(0, 2)`

Show Worked Solution

`text(Gradient of)\ \ y = 5x − 3\ text(is)\ 5.`

`y` `= 2e^x + 3x`
`(dy)/(dx)` `= 2e^x + 3`

 
`text(Find)\ \ x\ \ text(when)\ \ (dy)/(dx) = 5`

`5` `= 2e^x + 3`
`2e^x` `= 2`
`e^x` `= 1`
`x` `= 0`

 
`text(When)\ \ x = 0`

`y` `= 2e^0 + (3 xx 0)`
  `= 2`

 
`:.P\ \ text{has coordinates (0, 2)}`

Plane Geometry, 2UA 2005 HSC 5b

The diagram shows a parallelogram `ABCD` with `∠DAB = 120^@`. The side `DC` is produced to `E` so that `AD = BE`.

Prove that `ΔBCE` is equilateral.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`BC` `= AD\ text{(opposite sides of parallelogram}\ ABCD)`
`∠BCD` `= 120^@\ text{(opposite angles of parallelogram}\ ABCD)`
`∠BCE` `= 60^@\ (∠DCE\ text{is a straight angle)}`
`∠CEB` `= 60^@\ text{(base angles of isosceles}\ \Delta BCE)`
`∠CBE` `= 60^@\ text{(angle sum of}\ ΔBCE)`

 
`:.ΔBCE\ text(is equilateral)`

Calculus, 2ADV C3 2005 HSC 4b

A function  `f(x)`  is defined by  `f(x) = (x + 3)(x^2- 9)`.

  1. Find all solutions of  `f(x) = 0`  (2 marks)

  2. Find the coordinates of the turning points of the graph of  `y = f(x)`, and determine their nature.  (3 marks)

  3. Hence sketch the graph of  `y = f(x)`, showing the turning points and the points where the curve meets the `x`-axis.  (2 marks)

  4. For what values of `x` is the graph of  `y = f(x)`  concave down?  (1 mark)

Show Answers Only
  1. `−3 or 3`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(See worked solutions)`
  4. `x < −1`
Show Worked Solutions
i.    `f(x)` `= (x + 3)(x^2 − 9)`
    `= (x + 3)(x +3)(x − 3)`
  `:. f(x)` `= 0\ text(when)\ \ x=–3\ text(or)\ 3`

 

ii.   `f (x)` `= (x +3)(x^2 − 9)`
    `= x^3 − 9x + 3x^2 − 27`
    `= x^3 + 3x^2 − 9x − 27`
  `f′(x)` `= 3x^2 + 6x − 9`
  `f″(x)` `= 6x + 6`

 

`text(S.P.’s  when)\ \ f′(x) = 0`

`3x^2 + 6x − 9` `= 0`
`3(x^2 + 2x − 3)` `= 0`
`3(x − 1)(x + 3)` `= 0`

 

`text(At)\ x =1`

`f(1)` `= (4)(−8)=−32`
 `f″(1)` `= 6 + 6=12>0`
`:.\ text(MIN at)\ (1, −32)` 

 

`text(At)\ x = −3`

`f(-3)` `= 0`
`f″(−3)` `= (6 xx −3) + 6 = −12 <0`
`:.\ text(MAX at)\ (−3, 0)`

 

iii.   Geometry and Calculus, 2UA 2005 HSC 4b Answer

 

iv.  `f(x)\ \ text(is concave down when)`

`f″(x)` `< 0`
`6x + 6` `< 0`
`6x` `< −6`
`x` `< −1`

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` 

Trigonometry, 2ADV T1 2005 HSC 3b

The lengths of the sides of a triangle are 7 cm, 8 cm and 13 cm.

  1. Find the size of the angle opposite the longest side.  (2 marks)

  2. Find the area of the triangle.  (1 marks)

Show Answers Only
  1. `120^@`
  2. `14sqrt3\ text(cm)`
Show Worked Solution

i.

 Trig Ratios, 2UA 2005 HSC 3b Answer1 

`∠ABC\ \ text(is opposite the longest side)`

`text(Using the cosine rule)`

`cos\ ∠ABC` `= (7^2 + 8^2 −13^2)/(2 xx 7 xx 8)`
  `= text(−)1/2`

 
`text(S)text(ince cos)\ 60^@ = 1/2\ text(and cos is negative)`

`text(in 2nd quadrant,)`

`∠ABC` `= 180− 60`
  `= 120^@`

 

ii.  `text(Using the sine rule)`

`text(Area)\ ΔABC` `= 1/2\ ab\ sin\ C`
  `= 1/2 xx 7 xx 8\ sin 120^@`
  `= 28 xx sqrt3/2`
  `= 14sqrt3\ text(cm)^2`

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

Trigonometry, 2ADV T3 2006 HSC 7b

A function  `f(x)`  is defined by  `f(x) = 1 + 2 cos x`.

  1. Show that the graph of  `y = f(x)`  cuts the `x`-axis at  `x = (2 pi)/3`.   (1 mark)

  2. Sketch the graph of  `y = f(x)`  for  `-pi <= x <= pi`  showing where the graph cuts each of the axes.   (3 marks)

  3. Find the area under the curve  `y = f(x)` between  `x = -pi/2`  and  `x = (2 pi)/3`.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2.   
  3. `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`
Show Worked Solution

i.   `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x` `= 0`
`2 cos x` `=-1`
 `cos x` `= -1/2`

`:.  x = (2 pi)/3\ …\ text(as required)`

 

ii.   2UA HSC 2006 7b

 

iii.  `text(Area)` `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
  `= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
  `= [((2 pi)/3 + 2 sin­ (2 pi)/3) – ((-pi)/2 + 2 sin­ (-pi)/2)]`
  `= ((2 pi)/3 + 2 xx sqrt 3/2) – ((-pi)/2 +2(- 1))`
  `= (2 pi)/3 + sqrt(3) + pi/2 + 2`
  `= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`

Calculus, 2ADV C3 2006 HSC 5a

A function  `f(x)`  is defined by  `f(x) =2x^2(3 - x)`.

  1. Find the coordinates of the turning points of  `y =f(x)`  and determine their nature.  ( 3 marks)

  2. Find the coordinates of the point of inflection.  (1 mark)

  3. Hence sketch the graph of  `y =f(x)`, showing the turning points, the point of inflection and the points where the curve meets the `x`-axis.  (3 marks)

  4. What is the minimum value of  `f(x)`  for  `–1 ≤ x ≤4`?  (1 mark)

Show Answers Only
  1. `text(Min)\ (0, 0),\ text(Max)\ (2, 8)`
  2. `text(P.I. at)\ (1, 4)`
  3.  
  4. `-32`
Show Worked Solution
i.       `f(x)` `= 2x^2 (3 – x)`
  `= 6x^2 – 2x^3`
`f prime (x)` `= 12x – 6x^2`
`f″(x)` `= 12 – 12x`

 

`text(S.P.’s when)\ f′(x) = 0`

`12x – 6x^2` `= 0`
`6x(2 – x)` `= 0`

`x = 0 or 2`

`text(When)\ x = 0`

`f(0)` `= 0`
`f″(0)` `= 12 – 0 = 12 > 0`
`:.\ text(MIN at)\ (0, 0)`

 

`text(When)\ x = 2`

`f(2)` `= 2 xx 2^2 (3 – 2)` `= 8`
`f″(2)` `= 12 – (12 xx 2)` `= -12 < 0`
`:.\ text(MAX at)\ (2, 8)`

 

ii.  `text(P.I. when)\ f″(x) = 0`

`12 – 12x` `= 0`
`12x` `= 12`
`x` `= 1`
`f″(0.5)` `=6>0`
`f″(1.5)` `=-6<0`

`text(S)text(ince concavity changes)\ \ =>\  text(P.I. exists)` 

`f(1)` `= 2 xx 1^2(3 – 1)`
  `= 4`

`:.\ text(P.I. at)\ (1, 4)`

 

iii.  `f(x)\ text(meets)\ x text(-axis when)\ f(x) = 0`

`2x^2 xx (3 – x) = 0`

`x = 0 or 3`

2UA HSC 2006 5a

 

(iv)  `text(The graph clearly shows that in the given range)`

`-1<= x<=4,\ text(the minimum will occur when)\ x = 4`

`:.\ text(Minimum` `= 2 xx 4^2 (3 – 4)`
  `= -32`

Quadratics, 2UA 2005 HSC 1f

Find the coordinates of the focus of the parabola  `x^2 = 8(y − 1)`.  (2 marks)

Show Answers Only

`(0, 3)`

Show Worked Solution

`x^2 = 8(y − 1)`

`text(Vertex)` `= (0, 1)`
`4a` `= 8`
`a` `= 2`

Algebra, 2UA 2005 HSC 1f Answer

`:.\ text(Focus has coordinates)\ \  (0,3).`

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 S1 2005 HSC 27d

Nine students were selected at random from a school, and their ages were recorded.
 

2UG-2005-27d 

  1. What is the sample standard deviation, correct to two decimal places?   (2 marks)

  2. Briefly explain what is meant by the term standard deviation.   (1 mark)

Show Answers Only
  1. `text{1.69  (to 2 d.p.)}`
  2. `text(Standard deviation is a measure of how much)`

     

    `text(members of a data group differ from the mean)`

     

    `text(value of the group)`

Show Worked Solution

i.  `text(Sample standard deviation)`

`= 1.6914…\ text{(by calculator)}`

`= 1.69\ \ \ text{(to 2 d.p.)}`

 

ii.  `text(Standard deviation is a measure of how much)`

`text(members of a data group differ from the mean)`

`text(value of the group.)`

Measurement, STD2 M6 2005 HSC 27c

2UG-2005-27c
 

The bearing of `C` from `A` is 250° and the distance of `C` from `A` is 36 km.

  1. Explain why  `theta`  is 110°.   (1 mark)

  2. If  `B`  is 15 km due north of  `A`, calculate the distance of  `C`  from  `B`, correct to the nearest kilometre.   (3 marks)

Show Answers Only
  1. `110^@`
  2. `text{43 km (nearest km)}`
Show Worked Solution

i.  `text(There is 360° about point)\ A`

`:.theta + 250^@` `= 360^@`
`theta` `= 110^@`

 

ii.   
`a^2` `= b^2 + c^2 − 2ab\ cos\ A`
`CB^2` `= 36^2 + 15^2 − 2 xx 36 xx 15 xx cos\ 110^@`
  `= 1296 + 225 −(text(−369.38…))`
  `= 1890.38…`
`:.CB` `= 43.47…`
  `= 43\ text{km  (nearest km)}`

Measurement, 2UG 2005 HSC 27b

This diagram represents Earth. `O` is at the centre, and `A` and `B` are points on the surface.

2UG-2005-27b

Calculate the distance from `A` to `B` along the great circle through `A` and `B`.
Give your answer in to the nearest km.   (2 marks)

Show Answers Only

`text{4803 km  (nearest km)}`

Show Worked Solution

`A : 35^@text(N)\ 20^@text(E)\ \ \ \ B:8^@text(S)\ 20^@text(E)`

HSC 2005 27b

`text(Angular difference)` `= 35^@ + 8^@`
  `= 43^@`

 

`:.text(Distance from)\ A\ text(to)\ B`

`= 43/360 xx 2 xx pi xx r`

`= 43/360 xx 2 xx pi xx 6400`

`= 4803.1…`

`= 4803\ text{km  (nearest km)}`

Data, 2UG 2005 HSC 27a

The area graph shows sales figures for Shoey’s shoe store.

2UG-2005-27a

  1. Approximately how many school shoes were sold in January?   (1 mark)
  2. For which month does the graph indicate that the same number of school shoes and business shoes was sold?   (1 mark)
  4. Identify ONE trend in this graph, and suggest a valid reason for this trend.   (2 marks)
Show Answers Only
  1. `15\ 000`
  2.  `text(April)`
  3. `text(Possible trends include)`
  4. `text(- Boot sales peak in the Winter months of Jun-Aug)`
  5. `\ \ text(because of the colder weather.)`
    `text(- School shoe sales peak in January as children prepare)`
    `\ \ text(to return to school for the new year.)`
Show Worked Solution

(i)   `text(School shoes sold)`

`≈ 18\ 000 − 3000`

`≈ 15\ 000`

 

(ii)  `text(April)`

 

(iii) `text(Possible trends include)`

`text(- Boot sales peak in the Winter months of Jun-Aug)`

`\ \ text(because of the colder weather.)`

`text(- School shoe sales peak in January as children prepare)`

`\ \ text(to return to school for the new year.)`

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 2005 HSC 26a

A sports car worth $150 000 is bought in December 2005.

In December each year, beginning in 2006, the value of the sports car is depreciated by 10% using the declining balance method of depreciation.

In which year will the depreciated value first fall below $120 000?   (2 marks)

Show Answers Only

`text(The value falls below $120 000 in the third year)`

`text{which will be during 2008.}`

Show Worked Solution

`text(Using)\ \ S = V_0(1-r)^n`

`text(where)\ \ V_0 = 150\ 000, r = text(10%)`

`text(If)\ \ n = 2,`

`S` `= 150\ 000(1-0.1)^2`
  `= 121\ 500`

 
`text(If)\ \ n= 3,`

`S` `= 150\ 000(1-0.1)^3`
  `= 109\ 350`

 

`:.\ text(The value falls below $120 000 in the third year)`

`text{which will be during 2008.}`

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`  

Measurement, 2UG 2004 HSC 26b

The location of Sorong is `text(1°S 131°E)` and the location of Darwin is `text(12°S 131°E)`.

  1. What is the difference in the latitudes of Sorong and Darwin?  (1 mark)
  2. The radius of Earth is approximately `text(6400 km.)`

  3. Show that the great circle distance between Sorong and Darwin is approximately `text(1200 km)`.  (2 marks)

 

Show Answers Only
  1. `11°`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Difference in latitudes)`

`= 12^@ – 1^@`

`= 11^@`

 

(ii)   `text(S)text(ince longitude is the same,)`

`text(Distance between Sarang and Darwin)`

`= 11/360 xx 2 pi r`

`= 11/360 xx 2 xx pi xx 6400`

`= 1228.7…\ text(km)`

`~~ 1200\ text(km … as required.)`

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)}`

Data, 2UG 2006 HSC 23d

The graph shows the amounts charged by Company `A` and Company `B` to deliver parcels of various weights.

2UG-2006-23d

  1. How much does Company `A` charge to deliver a `3` kg parcel?  (1 mark)
  2. Give an example of the weight of a parcel for which both Company `A` and Company `B` charge the same amount.  (1 mark)
  4. For what weight(s) is it cheaper to use Company `A`?  (2 marks)
  5. What is the rate per kilogram charged by Company `B` for parcels up to `8` kg?  (1 mark)
Show Answers Only
  1. `$6`
  2. `text(4 kg or 7 kg)`
  3. `x > 7`
  4. `text($1.50 per kg)`
Show Worked Solution

(i)   `$6`

 

(ii)  `text(4 kg or 7 kg)`

 

(iii)  `text(Let)\ x = text(weights where Company)\ A`

`text{is cheaper (kg).}`

`4 < x <= 6`

`x > 7`

 

(iv)   `text(Company)\ B\ text(charges $12 for 8 kg)`

`:.\ text(Rate)` `= 12/8`
  `= $1.50\  text(per kg)`

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 24a

2UG-2006-24a

List TWO ways in which this graph is misleading.  (2 marks)

Show Answers Only

`text(Reasons the graph is misleading include)`

`text(- the columns are a different widths/volumes)`

`text(- the vertical axis doesn’t start at zero)`

Show Worked Solution

`text(Reasons the graph is misleading include)`

`text(- the columns are a different widths/volumes)`

`text(- the vertical axis doesn’t start at zero)`

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…)`

Measurement, STD2 M6 2005 HSC 25b

2UG-2005-25b

  1. Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle.  (1 mark)

  2. Calculate the size of `∠ABC` to the nearest minute.  (2 marks)

Show Answers Only
  1. `text(Proof)`
  2. `67°23^{′}`
Show Worked Solution

i.   `ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2`

`a^2 + b^2` `= 5^2 + 12^2`
  `= 169`
  `= 13^2`
  `= c^2…\ text(as required.)`

MARKER’S COMMENT: Know your calculator process for producing an angle in minutes/seconds. Note >30 “seconds” rounds up to the higher “minute”.

 
ii.  `sin ∠ABC = 12/13`

`:.∠ABC` `= 67.38…°`
  `=67°22^{′}48^{″}`
  `= 67°23^{′}\ \ \ text{(nearest minute)}`

Statistics, STD2 S1 2005 HSC 24d

The sector graph shows the proportion of people, as a percentage, living in each region of Sumcity. There are 24 000 people living in the Eastern Suburbs.
 

2UG-2005-24d1
 

  1. Show that the total number of people living in Sumcity is  160 000.  (1 mark)

Jake used the information above to draw a column graph.
 

2UG-2005-24d2

  1. The column graph height is incorrect for one region.

     

    Identify this region and justify your answer.   (2 marks)

Show Answers Only
  1. `160\ 000`
  2. `text(Western Suburbs population)`
Show Worked Solution

i.   `text(Let the population of Sumcity =)\ P`

`text(15%)× P` `= 24\ 000`
`:.P`  `= (24\ 000)/0.15` 
  `= 160\ 000\ …\ text(as required)` 

 

ii.  `text(Western Suburbs population)`

`= text(10%) × 160\ 000`

`= 16\ 000`

`text(The column graph has this population as)`

`text(12 000 people which is incorrect.)`

Algebra, STD2 A1 2005 HSC 24b

The formula  `D = (2A)/15`  is used to calculate the dosage of Hackalot cough medicine to be given to a child.

    • D is the dosage of Hackalot cough medicine in millilitres (mL).
    • A is the age of the child in months.
  1. If George is nine months old, what dosage of Hackalot cough medicine should he be given?  (1 mark)

The correct dosage of Hackalot cough medicine for Sam is 4 mL.

  1. What is the difference in the ages of Sam and George, in months?  (3 marks)

Show Answers Only
  1. `text(1.2 mL)`
  2. `30`
Show Worked Solution
i. `D` `= (2A)/15`
    `= (2 × 9)/15`
    `=1.2\ text(mL)`

 

`:.\ text(George should be given a dosage of 1.2 mL)`

 

ii.   `text(Find)\ A\ text(when)\ D = text(4 mL)`

`4` `= (2 × A)/15`
 `2A` `= 60`
`A`  `= 30` 

 

`:.\ text(Sam is 30 months old and is 21 months)`

`text(older than George.)`

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`

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`

Data, 2UG 2004 HSC 24a

The following graphs have been constructed from data taken from the Bureau of Meteorology website. The information relates to a town in New South Wales.

The graphs show the mean 3 pm wind speed (in kilometres per hour) for each month of the year and the mean number of days of rain for each month (raindays).

2004 24a

  1. What is the mean 3 pm wind speed for September?  (1 mark)
  2. Which month has the lowest mean 3 pm wind speed?  (1 mark)
  3. In which three-month period does the town have the highest number of raindays?  (1 mark)
  4. Briefly describe the pattern relating wind speed with the number of raindays for this town. Refer to specific months.  (2 marks)

 

Show Answers Only
  1. `text(15 km/h)`
  2. `text(February)`
  3. `text(Jan – Mar)`
  4. `text(The mean number of rain days tends to be higher)`
    `text(when the wind speed is lower and vice versa.)`
    `text(For example, the highest number of mean rain)`
    `text(days is in Feb, which is also the month of the lowest)`
    `text(mean wind speed.)`

 

 

Show Worked Solution

(i)    `text(15 km/h)`

(ii)   `text(February)`

(iii)  `text(Jan – Mar)`

(iv)  `text(The mean number of rain days tends to be higher)`

`text(when the wind speed is lower and vice versa.)`

`text(For example, the highest number of mean rain)`

`text(days is in Feb, which is also the month of the lowest)`

`text(mean wind speed.)`

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.)}`

Probability, STD2 S2 2005 HSC 16 MC

On a television game show, viewers voted for their favourite contestant. The results were recorded in the two-way table.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textbf{Male viewers} & \textbf{Female viewers} \\
\hline
\rule{0pt}{2.5ex}\textbf{Contestant 1}\rule[-1ex]{0pt}{0pt} & 1372 & 3915\\
\hline
\rule{0pt}{2.5ex}\textbf{Contestant 2}\rule[-1ex]{0pt}{0pt} & 2054 & 3269\\
\hline
\end{array}

One male viewer was selected at random from all of the male viewers.

What is the probability that he voted for Contestant 1?

  1. `1372/(10\ 610)`
  2. `1372/5287`
  3. `1372/3426`
  4. `1372/2054`
Show Answers Only

`C`

Show Worked Solution

`text(Total male viewers)\ = 1372 + 2054= 3426`

  
`P\ text{(Male viewer chosen voted for C1)}`

`= text(Males who voted for C1)/text(Total male viewers)`

`= 1372/3426`
 

`=>  C`

Financial Maths, STD2 F4 2005 HSC 15 MC

A car bought for  $50 000  is depreciated using the declining balance method.

Which graph best represents the salvage value of the car over time?

 

2UG-2005-15abMC

2UG-2005-15cdMC

Show Answers Only

`D`

Show Worked Solution

`text(Declining Balance Method means that the salvage value)`

`text(of the car drops the most value in the 1st year and then)`

`text(drops less value each following year.)`

`=>  D`

Algebra, STD2 A1 2005 HSC 14 MC

Using the formula  `d = 5t^3 - 2`, Marcia tried to find the value of  `t`  when `d = 137`.

Here is her solution. She has made one mistake.
 

2UG-2005-14MC

Which line does NOT follow correctly from the previous line?

  1. `text(Line)\ A`
  2. `text(Line)\ B`
  3. `text(Line)\ C`
  4. `text(Line)\ D`
Show Answers Only

`B`

Show Worked Solution
`d` `= 5t^3 – 2`  
 `137` `= 5t^3 – 2 \ \ \ ` ` …text( Line A)`
 `139` `= 5t^3` `…text( Line B)`

 

`:.\ text(Line)\ B\ text(doesn’t follow on correctly.)`

`=>  B`

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`

Financial Maths, STD2 F4 2004 HSC 25a

Tai uses the declining balance method of depreciation to calculate tax deductions for her business. Tai’s computer is valued at $6500 at the start of the 2003 financial year. The rate of depreciation is 40% per annum.

  1. Calculate the value of her tax deduction for the 2003 financial year.  (1 mark)

  2. What is the value of her computer at the start of the 2006 financial year?  (2 marks)

Show Answers Only
  1. `$2600`
  2. `$1404`
Show Worked Solution
i.  `text(Tax deduction)` `= 40 text(%) xx $6500`
  `= $2600`

 

ii. `text(Using)\  S = V_0(1 – r)^n,`

`text(Value at the start of 2006 FY)`

`= 6500(1 – 0.4)^3`

`= $1404`

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)`