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Statistics, 2ADV S3 2023 HSC 23

A random variable is normally distributed with a mean of 0 and a standard deviation of 1 . The table gives the probability that this random variable lies below `z` for some positive values of `z`.

The probability values given in the table are represented by the shaded area in the following diagram.
 

The weights of adult male koalas form a normal distribution with mean `mu` = 10.40 kg, and standard deviation `sigma` = 1.15 kg.

In a group of 400 adult male koalas, how many would be expected to weigh more than 11.93 kg?  (4 marks)

Show Answers Only

`37\ text{koalas*}`

`text{*36 or 36.72 koalas would also receive full marks}`

Show Worked Solution
`ztext{-score (11.93)}` `=(x-mu)/sigma`  
  `=(11.93-10.4)/1.15`  
  `=1.330`  

 
`Ptext{(Koala weighs > 11.93 kg)}\ = P(z>1.330)`

`text{Using the table:}`

`P(z>1.33)` `=1-0.9082`  
  `=0.0918`  

 

`:.\ text{Expected koalas > 11.93 kg}` `=0.0918 xx 400`  
  `=36.72`  
  `=37\ text{koalas*}`  

 
`text{*36 or 36.72 koalas would also receive full marks}`

Functions, 2ADV F2 2023 HSC 19

  1. Sketch the graphs of the functions  `f(x)=x-1`  and  `g(x)=(1-x)(3+x)`  showing the `x`-intercepts.  (2 marks)
        

  1. Hence, or otherwise, solve the inequality  `x-1<(1-x)(3+x)`.  (2 marks)

Show Answers Only

a.   
         

b.    `-4<x<1`

Show Worked Solution

a.   `g(x)\ text{cuts}\ xtext{-axis at 1 and}\ -3.`

`g(x)_max=g(-1)=4`

`text{Find intersection of graphs:}`

`(1-x)(3+x)` `=x-1`  
`3+x-3x-x^2` `x-1`  
`x^2+3x-4` `=0`  
`(x+4)(x-1)` `=0`  

 
`text{Intersections at:}\ (1,0), (-4,-5)`
 

 

b.    `text{From the graph:}`

`x-1<(1-x)(3+x)\ \ text{when}\ \ -4<x<1`

`text{Test}\ \ x=0:`

`0-1<(1-0)(3+0)\ \ =>\ \ -1<3\ \ text{(correct)}`

Calculus, 2ADV C3 2023 HSC 6 MC

The following table gives the signs of the first and second derivatives of a function  \(y=f(x)\)  for different values of \(x\).
 

\(x\) \(-2\ \ \) \(0\) \(2\)
\( f^{′}(x) \) \(+\) \(0\) \(+\)
\(f^{″}(x)\) \(-\) \(0\) \(+\)

 
Which of the following is a possible sketch of  \(y=f(x)\)?
 

Show Answers Only

\(C\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{Gradient is positive at}\ x=-2\ \text{and}\ 2\ \ (\text{Eliminate}\ B\ \text{and}\ D)\)

\(\text{SP and POI at}\ x=0\ \ (\text{Eliminate}\ A)\)

\(\Rightarrow C\)

Functions, 2ADV F1 2023 HSC 4 MC

The graph of a polynomial is shown.
 


  

Which row of the table is correct for this polynomial?
 

Show Answers Only

`B`

Show Worked Solution

`text{By elimination:}`

`text{Double root where graph SP touches}\ xtext{-axis}`

`:.\ C and D\ text{incorrect}`

`(x-c)^2\ text{occurs where}\ x>0`

`:. c>0\ (A\ text{is incorrect})`

`=>B`

Financial Maths, 2ADV M1 2023 HSC 21

The fourth term of a geometric sequence is 48 .

The eighth term of the same sequence is `3/16`.

Find the possible value(s) of the common ratio and the corresponding first term(s).  (3 marks)

Show Answers Only

`a=3072,\ r=1/4, or`

`a=-3072,\ r=-1/4`

Show Worked Solution

`T_4=ar^3=48\ …\ (1)`

`T_8=ar^7=3/16\ …\ (2)`

`(ar^7)/(ar^3)` `=(3/16)/48`  
`r^4` `=1/256`  
`r` `=+-1/4`  

 
`text{If}\ \ r=1/4`

`a(1/4)^3` `=48`  
`a/64` `=48`  
`a` `=3072`  

 
`text{If}\ \ r=-1/4,\ \ a=-3072`

Trigonometry, 2ADV T2 2023 HSC 20

Find all the values of `theta`, where  `0^@ <=theta <= 360^@`, such that

`sin(theta-60^@)=-sqrt3/2`  (3 marks)

Show Answers Only

`theta=0^@, 300^@ and 360^@`

Show Worked Solution

`sin60^@=sqrt3/2\ \ =>\ \ text{Base angle}\ =60^@`

`=>\ text{sin is negative in 3rd and 4th quadrants}`

`sin(theta-60^@)` `=180+60, 360-60`  
  `=240^@, 300^@`  

 
`theta-60^@=240^@\ \ =>\ \ theta=300^@`

`theta-60^@=300^@\ \ =>\ \ theta=360^@`

 
`text{Consider}\ theta = 0^@`

`sin(0-60^@)=sin(-60^@)=-sqrt3/2`

`:.theta=0^@, 300^@ and 360^@`

Statistics, 2ADV S2 2023 HSC 18

A university uses gas to heat its buildings. Over a period of 10 weekdays during winter, the gas used each day was measured in megawatts (MW) and the average outside temperature each day was recorded in degrees Celsius (°C).

Using `x` as the average daily outside temperature and `y` as the total daily gas usage, the equation of the least-squares regression line was found.

The equation of the regression line predicts that when the temperature is 0°C, the daily gas usage is 236 MW.

The ten temperatures measured were: 0°, 0°, 0°, 2°, 5°, 7°, 8°, 9°, 9°, 10°,

The total gas usage for the ten weekdays was 1840 MW.

In any bivariate dataset, the least-squares regression line passes through the point `(bar x,bar y)`, where `bar x` is the sample mean of the `x`-values and `bary` is the sample mean of the `y`-values.

  1. Using the information provided, plot the point `(bar x,bar y)` and the `y`-intercept of the least-squares regression line on the grid.  (3 marks)
     

 

  1. What is the equation of the regression line?  (2 marks)

  2. In the context of the dataset, identify ONE problem with using the regression line to predict gas usage when the average outside temperature is 23°C.  (1 mark)

Show Answers Only

a.    
       

b.    `y=-10.4x+236`

c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Show Worked Solution

a.    `barx=(0+0+0+2+5+7+8+9+9+10)/10=5^@text{C}`

`bary=1840/10=184`

`text{Regression line passes through:}\ (0,236) and (5,184)`
 

 

b.    `m=(y_2-y_1)/(x_2-x_1)=(184-236)/(5-0)=-10.4`

`text{Equation of line}\ m=-10.4\ text{passing through}\ (0,236):`

`(y-y_1)` `=m(x-x_1)`  
`y-236` `=-10.4(x-0)`  
`y` `=-10.4x+236`  

 
c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Trigonometry, 2ADV T1 2023 HSC 16

The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.

The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.

 

What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place.  (4 marks)

Show Answers Only

`23.8\ text{m}`

Show Worked Solution
`text{Arc}\ PQ` `=110/360 xx pi xx 2 xx 2.1`  
  `=4.03171… \ text{m}`  

 
`text{Consider}\ ΔOPQ:`
 
 

`sin 55^@` `=x/2.1`  
`x` `=2.1 xx sin 55^@`  
  `=1.7202…`  

 
`PQ=2x=3.440\ text{m}`

`:.\ text{Perimeter}` `=8+(2xx3.6)+4.031+(8-3.440)`  
  `=23.79…\ text{m}`  
  `=23.8\ text{m  (to 1 d.p.)}`  

Calculus, 2ADV C4 2023 HSC 13

Let `P(t)` be a function such that `(dP)/(dt)=3000 e^{2t}`.

When `t=0, P=4000`.

Find an expression for `P(t)`.  (2 marks)

Show Answers Only

`P(t)=1500e^{2t}+2500`

Show Worked Solution
`P(t)` `=int (dP)/(dt)\ dt`  
  `=int 3000e^{2t}\ dt`  
  `=1500e^{2t}+c`  

 
`text{When}\ t=0, P=4000`

`4000` `=1500e^0+c`  
`c` `=2500`  

 
`:.P(t)=1500e^{2t}+2500`

Probability, 2ADV S1 2023 HSC 12

The table shows the probability distribution of a discrete random variable.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 1 & 2 & 3 & 4 \\
\hline
\rule{0pt}{2.5ex} P(X = x) \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ 0.3\ \  & \ \ 0.5\ \  & \ \ 0.1\ \  & \ \ 0.1\ \  \\
\hline
\end{array}

  1. Show that the expected value `E(X)=2`.  (1 mark)

  2. Calculate the standard deviation, correct to one decimal place.  (2 marks)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `0.9`
Show Worked Solution
a.     `E(X)` `=0+1xx0.3+2xx0.5+3xx0.1+4xx0.1`
    `=0.3+1+0.3+0.4`
    `=2`

 

b.     `text{Var}(X)` `=E(X^2)-[E(X)]^2`
    `=(0+1^2xx0.3+2^2xx0.5+3^2xx0.1+4^2xx0.1)-2^2`
    `=(0.3+2+0.9+1.6)-4`
    `=0.8`

 

`:. sigma` `=sqrt(0.8)`  
  `=0.8944…`  
  `=0.9\ \ text{(to 1 d.p.)}`  

L&E, 2ADV E1 2023 HSC 8 MC

What is the solution of the equation `log _a x^3=b`, where a and b are positive constants?

  1. `x=b^(a/3)`
  2. `x=a^(b/3)`
  3. `x=b^a/3`
  4. `x=a^b/3`
Show Answers Only

`B`

Show Worked Solution
`log_a x^3` `=b`  
`3log_a x` `=b`  
`log_a x` `=b/3`  
`:.x` `=a^(b/3)`  

 
`=>B`

Financial Maths, STD2 F1 2023 HSC 37

The table shows personal income tax rates for different taxable incomes for a particular country.

Taxable income Tax payable
$0 − $11 000 Nil
$11 001 − $42 400 20 cents for each $1 over $11 000
$42 401 − $78 800 $6280 plus 33 cents for each $1 over $42 400
$78 801 − $108 400 $18 292 plus `X` cents for each $1 over $78 800
$108 401 and over $31 316 plus 48 cents for each $1 over $108 400

 
A person with a taxable income of $90 000 pays 25.8% of that income in tax (excluding any levies).

What is the value of `X` in the table?  (3 marks)

Show Answers Only

`44\ text{cents}`

Show Worked Solution

`text{Tax paid}\ =90\ 000 xx 0.258=$23\ 220`

`text{Equating with tax payable formula in the table:}`

`23\ 220` `=18\ 292+X(90\ 000-78\ 800)`  
`X(11\ 200)` `=23\ 220-18\ 292`  
`X` `=4928/(11\ 200)`  
  `=0.44\ text{dollars}`  
  `=44\ text{cents}`  

Algebra, STD2 A1 2023 HSC 36

The following formula can be used to calculate an estimate for blood alcohol content (`BAC`) for males.
 

`BAC_text{male}=(10N-7.5H)/(6.8M)`

`N` is the number of standard drinks consumed

`M` is the person's weight in kilograms

`H` is the number of hours of drinking
 

Cameron weighs 75 kg. His `BAC` was zero when he began drinking alcohol. At 9:00 pm, after consuming 3 standard drinks, his `BAC` was 0.02.

Using the formula, estimate at what time Cameron began drinking alcohol, to the nearest minute.  (4 marks)

Show Answers Only

`text{6:22 pm}`

Show Worked Solution
`BAC` `=(10N-7.5H)/(6.8M)`  
`0.02` `=(10xx3-7.5xxH)/(6.8 xx 75)`  
`0.02xx510` `=30-7.5H`  
`7.5H` `=30-10.2`  
`H` `=19.8/7.5`  
  `=2.64\ text{hours}`  
  `=2\ text{hours}\ 38\ text{minutes (nearest minute)}`  

 
`text{Time Cameron began drinking}`

`=\ text{9 pm less 2 h 38 m}`

`=\ text{6:22 pm}`

Mean mark 56%.

Measurement, STD2 M6 2023 HSC 35

The diagram shows triangle `ABC`.
 

Calculate the area of the triangle, to the nearest square metre.  (3 marks)

Show Answers Only

`147\ text{m}^2`

Show Worked Solution

`text{Using the sine rule:}`

`(CB)/sin60^@` `=12/sin25^@`  
`CB` `=sin60^@ xx 12/sin25^@`  
  `=24.590…`  

 
`angleACB=180-(60+25)=95^@\ \ text{(180° in Δ)}`
 

`text{Using the sine rule (Area):}`

`A` `=1/2 xx AC xx CB xx sin angleACB`  
  `=1/2 xx 12 xx 24.59 xx sin95^@`  
  `=146.98…`  
  `=147\ text{m}^2`  

Networks, STD2 N3 2023 HSC 31

A function centre employs staff so that all necessary tasks can be completed between the end of one function and the beginning of the next function.

The network diagram shows the time taken in hours for the tasks that need to be completed.
 


 

  1. Find the TWO critical paths.  (2 marks)

  2. The function centre wants to decrease the length of each critical path by 3 hours. They can do this by hiring more staff to do ONE of the tasks so it takes less time to complete.
  3. For which task should the centre hire more staff, and how long should that task take to ensure all tasks can be completed in 14 hours?  (2 marks) 

Show Answers Only

a.    `HIGC and HIK`

b.    `text{More staff should be hired for task}\ I.`

`text{By decreasing task}\ I\ text{by 3 hours (so it takes 4 hours), the}`

`text{critical path of the network reduces to 14 hours.}`

Show Worked Solution

a.    `text{Scanning both ways:}`
 

`text{Critical paths:}\ HIGC and HIK`

♦ Mean mark (a) 46%.

 
b.   `text{More staff should be hired for task}\ I.`

`text{By decreasing task}\ I\ text{by 3 hours (so it takes 4 hours), the}`

`text{critical path of the network reduces to 14 hours.}`

Mean mark (b) 52%.

Financial Maths, STD2 F4 2023 HSC 28

A plumber leases equipment which is valued at $60 000.

The salvage value of the equipment at any time can be calculated using either of the two methods of depreciation shown in the table.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Method of depreciation} \rule[-1ex]{0pt}{0pt} & \textit{Rate of depreciation} \\
\hline
\rule{0pt}{2.5ex} \text{Straight-line method} \rule[-1ex]{0pt}{0pt} & \text{\$3500 per annum} \\
\hline
\rule{0pt}{2.5ex} \text{Declining balance method} \rule[-1ex]{0pt}{0pt} & \text{12% per annum} \\
\hline
\end{array}

Under which method of depreciation would the salvage value of the equipment be lower at the end of 3 years? Justify your answer with appropriate mathematical calculations.   (3 marks)

Show Answers Only

`text{Straight-line method:}`

`S` `=V_0-Dn`  
  `=60\ 000-3500×3`  
  `=$49\ 500`  

 
`text{Declining-balance method:}`

`S` `=V_0(1-r)^n`  
  `=60\ 000(1-0.12)^3`  
  `=60\ 000(0.88)^3`  
  `=$40\ 888.32`  

 
`text{Salvage value is lower for the declining-balance method.}`

Show Worked Solution

`text{Straight-line method:}`

`S` `=V_0-Dn`  
  `=60\ 000-3500×3`  
  `=$49\ 500`  

 
`text{Declining-balance method:}`

`S` `=V_0(1-r)^n`  
  `=60\ 000(1-0.12)^3`  
  `=60\ 000(0.88)^3`  
  `=$40\ 888.32`  

 
`text{Salvage value is lower for the declining-balance method.}`

Measurement, STD2 M6 2023 HSC 27

The diagram shows the location of three places `X`, `Y` and `C`.

`Y` is on a bearing of 120° and 15 km from `X`.

`C` is 40 km from `X` and lies due west of `Y`.

`P` lies on the line joining `C` and `Y` and is due south of `X`.
  

  1. Find the distance from `X` to `P`.  (2 marks)

  2. What is the bearing of `C` from `X`, to the nearest degree?  (2 marks)

Show Answers Only
  1. `7.5\ text{km}`
  2. `259^@`
Show Worked Solution

a.    `text{In}\ ΔXPY:`

`anglePXY=180-120=60^@`

`cos 60^@` `=(XP)/15`  
`XP` `=15 xx cos 60^@`  
  `=7.5\ text{km}`  

 
b.    `text{In}\ ΔXPC:`

`text{Let}\ \ theta = angleCXP`

`cos theta` `=7.5/40`  
`theta` `=cos^{-1}(7.5/40)`  
  `=79.193…`  
  `=79^@\ \ text{(nearest degree)}`  

 

`text{Bearing}\ C\ text{from}\ X` `=180+79`  
  `=259^@`  

♦ Mean mark (b) 39%.

Measurement, STD2 M7 2023 HSC 26

Kim is building a path around a garden at the back of a house, as shown. The path is 0.5 m wide.

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

  2. Kim is mixing some concrete for the path. The concrete mix is made up of crushed rock, sand and cement in the ratio of 4 : 2 : 1 by weight.
  3. Kim needs 2.1 tonnes of concrete in the correct ratio.
  4. Calculate how many 15 kg bags of cement Kim needs to buy.   (3 marks)

Show Answers Only
  1. `6.5\ text{m}^2`
  2. `20\ text{bags}`
Show Worked Solution

a.    `text{Area outer rectangle}\ = 3xx8=24\ text{m}^2`

`text{Area garden}\ = 2.5xx7=17.5\ text{m}^2`

`A_text{path}` `=24-17.5`  
  `=6.5\ text{m}^2`  

 
b.    `text{7 parts = 2.1 tonnes}`

`text{1 part}\ = 2.1/7=0.3\ text{tonnes}\ =300\ text{kgs}`

`text{Rock}:text{Sand}:text{Cement} = 4:2:1 = 1200:600:300`

`=>\ text{300 kgs of cement are required}`

`:.\ text{Bags of cement}` `=300/15`  
  `=20\ text{bags}`  

Financial Maths, STD2 F5 2023 HSC 25

A table of future value interest factors for an annuity of $1 is shown.
 

  1. Micky wants to save $450 000 over the next 10 years.
  2. If the interest rate is 6% per annum compounding annually, how much should Micky contribute each year? Give your answer to the nearest dollar.  (2 marks)

  3. Instead, Micky decides to contribute  $8535 every three months for 10 years to an annuity paying 6% per annum, compounding quarterly.
  4. How much will Micky have at the end of 10 years?  (3 marks)

Show Answers Only

  1. `$34\ 140`
  2. `$463\ 177.38`

Show Worked Solution

a.    `text{Applicable interest rate}\ =6%`

`text{Compounding periods}\ =10xx1=10`

`=>\ text{Factor}\ = 13.181`

`:.\ text{Contribution (annual)}` `=(450\ 000)/13.181`  
  `=$34\ 140`  

 
b.    `text{Applicable interest rate}\ =(6%)/4=1.5%\ text{per quarter}`

`text{Compounding periods}\ =10xx4=40`

`=>\ text{Factor}\ = 54.268`

`text{Total (after 10 years)}` `=8535 xx 54.268`  
  `=$463\ 177.38`  
Mean mark (b) 53%.
 

Measurement, STD2 M1 2023 HSC 24

The diagram shows the cross-section of a wall across a creek. 
 


 
  1. Use two applications of the trapezoidal rule to estimate the area of the cross-section of the wall.  (2 marks)

  2. The wall has a uniform thickness of 0.80 m. The weight of 1 m³ of concrete is 3.52 tonnes.  
  3. How many tonnes of concrete are in the wall? Give the answer to two significant figures.  (3 marks)

Show Answers Only
  1. `18\ text{m}^2`
  2. `text{51 tonnes}`
Show Worked Solution

a.    `h=8.0/2=4`

`A` `~~h/2[1.9+1.7+2(2.7)]`  
  `~~4/2(9)`  
  `~~18\ text{m}^2`  

 
b.   `V_text{wall}=18 xx 0.8=14.4\ text{m}^3`

`text{Mass of concrete}` `=14.4 xx 3.52`  
  `=50.688`  
  `=51\ text{tonnes (2 sig.fig.)}`  
♦ Mean mark (b) 46%.
 

Probability, STD2 S2 2023 HSC 23

One hundred tickets are sold in a raffle which offers two prizes. Hazel buys five of the tickets.

A ticket is drawn at random for the first prize. A second ticket is drawn from the remaining tickets for the other prize.

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

Show Answers Only

`1/495`

Show Worked Solution

`P(W_1, W_2) = 5/100 xx 4/99=1/495`

Algebra, STD2 A4 2023 HSC 22

The braking distance of a car, in metres, is directly proportional to the square of its speed in km/h, and can be represented by the equation

`text{braking distance}\ = k xx text{(speed)}^2`

where `k` is the constant of variation.

The braking distance for a car travelling at 50 km/h is 20 m.

  1. Find the value of `k`.  (2 marks)

  2. What is the braking distance when the speed of the car is 90 km/h?  (1 mark)

Show Answers Only

a.    `k=0.008`

b.    `64.8\ text{m}`

Show Worked Solution

a.  `text{braking distance}\ = k xx text{(speed)}^2`

`20` `=k xx 50^2`  
`k` `=20/50^2`  
  `=0.008`  

 
b.    `text{Find braking distance}\ (d)\ text{when speed = 90 km/h:}`

`d` `=0.008 xx 90^2`  
  `=64.8\ text{m}`  

Algebra, STD2 A4 2023 HSC 21

Electricity provider \(A\) charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40.

  1. Complete the table showing Provider \(A\)'s monthly charges for different levels of electricity usage.  (1 mark)
     
\(\textit{Electricity used in a
 month (kWh)}\)		 \(\ \ 0\ \ \) \(400\) \(1000\)
\(\textit{Monthly charge (\$)}\) \(40\)   \(290\)
 
Provider \(B\) charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider \(B\)'s charges vary with the amount of electricity used in a month.
 

 
  1. On the grid above, graph Provider \(A\)'s charges from the table in part (a).   (1 mark)
  2. Use the two graphs to determine the number of kilowatt hours per month for which Provider \(A\) and Provider \(B\) charge the same amount.   (1 mark)

  3. A customer uses an average of 800 kWh per month.
  4. Which provider, \(A\) or \(B\), would be the cheaper option and by how much?   (2 marks)

Show Answers Only

a. 

\(\textit{Electricity used in a
 month (kWh)}\)		 \(\ \ 0\ \ \) \(400\) \(1000\)
\(\textit{Monthly charge (\$)}\) \(40\) \(140\) \(290\)

b.    
         

c.    \(\text{400 kWh}\)

d.    \(\text{Provider}\ A\ \text{is cheaper by \$40.}\)

Show Worked Solution

a. 

\(\textit{Electricity used in a
 month (kWh)}\)		 \(\ \ 0\ \ \) \(400\) \(1000\)
\(\textit{Monthly charge (\$)}\) \(40\) \(140\) \(290\)

b. 
          
 

c.    \(A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}\)

\(\therefore\ \text{Same charge at 400 kWh}\)
 

d.    \(\text{Cost at 800 kWh:}\)

\(\text{Provider}\ A: \ 40 + 0.25 \times 800 = $240\)

\(\text{Provider}\ B: \ 0.35 \times 800 = $280\)

\(\therefore \text{Provider}\ A\ \text{is cheaper by \$40.}\)

Measurement, STD2 M7 2023 HSC 11 MC

A bag contains 150 jelly beans. Some of them are red and the rest are blue. The ratio of red to blue jelly beans is `2:3`.

Sophie eats 10 of each colour.

What is the new ratio of red to blue jelly beans?

  1. `2:3`
  2. `4:9`
  3. `5:8`
  4. `11:17`
Show Answers Only

`C`

Show Worked Solution

`text{Original ratio}\ = 2:3`

`text{5 parts = 150}\ \ =>\ \ text{1 part}\ = 150/5=30`

`text{Original ratio (by number)}\ = 2 xx 30:3 xx 30 = 60:90`

`text{After eating 10 of each colour:}`

`text{Ratio}\ = 50:80 = 5:8`

`=>C`

Financial Maths, STD2 F4 2023 HSC 10 MC

An amount of $25 000 is invested for six years. Interest is earned at a rate of 8% per annum, compounding quarterly.

Which expression gives the value of the investment after 6 years, in dollars?

  1. `25\ 000 xx 1.02^{24}`
  2. `25\ 000 xx 1.02^{6}`
  3. `25\ 000 xx 1.08^{24}`
  4. `25\ 000 xx 1.08^{6}`
Show Answers Only

`A`

Show Worked Solution

`text{Interest rate}\ = 8/2=2%\ text{per quarter}`

`text{Compounding periods}\ = 6xx4=24`

`:.FV=25\ 000 xx 1.02^{24}`

`=>A`

Measurement, STD2 M2 2023 HSC 7 MC

City `A` is at latitude 34°S and longitude 151°E. City `B` is 72° north of City `A` and 25° west of City `A`.

What are the latitude and longitude of City `B`?

  1. 16°N, 126°E
  2. 16°N, 176°E
  3. 38°N, 126°E
  4. 38°N, 176°E
Show Answers Only

`C`

Show Worked Solution

`text{Latitude of city}\ A: \ -34+72=38°text{N}`

`text{Longitude of city}\ A: \ 151-25=126°text{E}`

`=>C`

Measurement, STD2 M7 2023 HSC 5 MC

Four petrol pumps are shown, each with the amount of petrol purchased and its cost.

Which one represents the best value?
 

Show Answers Only

`B`

Show Worked Solution

`text{Calculate cost per litre of each option}`

`text{Option}\ A:\ $46.70 -: 24= $1.946\ text{per litre}`

`text{Option}\ B:\ $48.50 -: 25= $1.940\ text{per litre}`

`text{Option}\ C:\ $52.30 -: 26= $2.01\ text{per litre}`

`text{Option}\ D:\ $54.80 -: 27= $2.03\ text{per litre}`

`=>B`

Statistics, STD2 S4 2023 HSC 3 MC

The number of bees leaving a hive was observed and recorded over 14 days at different times of the day.
 

Which Pearson's correlation coefficient best describes the observations?

  1. `-0.8`
  2. `-0.2`
  3. `0.2`
  4. `0.8`
Show Answers Only

`D`

Show Worked Solution

`text{Correlation is positive and strong.}`

`text{Best option:}\ r=0.8`

`=>D`

NOTE: Inputting all data points into a calculator is unnecessary and time consuming here.

BIOLOGY, M2 2013 HSC 35d

Analyse how the use of isotopes has contributed to tracing biochemical pathways in plants.  (6 marks)

Show Answers Only
  • Scientists are able to synthesise chemicals with implanted radio-isotopes which organisms use identically to the natural chemicals.
  • Autoradiography is the use of x-ray or photographic film to detect these radioactive materials. They produce permanent record of positions and relative intensities which scientists can analyse.
  • \(\ce{^3H}\) can be used to track the transport of \(\ce{H}\) across the thylakoid membrane.
  • \(\ce{^{32}P}\) can be used to show that water moves in xylem vessels. When roots were surrounded by water containing \(\ce{^{32}P}\), it shown it was taken up by the roots, then through the xylem vessels into other plant organs.
Show Worked Solution
  • Scientists are able to synthesise chemicals with implanted radio-isotopes which organisms use identically to the natural chemicals.
  • Autoradiography is the use of x-ray or photographic film to detect these radioactive materials. They produce permanent record of positions and relative intensities which scientists can analyse.
  • \(\ce{^3H}\) can be used to track the transport of \(\ce{H}\) across the thylakoid membrane.
  • \(\ce{^{32}P}\) can be used to show that water moves in xylem vessels. When roots were surrounded by water containing \(\ce{^{32}P}\), it shown it was taken up by the roots, then through the xylem vessels into other plant organs.

BIOLOGY, M1 2013 HSC 19 MC

An enzyme was extracted from a mammal. The graph shows the rate at which bubbles are produced in a reaction at 36°C using the extracted enzyme.
 

Which of the following graphs would show the results if the enzyme reaction were carried out at 18°C ?
 

Show Answers Only

\(D\)

Show Worked Solution
  • The reduction in temperature will reduce the reaction rate. Option D is the only graph which accurately represents this change in relevance to the original.

\(\Rightarrow D\)

BIOLOGY, M2 2013 HSC 10 MC

What does the structure of arteries allow them to do?

  1. Transport oxygen rich blood
  2. Withstand high blood pressure
  3. Release carbon dioxide to the lungs
  4. Remove nitrogenous waste via the kidneys
Show Answers Only

\(B\)

Show Worked Solution
  • Arteries have thick muscular walls which allow them to withstand the higher blood pressure when it is pumped from the heart.

\(\Rightarrow B\)

BIOLOGY, M2 2014 HSC 35a

  1. Name a scientist from the 17th or 18th century who contributed to the development of ideas on the structure and function of plants.   (1 mark)

  2. Outline how the findings from a 17th or 18th century experiment informed scientists about plant structure and/or function.   (3 marks)

Show Answers Only

i.    Scientists could include one of:

  • Van Helmont, Hales, Priestley, Ingen-Housz, Senebier, Saussure 

ii.   Priestly’s Experiment

  • In 1771, Joseph Priestly ignited a candle in a jar with some mint, in which he ignited the candle and then it went out. After a few trials, the candle wouldn’t ignite as all the oxygen in the air had been used (note that oxygen was not yet discovered). However after 27 days, by igniting the candle with a mirror and sunlight as to not open the jar, it reignited.
  • Similarly, by putting mice into a jar with and without a sprig of mint, the jar with the mint allowed the mouse to survive much longer.
  • These observations lead to questions about plant structure, and how they were somehow able to ‘restore air’ that is used by animals breathing and combustion reactions. 
Show Worked Solution

i.    Scientists could include one of:

  • Van Helmont, Hales, Priestley, Ingen-Housz, Senebier, Saussure 

ii.   Priestly’s Experiment

  • In 1771, Joseph Priestly ignited a candle in a jar with some mint, in which he ignited the candle and then it went out. After a few trials, the candle wouldn’t ignite as all the oxygen in the air had been used (note that oxygen was not yet discovered). However after 27 days, by igniting the candle with a mirror and sunlight as to not open the jar, it reignited.
  • Similarly, by putting mice into a jar with and without a sprig of mint, the jar with the mint allowed the mouse to survive much longer.
  • These observations lead to questions about plant structure, and how they were somehow able to ‘restore air’ that is used by animals breathing and combustion reactions. 

BIOLOGY, M3 2014 HSC 7 MC

Which of the following is essential in any model of natural selection?

  1. High reproductive rates
  2. Random selection of prey
  3. A population of predators
  4. Differences in the population
Show Answers Only

\(D\)

Show Worked Solution
  • For natural selection to take effect there must be differences within the population in which a factor will favour a certain characteristic.

\(\Rightarrow D\)

BIOLOGY, M3 2015 HSC 30

The graph shows the history of the relative numbers of three varieties of bird \((X, Y\) and \(Z\) ) within a bird species, on a remote island in the Pacific Ocean.

The bird species arrived on the island in a migration event. Before migration, the bird species was not present on the island.

The graph record of bird numbers on the island is divided into two sections (1 and 2). Over the time data were recorded, the environment of the island did not change.
 

  1. What bird varieties originally migrated to the island?   (1 mark)

  2. Using the Darwin/Wallace theory of evolution, and making reference(s) to the data in the graph, explain the changes to the population of each variety of bird in
  3.   i. section (1) of the graph.  (2 marks)

  4.  ii. section (2) of the graph.  (4 marks)

Show Answers Only

a.    Varieties \(Z\) and \(Y\).

b.i.  In section (1):

  • Variety \(Z\) is naturally selected as its numbers grow over time until they plateau when it reaches the limits of resources the environment can provide.
  • Variety \(Y\) is not naturally selected as it is not suited to the environment and becomes extinct as its numbers fall over time to zero.

b.ii. In section (2):

  • Variety \(X\) develops, likely from mutation.
  • Variety \(Z\) faces competition from variety \(X\) as the numbers of variety \(Z\) quickly fall as the numbers of variety \(X\) quickly grow. This  shows that variety \(X\) possesses characteristics which vary from \(Z\) and make it more suited to the island.
  • Variety \(Z\) population remains at low levels due to the difficulty of competing for resources with variety \(X\), yet they do not go extinct but instead become a minority variety in the species.
  • The numbers of variety \(X\) plateau off at a lower level than variety \(Z\) as it reaches the limits of resources the environment can provide for both variety \(Z\) and \(X\).
Show Worked Solution

a.    Varieties \(Z\) and \(Y\). 

b.i.  In section (1):

  • Variety \(Z\) is naturally selected as its numbers grow over time until they plateau when it reaches the limits of resources the environment can provide.
  • Variety \(Y\) is not naturally selected as it is not suited to the environment and becomes extinct as its numbers fall over time to zero. 
♦ Mean mark (b)(i) 47%.

b.ii. In section (2):

  • Variety \(X\) develops, likely from mutation.
  • Variety \(Z\) faces competition from variety \(X\) as the numbers of variety \(Z\) quickly fall as the numbers of variety \(X\) quickly grow. This  shows that variety \(X\) possesses characteristics which vary from \(Z\) and make it more suited to the island.
  • Variety \(Z\) population remains at low levels due to the difficulty of competing for resources with variety \(X\), yet they do not go extinct but instead become a minority variety in the species.
  • The numbers of variety \(X\) plateau off at a lower level than variety \(Z\) as it reaches the limits of resources the environment can provide for both variety \(Z\) and \(X\).
♦ Mean mark (b)(ii) 47%.

BIOLOGY, M4 2018 HSC 34aii

Using an example, outline a limitation of only using the fossil record to interpret the past.  (2 marks)

Show Answers Only
  • Using just a fossil record to date events may lead to inaccurate results if not compared to another dating method.
  • For example, radioactive dating of fossils from campsites gives the date for colonisation of Australia at 40,000 years ago while thermoluminescence data, a process used to date certain crystalline minerals, puts the date at 50-60,000 years.
Show Worked Solution
  • Using just a fossil record to date events may lead to inaccurate results if not compared to another dating method.
  • For example, radioactive dating of fossils from campsites gives the date for colonisation of Australia at 40,000 years ago while thermoluminescence data, a process used to date certain crystalline minerals, puts the date at 50-60,000 years.

BIOLOGY, M2 2015 HSC 36e

'Science has been used to solve problems in the investigation of photosynthesis, and so has provided information of benefit to society.' 

Justify this statement with reference to the scientific knowledge behind radioactive tracers for the study of photosynthesis.   (7  marks)

Show Answers Only
  • The intricate nature of photosynthesis has posed many barriers when scientists attempt to study it. To overcome this scientists have used radioactive tracers, synthetic chemicals which are taken up by the plant and act as normal organic chemicals except they contain a radioactive component.
  • Radioactive atoms release radiation that can be seen by technologies like X-ray film, geiger counters etc. The pathway of a radioactive substance through the living thing can be followed, as can the biochemical pathways that the molecule/atom is involved in.
  • \(\ce{C^{14}O2}\) and \(\ce{H2O^{18}}\) are both radioactive tracers that can be used to study photosynthesis.
  • If plants are surrounded by \(\ce{C^{14}O2 (g)}\) the radioactivity is soon seen in starch granules in the leaves of the plant. This shows that the starch is formed from the \(\ce{CO2}\) in the air, and is composed of carbon atoms from the air. None of the \(\ce{C^{14}}\) in the \(\ce{C^{14}O2 (g)}\) taken in by the plant is lost.
  • If plants are watered with \(\ce{H2O^{18} (l)}\) the radioactivity is seen in the \(\ce{O2}\) that the plant releases into the air around the plant and not in molecules constructed by photosynthesis contained within the leaf.
  • Therefore in photosynthesis the water is split and the oxygen released into the atmosphere, the \(\ce{H}\) incorporated into the plant within intermediate molecules in a biochemical pathway, and then finally into a starch molecule.
  • This knowledge is of benefit to society because we need to find ways of reducing the carbon in the atmosphere because of excess use of fossil fuel combustion and its resultant climate change. We can understand that land clearing with its removal of photosynthetic species will exacerbate the build up of carbon in the atmosphere because of the loss of photosynthesis it causes.
  • Society is also concerned about the need to generate oxygen such as in the context of massive amounts of fossil fuel combustion also removing oxygen from the atmosphere. Understanding that plants release oxygen in photosynthesis is part of the offsets for fossil fuel use in re-forestation projects as the carbon is locked up in the plant and oxygen is released into the atmosphere.
Show Worked Solution
  • The intricate nature of photosynthesis has posed many barriers when scientists attempt to study it. To overcome this scientists have used radioactive tracers, synthetic chemicals which are taken up by the plant and act as normal organic chemicals except they contain a radioactive component.
  • Radioactive atoms release radiation that can be seen by technologies like X-ray film, geiger counters etc. The pathway of a radioactive substance through the living thing can be followed, as can the biochemical pathways that the molecule/atom is involved in.
  • \(\ce{C^{14}O2}\) and \(\ce{H2O^{18}}\) are both radioactive tracers that can be used to study photosynthesis.
  • If plants are surrounded by \(\ce{C^{14}O2 (g)}\) the radioactivity is soon seen in starch granules in the leaves of the plant. This shows that the starch is formed from the \(\ce{CO2}\) in the air, and is composed of carbon atoms from the air. None of the \(\ce{C^{14}}\) in the \(\ce{C^{14}O2 (g)}\) taken in by the plant is lost.
  • If plants are watered with \(\ce{H2O^{18} (l)}\) the radioactivity is seen in the \(\ce{O2}\) that the plant releases into the air around the plant and not in molecules constructed by photosynthesis contained within the leaf.
  • Therefore in photosynthesis the water is split and the oxygen released into the atmosphere, the \(\ce{H}\) incorporated into the plant within intermediate molecules in a biochemical pathway, and then finally into a starch molecule.
  • This knowledge is of benefit to society because we need to find ways of reducing the carbon in the atmosphere because of excess use of fossil fuel combustion and its resultant climate change. We can understand that land clearing with its removal of photosynthetic species will exacerbate the build up of carbon in the atmosphere because of the loss of photosynthesis it causes.
  • Society is also concerned about the need to generate oxygen such as in the context of massive amounts of fossil fuel combustion also removing oxygen from the atmosphere. Understanding that plants release oxygen in photosynthesis is part of the offsets for fossil fuel use in re-forestation projects as the carbon is locked up in the plant and oxygen is released into the atmosphere.

BIOLOGY, M2 2016 HSC 36e

'Over the past 400 years, the development of our knowledge of the chemical transformations occurring both inside and outside plants has led to our current understanding of photosynthesis.'

Evaluate this statement with reference to the experiments of TWO named scientists.  (7 marks)

Show Answers Only
  • In 1774 Joseph Priestley made observations about changes to the air around a living plant. He observed a candle would burn longer within a bell jar of air if a plant were included under the bell jar.
  • He deduced that the burning of the candle was using up something in the air and thus had a limited burning time in a fixed volume of air. He further deduced the plant was reversing that change and restoring the air to allow the candle to burn longer.
  • Priestley later described a gas (oxygen) used up from air in combustion of the candle, yet released by plants. Adding oxygen to the outside air mixture can be termed a chemical transformation of air done by plants. This led to the understanding that photosynthesis produces oxygen.

Data collected inside the plant

  • The experiments by Calvin in the 1940s involved an illuminated flattened glass vessel, wide but thin, containing microscopic algae in solution. The algae were given a pulse of radioactive carbon dioxide \(\ce{(^{14}CO2)}\), injected into a stream of air into the vessel.
  • Samples of Chlorella were then released at intervals (3, 5, 10 seconds and then 15 second periods) into boiling alcohol to kill the algae and stop the progress of biochemical reactions containing the \(\ce{(^{14}CO2)}\).
  • Compounds that the radioactive carbon had reached at a particular moment were determined by two-dimensional paper chromatography and autoradiography after the chlorella cells were broken up.
  • New chemicals formed were deduced with the same results with chromatography of standard chemicals. For example, phosphoglycerate was identified as the first metabolite in the carbon cycle that was then changed into the glyceraldehyde phosphate.
  • In this way the series of chemical transformations of carbon compounds in the light-independent reactions of photosynthesis could be followed.
  • The work of scientists studying both the external and internal environments was essential in developing our understanding of photosynthesis.
Show Worked Solution
  • In 1774 Joseph Priestley made observations about changes to the air around a living plant. He observed a candle would burn longer within a bell jar of air if a plant were included under the bell jar.
  • He deduced that the burning of the candle was using up something in the air and thus had a limited burning time in a fixed volume of air. He further deduced the plant was reversing that change and restoring the air to allow the candle to burn longer.
  • Priestley later described a gas (oxygen) used up from air in combustion of the candle, yet released by plants. Adding oxygen to the outside air mixture can be termed a chemical transformation of air done by plants. This led to the understanding that photosynthesis produces oxygen.

Data collected inside the plant

  • The experiments by Calvin in the 1940s involved an illuminated flattened glass vessel, wide but thin, containing microscopic algae in solution. The algae were given a pulse of radioactive carbon dioxide \(\ce{(^{14}CO2)}\), injected into a stream of air into the vessel.
  • Samples of Chlorella were then released at intervals (3, 5, 10 seconds and then 15 second periods) into boiling alcohol to kill the algae and stop the progress of biochemical reactions containing the \(\ce{(^{14}CO2)}\).
  • Compounds that the radioactive carbon had reached at a particular moment were determined by two-dimensional paper chromatography and autoradiography after the chlorella cells were broken up.
  • New chemicals formed were deduced with the same results with chromatography of standard chemicals. For example, phosphoglycerate was identified as the first metabolite in the carbon cycle that was then changed into the glyceraldehyde phosphate.
  • In this way the series of chemical transformations of carbon compounds in the light-independent reactions of photosynthesis could be followed.
  • The work of scientists studying both the external and internal environments was essential in developing our understanding of photosynthesis.

BIOLOGY, M2 2018 HSC 9 MC

Sunken stomata can be found in the leaves of some Australian plants. A section of such a leaf is shown.
 

How do sunken stomata assist the plant to conserve water in a dry environment?

  1. They trap moist air, reducing humidity.
  2. They prevent entry of gases into the leaf.
  3. They accumulate moist air, reducing transpiration.
  4. They increase the surface area available for transpiration.
Show Answers Only

\(C\)

Show Worked Solution
  • A sunken stomata can accumulate moist air between it and the cuticle, hence reducing transpiration and water loss.

\(\Rightarrow C\)

BIOLOGY, M1 2015 HSC 5 MC

The diagram shows a model for enzyme action.
 

Which letter indicates the substrate in this diagram?

  1. \(W\)
  2. \(X\)
  3. \(Y\)
  4. \(Z\)
Show Answers Only

\(A\)

Show Worked Solution
  • \(W\) splits into 2 components, \(Y\) and \(Z\), indicating that this is the substrate being split into the products, while the molecule that remains in contact, \(X\) is the enzyme.

\(\Rightarrow A\)

BIOLOGY, M1 2016 HSC 5 MC

The graph shows the relationship between substrate concentration and reaction rate in an enzyme-catalysed reaction.
  

Why does the reaction rate NOT continue to increase after point \(X\)?

  1. The enzyme has been denatured.
  2. The substrate has been denatured.
  3. The enzyme is no longer acting as a catalyst.
  4. All the enzyme active sites are being occupied.
Show Answers Only

\(D\)

Show Worked Solution
  • The reason substrate concentration does not increase but reaction rate is still at it’s peak is due to all the enzyme active sites being occupied.

\(\Rightarrow D\)

BIOLOGY, M1 2017 HSC 36a

  1. Identify TWO reasons for studying photosynthesis.  (2 marks)

  2. What are the functions of the products of photosynthesis?  (2 marks)

Show Answers Only

i.     Studying the process of photosynthesis:

  • Leads to an understanding of how oxygen is produced.
  • Leads to an understanding of how raw materials can be produced for a range of human needs, such as in medicinal and agricultural fields. 

ii.    Functions of photosynthesis products:

  • The products of photosynthesis are oxygen and sugar.
  • They are used in cellular respiration in all cells which contain mitochondria, in both animals and plants.

Show Worked Solution

i.     Studying the process of photosynthesis:

  • Leads to an understanding of how oxygen is produced.
  • Leads to an understanding of how raw materials can be produced for a range of human needs, such as in medicinal and agricultural fields. 

ii.    Functions of photosynthesis products:

  • The products of photosynthesis are oxygen and sugar.
  • They are used in cellular respiration in all cells which contain mitochondria, in both animals and plants.

BIOLOGY, M1 2017 HSC 36e

Analyse the impact of the development of the electron microscope on the understanding of chloroplast structure and function.  (7 marks)

Show Answers Only
  • Using light microscopes, scientists were able to view and identify chloroplasts. However, it wasn’t until the development of the electron microscope with its greater magnification and resolution, that scientists were able to view a chloroplast’s internal structure.
  • Structures such as the grana, stroma and thylakoids could then be identified. The role of each in the process of photosynthesis could then be studied.
  • Thylakoids are flattened, hollow discs which are arranged in stacks called grana. The stacking of the layers into grana increases stability and surface area for the capture of light.
  • The membranes of these thylakoids contain chlorophyll and are the site for the light-dependent reactions of photosynthesis.
  • The space outside the thylakoid is called the stroma, which is an aqueous fluid present within the inner membrane of the chloroplast. It contains DNA, ribosomes, lipid droplets and starch granules. This is where the light independent reactions, the Calvin cycle, takes place.
  • The functions described would not have been linked to the internal structures of the chloroplast without the development of an electron microscope.
Show Worked Solution
  • Using light microscopes, scientists were able to view and identify chloroplasts. However, it wasn’t until the development of the electron microscope with its greater magnification and resolution, that scientists were able to view a chloroplast’s internal structure.
  • Structures such as the grana, stroma and thylakoids could then be identified. The role of each in the process of photosynthesis could then be studied.
  • Thylakoids are flattened, hollow discs which are arranged in stacks called grana. The stacking of the layers into grana increases stability and surface area for the capture of light.
  • The membranes of these thylakoids contain chlorophyll and are the site for the light-dependent reactions of photosynthesis.
  • The space outside the thylakoid is called the stroma, which is an aqueous fluid present within the inner membrane of the chloroplast. It contains DNA, ribosomes, lipid droplets and starch granules. This is where the light independent reactions, the Calvin cycle, takes place.
  • The functions described would not have been linked to the internal structures of the chloroplast without the development of an electron microscope.

BIOLOGY, M1 2017 HSC 16 MC

Which row in the table correctly identifies the features of the named transport mechanism?
 

Show Answers Only

\(B\)

Show Worked Solution

By Elimination

  • Diffusion moves molecules from high to low concentration (Eliminate A).
  • Active transport uses energy to move molecules against the concentration gradient, therefore from low to high (Eliminate C).
  • Passive transport refers to the diffusion of simple molecules which can ‘squeeze between’ the phospholipid molecules of the membrane. Sugars are too large to do this and require carrier or channel proteins. (Eliminate D).

\(\Rightarrow B\)

BIOLOGY, M1 2018 HSC 23

  1. A student plans to investigate the effect of light intensity on transpiration in plants. 
  2. Complete the table, identifying features that this student should include to ensure valid experimental design.  (3 marks)

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Dependent variable}\rule[-1ex]{0pt}{0pt} &  \text{•}  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Control}\rule[-1ex]{0pt}{0pt} &  \text{•} \\
\hline
\rule{0pt}{2.5ex}\textit{Variables to be kept constant}\rule[-1ex]{0pt}{0pt} &   \text{•} \\  &   \text{•} \\
\hline
\end{array}

  1. Explain ONE mechanism for the movement of materials in xylem vessels.  (2 marks)

Show Answers Only

a.   

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Dependent variable}\rule[-1ex]{0pt}{0pt} &  \text{• Amount of water lost}  \\
\hline
\rule{0pt}{2.5ex}\textit{Control}\rule[-1ex]{0pt}{0pt} &  \text{• Plant kept in the dark} \\
\hline
\rule{0pt}{2.5ex}\textit{Variables to be kept constant}\rule[-1ex]{0pt}{0pt} &   \text{• Plant type} \\  &   \text{• Temperature} \\
\hline
\end{array}
 

b.    Water molecules have chemical properties which make them cohesive.

  • Due to this, when water evaporates or transpires, other water molecules are pulled up the xylem vessels.
  • This is one of the core mechanisms that drives the movement of water from roots to leaves.

Show Worked Solution

a.   

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Dependent variable}\rule[-1ex]{0pt}{0pt} &  \text{• Amount of water lost}  \\
\hline
\rule{0pt}{2.5ex}\textit{Control}\rule[-1ex]{0pt}{0pt} &  \text{• Plant kept in the dark} \\
\hline
\rule{0pt}{2.5ex}\textit{Variables to be kept constant}\rule[-1ex]{0pt}{0pt} &   \text{• Plant type} \\  &   \text{• Temperature} \\
\hline
\end{array}

♦ Mean mark (a) 54%.

 b.    Water molecules have chemical properties which make them cohesive.

  • Due to this, when water evaporates or transpires, other water molecules are pulled up the xylem vessels.
  • This is one of the core mechanisms that drives the movement of water from roots to leaves. 

CHEMISTRY, M3 2009 HSC 9 MC

One test used for random breath testing in NSW involved crystals of potassium dichromate reacting with ethanol. In this reaction the orange dichromate ion, \(\ce{Cr2O7}^{2-}\), changes to the green chromium ion, \(\ce{Cr^3+}\).

Which statement is true for this reaction?

  1. Chromium has lost electrons and reached a lower oxidation state.
  2. Chromium has lost electrons and reached a higher oxidation state.
  3. Chromium has gained electrons and reached a lower oxidation state.
  4. Chromium has gained electrons and reached a higher oxidation state.
Show Answers Only

\(C\)

Show Worked Solution
  • Let \(x\) equal the oxidation number of Cr in \(\ce{Cr2O7}^{2-}\).
  •    \(2x+ 7 \times -2 = -2\ \ \Rightarrow \ \ 2x= 12\ \ \Rightarrow \ \ x=6\)
  • The Chromium has been reduced as the oxidation number has decreased from 6 to 3, thus it gains electrons and the oxidation state is lower.

\(\Rightarrow C\)

CHEMISTRY, M3 2012 HSC 13-14 MC

Use the information provided to answer Questions 13 and 14.

\begin{array} {|l|}
\hline \text{This equation represents a common redox reaction.} \\ \ \ \ \ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6Fe^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 6Fe^{3+}(aq) + 7H2O(l)} \\
\hline \end{array}

Question 13

What is the oxidising agent in the reaction?

  1. \(\ce{H^+}\)
  2. \(\ce{Cr^3+}\)
  3. \(\ce{Fe^2+}\)
  4. \(\ce{Cr2O7^2-}\)

 
Question 14

What is the value of \(\ce{E}_{\text {cell }}^{\ominus}\) for the reaction?

  1. 0.59 V
  2. 0.92 V
  3. 1.90 V
  4. 2.13 V
Show Answers Only

\(\text{Question 13:}\ D\)

\(\text{Question 14:}\ A\)

Show Worked Solution

Question 13

  • The oxidising agent is the chemical that undergoes reduction. The chromium changes its oxidation number from \(+6\) to \(+3\) indication reduction.
  • Thus it is the dichromate ion which undergoes reduction.

\(\Rightarrow D\)
 

Question 14

From the list of standard potentials:

  • The reduction voltage of dichromate ions is 1.36 \(\text{V}\) 
  • The oxidation of \(\ce{Fe^2+}\) ions is -0.77 \(\text{V}\) 
  •    \(1.36 + -0.77= 0.59\ \text{V}\) 

\(\Rightarrow A\)

CHEMISTRY, M3 2015 HSC 7 MC

A diagram of a simple cell is shown.

Which of the following occurs when the cell is in operation?

  1. Silver ions are formed in solution.
  2. The copper electrode loses electrons.
  3. Electrons travel through the electrolyte.
  4. The copper electrode increases in mass.
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\(B\)

Show Worked Solution
  • Copper is a more active metal than silver.
  • Thus the copper electrode will be the anode and the silver electrode will be the cathode, and in REDOX reactions, electrons flow from the anode to the cathode.
  • Hence the copper electrode will lose electrons.

\(\Rightarrow B\)

CHEMISTRY, M3 2015 HSC 4 MC

What happens to \(\ce{Fe^2+} \) in the following reaction?

\( \ce{Sn^4+} + \ce{Fe^2+} \rightarrow \ce{Sn^3+} + \ce{Fe^3+} \)

  1. It undergoes oxidation and gains electrons.
  2. It undergoes reduction and gains electrons.
  3. It undergoes oxidation and loses electrons.
  4. It undergoes reduction and loses electrons.
Show Answers Only

\(C\)

Show Worked Solution
  • The increase from \(2^+\) to \(3^+\) indicates oxidation.
  • During oxidation, the species loses electrons.

\(\Rightarrow C\)

CHEMISTRY, M3 2016 HSC 10* MC

Which of the following are the balanced products of the following reaction:

\(\ce{H3PO4 + 3NaOH \rightarrow ?}\)

  1. \( \ce{Na3PO4 + 3H2O}\)
  2. \( \ce{NaPO4 + 3H2O} \)
  3. \( \ce{Na3PO4 + H2O} \)
  4. \( \ce{3NaPO4 + 3H2O} \)
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\(A\)

Show Worked Solution
  • An acid base reaction result in the formation of a salt (\ce{Na3PO4}) and water (\ce{H2O}\) and 3 (\ce{H2O}\) are required to balance the equation.

\(\Rightarrow A\)

CHEMISTRY, M2 2004 HSC 16b

Calculate the mass of solid sodium hydrogen carbonate required to make 250 mL of 0.12 mol L\(^{-1}\) solution.  (2 marks)

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\(2.52\ \text{g} \)

Show Worked Solution

\(\ce{n(NaHCO3\ \text{in solution})\ = 0.12 \times \dfrac{250}{1000} = 0.03\ \text{mol}}\)

\(\ce{MM(NaHCO3) = 22.99 + 1.008 + 12.01 + 3 \times 16.00 = 84.0\ \text{g mol}^{-1}} \)

\(\ce{m(NaHCO3) = n \times MM = 0.03 \times 84.0 = 2.52\ \text{g}} \)

CHEMISTRY, M2 2006 HSC 18

A student studying the mass change that occurs during fermentation added glucose, water and yeast to a flask and stoppered the flask with some cotton wool.

The student measured the mass of the flask daily for seven days. The table shows the data collected.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \ \textit{Day}\ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \textit{Mass}\ \text{(g)}\ \ \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 381.05\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 376.96\\
\hline
\rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt} & 373.42\\
\hline
\rule{0pt}{2.5ex} 4 \rule[-1ex]{0pt}{0pt} & 370.44\\
\hline
\rule{0pt}{2.5ex} 5 \rule[-1ex]{0pt}{0pt} & 370.42\\
\hline
\rule{0pt}{2.5ex} 6 \rule[-1ex]{0pt}{0pt} & 370.40\\
\hline
\rule{0pt}{2.5ex} 7 \rule[-1ex]{0pt}{0pt} & 370.39\\
\hline
\end{array}

  1. Calculate the moles of \(\ce{CO2}\) released between days 1 and 7.  (1 mark)

  2. Calculate the mass of glucose that underwent fermentation between days 1 and 7. Include a balanced chemical equation in your answer.  (3 marks)

Show Answers Only

a.    59.37 moles

b.    5395.92 g

Show Worked Solution

a.    \(\ce{MM (CO2) = 12.01 + 2 \times 16.0 = 44.01 g mol^{-1}}\)

\(\ce{Mass CO2 released = 2613.08 g}\)

\[\ce{n(CO2 released) = \frac{2613.08}{44.01} = 59.37 moles}\]  

b.    \(\ce{C6H12O6 \rightarrow 2C2H6O + 2CO2}\)

  • The moles of \(\ce{CO2}\) released and the reaction’s molar ratio can be used to calculate the mass of glucose that underwent fermentation.
  • Molar ratio between glucose and carbon dioxide = \(1:2\)
  •    \(\ce{n(C6H12O6) = \dfrac{1}{2} \times 59.37 = 29.685\ \text{mol}} \)
  •    \(\ce{MM(C6H12O6) = 180.56\ \text{g mol}^{-1}} \)
  •    \(\ce{m(C6H12O6) = 29.685 \times 180.56 = 5395.92\ \text{g}} \)

CHEMISTRY, M6 2012 HSC 30

A chemist analysed aspirin tablets for quality control. The initial step of the analysis was the standardisation of a \(\ce{NaOH}\) solution. Three 25.00 mL samples of a 0.1034 mol L\(^{-1}\) solution of standardised \(\ce{HCl}\) were titrated with the \( \ce{NaOH} \) solution. The average volume required for neutralisation was 25.75 mL.

  1. Calculate the molarity of the \(\ce{NaOH}\) solution.  (2 marks)

Three flasks were prepared each containing a mixture of 25 mL of water and 10 mL of ethanol. An aspirin tablet was dissolved in each flask. The aspirin in each solution was titrated with the standardised \(\ce{NaOH}\) solution according to the following equation:

\(\ce{C9H8O4(aq) + NaOH(aq) \rightarrow C9H7O4Na(aq) + H2O(l)}\)

The following titration results were obtained.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Tablet}\rule[-1ex]{0pt}{0pt} & \textit{Volume}\ \text{(mL)}\\
\hline
\rule{0pt}{2.5ex}\text{1}\rule[-1ex]{0pt}{0pt} & 16.60\\
\hline
\rule{0pt}{2.5ex}\text{2}\rule[-1ex]{0pt}{0pt} & 16.50\\
\hline
\rule{0pt}{2.5ex}\text{3}\rule[-1ex]{0pt}{0pt} & 16.55\\
\hline
\end{array}

  1. Calculate the average mass (mg) of aspirin per tablet.  (3 marks)
Show Answers Only

a.    \(\ce{0.1004 mol L^{-1}}\)

b.    \(\ce{299.2 mg}\)

Show Worked Solution

a.    \(\ce{n(HCl) = c \times V = 0.1034 \times 0.02500 = 2.585 \times 10^{-3} moles}\)

\(\ce{n(HCl) = n(OH^{-})}\)

\[\ce{[OH-] = \frac{2.585 \times 10^{-3}}{0.02575} = 0.1004 mol L^{-1}}\]  

b.    \(\ce{n(HCl) = c \times V = 0.1004 \times 0.01655 = 1.661 \times 10^{-3} moles}\)

\(\ce{n(HCl) = n(C9H8O4) = 1.661 \times 10^{-3} moles}\)

\(\ce{MM(C9H8O4) = 9 \times 12.01 + 8 \times 1.008 + 4 \times 16.00 = 180.154 g}\)

\(\ce{Average mass of C9H8O4  per tablet}\)

\(\ce{= n \times MM = 1.661 \times 10^{-3} \times 180.154 = 0.2992 g = 299.2 mg}\)

CHEMISTRY, M2 2013 HSC 28

A student attempted to determine the concentration of a hydrochloric acid solution. The following steps were performed.

Step 1. A conical flask was rinsed with water.

Step 2. A 25.0 mL pipette was rinsed with water.

Step 3. The student filled the pipette with a standard sodium carbonate solution to the level shown in the diagram.

Step 4. The standard sodium carbonate solution in the pipette was transferred to the conical flask. The student ensured that all of the sodium carbonate solution was transferred to the conical flask by blowing through the pipette. Three drops of an appropriate indicator were added to the conical flask.

Step 5. A burette was rinsed with the hydrochloric acid solution and then filled with the acid. The student then carried out a titration to determine the concentration of the hydrochloric acid solution.

In steps 2,3 and 4 above the student did not follow acceptable procedures.

  1. Identify the mistake the student made in step 4 and propose a change that would improve the validity of the result.  (2 marks)
  2. Explain the effect of the mistakes made in steps 2 and 3 on the calculation of the concentration of the hydrochloric acid solution.  (3 marks)
Show Answers Only

a.   Mistake: blowing through the pipette

  • Proposed change: student should have touched the end of the pipette to the surface of flask to draw out the liquid.

b.   Mistake (step 2): rinsing the pipette with water

  • This would decrease the number of moles of \(\ce{Na2CO3}\) it contains.

Mistake (step 3): not filling to the gradation mark

  • By not filling to the mark, the pipette would contain fewer moles of \(\ce{Na2CO3}\).
  • Hence, a lower volume of the \(\ce{HCl}\) would be added from the burette, but the student would think that there were more moles of \(\ce{HCl}\) in this volume.
  • As a result, the calculated concentration of the acid solution would be higher than the actual concentration.
Show Worked Solution

a.   Mistake: blowing through the pipette

  • Proposed change: student should have touched the end of the pipette to the surface of flask to draw out the liquid.

b.   Mistake (step 2): rinsing the pipette with water

  • This would decrease the number of moles of \(\ce{Na2CO3}\) it contains.

Mistake (step 3): not filling to the gradation mark

  • By not filling to the mark, the pipette would contain fewer moles of \(\ce{Na2CO3}\).
  • Hence, a lower volume of the \(\ce{HCl}\) would be added from the burette, but the student would think that there were more moles of \(\ce{HCl}\) in this volume.
  • As a result, the calculated concentration of the acid solution would be higher than the actual concentration.
♦ Mean mark (b) 54%.

CHEMISTRY, M6 2012 HSC 28

A solution was made by mixing 75.00 mL of 0.120 mol L\(^{-1}\) hydrochloric acid with 25.00 mL of 0.200 mol L\(^{-1}\) sodium hydroxide.

What is the pH of the solution?  (3 marks)

Show Answers Only

\(\ce{pH = 1.4}\)

Show Worked Solution

\(\ce{HCl + NaOH \rightarrow NaCl + H2O}\)

\(\ce{n(H3O+) = c \times V = 0.120 \times 0.07500 = 0.00900 moles}\)

\(\ce{n(OH) = c \times V = 0.02500 \times 0.200 = 0.00500 moles}\)

\(\ce{n(H3O+ excess) = 9 \times 10^{-3}-5 \times 10^{-3} = 4 \times 10^{-3} moles}\)

\[\ce{[H3O+] = \frac{4 \times 10^{-3}}{0.100} = 4.00 \times 10^{-2}}\]

\(\ce{pH = -log_10[H3O+] = -log_10 4.00 \times 10^{-2} = 1.4}\)