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Rates of Change, SMB-010

Moses finds that for a Froghead eel, its mass is directly proportional to the square of its length.

An eel of this species has a length of 72 cm and a mass of 8250 grams.

What is the expected length of a Froghead eel with a mass of 10.2 kg? Give your answer to one decimal place.  (3 marks)

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`80.1\ text{cm}`

Show Worked Solution

`text(Mass) prop text(length)^2`

`m = kl^2`
  

`text(Find)\ k:`

`8250` `= k xx 72^2`
`k` `= 8250/72^2`
  `= 1.591…`

 
`text(When)\ \ l\ \ text(when)\ \ m = 10\ 200:`

`10\ 200` `= 1.591… xx l^2`
`l^2` `= (10\ 200)/(1.591…)`
`:. l` `= 80.058…`
  `= 80.1\ text{cm  (to 1 d.p.)}`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-40-a prop other

Rates of Change, SMB-009

The number of trees that can be planted along the fence line of a paddock varies inversely with the distance between each tree.

There will be 108 trees if the distance between them is 5 metres.

  1. How many trees can be planted if the distance between them is 6 metres?  (2 marks)

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  2. What is the distance between the trees if 120 trees are planted?  (1 mark)

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Show Answers Only
  1. `90`
  2. `4.5\ text(metres)`
Show Worked Solution

i.   `t prop 1/d`

`t` `= k/d`
`108` `= k/5`
`k` `= 540`

 
`text(Find)\ t\ text(when)\ d = 6:`

`t` `= 540/6`
  `= 90`

 

ii.   `text(Find)\ d\ text(when)\ t = 120:`

`120` `= 540/d`
`d` `= 540/120`
  `= 4.5\ text(metres)`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-30-a prop 1/b

Rates of Change, SMB-008

It is known that the quantity of steel produced in tonnes `(S)`, is directly proportional to the tonnes of iron ore used in the process `(I)`.

If 16 tonnes or iron ore produces 10 tonnes of steel, calculate the tonnes of iron ore required to produce 48 tonnes of steel.  (3 marks)

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`76.8\ text{tonnes}`

Show Worked Solutions

`S prop I\ \ =>\ \ S=kI`

`text(Find)\ k\ text{given}\ S=10\ text{when}\ I=16:`

`10` `=k xx 16`
`k` `=10/16`
  `=0.625`

 
`text{Find}\ I\ text{when}\ S=48:`

`48` `=0.625 xx I`
`:. I` `=48/0.625`
  `=76.8\ text{tonnes}`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-10-a prop b

Rates of Change, SMB-007

It is known that a quantity `P` kgs is proportional to the reciprocal of another quantity `Q` kgs such that `P prop 1/Q`.

If  `P=12` when `Q=20`, calculate the estimated quantity of `Q` when `P=45` kgs, to the nearest gram.  (3 marks)

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`5333\ text{g}`

Show Worked Solutions

`P prop 1/Q\ \ =>\ \ P=k/Q`

`text(Find)\ k\ text{given}\ P=12\ text{when}\ Q=20:`

`12` `=k/20`
`:. k` `=12 xx 20`
  `=240`

 
`text{Find}\ Q\ text{when}\ P=45:`

`45` `=240/Q`
`:. Q` `=240/45`
  `=5.3333\ text{kg}`
  `=5333\ text{g}`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-30-a prop 1/b

Rates of Change, SMB-006

The stopping distance of a car on a certain road, once the brakes are applied, is directly proportional to the square of the speed of the car when the brakes are first applied.

A car travelling at 70 km/h takes 58.8 metres to stop.

How far does it take to stop if it is travelling at 105 km/h?  (3 marks)

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`132.3\ text(metres)`

Show Worked Solution

`text(Let)\ \ d\ text(= stopping distance)`

`d prop s^2`

`d = ks^2`
 

`text(Find)\ k,`

`58.8` `= k xx 70^2`
`k` `= 58.8/(70^2)`
  `= 0.012`

 
`text(Find)\ d\ \ text(when)\ s = 105:`

`d` `= 0.012 xx 105^2`
  `= 132.3\ text(metres)`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-40-a prop other

Rates of Change, SMB-005

Fuifui finds that for Giant moray eels, the mass of an eel `(M)` is directly proportional to the cube of its length `(l)`.

An eel of this species has a length of 15 cm and a mass of 675 grams.

What is the expected length of a Giant moray eel with a mass of 3.125 kg? Give your answer to one decimal place.  (3 marks)

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`25\ text{cm}`

Show Worked Solution

`M prop l^3`

`M = kl^3`
 

`text(Find)\ k:`

`675` `= k xx 15^3`
`k` `= 675/15^3`
  `= 0.2`

 
`text(Find)\ \ l\ \ text(when)\ \ M = 3125:`

`3125` `= 0.2 xx l^3`
`l^3` `= 3125/0.2`
`:. l` `= root3(15\ 625)`
  `= 25\ text{cm}`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-40-a prop other

Rates of Change, SMB-004

Jacques is a marine biologist and finds that the mass of a crab `(M)` is directly proportional to the cube of the diameter of its shell `(d)`.

If a crab with a shell diameter of 15 cm weighs 680 grams, what will be the diameter of a crab that weighs 1.1 kilograms? Give your answer to 1 decimal place.  (3 marks)

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`17.6\ text(cm)`

Show Worked Solution
`M` `prop d^3`  
`M` `= kd^3`  

 
`text(When)\ \ M=680, \ d=15`

`680` `=k xx 15^3`  
`k` `=0.201481…`  

 
`text(Find)\ \ d\ \ text(when)\ \ M=1100:`

`1100` `=0.20148… xx d^3`  
`d` `=root3(1100/(0.20148…))`  
  `=17.608…`  
  `=17.6\ text{cm  (to 1 d.p.)}`  

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-40-a prop other

Rates of Change, SMB-003

The current of an electrical circuit, measured in amps (A), varies inversely with its resistance, measured in ohms (R).

When the resistance of a circuit is 28 ohms, the current is 3 amps.

What is the current when the resistance is 8 ohms? (2 marks)

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`10.5`

Show Worked Solution
`A` `prop 1/R`  
`A` `= k/R`  

 
`text(When)\ \ A=3, \ R=28`

`3` `=k/28`  
`k` `=84`  

 
`text(Find)\ \ A\ \ text(when)\ \ R=8:`

`A` `=84/8`  
  `=10.5`  

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-30-a prop 1/b

Rates of Change, SMB-002

It is known that at a constant speed, the distance travelled in kilometres `(d)` is directly proportional to the time of travel in hours `(t)`, or  `d prop t`.

  1. If `d=75` when `t=5`, calculate the constant of variation `k`.  (2 marks)

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  2. In the context of this question, what does the value of `k` represent?  (1 mark)

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i.    `k=15`

ii.   `text{Speed}`

Show Worked Solutions

i.    `d prop t`

`d=kt`

`text(Find)\ k\ text{given}\ d=75\ text{when}\ t=5:`

`75` `=k xx 5`
`:. k` `=75/5`
  `=15`

 
ii.
   `k\ text{represents the speed.}`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-10-a prop b

Rates of Change, SMB-001

It is known that a quantity `y` is inversely proportional to another quantity `x`.

If  `y=3` when `x=1.8`, calculate the constant of variation `k`.  (2 marks)

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`k=5.4`

Show Worked Solutions

`y prop 1/x`

`y=k/x`

`text(Find)\ k\ text{given}\ y=3\ text{when}\ x=1.8:`

`3` `=k/1.8`
`:. k` `=3xx1.8`
  `=5.4`

Filed Under: Variation and Rates of Change Tagged With: num-title-ct-patha, smc-4239-30-a prop 1/b

Algebra, STD1 A3 2019 HSC 9 MC

The container shown is initially full of water.
 

Water leaks out of the bottom of the container at a constant rate.

Which graph best shows the depth of water in the container as time varies?
 

A. B.
C. D.
Show Answers Only

`D`

Show Worked Solution

`text(Depth will decrease slowly at first and accelerate.)`

`=> D`

Filed Under: Non-Linear: Inverse and Other Problems (Std 2), Quadratics, Variation and Rates of Change Tagged With: Band 5, num-title-ct-pathb, num-title-qs-hsc, smc-4239-60-Variable rates of change, smc-795-20-Other Relationship

Functions, 2ADV F1 2022 HSC 12

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.    (2 marks)

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  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}`.   (2 marks)
     

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M &  &  &  \\
\hline \end{array}

 
                   

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a.    `M=180/T`

 b.    

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}       

 

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`
 


♦ Mean mark (a) 49%.

b.   

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}

Filed Under: Direct and Inverse Variation (Adv-2027), Further Functions and Relations (Y11), Variation and Rates of Change Tagged With: 2adv-std2-common, Band 4, Band 5, common-content, num-title-ct-patha, num-title-qs-hsc, smc-4239-30-a prop 1/b, smc-6383-30-prop 1/(kx^n), smc-987-30-Reflections and Other Graphs, smc-987-60-Proportional

Algebra, STD2 A2 2019 HSC 34

The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.
 

  1. Write the direct variation equation relating British pounds to Australian dollars in the form  `p = md`. Leave `m` as a fraction.  (1 mark)

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  2. The relationship between Japanese yen `(y)` and Australian dollars `(d)` on the same day is given by the equation  `y = 76d`.

     

    Convert 93 100 Japanese yen to British pounds.  (2 marks)

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  1. `p = 4/7 d`
  2. `93\ 100\ text(Yen = 700 pounds)`
Show Worked Solution

a.   `m = text(rise)/text(run) = 4/7`

♦ Mean mark 42%.

`p = 4/7 d`

 

b.   `text(Yen to Australian dollars:)`

`y` `=76d`
`93\ 100` `= 76d`
`d` `= (93\ 100)/76`
  `= 1225`

 
`text(Aust dollars to pounds:)`

`p` `= 4/7 xx 1225`
  `= 700\ text(pounds)`

 
`:. 93\ 100\ text(Yen = 700 pounds)`

Filed Under: Applications: Currency, Fuel and Other Problems (Std 2), Applications: Currency, Fuel and Other Problems (Std2-2027), Linear Functions (Adv-2027), Linear Functions (Y11), Variation and Rates of Change Tagged With: Band 4, Band 5, common-content, num-title-ct-patha, num-title-qs-hsc, smc-4239-70-Currency convert, smc-6214-20-Other Real World Applications, smc-6249-40-Graphical Solutions, smc-6256-10-Currency Conversion, smc-793-10-Currency Conversion, smc-985-20-Other Linear Applications

Algebra, STD2 A4 2019 HSC 33

The time taken for a car to travel between two towns at a constant speed varies inversely with its speed.

It takes 1.5 hours for the car to travel between the two towns at a constant speed of 80 km/h.

  1. Calculate the distance between the two towns.  (1 mark)

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  2. By first plotting four points, draw the curve that shows the time taken to travel between the two towns at different constant speeds.  (3 marks)

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  1. `120\ text(km)`
  2.  
Show Worked Solution
a.    `D` `= S xx T`
    `= 80 xx 1.5`
    `= 120\ text(km)`

 
b. 
 

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ s\ \  \rule[-1ex]{0pt}{0pt} & 20 & 40 & 60 & 80 \\
\hline
\rule{0pt}{2.5ex} t \rule[-1ex]{0pt}{0pt} & 6 & 3 & 2 & 1.5 \\
\hline
\end{array}

Filed Under: Non-Linear: Inverse and Other Problems (Std 2), Variation and Rates of Change Tagged With: Band 3, Band 4, num-title-ct-patha, num-title-qs-hsc, smc-4239-30-a prop 1/b, smc-795-10-Inverse

Algebra, STD2 A2 2007 HSC 24c

Sandy travels to Europe via the USA. She uses this graph to calculate her currency conversions.
  
  
 

  1. After leaving the USA she has US$150 to add to the A$600 that she plans to spend in Europe.

     

    She converts all of her money to euros. How many euros does she have to spend in Europe?    (3 marks)

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  2. If the value of the euro falls in comparison to the Australian dollar, what will be the effect on the gradient of the line used to convert Australian dollars to euros?   (1 mark)

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  1. `480\ €`
  2. `text(If the value of the euro falls against the)`
  3.  

    `text(A$, then 1 A$ will buy more euros than)`

  4.  

    `text(before and the gradient used to convert)`

  5.  

    `text{the currencies will steepen (increase).}`

Show Worked Solution

i.   `text(From graph:)`

`75\ text(US$)` `=\ text(100 A$)`
`=> 150\ text(US$)` `=\ text(200 A$)`

 
`:.\ text(Sandy has a total of 800 A$)`
 

`text(Converting A$ to €:)`

`text(100 A$)` `= 60\ €`
`:.\ text(800 A$)` `= 8 xx 60`
  `= 480\ €`

 

ii.   `text(If the value of the euro falls against the)`

`text(A$, then 1 A$ will buy more euros than)`

`text(before and the gradient used to convert)`

`text{the currencies will steepen (increase).}`

Filed Under: AM2 - Linear Relationships (Prelim), Applications: Currency, Fuel and Other Problems (Std 1), Applications: Currency, Fuel and Other Problems (Std 2), Applications: Currency, Fuel and Other Problems (Std2-2027), Variation and Rates of Change Tagged With: Band 4, num-title-ct-patha, num-title-qs-hsc, smc-1119-10-Currency Conversion, smc-4239-70-Currency convert, smc-6256-10-Currency Conversion, smc-793-10-Currency Conversion

Algebra, STD2 A4 2007 HSC 15 MC

If pressure (`p`) varies inversely with volume (`V`), which formula correctly expresses  `p`  in terms of  `V`  and  `k`, where  `k`  is a constant?

  1. `p = k/V`
  2. `p = V/k`
  3. `p = kV`
  4. `p = k + V`
Show Answers Only

`A`

Show Worked Solution

`p prop 1/V`

`p = k/V`

`=>  A`

Filed Under: Inverse, Non-Linear: Inverse and Other Problems (Std 2), Variation and Rates of Change Tagged With: Band 5, num-title-ct-patha, num-title-qs-hsc, smc-4239-30-a prop 1/b, smc-795-10-Inverse, smc-795-40-Proportional

Algebra, STD2 A2 2014 HSC 26f

The weight of an object on the moon varies directly with its weight on Earth.  An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon.

A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth.   (2 marks)

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 `14\ 694\ text(kg)`

Show Worked Solution

`W_text(moon) prop W_text(earth)`

`=> W_text(m) = k xx W_text(e)`

`text(Find)\ k\ text{given}\  W_text(e) = 84\ text{when}\ W_text(m) = 14`

`14` `= k xx 84`
`k` `= 14/84 = 1/6`

 

`text(If)\ W_text(m) = 2449\ text(kg),\ text(find)\ W_text(e):`

`2449` `= 1/6  xx W_text(e)`
`W_text(e)` `= 14\ 694\ text(kg)`

 

`:.\ text(Landing craft weighs)\ 14\ 694\ text(kg on earth)`

Filed Under: Applications: Currency, Fuel and Other Problems (Std 1), Applications: Currency, Fuel and Other Problems (Std 2), Direct Variation (Std2-2027), Other Linear Modelling, Variation and Rates of Change Tagged With: Band 4, num-title-ct-patha, num-title-qs-hsc, smc-1119-30-Other Linear Applications, smc-1119-50-Proportional, smc-4239-10-a prop b, smc-6249-30-Algebraic Solutions, smc-793-30-Other Linear Applications, smc-793-50-Proportional

Algebra, STD2 A4 2011 HSC 28a

The air pressure, `P`, in a bubble varies inversely with the volume, `V`, of the bubble. 

  1. Write an equation relating `P`, `V` and `a`, where `a` is a constant.    (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  2. It is known that `P = 3` when `V = 2`.

     

    By finding the value of the constant, `a`, find the value of `P` when `V = 4`.    (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

  3. Sketch a graph to show how `P` varies for different values of `V`.

     

    Use the horizontal axis to represent volume and the vertical axis to represent air pressure.   (2 marks)

    --- 8 WORK AREA LINES (style=lined) ---


Show Answers Only
  1. `P = a/V`
  2. `P = 1 1/2`
  3.  
     
Show Worked Solution
♦ Mean mark (i) 39%
COMMENT: Expressing the proportional relationship `P prop 1/V` as the equation `P=k/V` is a core skill here.
i. `P` `prop 1/V`
    `= a/V`

 

ii. `text(When)\ P=3,\ V = 2`
`3` `= a/2`
`a` `=6`

 

`text(Need to find)\ P\ text(when)\ V = 4`  

♦ Mean mark (ii) 47%
`P` `=6/4`
  `= 1 1/2`

  

♦♦ Mean mark (iii) 26%
COMMENT: An inverse relationship is reflected by a hyperbola on the graph.
iii.

Filed Under: Inverse, Non-Linear: Inverse and Other Problems (Std 2), Variation and Rates of Change Tagged With: Band 5, num-title-ct-patha, num-title-qs-hsc, smc-4239-30-a prop 1/b, smc-795-10-Inverse, smc-795-40-Proportional

Algebra, STD2 A4 2009 HSC 28c

The height above the ground, in metres, of a person’s eyes varies directly with the square of the distance, in kilometres, that the person can see to the horizon.

A person whose eyes are 1.6 m above the ground can see 4.5 km out to sea.

How high above the ground, in metres, would a person’s eyes need to be to see an island that is 15 km out to sea? Give your answer correct to one decimal place.   (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

 `17.8\ text(m)\ \ text{(to 1 d.p.)}`

Show Worked Solution
♦♦ Mean mark 22%
CRITICAL STEP: Reading the first line of the question carefully and establishing the relationship `h=k d^2` is the key part of solving this question.

`h prop d^2`

`h=kd^2`

`text(When)\ h = 1.6,\ d = 4.5`

`1.6` `= k xx 4.5^2`
`:. k` `= 1.6/4.5^2`
  `= 0.07901` `…`

 

`text(Find)\ h\ text(when)\ d = 15`

`h` `= 0.07901… xx 15^2`
  `= 17.777…`
  `= 17.8\ text(m)\ \ \ text{(to 1 d.p.)}`

Filed Under: Exponential/Quadratic (Projectile), Non-Linear: Exponential/Quadratics (Std 2), Variation and Rates of Change Tagged With: Band 5, num-title-ct-patha, num-title-qs-hsc, smc-4239-40-a prop other, smc-830-20-Quadratics, smc-830-60-Proportional

Algebra, STD2 A1 2009 HSC 16 MC

The time for a car to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?
 

Show Answers Only

`A`

Show Worked Solution
`T` `prop 1/S`
`T` `= k/S`

 
`text{By elimination:}`

`text(As   S) uarr text(, T) darr => text(cannot be B or D)`

♦ Mean mark 38%

`text(C  is incorrect because it graphs a linear relationship)`

`=>  A`

Filed Under: Applications: BAC, Medication and D=SxT (Std 2), Inverse, Linear Equations and Basic Graphs (Std 2), Non-Linear: Inverse and Other Problems (Std 2), Safety: D=ST & BAC, Variation and Rates of Change Tagged With: Band 5, num-title-ct-patha, num-title-qs-hsc, smc-4239-30-a prop 1/b, smc-6235-20-Speed Distance Time, smc-6255-40-Other, smc-791-20-Speed Distance Time, smc-792-40-Other, smc-795-10-Inverse

Algebra, STD2 A4 2010 HSC 13 MC

The number of hours that it takes for a block of ice to melt varies inversely with the temperature. At 30°C it takes 8 hours for a block of ice to melt.

How long will it take the same size block of ice to melt at 12°C?  

  1. 3.2 hours
  2. 20 hours
  3. 26  hours
  4. 45 hours
Show Answers Only

`B`

Show Worked Solution
 
♦ Mean mark 50% 

`text{Time to melt}\ (T) prop1/text(Temp) \ => \ T=k/text(Temp)`

`text(When) \ T=8, text(Temp = 30)`

`8` `=k/30`
`k` `=240`

  

`text{Find}\ T\ text{when  Temp = 12:}`

`T` `=240/12`
  `=20\ text(hours)`

 
`=>  B`

Filed Under: Inverse, Non-Linear: Inverse and Other Problems (Std 2), Variation and Rates of Change Tagged With: Band 5, num-title-ct-patha, num-title-qs-hsc, smc-4239-30-a prop 1/b, smc-795-10-Inverse, smc-795-40-Proportional

Algebra, STD2 A2 2012 HSC 13 MC

Conversion graphs can be used to convert from one currency to another.  
  

 
  

Sarah converted  60  Australian dollars into Euros. She then converted all of these Euros
into New Zealand dollars.

How much money, in New Zealand dollars, should Sarah have? 

  1. $26  
  2. $45
  3. $78
  4. $135
Show Answers Only

`C`

Show Worked Solution

`text(Using the graphs)`

`$60\ text(Australian)` `=46\ text(Euro)`
`46 \ text(Euro)` `=$78\  text(New Zealand)`

 
`=>  C`

Filed Under: AM2 - Linear Relationships (Prelim), Applications: Currency, Fuel and Other Problems (Std 1), Applications: Currency, Fuel and Other Problems (Std 2), Applications: Currency, Fuel and Other Problems (Std2-2027), Variation and Rates of Change Tagged With: Band 2, num-title-ct-patha, num-title-qs-hsc, smc-1119-10-Currency Conversion, smc-4239-70-Currency convert, smc-6256-10-Currency Conversion, smc-793-10-Currency Conversion

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