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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), Direct Variation and Currency Conversion (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-6249-30-Graphical Solutions, smc-6249-50-Currency Conversion, smc-793-10-Currency Conversion

L&E, 2ADV E1 2013 HSC 9 MC

What is the solution of   `5^x=4`?

  1. `x=(log_2 4)/5`
  2. `x=4/(log_2 5)`
  3. `x=(log_2 4)/(log_2 5)`
  4. `x=log_2(4/5)`

 

Show Answer Only

 `C`

Show Worked Solutions
`5^x` `=4`
`log_2 5^x` `=log_2 4`
`x  log_2 5` `=log_2 4`
`:.x` `=(log_2 4)/(log_2 5)`

 
`=>C`

Filed Under: Log/Index Laws and Equations (Adv-2027), Log/Index Laws and Equations (Y11), Log/Index laws and Other Equations, Logarithms Tagged With: Band 4, num-title-ct-patha, num-title-qs-hsc, smc-4243-50-Change of base, smc-6455-20-Logs - Power Rule, smc-963-20-Log - power rule

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