SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

v1 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 63 kg on Earth weighs only 9 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)

Show Answers Only

 \(17\ 143\ \text{kg}\)

Show Worked Solution

\(W_{\text{moon}}\propto W_{\text{earth}}\)

\(\rightarrow\  W_{\text{m}}=k\times W_{\text{e}}\)

\(\text{Find}\ k,\ \text{given}\ W_{\text{e}}=63\ \text{when}\ W_{\text{m}}=9\)

\(9\) \(=k\times 63\)
\(k\) \(=\dfrac{9}{63}=\dfrac{1}{7}\)

  
\(\text{If}\ W_{\text{m}}=2449\ \text{kg, find}\ W_{\text{e}}:\)

\(2449\) \(=\dfrac{1}{7}\times W_{\text{e}}\)
\(W_{\text{e}}\) \(=7\times 2449=17\ 143\)

  
\(\text{Landing craft weighs}\ 17\ 143\ \text{kg on earth}\)

v1 Algebra, STD2 A2 SM-Bank 2

The cost of apples per kilogram, \(C\), varies directly with the weight of apples purchased, \(w\).

If 12 kilograms costs $56.64, calculate the cost of 4.5 kilograms of apples.  (2 marks)

Show Answers Only

\($21.24\)

Show Worked Solution

\(C\ \propto \ w\)

\(C=kw\)

\(\text{When}\ C=$56.64\ \text{kg},\ w=12\ \text{kg}\)

\(56.64\) \(=k\times 12\)
\(k\) \(=\dfrac{56.64}{12}\)
  \(=$4.72\)

 

\(\text{When}\ \ w=4.5\ \text{kg,}\)

\(C\) \(=4.72\times 4.5\)
  \(=$21.24\)

v1 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)

  2. The relationship between Japanese yen \((y)\) and Australian dollars \((d)\) on the same day is given by the equation  \(y=84d\).

     

    Convert \(107\ 520\) Japanese yen to British pounds.  (2 marks)

Show Answers Only
  1. \(p=\dfrac{5}{8}d\)
  2. \(107\ 520\ \text{yen = 800 pounds}\)
Show Worked Solution

a.   \(m=\dfrac{\text{rise}}{\text{run}}=\dfrac{5}{8}\)

\(p=\dfrac{5}{8}d\)


♦ Mean mark 42%.

b.   \(\text{Yen to Australian dollars:}\)

\(y\) \(=84d\)
\(107\ 520\) \(=84d\)
\(d\) \(=\dfrac{107\ 520}{84}\)
  \(= 1280\ $\text{A}\)

 
\(\text{Australian dollars to pounds:}\)

\(p\) \(=\dfrac{5}{8}\times 1280\)
  \(=800\ \text{pounds}\)

  
\(\therefore\ 107\ 520\ \text{yen = 800 pounds}\)

v1 Algebra, STD1 A2 2020 HSC 20

The height of a bundle of photographic paper (\(H\) mm) varies directly with the number of sheets (\(N\)) of photographic paper that the bundle contains.

This relationship is modelled by the formula  \(H=kN\), where  \(k\)  is a constant.

The height of a bundle containing 150 sheets of photographic paper is 2.7 centimetres.

  1. Show that the value of  \(k\)  is 0.18.   (1 mark)

  2. A bundle of photographic paper has a height of 36 centimetres. Calculate the number of sheets of photographic paper in the bundle.   (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(2000\ \text{sheets}\)
Show Worked Solution

a.    \(H=2.7\ \text{cm }=27\ \text{mm, when}\  N=150:\)

\(H\) \(=kN\)
\(2.7\) \(=k\times 150\)
\(\therefore\ k\) \(=\dfrac{2.7}{150}\)
  \(=0.18\)

  

b.     \(\text{Find}\ \ N \ \text{when} \ \ H=36\ \text{cm}=360\ \text{mm:}\)

\(360\) \(=0.18\times N\)
\(\therefore\ N\) \(=\dfrac{360}{0.18}\)
  \(=2000\ \text{sheets}\)

♦ Mean mark 50%.