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Algebra, STD2 EQ-Bank 24

It is known that a quantity \(N\) varies directly with another quantity \(Q\).

The relationship can be modelled by the equation  \(N= k \times Q\), where \(k\) is a constant.

If \(N = 18\)  when  \(Q=4:\)

  1. Show the value of  \(k=4.5\).   (1 mark)

  2. Hence find the value of \(Q\) when  \(N=63\).  (1 mark)

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a.    \(\text{Using}\ \ N=k \times Q:\)

\(18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5\)

b.    \(Q=14\)

Show Worked Solution

a.    \(\text{Using}\ \ N=k \times Q:\)

\(18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5\)
 

b.    \(\text{Find}\ Q\ \text{when}\ \ N=63:\)

\(63\) \(=4.5 \times Q\)  
\(Q\) \(=\dfrac{63}{4.5}=14\)  

Algebra, STD2 EQ-Bank 25

Two quantities,  \(M\)  and  \(t\),  have a relationship such that  \(M\)  varies directly with  \(t\)  and   \(M=2.4\)   when  \(t=8\).

Find the value of  \(t\)  when  \(M=5.1\).   (2 marks)

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\(t=17\)

Show Worked Solution

\(M \propto t \ \ \Rightarrow \ \ M=kt\)

\(\text{Given} \ \ M=2.4 \ \ \text{when} \ \ t=8:\)

\(2.4=8k \ \ \Rightarrow \ \ k=\dfrac{2.4}{8}=0.3\)

\(\text{When} \ \ M=5.1:\)

\(5.1\) \(=0.3 \times t\)  
\(t\) \(=\dfrac{5.1}{0.3}=17\)  

Algebra, STD2 EQ-Bank 26

\(F\) varies directly with  \(m\)  and  \(F=14\)  when  \(m=3.5\).

Find the value of  \(m\)  when  \(F=50\).   (2 marks)

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\(m=12.5\)

Show Worked Solution

\(F \propto m \ \ \Rightarrow \ \ F=km\)

\(\text{Given} \ \ F=14 \ \ \text{when} \ \ m=3.5:\)

\(14=3.5k \ \ \Rightarrow \ \ k=4\)

\(\text{When} \ \ F=50:\)

\(50\) \(=4 \times m\)  
\(m\) \(=\dfrac{50}{4} = 12.5\)  

Algebra, STD2 EQ-Bank 15 MC

\(d\) varies directly with \(h\) and it is known that  \(d=1.2\)  when  \(h=5\).

The value of \(h\) when  \(d=48\)  is

  1. \(11.52\)
  2. \(57.6\)
  3. \(200\)
  4. \(240\)
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\(C\)

Show Worked Solution

\(d \propto h \ \ \Rightarrow \ \ d=k \times h\)

\(\text{Find \(k\) given \(\ d=1.2\)  when \(\ h=5\):}\)

\(1.2\) \(=5 \times k\)  
\(k\) \(=\dfrac{1.2}{5}=0.24\)  

 
\(\text{Find} \ h \ \text{when} \ \ d=48:\)

\(48\) \(=0.24 \times h\)  
\(h\) \(=\dfrac{48}{0.24}=200\)  

 
\(\Rightarrow C\)

Algebra, STD2 EQ-Bank 11 MC

\(Q\) varies directly with \(r\) and  \(Q=2\) when  \(r=16\).

The value of \(r\) when  \(Q=13\)  is

  1. \(1.625\)
  2. \(26\)
  3. \(76\)
  4. \(104\)
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\(D\)

Show Worked Solution

\(Q \propto r \ \Rightarrow \ Q=kr\)

\(\text{Find \(k\) given \(\ Q=2\) given \(\ r=16\):}\)

\(2=16 \times k \ \Rightarrow \ k=\dfrac{1}{8}\)

\(\text{Find \(r\) when \(\ Q=13:\)}\)

\(13\) \(=\dfrac{1}{8} \times r\)  
\(r\) \(=13 \times 8=104\)  

 
\(\Rightarrow D\)

Algebra, STD1 A2 2025 HSC 16

The mass \((M )\) of a box with a square base, in grams, is directly proportional to the area of its base, in cm².
 

A box with a square base of side length 5 cm has a mass of 500 g.

What is the mass of a similar box with a square base of side length 3 cm?    (3 marks)

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\(M=180\ \text{grams}\)

Show Worked Solution

\(\text{Area of square base }= s^2\)

\(M \propto s^2\ \ \Rightarrow \ \ M=k\times s^2\)

\(\text{Find \(k\) given \(\ s=5\ \) when \(\ M=500\):}\)

\(500\) \(=k\times 5^2\)
\(25k\) \(=500\)
\(k\) \(=\dfrac{500}{25}=20\)

 
\(\text{Find \(M\) when  \(s=3\):}\)

\(M=20\times 3^2=180\ \text{grams}\)


♦♦♦ Mean mark 21%.