Aussie Maths & Science Teachers: Save your time with SmarterEd
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 \(Q=4:\)
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a. \(\text{Using}\ \ N=k \times Q:\)
\(18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5\)
b. \(Q=14\)
a. \(\text{Using}\ \ N=k \times Q:\)
\(18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5\)
b. \(\text{Find}\ Q\ \text{when}\ \ N=63:\)
| \(63\) | \(=4.5 \times Q\) | |
| \(Q\) | \(=\dfrac{63}{4.5}=14\) |
The amount of water (\(W\)) in litres used by a garden irrigation system varies directly with the time (\(t\)) in minutes that the system operates.
This relationship is modelled by the formula \(W=kt\), where \(k\) is a constant.
The irrigation system uses 96 litres of water when it operates for 24 minutes.
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a. \(W=kt\)
\(\text{When } W = 96 \text{ and } t = 24:\)
| \(96\) | \(=k \times 24\) |
| \(k\) | \(=\dfrac{96}{24}=4\ \text{(as required)}\) |
b. \(162.5\ \text{minutes}\)
a. \(W=kt\)
\(\text{When } W = 96 \text{ and } t = 24:\)
| \(96\) | \(=k \times 24\) |
| \(k\) | \(=\dfrac{96}{24}=4\ \text{(as required)}\) |
b. \(W = 4t\)
\(\text{When } W = 650:\)
| \(650\) | \(=4\times t\) |
| \(t\) | \( =\dfrac{650}{4}=162.5\ \text{minutes}\) |
\(\therefore\ \text{The irrigation system can operate for 162.5 minutes.}\)
The cost (\(C\)) of copper wire varies directly with the length (\(L\)) in metres of the wire.
This relationship is modelled by the formula \(C = kL\), where \(k\) is a constant.
A 250 metre roll of copper wire costs $87.50.
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a. \(C=kL\)
\(\text{When } C = 87.50 \text{ and } L = 250:\)
| \(87.50\) | \(=k \times 250\) |
| \(k\) | \(=\dfrac{87.50}{250}\) |
| \(k\) | \(=0.35\ \text{(as required)}\) |
b. \(400\ \text{m}\)
a. \(C=kL\)
\(\text{When } C = 87.50 \text{ and } L = 250:\)
| \(87.50\) | \(=k \times 250\) |
| \(k\) | \(=\dfrac{87.50}{250}\) |
| \(k\) | \(=0.35\ \text{(as required)}\) |
b. \(C = 0.35L\)
\(\text{When } C = 140:\)
| \(140\) | \(=0.35\times L\) |
| \(L\) | \( =\dfrac{140}{0.35}=400\ \text{m}\) |
\(\therefore\ \text{The builder can purchase 400 metres of wire.}\)
The distance (\(d\)) a spring stretches varies directly with the force (\(F\)) applied to it.
When a force of 18 newtons is applied, the spring stretches 27 mm.
What is the value of the constant of variation (\(k\))?
\(D\)
\(d \propto F\)
\(d=kF\)
\(\text{When } d = 27 \text{ and } F = 18:\)
| \(27\) | \(=k \times 18\) |
| \(k\) | \(=\dfrac{27}{18}=\dfrac{3}{2}\) |
\(\Rightarrow D\)
The cost (\(C\)) of printing flyers varies directly with the number of flyers (\(n\)) printed.
A printing company charges $45 to print 300 flyers.
What is the value of the constant of variation (\(k\))?
\(B\)
\(C \propto n\)
\(C=kn\)
\(\text{When } C = 45 \text{ and } n = 300:\)
| \(45\) | \(=k \times 300\) |
| \(k\) | \(=\dfrac{45}{300}=\dfrac{3}{20}\) |
\(\Rightarrow B\)
The weight of a bundle of A4 paper (`W` kg) varies directly with the number of sheets (`N`) of A4 paper that the bundle contains.
This relationship is modelled by the formula `W = kN`, where `k` is a constant.
The weight of a bundle containing 500 sheets of A4 paper is 2.5 kilograms.
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a. `text{Show Worked Solutions}`
b. `240 \ text{sheets}`
a. `W = 2.5\ text{kg when} \ N = 500:`
| `2.5` | `= k xx 500` |
| `therefore \ k` | `= frac{2.5}{500}= 0.005` |
b. `text{Find}\ \ N \ \ text{when} \ \ W = 1.2\ text{kg:}`
| `1.2` | `= 0.005 xx N` |
| `therefore N` | `= frac{1.2}{0.005}= 240 \ text{sheets}` |