(a) Solve the simultaneous equation: \[ \log_{10} x + \log_{10} y = 4 \] \[ \log_{10} x + 2\log_{10} y = 3 \] (b) The time, \(t\), taken to buy fuel at a pe...
Question 1 Report
(a) Solve the simultaneous equation:
\[ \log_{10} x + \log_{10} y = 4 \]
\[ \log_{10} x + 2\log_{10} y = 3 \]
(b) The time, \(t\), taken to buy fuel at a petrol station varies directly as the number of vehicles \(V\) on queue and jointly varies inversely as the number of pumps \(P\) available in the station. In a station with 5 pumps, it took 10 minutes to fuel 20 vehicles. Find:
(i) the relationship between \(t\), \(P\) and \(V\); (ii) the time it will take to fuel 50 vehicles in the station with 2 pumps; (iii) the number of pumps required to fuel 40 vehicles in 20 minutes.
Answer Details
(a) Let \(\log_{10}x = u\) and \(\log_{10}y = v\).
\[u + v = 4 \quad\text{...(1)}\qquad u + 2v = 3 \quad\text{...(2)}\]
Subtract (1) from (2): \(v = -1\). Then \(u = 4 - (-1) = 5\).
\[\log_{10}x = 5 \Rightarrow x = 10^5 = 100000, \qquad \log_{10}y = -1 \Rightarrow y = 10^{-1} = 0.1\]
(b) \(t\) varies directly as \(V\) and inversely as \(P\): \(t = \dfrac{kV}{P}\).
With \(P = 5,\, V = 20,\, t = 10\): \(10 = \dfrac{k(20)}{5} = 4k \Rightarrow k = 2.5\).
(i) \(t = \dfrac{2.5V}{P}\) (equivalently \(t = \dfrac{5V}{2P}\)).
(ii) \(V = 50,\, P = 2\): \(t = \dfrac{2.5(50)}{2} = 62.5\) minutes.
(iii) \(V = 40,\, t = 20\): \(20 = \dfrac{2.5(40)}{P} \Rightarrow 20 = \dfrac{100}{P} \Rightarrow P = 5\) pumps.
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