(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.
Explanation
(a) \(\log_{10} x + \log_{10} y = 4 ... (1)\) \(\log_{10} x + 2 \log_{10} y = 3 ...(2)\) \((1) \implies \log_{10} (xy) = 4 \therefore xy = 10^{4} ... (3)\) \((2) \implies \log_{10} (xy^{2}) = 3 \therefore xy^{2} = 10^{3} ... (4)\) Divide (4) by (3) : \(y = 10^{-1} = 0.1\) Put y in (3), we have \(\frac{x}{10} = 10^{4} \implies x = 10^{5}\) \((x, y) = (10^{5}, 10^{-1})\) (b) (i) \(t \propto \frac{V}{P}\) \(t = \frac{kV}{P}\) when t = 10, V = 20 and P = 5. \(10 = \frac{20k}{5} \implies 50 = 20k\) \(k = \frac{5}{2}\) \(\therefore t = \frac{5V}{2P}\) (ii) When V = 50, P = 2, we have \(t = \frac{5(50)}{2(2)} = \frac{250}{4}\) = \(62.5 minutes\) (iii) when V = 40, t = 20 minutes \(20 = \frac{5(40)}{2P} \implies 40P = 200\) \(P = \frac{200}{40} = 5 pumps\)
