(a) Solve the simultaneous equation : \(\log_{10} x + \log_{10} y = 4\) \(\log_{10} x
(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
\(20 = \frac{5(40)}{2P} \implies 40P = 200\)
\(P = \frac{200}{40} = 5 pumps\)