(a) Solve the simultaneous equation : (log_{10} x + log_{10} y = 4)
(log_{10} x + 2log_{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)