Question
A man travels a specific distance at a given speed,
completing the journey in a set amount of time. If he had reduced his speed by 10 km/h, it would have taken him an additional 2 hours to cover the same distance. Further, if he had decreased his speed by another 10 km/h (a total reduction of 20 km/h from the original speed), the journey would have taken an additional 3 hours beyond the initial 2-hour delay. Determine the total distance he traveled.Solution
Let, distance be βdβ km So, d/(s β 10) β d/s = 2 Or, 10d = 2s(s β 10) Or, d = 2 Γ s(s β 10)/10 -----(i) And, d/(s β 20) β d/s = 5 Or, 20d = 5 Γ s(s β 20) Or, d = 5 Γ s(s β 20)/20 --------(ii) From (i) and (ii), Let S1 = 10 km/h S2 = 20 km/h T1 = 2 hours T2 = 5 hours Let original speed be S km/h [(S β 10) Γ 2]/10 = [(S β 20) Γ 5]/20 4(S β 10) = 5(S β 20) 4S β 40 = 5S β 100 S = 60 km/h Required distance = 60 Γ (60 β 10)/10 Γ 2 = 60 Γ 5 Γ 2 = 600 km
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