An express train covers a certain distance at a uniform speed. If its speed was increased by 16 km/h, it would take 8 hours less than the scheduled time. If its speed was decreased by 16 km/h, it would take 12 hours more than the scheduled time. Find the length of the journey.
3840 km
Let the usual speed be v km/h and the scheduled time be t hours, so the distance is vt km and stays the same in every case.
Faster case: (v + 16)(t – 8) = vt gives –8v + 16t – 128 = 0, i.e. v = 2t – 16.
Slower case: (v – 16)(t + 12) = vt gives 12v – 16t – 192 = 0, i.e. 3v – 4t = 48.
Substituting v = 2t – 16: 3(2t – 16) – 4t = 48, so 2t = 96, t = 48 hours and v = 2 × 48 – 16 = 80 km/h.
Distance = 80 × 48 = 3840 km.
Check: at 96 km/h the time is 3840 ÷ 96 = 40 h (8 h less than 48 h), and at 64 km/h it is 3840 ÷ 64 = 60 h (12 h more than 48 h), as stated.
Hence, 3840 km.
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