A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

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Aditya Kumar

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distance=360 km time difference b/w both speed = 48/60 hr let speed of train is x km/hr then time taken by train when speed is x km/hr = 360/x hr time taken by train when speed is x+5 km/hr = 360/(x+5) hr A/C to question 360/x = 360/(x+5) +48/60 after solving this equation you will get the original speed...
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distance=360 km time difference b/w both speed = 48/60 hr let speed of train is x km/hr then time taken by train when speed is x km/hr = 360/x hr time taken by train when speed is x+5 km/hr = 360/(x+5) hr A/C to question 360/x = 360/(x+5) +48/60 after solving this equation you will get the original speed of train i.e equal to 45 km/hr read less
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Let original speed of train = x km/h We know, Time = distance/speed First case: Time taken by train = 360/x hour Second case: Time taken by train its speed increase 5 km/h = 360/(x + 5) Time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour 360/x - 360/(x +5) = 48/60...
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Let original speed of train = x km/h We know, Time = distance/speed First case: Time taken by train = 360/x hour Second case: Time taken by train its speed increase 5 km/h = 360/(x + 5) Time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour 360/x - 360/(x +5) = 48/60 = 4/5 360 {1/x - 1/(x +5)} = 4/5 360 ×5/4 {5/(x²+5x)}=1 450 x 5 = x² + 5x x²+5x - 2250 = 0 x = {-5±√(25+9000)}/2 = (-5 ±√(9025))/2 =(-5 ± 95)/2 = -50, 45 But x ≠ -50 because speed doesn't negative, So, x = 45 km/h Hence, original speed of train = 45 km/h read less
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