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Let the usual speed and time be 'x' km/hr and 't' minutes, respectively. ATQ, x X 0.75 X {(t + 53)/60} = x X (t/60) Or, 75/100 x (t + 53) = t Or, 3t + 159 = 4t So, 't' = 159 Alternate solution To cover a certain distance, ratio of time taken is reciprocal of ratio of speeds Since, speed is decreased to 75% Therefore, ratio of usual speed to that of decreased speed = 100:75 = 4:3 Or, ratio of usual time taken to the time taken with decreased speed = 3:4 Required time taken = 53 X {3/(4 - 3)} = 159 minutes
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