A school bus transported am excursion party to picnic spot 150 km away . While returning , it was raining and the bus had to reduce its speed by 5 km/hr , and it took 1 hour longer to make the return trip . Fine the time taken to return
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Speed = Distance / Time
For the journey to the picnic spot,
s=(150) / t (1)
For the journey from the picnic spot,
s-5=(150) / (t+1)
s= [ (150) / (t+1) ] + 5 (2)
Now equating (1) & (2)
(150) / t = [ (150) / (t+1) ] + 5
(150) / t = (150 + 5x(t+1)) / (t+1)
150x(t+1) = 150xt + 5x(t+1)xt
150t + 150 = 150t + 5x(t²+t)
150 = 5t² + 5t
5t² + 5t - 150 = 0
t² + t - 30 = 0
So, t = 5 & t = -6
Hence t = 5 Hours.
For the journey to the picnic spot,
s=(150) / t (1)
For the journey from the picnic spot,
s-5=(150) / (t+1)
s= [ (150) / (t+1) ] + 5 (2)
Now equating (1) & (2)
(150) / t = [ (150) / (t+1) ] + 5
(150) / t = (150 + 5x(t+1)) / (t+1)
150x(t+1) = 150xt + 5x(t+1)xt
150t + 150 = 150t + 5x(t²+t)
150 = 5t² + 5t
5t² + 5t - 150 = 0
t² + t - 30 = 0
So, t = 5 & t = -6
Hence t = 5 Hours.
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