To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?
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Correct Answer: Option C
Explanation:
\(speed = \frac{distance}{time}\\
72 = \frac{60}{time}\\
t = \frac{60}{72} = \frac{5}{6}hr\)
time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\)
Time required for the journey
\(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\
speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)
\(speed = \frac{distance}{time}\\
72 = \frac{60}{time}\\
t = \frac{60}{72} = \frac{5}{6}hr\)
time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\)
Time required for the journey
\(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\
speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)