Discussion
After covering a distance of 30 km with a uniform speed there is some defect in the train engine and therefore, its speed is reduced to \(\frac{4}{5}\)th of its original speed. Consequently, the train reaches its destination late by 45 minutes. Had it happened after covering 18 km more, the train would have reached 9 minutes earlier. Find the distance of the journey.
Answer: C
Let the original speed be v and the distance D. After the fault the speed is 4v/5, so the extra time on the remaining distance is one fourth of the original time for it. The delay is 45 minutes when the fault comes after 30 km and 36 minutes when it comes 18 km later, after 48 km.
\[ \frac{D-30}{4v} = \tfrac{3}{4} \]
\[ \frac{D-48}{4v} = \tfrac{3}{5} \]
\[ \frac{18}{4v} = \tfrac{3}{20} \Rightarrow v = 30 \]
Then (D - 30) / 120 = 3/4, so D - 30 = 90 and D = 120 km. The other options do not give a delay of 45 minutes at 30 km per hour.
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