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A train covers a distance of $$193\frac{1}{3}$$ km in $$4\frac{1}{4}$$ hours with one stoppage of 10 minutes, two of 5 minutes and one of 3 minutes on the way. The average speed of the train is :
Answer & Solution
Correct Answer:
Option
B
Actual time taken for the journey :
= 4 hrs 15 min - (10 + 2 × 5 + 3) min
= 4 hrs 15 min - 23 min
= 3 hrs 52 min
= $$3\frac{26}{30}$$ hrs
= $$\frac{116}{30}$$ hrs
∴ Average speed :
$$\eqalign{ & = \left( {\frac{{580}}{3} \times \frac{{30}}{{116}}} \right){\text{ km/hr}} \cr & = 50{\text{ km/hr}} \cr} $$
= 4 hrs 15 min - (10 + 2 × 5 + 3) min
= 4 hrs 15 min - 23 min
= 3 hrs 52 min
= $$3\frac{26}{30}$$ hrs
= $$\frac{116}{30}$$ hrs
∴ Average speed :
$$\eqalign{ & = \left( {\frac{{580}}{3} \times \frac{{30}}{{116}}} \right){\text{ km/hr}} \cr & = 50{\text{ km/hr}} \cr} $$
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