# The average speed of a train in the return journey is 33.33% more than that in the onward Journey. The train halts for two hours on reaching the desti

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The average speed of a train in the return journey is 33.33% more than that in the onward Journey. The train halts for two hours on reaching the destination. The total distance covering the complete to and for journey is 1200 km in 16 hours. If the speed of train in onward journey is 50% more than the speed of the car. Then, find the time taken by car to cover the distance of 625 km?

1. 12.5 hours
2. 12 hours
3. 10.5 hours
4. 15.5 hours
5. None of these

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Correct Answer - Option 1 : 12.5 hours

Given:

Speed of the train in onward journey = v km/hr

Speed of the train in return journey = 4v/3 km/hr

Total halt = 2 hr

Total time = 16 hrs

Total distance = 1200 km

Speed of train in onward journey = 150% of Speed of Car

Formula Used:

Distance = Speed × Time

Calculation:

Suppose the speed of the train in onward = v km/hr

Speed of train in return = 4v/3 km/hr

According to question –

Total time taken = (16 – 2) hrs

⇒ 14 hrs

Average Speed = (Total Distance)/(Total Time)

⇒ (2V1V2)/(V1 + V2)

⇒ (2 × v × 4v/3)/(v + 4v/3)

⇒ (8v2/3)/(7v/3)

⇒ 8v/7

Average Speed = 1200/14

⇒ 8v/7 = 1200/14

⇒ v = 75 km/hr

The speed of the train in onward = 75 km/hr

According to question –

Suppose the speed of the car is x km/hr

⇒ 75 = 150% of x

⇒ 75 = 3x/2

⇒ x = 50 km/hr

The time taken by car to cover the distance of 625 km = 625/50 hr

⇒ 12.5 hours

∴ the time taken by the car to cover 625 km is 12.5 hours.