Bunuel wrote:
A train runs over a straight route from town A to town D. It is scheduled to depart town A at 7am and arrive at town D at 2pm, with 10 minute stops in towns B and C. The train’s top speed is 60mph. The entire length of the route is 320 miles. Will the train arrive on time?
(1) The train experiences a 30 minute delay in town B in addition to its scheduled stop.
(2) The train travels at its top speed for exactly 75% of the trip.
Kudos for a correct solution.
Total Distance = 320 Miles
Total Time = 7 hours (7 AM to 2 PM) = 7*60 = 420 minutes
Time of Break = 2*10 = 20 minutes
Travel time to cover 320 Miles = 420 - 20 = 400 Mins
Max. Speed = 60 mph
Min speed required throughout to travel 320 miles in 400 mins = 320*60/400 = 48 mph
Question : Is Time taken to cover distance of 320 miles = 400 minutes ?Statement 1: The train experiences a 30 minute delay in town B in addition to its scheduled stop.Time taken by train when it travels the complete distance at max speed = 320*60/60 = 320 mins
including 10 mins break, total time = 320+20 = 340 Mins
Maximum delay acceptable = 420 - 340 = 80 mins
i.e. a train can reach on time even with 80 mins delay if it travels at top speed for all the time
30 mins delay may cause delay or on time journey. Hence,
NOT SUFFICIENT
Statement 2: The train travels at its top speed for exactly 75% of the trip.75% of Trip = 320*75/100 = 240 miles
Time to travel 240 miles at Top speed = 240/60 = 4 hours = 240 mins
But Min speed is unknown which may or may NOT have caused any delay, Hence
NOT SUFFICIENT
Combining the two statements:240 mins exhausted to travel 240 miles
Remaining travel time = 400-240 = 160 mins
Remaining distance = 320-240 = 80 miles
Speed required to travel remaining distance in remaining time to be on schedule = 80*60/160 = 30 miles/hr
But Minimum Speed is unknown. Hence
NOT SUFFICIENT
Answer: Option E