Jake left point A for point B. 2 hours and 15 minutes later, Paul left A for B and arrived at B at the same time as Jake. Had both of them started simultaneously from A and B travelling towards each other, they would have met in 120 minutes. How much time (hours) did it take for the slower one to travel from A to B if the ratio of speeds of the faster to slower is 3:1?
202.5 minutes
205.4 minutes
208.5 minutes
201.5 minutes
Explanation:
It seems there is some problem with this question.
Let the distance between A and B is D km. As Paul is faster, take the speeds of Jake and Paul
are s and 3s kmph.
As the speeds are in the ratio of 1 : 3, times taken by them should be 3 : 1.
Take the times taken by them are 3x , x. But We know that 3x - x = 2 hour 15 min. So 2x = 9/4 hours, x = 9/8 hours.
So time taken by the slower one (Jake) takes 3x time = 3 x 9/8 = 27/8 hours = 202.5 minutes.
It seems there is some problem with this question.
Let the distance between A and B is D km. As Paul is faster, take the speeds of Jake and Paul
are s and 3s kmph.
As the speeds are in the ratio of 1 : 3, times taken by them should be 3 : 1.
Take the times taken by them are 3x , x. But We know that 3x - x = 2 hour 15 min. So 2x = 9/4 hours, x = 9/8 hours.
So time taken by the slower one (Jake) takes 3x time = 3 x 9/8 = 27/8 hours = 202.5 minutes.