A and B have to go from X to Y which are at a distance of 230 kms from each other. A drives at a constant speed of 100 kms/hr for the first 115 kms and then at an average speed of 50 kms/hr for the remaining distance. B starts with an initial speed of 50 kms/hr at X and drives with constant acceleration such that when he reaches Y, his speed is 100 kms/hr. Which of the following statement is true?
- A reaches Y first
- B reaches Y first
- A and B reach Y together
- Cannot calculate from given data
- None of these
The Usual Method
A covers half of the distance at 100 kms/hr and the remaining half at 50 kms/hr. Hence average speed of A = 66.67 kms/hr
Average speed of B for the whole journey = 75 kms/hr
Average speed of B > Average speed of A,
B will reach Y before A does.
Estimated Time to arrive at the answer = 45 seconds.
Note that speed of B follows Arithmetic mean and that of A follows Harmonic mean.
We know that for a given set of numbers, Harmonic Mean is always < Arithmetic Mean
Hence speed of A < speed of B, so B reaches Y before A.
Estimated Time to arrive at the answer = 10 seconds.