theperfectgentleman wrote:

A train after travelling 50 km meets with an accident and as a result, it proceeds at 3/4 of its former speed

and arrives at its destination 35 minutes late. Had the accident occurred 24 km further the train would have reached its

destination 25 minutes late. The original speed of the train is:

(1) 48 km/h

(2) 36 km/h

(3) 54 km/h

(4) 58 km/h

(5) 50 km/h

The algebraic solution for this question is pretty long and inefficient. Tried the following logic and my answer is wrong.

Case 1: Accident @ 50 Kms

-----------A1-------------

Case 2: Accident

24 Kms

----A2--------------------

Difference in time = 10 mins

Difference between A1 and A2 = 26 Kms

Ratio of speeds between case 1 and case 2 is 1:3/4 or, 4:3

Ratio of time (Inverse) 3:4

Time saved is 4-3 = 1 part; 1 part leads to difference of 10 mins

Time taken in case 1,

3 parts = 3*10= 30 mins

Original speed of train: 26/(30/60) = 42Kmph

Can someone highlight what is wrong in my approach?