Question

Two trains travel from A to B and B to A in 6 and 10 hours respectively. They start simultaneously at 10:00 a.m. If the first train meets with an accident and thereafter reduces its speed to half of the original speed, it meets with the second train at 2 p.m. At what time did the accident occurred?
a)1:20p.m. b)1:12 p.m. c)1:35 p.m. d)1:30 p.m
Please solve

Answer 1

Answer:

1 : 12 PM

Step-by-step explanation:

Let say Distance between Station = 60D km

Speed of A = 60D/6  = 10D km/Hr

Speed of B = 60D/10 = 6D km/Hr

Let say after x hr of 10 am accident happened

then for x hrs with speed of 10D km/hr

10-2:00 is 4 hrs and hence for 4-x hrs with 10D/2 = 5 D km/Hr speed

Distance covered by A =  10D * x  +  (5D) *(4-x)

= 10DX + 20D - 5DX

= 5DX + 20D

Distance covered by in 4 hrs = 6D * 4 = 24 D

Sum of both distance =  60D

5DX + 20D + 24D = 60D

=> 5DX = 16D

=> X = 3.2

3.2 hrs = 3 hrs 12 mins

3 hrs 12 min after 10 AM =  1 : 12 PM

option B is correct