A slow train leaves Ferris Creek at 10:15 am and arrives in Sawyers Siding at 1:15 pm. On the same morning, an express train leaves Ferris Creek at 10:45 am and arrives in Sawer Siding at 12:45 pm. At which time does the express train pass the slow train if each is travelling at constant speed?
Options:
a. 12:05pm b. 12:10 pm c. 11:45am d. 12:30 pm
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Answers
Answer:
Now!
Explanatory Answer
Method of solving this CAT Question from Speed Time Distance
Speed Time Distance - Finding the time taken
From question, XZ = 3d, YZ = 2d
As their speeds are in ratio 4:3
Let their speeds be 4V and 3V
Time (T to reach Z) = 3d4v hrs ---(1)
Time (S to reach Z) = 2d3v hrs ---(2)
We need to find the time taken for T To travel from X to Y
Time = 5d4v
Train S started one hour late than Train T
=> 3d4v - 1 = 2d3v
Let dv = k
34k - 23k = 1
k = 12
So, 5d4v = 5×124 = 15 hours
The question is "Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y? [TITA]"
Hence, the answer is 15 hours