Re: An express train leaves New York at 3pm and reaches Boston at 6pm.
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17 Jul 2023, 07:09
An express train leaves New York at 3pm and reaches Boston at 6pm. A slow train leaves Boston at 1.30pm and arrives at New York at 6pm. If both trains travel at constant speeds, at what time do they meet?
There are 2 trains into consideration
Express train (\(T^E\))
Slow Train (\(T^S\))
Now Start by assuming the distance to be same as D.
Scenario 1.
\(T^E\) starts at 3 pm and ends journey of distance D at 6pm
Therefore effectively taking 3 hrs to cover a distance of D.
Speed (\(T^E\)) = D/3 kmph.
Scenario 2 :
\(T^S\) starts journey from opposite end at Boston and completes distance of D in 4.5 Hrs.
Hence, Speed of \(T^S\) = D/4.5 = 10/45 D = 2/9 D in Kmph
Now we are asked that when will they meet.
First note that \(T^S\) is moving at a time earlier than \(T^E\).
the time difference is that \(T^S\) moves at 1.30 pm while \(T^E\) moves at 3.00 pm
making a effective time difference of 1.5 hrs between there movement. in this duration \(T^S\) travels S*T = 10/45*D * 1.5 Km = D/3 km
So D - D/3 = 2D/3 is left.
Now at 3.00 pm \(T^E\) also starts its movement in the opposite direction as to \(T^E\).
As the time bracket is same, we use relative speed = speed \(T^S\) + speed \(T^E\) = D/3 + 2D/9 = 5D/9 Kmph
We also know that the distance left is 2D/3 do Time in which they will meet = D/S = 2D/3/5D/9= 6/5 hrs = 1.20 hrs or 1hr + 0.2*60 mins = 1hr + 12 mins.
This time is from the original 3.00 pm.
So IMO they will meet at 3.00 pm + 1hr + 12 mins = 4.12 pm.
Hence B