Manaswita wrote:
Train A leaves New York at 9am Eastern Time on Monday, headed for Los Angeles at a constant rate of 40 miles per hour. On the same 3,000-mile stretch of track, Train B leaves Los Angeles at noon Eastern Time on Monday, traveling to New York at its constant rate of 60 miles per hour, with one exception: Train B is delayed for exactly 2 hours in Las Vegas (approximately 200 miles from Los Angeles) due to track maintenance. Assuming the time it takes for Train B to decelerate and re-accelerate when stopping and resuming in Las Vegas is negligible, at what time (Eastern Time) will the two trains meet?
A) 3:00pm on Tuesday
B) 4:30pm on Tuesday
C) 5:00pm on Tuesday
D) 6:00pm on Tuesday
E) 7:30pm on Tuesday
There are various ways to do this problem.
Method 1: We need to find at what time they meet. Say they meet t hrs after train A starts at 9 AM EST.
Train A runs for t hrs. Train B runs for (t - 5) hrs since it starts 3 hrs late and takes a break of 2 hrs in between (It doesn't matter whether the break was taken 10 mins into the journey, 1 hr or 2 hrs as long as it was early enough so that the break was over before they met).
Together they covered 3000 miles.
40 * t + 60*(t - 5) = 3000
t = 33 hrs
That gives us 6 pm Tuesday.
Method 2: Explanation of Bunuel's method above that you (@teenavijay) are referring toIt doesn't matter WHERE the break was taken as long as it was early enough. For the rest of time, train B ran at the steady speed of 60 mph. So let's assume that the break was taken at the starting point itself such that train B started 5 hrs after train A. Think about it logically:
I start from home towards your home at 9 AM and walk consistently.
In one case, you start from your home at 12 noon, walk for 5 mins, then rest for 2 hrs and then walk consistently towards me. We meet at 6 PM.
In another case, you start from your home at 12 noon, walk for 10 mins, then rest for 2 hrs and then walk consistently towards me.
In yet another case, you start from your home at 12 noon, walk for 1 hr, then rest for 2 hrs and then walk consistently towards me.
Are the cases different? No. What part of the journey you covered at what time is irrelevant. You covered that distance at your consistent speed.
Hence, we can shift the break to the starting point and solve from there.
Now the question becomes:
Train A leaves New York at 9am Eastern Time on Monday, headed for Los Angeles at a constant rate of 40 miles per hour. On the same 3,000-mile stretch of track, Train B leaves Los Angeles at 2 PM Eastern Time on Monday, traveling to New York at its constant rate of 60 miles per hour. At what time (Eastern Time) will the two trains meet?
From 9 AM to 2 PM, in 5 hrs, train A travels 5 * 40 = 200 miles.
So total distance they need to travel from 2 PM onwards is 3000 - 200 = 2800 miles
Time taken = 2800/(40 + 60) = 28 hrs
28 hrs after 2 PM Monday again gives us 6 PM Tuesday.
Method 3: We can map their journey. Train A simply travels at 40 mph starting at 9 AM.
Train B starts at 12 noon at 60 mph and travels 200 miles. It reaches Vegas after 200/60 = 20/6 hrs.
It waits there for 2 hrs so after 20/6 + 2 = 16/3 hrs, both trains travel interrupted towards each other.
In these (3+16/3) hrs, train A covered (3 + 16/3)*40 = 120 + 640/3 miles.
So after (3+16/3) hrs, distance between them is 3000 - 200 - 120 - 640/3 = 2680 - 640/3 miles
Time taken = Distance between them/Relative speed = (2680 - 640/3) / (40 + 60) = (268/10 - 64/30) hrs
Hence total time taken = 3 + 16/3 + (268/10 - 64/30) hrs = 33 hrs
That gives us 6 pm Tuesday.
Hope you see that the intent of this question is to make you realise that the time of break is irrelevant as long as it is early enough. Method 3 is a nightmare.