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1.

Train X leaves Dhaka at 10:00 AM and travels East at a constant speed of x miles per hour . If another Train Y leaves Dhaka at 11:30 AM and travels east along the same tracks at speed 4x/3 , then at what time will Train Y catch Train X ?

  • A. 4 PM same day
  • B. 4 PM same day
  • C. 4 PM same day
  • D. 4 PM same day

Answer: Option A

Explanation:

For rate - time - distance questions, try to reduce the question to as few unknown quantities as possible. Instead of a separate r, t, and d for each train, look for commonalities:
Both trains travel the same distance, so there's only one "d"
We don't know the time, but we know that Train Y took less time, since it left an hour and half later. So if Train X took t hours to travel some distance, Train Y only took (t - 1.5) hours.
The rates aren't given, but we know that train Y was 4/3 the rate of train X.
Let's set it up. Since rate*time = distance and the distances are equal, we can set Train X's rate * time equal to Train Y's rate * time
xt = 43x(t−1.5)xt = 43x(t−1.5)
Working through the problem, we get
xt = 43xt−4/3∗1.5xt = 43xt−4/3∗1.5 here it can be helpful to recognize that 1.5 is just 3232, and cancel out the 3's: 43∗32 = 243∗32 = 2
xt = 43xt−2xt = 43xt−2
Subtract xt from both sides, and add 2 to both sides, and we get: 2 = 13t2 = (1/3)t
6 = t
So, adding 6 hours to Train X's starting time, we get 4:00 pm as the time that train Y will catch train X.

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