Quant Boosters - Rajesh Balasubramanian - Set 4



  • Now, let us try to jot down the points in a slightly different format and see if we can make any inferences. Let us assume speed of train P = p, speed of train Q = q, length of train P = L, length of train Q = M

    Take statements 1 and 3

    Train P crosses the stationmaster A entirely in 20 seconds (enters station at 8:00:00 and last carriage passes at 8:00:20) => Length of train = 20p => L = 20p

    Statement 5 tells us that engine of train Q crosses station master A at 8:00:55. Train Q enters the station at 8:00:10, so the train takes 45 seconds to cross the entire station. Or, it takes 45 seconds to travel 900 meters => Speed to train Q, q = 20m/s.

    Statement 3 tells us that the two trains cross each other at 8:00:22. This implies train P has traveled for 22 seconds since entering the station and train Q has traveled for 12 seconds since entering the station before they cross each other. The cumulative distance traveled by the two trains should be equal to the length of the station = 900m.

    => 900 = 22p + 12q

    q = 20m/s => p = 30m/s.

    So, train P_ takes 900/p = 30 seconds to cross the station. So, engine of train P will cross stationmaster B at 8:00:30.

    Statement 4 states that the last carriage of Q went past 22 seconds after the engine of P went by. Or, the last carriage of Q went by at 8:00:52.

    Engine of train Q went by at 8:00:10, last carriage went by at 8:00:52, or train Q took 42 seconds to cross station master B. Train Q travels at 20m/s. => Length of train Q = 840m

    Now, to the questions

    Q1) What is the length of train Q?

    Train Q is 840m long

    Q2) At what time do the rear ends of the two trains cross each other?

    The engines cross each other at 8:00:22. The relative speed of the two trains = 20+30 = 50m/s. The relative distance traveled by two trains from the time the engines cross each other to the times the rear ends cross each other = Sum of the two lengths = 600 + 840 =1440. Time taken = 1440/50 = 28.8 seconds past 8:00:22, or at 8:00:50.8 seconds.

    Q3) How far from station master A do the rear ends of the two trains cross each other?

    At 8:00:50.8, P would have traveled 50.8 * 30m/s post entering the station. Or, train P would have traveled 1524m. The rear end of train P would be at a point 1524-600m = 924m from stationmaster A (or 24 meters from stationmaster B and outside the station)

    Q4) You are told that the two trains enter the station at the same times mentioned and the length of the two trains are unchanged. Furthermore, train P continues to travel at the same speed (as computed above). At what minimum speed should train Q travel such that the rear ends of the two trains cross each other at a point within the length of the platform?

    Rear end of train P crosses the station completely at 8:00:50. (Train P takes 30 seconds to travel the station and 20 seconds to travel a distance equal to its length). Train Q should have traveled 840m by this time. => Train Q should travel 840 within 40 seconds.

    Or, minimum speed of Q = 840/40 = 21m/s.



  • we can do this another way too:
    The time in which antony covers 20 m, amar will cover 20 + 12 - 14 = 18 m
    => (20/antony) = (18/amar)
    => amar = (18/20) antony = (9/10) antony
    Choice a .



  • As C is at some fixed distance say x from B

    shouldnt the distance at time t include x also?

    D= sqrt((200-40t)^2+ (x-20t)^2)???



  • In every meet their combined distance travelled is 100 m.

    Let there be "a" meets.

    =>[(100a) * 5/(5+3)]=300
    Which gives a=4.8

    Though it cant be decimal.

    But What is wrong with this approach????

    As there is a similar question in gyan room-arithmetic shortcuts...and i hv applied the same approach here..



  • @rajesh_balasubramanian Hello. I was just looking for an online CAT coaching and stumbled upon this. The answer seems to be 7. And i don't think 10 can be a value for B. Eg: Take the Track length as 300m. A will complete it in 50secs. And B will take 75 secs to overlap A for the first time(300/4). I can email you my solution if you'd like to give it a look.
    Thank you.


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