CAT Question Bank - Time, Speed & Distance

  • Q20) Sachin and Sunil started with a speed of 40 km/hr and 18 km/hr respectively on their bikes for Patna at 8 a.m. from the same place. Sunil completed the entire journey with the same speed with which he had started. At 9:30 a.m., Sachin was stuck in traffic and had to reduce his speed. So in the next 30 minutes, he could cover only 15 km and covered the remaining distance with a speed equal in magnitude to the total time taken by Sunil in hours to cover the entire journey and reached Patna at the same time as Sunil and in broad daylight. At what time did both of them reach Patna?

  • Q21) Vijay went from A to B at a certain speed and came back to A from B at a speed that was 4 times the speed at which he went from A to B. The entire journey took 3 hr. Had Vijay come back at a speed that was only half the speed at which he actually returned, how long would it have taken him for the entire journey?

  • Q22) Two motorcyclst Ajay and vijay, start Simultaneously from a point S on an oval track and drive around the track in the same direction, with speeds of 29 kmph and 19 kmph resp. Every time ajay overtakes vijay (anywhere on track); both of them decrease there speed by 1kmph. If the length of the track is 1 Km. How many times do they meet at the starting point before vijay comes to rest?

  • Q23) A husband used to pick-up his wife at 4 pm and reach home at 6 pm. One day he got late and called his wife at 4 pm and asked her to start walking. He left at 4.10 pm and met her in between and was able to reach home at 6 pm itself. What is the ratio of speed of wife : husband ?

  • Q24) A man starts from point A and walks northwards at the speed of 6 km/hr. He then turns right and walks eastwards at the speed of 4 km/hr, turns right again and walks southwards at the speed of 8 km/hr and turns left and walks eastwards at the rate of 6 km/hr. The time he walks in various directions is inversely proportional to the speed with which he walks. If the total distance he covers is equal to 100 km, then find how far he is from the starting point.

  • Q25) Ramesh, Ganesh, Mahesh, Jignesh decide to run around a circular garden which is 2500 m long at respective speeds of 18, 9, 72 and 36 km/hr. If they start at the same time from the same point with Ramesh and Mahesh running in same direction and Ganesh and Jignesh in opposite directions when will all four meet again at the starting point?
    a) 16.5 min
    b) 800 sec
    c) 15 min
    d) 15.5 min
    e) All 4 won’t meet at the same time

  • Q26) R, C and S run on a linear racetrack. If R beats C by 20 seconds or 180 meters, C beats S by 40 seconds or 240 meters, what is the ratio of speeds of R and S?
    a) 2 : 1
    b) 1 : 2
    c) 3 : 2
    d) 4 : 3

  • Q27) By the schedule, students in a school will visit a farm and the farm's bus will arrive at the school at 7:00 AM to take the students. Because the bus has a trouble on the way from the farm to the school, the bus stops to repair. At 7:10 AM, students begin to walk to the farm. On the way, they meet the incoming repaired bus and get on it. In this way, they arrive at the farm 30 minutes later than the schedule. If the bus' speed is 6 times walking speed, how many minutes are used to repair the bus on the way?

  • Q28) On Monday, Lou drives his ford escort with 28-inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his car to 32-inch tires yet drives to work at the same average speed as on Monday. What is the percent change from Monday to Tuesday in the average number of revolutions that Lou's tires make per second?
    a) Decrease by 14.3%
    b) Decrease by 12.5%
    c) Increase by 14.3%
    d) Increase by 12.5%
    e) Cannot be determined with the given information

  • Q29) Two cars start simultaneously from cities A and B, towards B and A respectively, on the same route. Once the two cars reach their destinations they turn around and move towards the other city. The two cars continue shuttling in this manner for exactly 120 hours. If the speed of the car starting from A is 80 km/hr and the speed of car starting from B is 60 km/hr and the distance between the two cities is 240 km, find the number of times the two cars meet in this duration, after they start. (Assume that throughout the journey, cars from cities A and B travel at their respective uniform speeds)
    a. 26
    b. 30
    c. 25
    d. None of these

  • Q30) A and B start swimming simultaneously from opposite ends i.e Deep end n shallow end of a swimming pool with speeds 5m/s and 2 m/s. They swim n and after reaching opposite ends, they change the directions and continue swimming with their respective uniform speeds. When they meet for the 8th time, they meet at a distance of 10 meters from the shallow end. if the race is of the 200 rounds,
    a) How many times will they meet
    b) Find the length of the swimming pool

  • @rowdy-rathore 38 minutes

  • @rowdy-rathore 38 min correct

  • @rowdy-rathore could you explain how did you solve it?

  • @abhisd
    Let the total distance b/w home and school be d, and the point they meet (say M) is at a distance x from the school. Let t be the time taken to cover distance d if they travel in car. We will take speed of car as s and walking speed as (s/6).
    [H] -----< (d - x) >----< [M] >------< x >----- [S]
    we know, d/s = t --- (1)
    In the given situation, son walked till M (distance of x) and father drove a distance of (d - x) after they meet. And they reached 30 minutes late by doing so. Remember 10 min was lost in waiting, so 20 mins was lost while travelling.
    so, x/(s/6) + (d - x)/s = t + 20
    6x/s + d/s - x/6 = t + 20
    5x/s + d/s = t + 20 --- (2)
    (2) - (1) = 5x/s = 20
    x = 4s
    So to cover distance x by walking it will take x/(s/6) = 4s/(s/6) = 24 minutes. Car would take 24/6 = 4 minutes to cover this part and hence car has saved 4 + 4 = 8 minutes (onward and return) because of the son walking this part.
    Total time lost = Time lost in repair - Time saved by walking
    30 minutes = Time lost in repair - 8
    Time lost in repair = 38 minutes.

    Alternatively, a much faster thought process by Chandra sir (Takshzila)

    If son walked for 6t minutes, then the meeting of car and son happens at (10 + 6t) mins past 5:00 am.
    In the event of no fault, car would have been at same point, going back, at t mins past 7:00 am.
    Thus, (10 + 6t) - t = 30 i.e. t = 4
    The bus is 30 mins late after having saved 4 + 4 = 8 mins by not travelling the distance walked by students.
    Repair took 38 minutes.

  • @rowdy-rathore

    9 hours...................................

  • @rowdy-rathore


  • @rowdy-rathore

    d ..............70

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