New Upcoming Legal Jobs Free Workshop March 11 at 5 PM. Register here

    Question

    Three persons named P, Q, R started their journey at

    9.30 AM, 11.30 AM, 2 PM respectively from A to B. The speed of Q is √625 m/sec. Q reached the destination at 7.30 pm on the same day. Q is fastest & R is not the slowest. The speed of P is 20% less than the speed of Q, then in what time P will reach the destination? i. 6 hours more than Q ii. P will reach B at 7.30 pm iii. The total reaching time of P and Q is 18 hrs
    A Only i Correct Answer Incorrect Answer
    B Only ii Correct Answer Incorrect Answer
    C Only ii and iii Correct Answer Incorrect Answer
    D All i, ii and iii Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    Speed of Q = √625 m/sec = 25 m/sec = 25 x 18/5 = 90 km/hr Time taken to reach point B by Q = 8 hrs (11.30 am – 7:30 pm) Distance between A and B = 90 x 8 = 720 km Speed of P = 0.8 x 90 = 72 km/hr Now, From i: Time taken by P to cover the distance = 720/72 = 10 hrs Difference of time is 2 hrs, so i is not true. From ii: P started journey at 9:30 AM, so he will reach by 7:30 pm (10 hrs) So , ii is true. From iii: Total time  =  8 + 10 = 18 hrs iii is true.

    Practice Next

    Relevant for Exams: