Question Bank - 100 CAT level questions on Time, Speed and Distance topic



  • Q100) Rishi Kapoor can swim a certain course against the river flow in 84 minutes; he can swim the same course with the river flow in 9 minutes less than he can swim in still water. How long would he take to swim the course with the river flow?



  • total steps be T
    T/N -T/(N+4) =6
    3N^2+12N-2T=0 N=-12+rt (144+24T) /6 --(1)
    and T/(G-4)-T/G = 6
    3G^2 -12G-2T=0
    G= 12+rt (144+24T) /6 ---(2)
    relative speed = N+G
    so T= 4N+4G --(3)
    using 1 and 2,
    T = 4 * rt (144+24T)/3
    9T^2 = 16 (144+24T)
    9T^2 - 384T-2304 = 0
    T = (384 + 480)/18 = 48 steps



  • P's distance be Y, Q's be X
    Y/50 = (X-50)/25
    Y=2X-100
    say Y=100, X=100
    distance at any time=d
    d^2 = (100-50t)^2 + (100-25t)^2
    diff wrt t for minima
    (100-50t)*2 +(100-25t) = 0
    t=12/5
    d^2 = 20^2 + 40^2 = 2000 so d= 20 rt 5



  • Train speed= T, men speed: v1, v2. length L.
    so L/(T-v1)=20 and L/(T+v2) = 18
    so 20T-20v1=18T+18v2
    T= 9v2+10v1 --(1)
    Ditsane between the men = 600 (T+v2)
    Distance covered in these 10 mins = 600(v1+v2)
    so reqd time = {600(T+v2)-600(v1+v2)}/(v1+v2)
    = (6000-600)= 5400 sec= 90 mins ( using 1)



  • A=2x, B=x
    length= 2pi*r = 49pi = 154 m.
    so given, 154/3x = 14 x=11/3 so 2x=22/3



  • @mbatious answer 35?





  • @mbatious 9 hour 54 min is it?



  • @mbatious 6:24 am?



  • Let the distance between Home and School be D and the uniform speed during onward journey be S.
    While coming back, 2D/7 was covered with a speed of 2S/7
    Remaining distance = 5D/7 and 3/5th of this is 3D/7 which is covered with a speed of 3S/7.
    He finally reduces the speed to 3S/7 - 12S/42 = S/7 (as 66.67% is equivalent to 2/3rd) and cover the remaining distance, 2D/7.

    Average speed for the total journey = 14 kmph
    2D/14 = D/S + (2D/7) * (7/2S) + (3D/7) * (7/3S) + (2D/7) * (7/S)
    D/7 = D/S + D/S + D/S + 2D/S
    1/7 = 5/S
    S = 35 kmph



  • @mbatious option d



  • @mbatious 180



  • @mbatious pls provide answer key. asap. so that it is easier for me to prepare.



  • pawan's speed = 4x m/s
    escalator speed = y m/s
    400x - 100y = D
    rishabh's speed = 3x m/s
    120x + 40y = D
    400x - 100y = 120x + 40y
    280x = 140y
    ratio = 2:1
    it means P has to cover twice of step
    P's speed = 4x
    E's speed = 2x
    R's speed = 3x
    so , distance = 154/2 = 77
    77/(2x+5x) = 11/x sec
    so they will meet after 11/x secs that means (11/x) * 2x = 22 steps from bottom



  • D = 40 m
    T = 5 m/ sec
    R = 8 m/s
    1st meet at 25/3 s
    distance by R = 8 * 25/3 = 200/3 = 40 + 2 * 40/3
    had both started from same point then they would meet at exactly 3 points
    2 * 40/3 , 1 * 40/3 and 3 * 40/3
    Time difference between 1st meet and 3rd meet = 2 * 40/3 = 80/3
    total 25/3 + 80/3 = 35



  • @mbatious Could you please provide answer for this



  • @Akshata

    A good set which can give you a headache under exam pressure and if treated with calm, can fetch you 4 correct answers in some 5 minutes.

    First we know distance PM and MQ are in the ratio 1 : 2
    So lets consider PQ = 3d => PM = d and MQ = 2d
    Now let say Kedar took t minutes to cover PM.
    So, A would take t + 50 and B would take t + 25 (as they all started at different time and reached simultaneously) to cover PM.
    So time taken by K to cover PQ = twice the time he took to cover PM = 2t
    Similarly B will take 2t + 50

    K reached Q and ran back 20 more mins to meet with A and B again at S.
    So total time travelled = 2t + 20
    So B will take (2t + 50) - (2t + 20) = 30 minutes to reach Q.
    => B will take 30 minutes to cover what K covered in 20 minutes.
    Ratio of speed = 2 : 3
    (d/t) / (d/t+25) = 2/3
    (t + 25)/t = 2/3
    t = 50

    Question 2:
    So K will take t = 50 minutes to reach M.
    So All will reach M at 7.50 + 50 minutes = 8.40 am

    Question 4:
    If speed of K is 12 kmph = 1/5 km/minute
    Kedar has took t + 2t = 3t = 150 minutes to reach PQ
    So distance PQ = 150/5 = 30 KM

    Question 1:
    Total distance is 30 KM.
    Distance SQ = distance covered by K in 20 minutes = 20/5 = 4 km.
    Speed of A (before increasing it) = 10km/(100 minutes) = 1/10 km/minute
    When K took 100 minutes to reach Q from M, and in this time A would have covered 100/10 = 10 km from M.
    Now as A should cover another 20 - 10 - 4 = 6 km in 20 minutes to reach S with all others.
    A would have increased it speed to 6/20 km/min = 18 kmph.

    Question 3:
    Speed of A = 1/10
    Speed of K = 1/5
    When they meet, together they would have covered a total distance of 2MQ = 2 * 20 = 40 KM
    As ratio of speed is 1 : 2, distance will also be covered in same ratio
    So K will cover 80/3 and A will cover 40/3.
    Time taken by A to cover 40/3 = (40/3) / (1/10) = 400/3 = 133 minutes + 20 seconds
    So K will overtake A again after 8.40 + 133 minutes + 20 seconds = 10:53:20 AM!

    Happy Learning!



  • @mbatious it will be 35



  • @mbatious what is the answer to this team?



  • @mbatious 6mins.


Log in to reply