Time, Speed and Distance - Solved Questions - Part 1


  • Converted IIM Indore call | Mentor for Banking/RBI/SSC exams


    Concepts:

    If a body travels d1,d2..dn distances with speeds s1,s2..sn in time t1,t2…tn respectively, then the average speed of the body through the total distance is given by
    Average speed = Total distance travelled / Total time taken
    = (d1+d2…..dn)/(t1+t2+….tn)
    = (d1+d2+…..dn)/(d1/s1+d2/s2+….dn/sn)

    While travelling a certain distance d, if a man changes his speed in the ratio m:n, then the ratio of time taken becomes n:m.

    If a certain distance (d) say from A to B is covered at ‘a’ km/hr and same distance is covered from B to A in ‘b’ km/hr then average speed during the whole journey is given by
    Average speed = 2ab/(a+b) km/hr

    If two persons A and B start at the same time in opposite directions from two points and arrive at the two points in ‘a’ and ‘b’ hours respectively after having met, then
    A’s Speed/B’s speed = root(b)/root(a)

    Time taken by a moving object ‘x’ meters long in passing a stationary object of negligible length from the time they meet is same as the time taken by moving object to cover x meters with its own speed.

    Time taken by a moving object x meters long in passing a stationary object y meters long from the time they meet, is same as the time taken by moving object to cover x+y meters which its own speed.

    If two objects of length ‘x’ and ‘y’ meters move in the same direction at ‘a’ and ‘b’ m/s then the time taken to cross each other from the time they meet = Sum of their length/Relative speed
    = (x+y)/(a-b) if a>b -> else (x+y)/(b-a)

    If two objects of length ‘x’ and ‘y’ meters move in the opposite direction at ‘a’ and ‘b’ m/s then the time taken to cross each other from the time they meet = Sum of their length/Relative speed = (x+y)/(a+b)

    Questions

    Travelling at a speed of 4 km/hr from his home Rakesh reaches his office 45 minutes late. Next day Rakesh travels at a speed of 5 km/hr from his home and reaches his office on time. What is the distance between his home and office?

    Let D is the distance between his home and office.
    D/4 - D/5 = 3/4
    5D - 4D = 20 * 3/4 = 15
    D = 15 KM

    The ratio between the speeds of two trains is 4:5. If the first train runs 200 km in 2 hours, then the speed of the second train is

    Speed of first train = 200/2 = 100 KM/HR
    Hence speed of second train = 100 * 5/4 = 125 KM/HR

    A person goes from x to y at 20 kmph and comes back to x at 30 KMPH . Find his average speed?

    Average speed = 2ab/a+b
    a = 20kmph b= 30 kmph
    Average speed = 2 * 20 * 30/50 = 24 kmph

    A man performs (4/7)th of the total journey by train, (5/21) th by car and remaining 8 km on foot. His total journey is?

    Let Total journey is X km
    4x/7 + 5x/21 +8 = x
    12x+5x-21x = -8 *21
    -4x = -8 * 21
    x = 42 KM

    If a boy goes to school at 6km/hr and returns home at 4Km/hr. Find his average speed.

    Average speed = 2ab/(a+b)
    a = 6 Km/hr b = 4km/hr
    Average speed = 2 * 6 * 4/10 = 4.8 Km/hr

    A man starts from A to B, another from B to A at the same time. After passing each other they complete their journeys in 3 1/3 and 4 4/5 hours respectively. Find the speed of second man if the speed of the first is 12km/hr.

    First man’s speed/Second man’s speed = root(4 4/5)/root(3 1/3)
    = root(24/5) / root(10/3) = root(72/50) = root(36/25) = 6/5
    Second man’s speed = 5 * 12/6 = 10 Km/hr

    I shall be 40 min late to reach my office if I walk from my house at 3Km/hr. I shall be 30 min early if I walk at 4 km/hr . Find the distance between my house and the office.

    Let the usual time taken be ‘t’ and speed be ‘x’
    Distance = xt = 3*(t+2/3) = 4(t-1/2)
    3t+2 = 4t-2
    t = 4 hours
    Distance = 3(4+2/3) = 3 * 14/3 = 14KM

    A train after travelling 50 KM from A meets with an accident and proceeds at (4/5)th of the former speed and reaches B 45 min late. Had the accident happened 20 KM further on it would have arrived 12 min sooner. Find the original speed and distance.

    Let original speed of train is S
    20/(4S/5) – 20/S = 12/60 = 1/5
    100-80 = 4S/5
    4S = 100
    S = 25 KM/HR
    Let the original distance is D
    D-50/20 - D-50/25 = 3/4
    5D-250-4D+200 = 75
    D = 125 KM

    A train travelling at 25KM/HR leaves delhi at 9 am and another leaves Delhi at 35 Km/hr at 2 PM in the same direction. How many kms from Delhi do they meet?

    At 2 PM first train must have travelled 5 * 25 = 125 KM
    Relative velocity = 10 KM/HR
    Time taken to meet = 125/10 = 12.5 hours from 2PM
    Distance = 35 * 12.5 = 437.5 km

    Two trains 121 meters long and 99 meters long are running in opposite direction, the first at 40 KM/HR and the second at 32 KM/HR. In what time will they completely clear each other from the moment they meet?

    Total distance to be travelled = 121+99 = 220 meters
    Relative speed = 72 km/hr = 72*5/18 = 20 m/s
    Time required to clear each other = 220/20 = 11 seconds

    A train moving at uniform speed takes 20 sec to pass a cyclist riding at 11 km/hr but only 9 sec to pass a post. Find the length of train.

    Let length of train is is ‘L’ km and speed ‘S’ Km/hr
    L/X-11 = 20/3600 = 1/180
    L/X = 9/3600 = 1/400
    X = 400 L
    180 L = X – 11
    180L = 400L-11
    220L = 11
    L = 11/220 = 1/20 KM = 1000/20 =50 meters


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