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Speed, Time And Distance Test - 2

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Speed, Time And Distance Test - 2
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  • Question 1
    3 / -1

    A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?

    Solution

    If the distance between house and office is d km
    Then time taken to go from home to office = d/3 hr (Time = Distance/Speed)
    and Time taken to come back from office to home= d/2 hr
    So total time taken = (d/3 +d/2) hour
    But according to question , time taken = 5 hour
    so we have
    d/3+d/2=5
    or (2d+3d)/6= 5
    or 5d/6=5
    or, d=5 x6/5= 6km
    So distance between home and office = 6km

  • Question 2
    3 / -1

    A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?

  • Question 3
    3 / -1

    A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?

    Solution

    Let the time in which he travelled on foot = x hour
    Time for travelling on bicycle = (9 - x) hr

    Distance = Speed * Time, and Total distance = 61 km
    So,
    4x + 9(9-x) = 61
    => 5x = 20
    => x = 4

    So distance traveled on foot = 4(4) = 16 km

  • Question 4
    3 / -1

    It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. What is the ratio of the speed of the train to that of the car?

    Solution

    Let the speed of the train be x km/hr and that of the car be y km/hr.
    It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car.
    120/x + 480/y = 8 or 1/x + 4/y = 1/15
    If 200 km is done by train and the rest by car, then 20 minutes more time is taken.
    200/x + 400/y = 25/3 or 1/x + 2/y  = 1/24
    Solving both the equations,
    x = 60 and y = 80
    Ratio is 3 : 4

  • Question 5
    3 / -1

    In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. What is the duration of the flight ?

    Solution

    option "D"

  • Question 6
    3 / -1

    If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?

    Solution

    Distance he could travelled/speed diff.
    = 20/(14-10)
    = 20/4
    = 5 hrs
    Now his actual speed was 10 km/h
    Total distance travelled by him = speed × time
    = 10 × 5
    = 50 km.
     

  • Question 7
    3 / -1

    A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?

    Solution

    Total Distance = 10 +12 = 22 km
    Total time = 10/12 + 12/10 = 5/6 + 6/5 = 25/30 + 36/30 = 61/30 hours
    Average speed = 22 /(61/30) = 660/61 = 10.8 km/ hour

  • Question 8
    3 / -1

    The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, What is the the speed of the first train?

  • Question 9
    3 / -1

    Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?

    Solution

    Let the distance be x
    Let the original speed be y
    Time originally taken = distance / speed
    That means = x/y
    Now the new speed is (6/7)y
    And new time is distance /new speed
    x/(6/7)y that can be written as 7x/6y
    Now new time - original time = 12 min
    7x/6y-x/y=12
    After subtracting we get
    x/y = 12×6
    x/y = 72min

  • Question 10
    3 / -1

     

    Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B and C?

    Solution

    Let track length be equal to T. When A completes a lap, let us assume B has run a distance of (t - d). At this time, C should have run a distance of (t - 2d).

    After 3 laps C is in the same position as B was at the end of one lap. So, the position after 3t - 6d should be the same as t - d. Or, C should be at a distance of d from the end of the lap. C will have completed less than 3 laps (as he is slower than A), so he could have traveled a distance of either t - d or 2t - d.

    => 3t - 6d = t - d
    => 2t = 5d
    => d = 0.4t

    The distances covered by A, B and C when A completes a lap will be t, 0.6t and 0.2t respectively. Or, the ratio of their speeds is 5 : 3 : 1.

    In the second scenario, 3t - 6d = 2t - d => t = 5d => d = 0.2t.

    The distances covered by A, B and C when A completes a lap will be t, 0.8t and 0.6t respectively. Or, the ratio of their speeds is 5 : 4 : 3.

    The question is " Find the ratio of the speeds of A, B and C?"
    The ratio of the speeds of A, B and C is either 5 : 3 : 1 or 5 : 4 : 3.

    Hence, the answer is 5 : 4 : 3

     

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