Speed, Time and Distance Questions for CAT with Answers PDF

Since Chandu is moving at a speed of 10 m/s and he has to cover 360 km or 360000 meters,
the time taken would be given by 360000/10 seconds = 36000 seconds = 36000/60 minutes = 600 minutes = 10 hours.

Total distance/Total time = 1590/15 = 106 kmph. 

The distance covered in the various phases of his travel would be: 10 km + 25 km + 50 km + 60 km.
Thus the total distance covered = 145 km in 2 hours 50 minutes → 145 km in 2.8333 hours → 51.18 kmph

By increasing his speed by 25%, he will reduce his time by 20%.
(This corresponds to a 6 minute drop in his time for travel—since he goes from being 10 minutes late to only 4 minutes late.)
Hence, his time originally must have been 30 minutes.
Hence, the required distance is 20 kmph × 0.5 hours = 10 km.

If the car does half the journey @ 30 kmph and the other half at 40 kmph it’s average speed can be estimated using weighted averages.
Since, the distance traveled in each part of the journey is equal, the ratio of time for which the car would travel would be inverse to the ratio of speeds.
Since, the speed ratio is 3:4, the time ratio for the two halves of the journey would be 4:3.
The average speed of the car would be: (30 × 4 + 40 × 3)/7 = 240/7 kmph. It is further known that the car traveled for 17.5 hours (which is also equal to 35/2 hours).
Thus, total distance = average speed × total time = (240 × 35)/(2 × 7) = 120 × 5 = 600 km

You can solve this question using the options. Option (a) fits the given situation best as if we take the distance as 12 km he would have taken 1 hour to go by car and 4 hours to come back walking—a total of 5 hours as given in the problem.

In 6 minutes, the car goes ahead by 0.6 km.
Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph, since, the pedestrian is walking at 2 kmph, hence, the net speed is 8 kmph.

At 40 kmph, Harsh would cover (200/60) × 40 km. = 400/3 km. = 133.33 km.
This represents the distance by which Vijay would be ahead of Harsh, when Vijay reaches the endpoint means in essence that Vijay must have travelled for 133.33/20 hours → 6.66 hours Hence, the distance is 60 × 6.66 = 400 km.

The required speed s would be satisfying the equation:
300/s – 300/(s + 5) = 2
Solving for s from the options it is clear that s = 25.

By increasing the speed by 33.33%, it would be able to reduce the time taken for travel by 25%.
But since this is just able to overcome a time delay of 30 minutes, 30 minutes must be equivalent to 25% of the time originally taken.
Hence, the original time must have been 2 hours and the original speed would be 750 kmph.
Hence, the new speed would be 1000 kmph.

The average speed would be given by:

The total time taken by the motorist would be 200/53.333 = 200 × 3/160 = 3.75 hours = 3 hours 45 minutes.
In the first half of the journey the motorist covers 1/4^th the distance @ 40kmph.
This means that he takes 50/40 = 1.25 hours = 1 hour 15 minutes in covering the first 50 kms.
This also means that he covers the remaining distance of 150 km in 2 hours 30 minutes → a speed of 60 kmph.
Hence, option (b) is correct.

The speed of the first car would be 60 kmph while the speed of the second car would be 40 kmph.
The relative speed of the two cars would be 100 kmph.
To cover 480 km they would take 480/100 = 4.8 hours → In 4.8 hours, the car traveling from A to B would have traveled 4.8 → 60 = 288 kms.

The distance would get divided in the ratio of speeds (since time is constant).
Thus, the distance ratio would be 5 : 7 and required distance = 5/12 × 600 = 250 km.

The diameter of the circle would be given by the hypotenuse of the right triangle with legs 600 and 800 respectively.
Hence, the required diameter = 1000 meters.

From the situation described in the first condition itself we can see that the speed of coming back has to be double the speed of going downstream.
Checking the options, only option (a) fits this condition i.e. Downstream speed = 2 × Upstream speed.
Hence, option (a) is correct.

When speed goes down to three fourth (i.e. 75%) time will go up to 4/3rd (or 133.33%) of the original time.
Since, the extra time required is 16 minutes, it should be equated to 1/3rd of the normal time.
Hence, the usual time required will be 48 minutes.

The train that leaves at 6 am would be 75 km ahead of the other train when it starts.
Also, the relative speed being 36 kmph, the distance from Mumbai would be:
(75/36) × 13 6 = 283.33 km

From the figure above we see that Shyam would have walked a distance of 4 + 4 + 4 = 12 km. (G to P₁, P₁ to G and G to P₂).

When his speed becomes 3/4th, his time would increase by 1/3rd. Thus, the normal time = 7.5 hrs. (since increased time = 2.5 hrs).