Test: Speed, Time & Distance - 4 - SSC CGL MCQ

Distance=600 km

Speed=100 km/hr

Time =100/600​=6hr

Number of stops in 600 km=(600/75​)−1 = 7

Time of stopping after every 75 km = 3  min

Total time of stopping = 7×3 = 21min

Total time to cover 600 km is 6 hr 21 min.

Since, the ratio of speeds is 3:5, the ratio of times would be 5:3. The difference in the times
would be 2 (if looked at in the 5:3 ratio context.) Further, since Ram takes 30 minutes longer, 2
corresponds to 30. Hence, using unitary method, 5 will correspond to 75 and 3 will correspond to
45 minutes. Hence at 10 kmph, Bharat would travel 7.5 km.

Let the two Trains pass after x hours,

then for Calcutta mail, we have

Distance travelled in x hours = 48x kms

Distance travelled after passing Bombay mail = 

Now, For Bombay Mail, we have

Distance travelled in x hours = 576Km

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).

Kalu’s speed = 3 m/s.
For 1200 m, Kalu would take 400 seconds and Sambhu would take 10 seconds less. Hence, 390
seconds.

Since the train travels at 60 kmph, it’s speed per minute is 1 km per minute. Hence, if it’s speed
with stoppages is 40 kmph, it will travel 40 minutes per hour.

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
Æ
145 km in 2.8333 hours Æ 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.

Solve this question using the values given in the options. Option (d) can be seen to fit the situation
given by the problem as it gives us the following chain of thought:
If the average speed of the faster train is 48 kmph, the average speed of the slower train would be
32 kmph. In this case, the time taken by the faster train (192/48 = 4 hours) is 2 hours lesser than
the time taken by the slower train (192/32 = 6 hours). This satisfies the condition given in the
problem and hence option (d) is correct.

Solve through options using trial and error. For usual speed 3 kmph we have:
Normal time Æ 2/3 hours = 40 minutes.
At 4 kmph the time would be 2/4 hrs, this gives us a distance of 10 minutes. Hence option (c) is
correct.

The length of the circular track would be equal to the circumference of the circle. In 2 minutes
thus, the cyclist covers 3.14 × 200 = 628 meters (using the formula for the circumference of a
circle).
Thus, the cyclist’s speed would be 628/2 = 314 meters/minute.

Assume a distance of 60 km in each stretch. Get the average speed by the formula. Total distance/
Total time = 180/10 = 18 kmph.