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# Test: Speed, Time And Distance- 4

## 15 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Test: Speed, Time And Distance- 4

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This mock test of Test: Speed, Time And Distance- 4 for Quant helps you for every Quant entrance exam. This contains 15 Multiple Choice Questions for Quant Test: Speed, Time And Distance- 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Speed, Time And Distance- 4 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Test: Speed, Time And Distance- 4 exercise for a better result in the exam. You can find other Test: Speed, Time And Distance- 4 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1

### The Sinhagad Express left Pune at noon sharp. Two hours later, the Deccan Queen started from Pune in the same direction. The Deccan Queen overtook the Sinhagad Express at 8 p.m. Find theaverage speed of the two trains over this journey if the sum of their average speeds is 70 km/h.

Solution:

The ratio of time for the travel is 4:3 (Sinhagad to Deccan Queen). Hence, the ratio of speeds
would be 3:4. Since, the sum of their average speeds is 70 kmph, their respective speeds would
be 30 and 40 kmph respectively. Use alligation to get the answer as 34.28 kmph.

QUESTION: 2

### Ram and Bharat travel the same distance at the rate of 6 km per hour and 10 km per hourrespectively. If Ram takes 30 minutes longer than Bharat, the distance travelled by each is

Solution:

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.

QUESTION: 3

### Two trains, Calcutta Mail and Bombay Mail, start at the same time from stations Kolkata and Mumbai respectively towards each other. After passing each other, they take 12 hours and 3 hoursto reach Mumbai and Kolkata respectively. If the Calcutta Mail is moving at the speed of 48 km/h,the speed of the Bombay Mail is

Solution:

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 QUESTION: 4

Walking at 3/4 of his normal speed, a man takes 2(1/2) hours more than the normal time. Find thenormal time.

Solution:

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).

QUESTION: 5

Sambhu beats Kalu by 30 metres in 10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

Solution:

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

QUESTION: 6

Without stoppage, a train travels a certain distance with an average speed of 60 km/h, and withstoppage, it covers the same distance with an average speed of 40 km/h. On an average, howmany minutes per hour does the train stop during the journey?

Solution:

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.

QUESTION: 7

Rishikant, during his journey, travels for 20 minutes at a speed of 30 km/h, another 30 minutes at aspeed of 50 km/h, and 1 hour at a speed of 50 km/h and 1 hour at a speed of 60 km/h. What is theaverage velocity?

Solution:

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

QUESTION: 8

Narayan Murthy walking at a speed of 20 km/h reaches his college 10 minutes late. Next time he increases his speed by 5 km/h, but finds that he is still late by 4 minutes. What is the distance ofhis college from his house?

Solution:

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.

QUESTION: 9

A motor car does a journey in 17.5 hours, covering the first half at 30 km/h and the second half at 40 km/h. Find the distance of the journey.

Solution:

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

QUESTION: 10

Manish travels a certain distance by car at the rate of 12 km/h and walks back at the rate of 3km/h. The whole journey took 5 hours. What is the distance he covered on the car?

Solution:

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.

QUESTION: 11

A car driver, driving in a fog, passes a pedestrian who was walking at the rate of 2 km/h in thesame direction. The pedestrian could see the car for 6 minutes and it was visible to him up to adistance of 0.6 km. What was the speed of the car?

Solution:

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.

QUESTION: 12

A journey of 192 km takes 2 hours less by a fast train than by a slow train. If the average speed of the slow train be 16 kmph less than that of fast train, what is the average speed of the faster train?

Solution:

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.

QUESTION: 13

If Arun had walked 1 km/h faster, he would have taken 10 minutes less to walk 2 kilometre. What is Arun’s speed of walking?

Solution:

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.

QUESTION: 14

A cyclist moving on a circular track of radius 100 metres completes one revolution in 2 minutes.What is the average speed of cyclist (approximately)?

Solution:

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.

QUESTION: 15

A car travels 1/3 of the distance on a straight road with a velocity of 10 km/h, the next 1/3 with a velocity of 20 km/h and the last 1/3 with a velocity of 60 km/h. What is the average velocity of thecar for the whole journey?

Solution:

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