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Test: Speed, Time & Distance - 1 - UPSC MCQ


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20 Questions MCQ Test CSAT Preparation - Test: Speed, Time & Distance - 1

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Test: Speed, Time & Distance - 1 - Question 1

A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 1

To find the speed of the man in km per hour, we have to convert the given distance to km and also convert the given time to hour.
First, let us see the conversion of the given distance into km. We know that 1km = 1000m.
So, to convert 600m to km, we can write, 600m = x km.
x = 600/1000 km  = 0.6km 

Now, let us convert the given time into hours. We know that 1 hour = 60 minutes.
So, to convert 5 minutes to hours, we can write, 5 minutes = y hours.
y = 5/60 hrs = 0.083 hrs

Now we know that speed can be found out using the formula, 
Speed = Distance/Time
           = 0.6/ 0.083
           = 6×1000 / 83 × 10
           = 6000/ 830
           = 7.2 km/h

Therefore, we get the speed of the man crossing a 600m long street in 5 minutes as 7.2km/h.
Hence, option (D) is the correct answer.

Test: Speed, Time & Distance - 1 - Question 2

Q. A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways, is:

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 2

Let the distance be x km. Then,
(Time taken to walls x km) + (Time taken to ride x km) = 23/4 hrs
⇒  (Time taken to walls 2x km) + Time taken to ride 2x km  =  23/2 hrs
but time taken to ride 2x km = 15/4 hrs 
Therefore, Time taken to walk 2x km = (23/2 - 15/4) hrs = 31/4 hrs =7 hr 45 min.

Test: Speed, Time & Distance - 1 - Question 3

Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 3

It is given that, excluding stoppages, the speed of a bus is 54 kmph.
⇒ Distance travelled by bus in 1 hour, excluding stoppages = 54 km.
Also, it is given that including stoppages the speed of the bus is 45 km/hr.
⇒ Distance travelled by the bus in 1 hour, including stoppages = 45 km.
(Distance travelled by bus in 1 hour excluding stoppages – distance travelled by bus in 1 hour including stoppages)
⇒ (54km−45km) ⇒ 9 km
Due to stoppages, it covers 9 km less in an hour.
Students can easily solve this question by using this trick.
Required time for the stoppage per hour,

Test: Speed, Time & Distance - 1 - Question 4

A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 4

Let time taken to travel the first half = x hr 
⇒ Time taken to travel the second half = (10 - x) hr 

∵ Distance = Time * Speed
Distance covered in the first half = 21x 
Distance covered in the the second half = 24(10 - x)

∵  Distance covered in the the first half = Distance covered in the the second half
⇒ 21x = 24(10 - x)
⇒ 45x = 240
⇒ x = 16/3

Total Distance = 2 * 21(16/3) = 224 Km [∵ multiplied by 2 as 21x was the distance of half way]

Test: Speed, Time & Distance - 1 - Question 5

A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 5

Time taken = 1 hr 40 min 48 sec = 1 hr (40 + 4/5) min = 1 + 51/75 hrs = 126/75 hrs
Let the actual speed be x km/hr
Then, (5/7) * x * (126/75) = 42
⇒ x = 42 * 7 * (75 / 5) * 126
⇒ x = 35 km/hr

Test: Speed, Time & Distance - 1 - Question 6

A friend is spotted by Lalloo at a distance of 200 m. When Lalloo starts to approach him, the friend also starts moving in the same direction as Lalloo. If the speed of his friend is 15 kmph, and that of Lalloo is 20 kmph, then how far will the friend have to walk before Lalloo meets him?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 6

Lalloo is unfortunate that the friend is moving away from him.

(Because the friend moves in same direction as Lalloo).

relative speed= 20- 15= 5,kmph. distance= 200 m.

Thus, Lalloo will meet his friend when he gains 200 m over him.

=> time required = distance / speed = 0.2/5 = 1/25 hrs.

=>Distance travelled by the friend in 1/25 hrs. (when Lalloo catches up him)

=> Time x Speed = 1/25 x 15 = 3/5 km = 600 m.

Test: Speed, Time & Distance - 1 - Question 7

A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 7

Relative speed = Speed of A + Speed of B = 2 + 3 = 5 rounds per hour
(∴ they walk in opposite directions)

⇒ They cross each other 5 times in 1 hour and 2 times in 1/2 hour

∵ Time duration from 8 a.m. to 9.30 a.m. = 1.5 hour

∴ They cross each other 7 times before 9.30 a.m

Test: Speed, Time & Distance - 1 - Question 8

Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 8

In this type of question, we need to get the relative speed between them.

The relative speed of the boys = 5.5 – 5 = 0.5 km/h

The distance between them is 8.5 km.
Time = Distance/Speed
Time = 8.5/0.5 = 17 hrs

Test: Speed, Time & Distance - 1 - Question 9

In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun's speed?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 9

Let Anil takes x hrs.
⇒ Arun takes x+2

If Arun doubles the speed, he will take x-1 hrs
⇒ He needs 3 hours less.
∵ Double speed means half time.
∴ Half of the time required by Arun to cover 30 km = 3 hours

⇒ Time required by Arun to cover 30 km = 6 hour
Thus, Arun's speed = 30/6 = 5 km/h

Test: Speed, Time & Distance - 1 - Question 10

A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 10

Time taken = Distance/Speed
Total time taken = (160/64) + (160/80) = 9/2 hrs
Average Speed = Distance/Time
∴ Average speed = 320 / (9/2) = 320 x (2/9) = 71.11 km/hr.

Test: Speed, Time & Distance - 1 - Question 11

A jeep travels a distance of 100 km at a uniform speed. If the speed of the jeep is 5 kmph more, then it takes 1 hour less to cover the same distance. The original speed of the jeep is:

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 11

The original speed of the jeep is 20 kmph.

Let the original speed of the jeep be x kmph.

Time taken to cover 100 km at original speed = 100/x hours.

Time taken to cover 100 km at increased speed = 100/(x+5) hours.

Given that the time taken to cover 100 km at increased speed is 1 hour less than the time taken to cover 100 km at original speed.

Therefore, 100/x - 100/(x+5) = 1

Solving for x, we get x = 20.

Hence, the original speed of the jeep is 20 kmph.

Test: Speed, Time & Distance - 1 - Question 12

Two athletes cover the same distance at the rate of 10 and 15 kmph respectively. Find the distance travelled when one takes 15 minutes longer than the other.

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 12

The distance travelled is 7.5 km.

Let the time taken by the athlete travelling at 10 kmph be t hours.

The time taken by the athlete travelling at 15 kmph is t+15/60 hours.

The distance travelled by both athletes is the same.

Therefore, 10t = 15(t+15/60)

Solving for t, we get t = 3/4 hours.

The distance travelled by both athletes is 10t = 10 * 3/4 = 7.5 km.

Test: Speed, Time & Distance - 1 - Question 13

If Sita walks at 5 kmph, she misses her train by 10 minutes. If she walks at 7 kmph, she reaches the station 10 minutes early. How much distance does she walk to the station?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 13

The distance to the station can be calculated as follows:

Let's denote the distance to the station as "d" (in km), and the time difference between the two cases as "t" (in minutes).

In the first case, Sita walks at 5 km/h and misses the train by 10 minutes. So the time it would take her to get to the train on time is: d/5 (in hours) + 10/60 (in hours) = d/5 + 1/6 (in hours).

In the second case, Sita walks at 7 km/h and arrives 10 minutes early. So the time it takes her to get to the train is: d/7 - 10/60 = d/7 - 1/6 (in hours).

Since these two times should be the same, we can equate them:

d/5 + 1/6 = d/7 - 1/6

Solving this equation for "d" gives:

d = 35/6 km = 5.8 km

So the correct answer is 5.8 km.

Test: Speed, Time & Distance - 1 - Question 14

A friend is spotted by Lalloo at a distance of 200 m. When Lalloo starts to approach him, the friend also starts moving in the same direction as Lalloo. If the speed of his friend is 15 kmph, and that of Lalloo is 20 kmph, then how far will the friend have to walk before Lalloo meets him?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 14

Lalloo is unfortunate that the friend is moving away from him.

(Because the friend moves in same direction as Lalloo).

relative speed= 20- 15= 5,kmph. distance= 200 m.

Thus, Lalloo will meet his friend when he gains 200 m over him.

=> time required = distance / speed = 0.2/5 = 1/25 hrs.

=> Distance travelled by the friend in 1/25 hrs. (when Lalloo catches up him)

=> Time x Speed = 1/25 x 15 = 3/5 km = 600 m

Test: Speed, Time & Distance - 1 - Question 15

A person travelled 120 km by steamer, 450 km by train and 60 km by horse. It took him 13 hours 30 minutes. If the speed of the train is 3 times that of the horse and 1.5 times that of the steamer, then what is the speed (in km/h) of the steamer?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 15

A person travelled 120 km by steamer, 450 km by train and 60 km by horse

Formula used: Time = Distance/Speed

Calculation:
Let speed of horse be x km/hr
Speed of train = 3x km/hr
Speed of train = 1.5 × speed of steamer  ⇒ 3x = 1.5 × speed of steamer
Speed of steamer = 2x

According to the question: (120/2x) + (450/3x) + (60/x) = [13 + (30/60)] ⇒ (60/x) + (150/x) + (60/x) = (27/2) ⇒ (270/x) = 27/2
Upon solving, we get
⇒ x = 20
 

Now, Speed of steamer = 2x = 20 × 2 = 40 km/hr
∴ The speed of steamer is 40km/hr.

Test: Speed, Time & Distance - 1 - Question 16

A train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 40 seconds. What is the length of the train in meters?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 16

When a train overtakes another object such as a motorbike, whose length is negligible compared to the length of the train, then the distance traveled by the train while overtaking the motorbike is the same as the length of the train.

The length of the train = distance traveled by the train while overtaking the motorbike
= relative speed between the train and the motorbike * time taken

In this case, as both the objects i.e., the train and the motorbike are moving in the same direction, the relative speed between them = difference between their respective speeds = 100 - 64 = 36 kmph.

Distance traveled by the train while overtaking the motorbike = 36 kmph * 40 seconds.

The final answer is given in meters and the speed is given in kmph and the time in seconds.

So let us convert the given speed from kmph to m/sec.

1 kmph = 5/18 m/sec

Therefore, 36 kmph = 36 * 5 /18 = 10 m/sec.

Relative speed = 10 m/sec. Time taken = 40 seconds.

Therefore, distance traveled = 10 * 40 = 400 meters.

Test: Speed, Time & Distance - 1 - Question 17

Find the time taken by two trains, one 180 m long and the other 270 m long, to cross each other, if they are running at speeds of 46 kmph and 54 kmph respectively. Consider both possible cases of motion.

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 17

Case I: Motion in same direction

=>Relative speed = 54 — 46 = 8 km/hr.

Distance to be covered = 180 + 270 = 450 m.

=>Time = 0.450/8 = 0.056 hrs = 202.5 sec.
 

Case II: Motion in opposite direction.

=> Relative Speed = 54 + 46=100 km/hr.

Distance to be covered = 180 + 270= 450 m.

=>Time = 0.450/100 = 0.00045 hrs = 16.2 sec.

Test: Speed, Time & Distance - 1 - Question 18

A distance is covered at a certain speed in a certain time. If the double of this distance is covered in four times the time, then what is the ratio of the two speeds?

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 18

Case I : Distance D Speed S1 Time D/S1

Case II : Distance 2D Speed S2 Time 4(D/S1)

=> Speed for case II = S2 = Distance/Time = 2D/(4D/S1) = S1/22/(4/1) = 1/2

Hence, speed for case I : speed for case II = S1:S= 1:1/2 = 2:1 => Option C is correct.

Test: Speed, Time & Distance - 1 - Question 19

Two men A and B walk from P to Q, a distance of 21 km, at 3 and 4 km an hour respectively. B reaches Q, returns immediately and meets A at R. The distance from P to R is

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 19

When B meets A at R, B has walked the distance PQ + QR and A the distance PR.

That is, both of them have together walked twice the distance from P to Q, i.e. 42 km.

Now, the speed of A and B are in the ratio 3 : 4, and they have walked 42 km.

Hence, the distance PR travelled by A = 3/7 of 42 km = 18 km.

Test: Speed, Time & Distance - 1 - Question 20

A ship 156 km from the shore springs a leak which admits 2.5 metric tons of water In 61/2 minutes. A quantity of 68 metric tons would suffice to sink it, but the pumps can throw out 15 metric tons in an hour. The average rate of sailing so that it just reaches the shore as it begins to sink should be

Detailed Solution for Test: Speed, Time & Distance - 1 - Question 20

In one minute, amount flowing in = 15/39 MT (MT stands for metric tons)

In one minute, amount thrown out = 15/60 = 1/4 MT

Effective rate of filling in one hour = (15/39 - 1/4) MT = 21/56 MT/Min

Time till it just begins to sink = 68 / (21/156)

Speed required = (156/505) = 0.3 km/min = 0.3 x 60 km/hr = 18 km/hr.

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