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

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.

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.

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,

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]

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

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.

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

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

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

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.

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.

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.

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.

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

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.

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.

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.

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:S₂ = 1:1/2 = 2:1 => Option C is correct.

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.