Mastering Speed Distance and Time (50 Essential MCQs for Success)

In competitive and entrance exams, questions related to speed distance and time are frequently asked. These topics are important in various fields, such as physics, mathematics, engineering, and transportation. Understanding the relationship between speed distance and time is essential to solving these types of questions. In this chapter, we will discuss some of the most important MCQs related to speed, distance, and time, which will help you to prepare for various competitive and entrance exams.

Assignment-1

Que 1: Ramesh reaches his office 16 minutes late at three fourth of his normal speed. Find the normal time taken by him to cover the distance between his home and his office.
(a) 30 min
(b) 40 min
(c) 45 min
(d) 48 min

Solution: Speed⇒ 1 : ¾ =4 : 3
Time⇒ 3 : 4
Time difference (△t)= 4-3=1 unit= 16 min
Normal time duration= 3 unit= 3×16=48 min
Option (d) is correct

Que 2: Ram and Shyam cover the same distance at the speed of 6 km/hr and 10 km/hr respectively. If Ram takes 30 minutes more than Shyam, find the distance covered by each.
(a) 10 km
(b) 7.5 km
(c) 8.33 km
(d) 9.6 km

Solution: △t= 30 min= ½ hr and △s= 10-6= 4km/h
d=(\frac{s_1 \times s_2}{\triangle s}) \times \triangle t.
Distance= 7.5 km
Option (b) is correct

Que 3: If Rohan reduces his speed by 5 km/h, he will take four hours more to reach the destination. If he increases his speed by 5 km/h, he will take 2 hours less to reach the destination. Find the normal time taken by him.
(a) 6 hr
(b) 5.6 hr
(c) 8 hr
(d) 10.4 hr

Que 4: 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 the average speed of the two trains over this journey if the sum of their average speeds is 70 km/h?
(a) 34.28 km/h
(b) 50 km/h
(c) 35 km/h
(d) 12 km/h

Solution: t₁: t₂= 8:6=4:3
s₁:s₂=3:4
7k=70
k=10
s₁= 30 km/h and s₂=40 km/h
Average speed= (2×30×40)/(30+40)=34.28 km/h
Option (a) is correct

Que 5: 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 hours to reach Mumbai and Kolkata respectively. If the Calcutta Mail is moving at the speed of 48 km/h, what is the speed of the Bombay Mail?
(a) 24 km/h
(b) 21 km/h
(c) 22 km/h
(d) 96 km/h

Solution: \frac{s_1}{s_2}=\sqrt{\frac{t_2}{t_1}}.
s₁=48, t₁=12, t₂=3
s₂=96 km/h
Option (d) is correct

Que 6: A train travels a certain distance without any stoppage at an average speed of 60 km/h. However, when it includes stoppages, it covers the same distance with an average speed of 40 km/h. On average, how many minutes per hour does the train stop during its journey?
(a) 20 min/h
(b) 10 min/h
(c) 15 min/h
(d) 12 min/h

Solution: Stoppage per hour= (60-40)/60= 1/3 hr= 20 min
Option (a) is correct

Que 7: During his journey, Rishikant travels for 20 minutes at a speed of 30 km/h, another 30 minutes at a speed of 50 km/h, followed by 1 hour at a speed of 50 km/h, and finally 1 hour at a speed of 60 km/h. What is the average velocity of Rishikant’s entire journey?
(a) 51.18 km/h
(b) 39 km/h
(c) 63 km/h
(d) 48 km/h

Solution: Total distance (d)= 10+25+50+60= 145 km
Total Time (t)= 2 hr 50 min= 2+5/6=17/6 hr
Average velocity=d/t= 51.18 km/h
Option (a) is correct

Que 8: Two persons A and B start moving toward each other just after A has fired. After 10 minutes A fired again. The time between which B hears these two shots is 9 minutes. If the speed of A is 18 km/h, then find the speed of B. The speed of sound is given as 1080 km/hr.
(a) 120 km/h
(b) 100 km/h
(c) 96 km/h
(d) 84 km/h

Solution: In 10 minutes A covered ⇒ Speed×Time= 18×1/6 = 3 km
Time taken by sound to cover this 3 km= 3000/300= 10 sec
This means that if B does not move, he will hear the gunshot in 9 minutes and 50 seconds.
B heard the gunshot after 9 minutes. So the travel time of B is 9 minutes and the travel time of sound from the new position of B to the original position of B is 50 seconds.
Note: 9 min 50 sec- 9 min= 50 sec
Distance covered by B in 9 min= Distance covered by sound in 50-sec
k×(9×60)= 300×50
k=250/9 m/sec
k= (250/9)×(18/5)= 100 km/h
Option (b) is correct

Que 9: What is the actual speed of Rohan’s car given that he takes a lunch break of 15 minutes at a restaurant located 80 km away from point A and moves 16.67% slower, causing him to reach point B 90 minutes late, but if he takes this same lunch break at a restaurant located 60 km further ahead, he will reach point B 75 minutes late?
(a) 56 km/h
(b) 50 km/h
(c) 48 km/h
(d) 40 km/h

Solution: 16.67%=1/6
Let, the initial speed be 6x. So, the reduced speed is 5x

Case 1: At the first restaurant (C)
Speed⇒ 6:5
Time⇒ 5:6
Time difference= 1 unit= 75 min
Note: Exclude the lunchtime of 15 min
The original time taken to cover the journey from C to B=5 unit=5×75= 375 min

Case 2: At the second restaurant (D)
Speed⇒ 6:5
Time⇒ 5:6
Time difference= 1 unit= 60 min
Note: Exclude the lunchtime of 15 min
The original time taken to cover the journey from D to B=5 unit=5×60= 300 min

Time taken to cover the journey from C to D= 375-300= 75 min= 75/60=5/4 hr

The original speed= 60/(5/4) =48 kmph
Option (c) is correct

Que 10: Rohan and Sohan are at points A and B respectively which are 126 km apart, and start driving toward each other and meet at point P. After reaching B and A respectively they start back to their previous position and in the meantime meet at point Q. Find the distance between P and Q if Rohan takes 9 hours from P to B and Sohan takes 16 hours from P to A?
(a) 22.4 km
(b) 24 km
(c) 36 km
(d) 40 km

Solution: Let, the speeds of Rohan and Sohan are x and y respectively
Time taken by Rohan from P to B (t₁)= 9 hr
Time taken by Sohan from P to A (t₂)= 16 hr
x: y=√t₂: √t₁= 4:3

Let, x=4k and y=3k
Time taken by both to reach P (First meeting)= t=√(t₁×t₂)=12 hr
Time taken by Rohan to complete the journey from A to B= 12+9=21 hr
Distance AB= Speed of Rohan ×Time= 4k ×21=84k
Note: 84k=126 km (Given)

Total time taken for the second meeting= Distance/Relative Speed= (3×84k)/7k= 36 hr
Since Sohan takes 28 hours from B to A so for the second meeting at Q, he has to cover 8 hr journey.

Distance between A to Q= Speed of Sohan×Time= 3k×8= 24k
Distance between A to P= Speed of Rohan×Time= 4k×12= 48k

PQ=AP-AQ= 24k

Given, 84k= 126
So, 24k= 36 km
Option (c) is correct.

Que 11: What was the speed of the car if a pedestrian walking at 2 km/h in the same direction could see the car for 6 minutes and it was visible up to a distance of 0.6 km in the fog?
(a) 30 km/h
(b) 20 km/h
(c) 12 km/h
(d) 8 km/h

Solution: Let the speed of car=x
Time=6 min =6/60= 1/10 hr
Relative speed= x-2
Distance= 0.6 km
d=st
0.6=(x-2)×1/10
x=8 km/h
Option (d) is correct.

Que 12: Ram and Shyam run a race of 2000 m. First, Ram gives Shyam a start of 200 m and beats him by 30 s. Next, Ram gives Shyam a start of 3 min and is beaten by 1000 metres. Find the time in minutes in which Ram and Shyam can run the race separately.
(a) 8,10
(b) 5,9
(c) 4,5
(d) 6,9

Solution: Let the speed of Ram and Shyam be r and s (in m/s) respectively
Ram runs 2000⇒ Shyam runs (1800-30×s)
Ram runs 1000⇒ Shyam runs (2000-180×s)
2000/1000= (1800-2s)/(2000-180s)
s= 20/3 m/s
Time taken by Ramesh to cover 2000 m= distance/speed= 300 sec= 5 min
Ratio of distances covered in same time= 2000: 1600= 5:4
r:s= 5:4
Ratio of time taken to cover the same distance (2000 m)= 4:5
Option (c) is correct.

Que 13: In a 600-metre race, Ajay beats Vijay by 60 metres, and in a 500-metre race, Vijay beats Anjay by 25 metres. By how many metres will Ajay beat Anjay in a 400-metre race?
(a) 48 m
(b) 56 m
(c) 52 m
(d) 58 m

Solution: 600 m race⇒ Ajay: Vijay= 600: 540= 10:9
500 m race⇒ Vijay: Anjay= 500: 475= 20:19
400 m race⇒ Ajay: Vijay: Anjay= 400: 360: 342
400-342= 58 m
Option (d) is correct.

Que 14: What is the speed of a man in still water if he can row 30 km upstream and 44 km downstream in a total of 10 hours, and also can row 40 km upstream and 55 km downstream in a total of 13 hours?
(a) 4 km/h
(b) 8 km/h
(c) 6 km/h
(d) 12 km/h

Solution: 30/u +44/d=10
40/u+55/d= 13
u=5, d=11
Speed of man= (d+u)/2=8 km/h
Option (b) is correct.

Que 15: A motorboat went downstream for 28 km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 minutes. Find the speed of the boat in still water and the speed of the river flow.
(a) 9 km/h, 3 km/h
(b) 8 km/h, 2 km/h
(c) 9 km/h, 6 km/h
(d) 12 km/h, 3 km/h

Solution: tₔ:tᵤ=1:2
d:u= 2:1=2k: k
B= (d+u)/2= 3k/2 and S=(d-u)/2= k/2
B:S= 3:1 ……………(Option-a)

Second Case: B=1.5k & S’=k
d’=2.5k & u’=0.5k
672 min= 11+1/5=56/5 hr
28/d’ +28/u’=56/5
k=6
B=9 km/h and S=3 km/h
Option (a) is correct.

Que 16: A boat sails down the river for 10 km and then returns up the river for 6 km. The speed of the river flow is 1 km/h. What should be the minimum speed of the boat to ensure the entire trip takes a maximum of 4 hours?
(a) 2 kmph
(b) 3 kmph
(c) 4 kmph
(d) 5 kmph

Solution: 10/d+6/u ≤4
d=b+1 and u=b-1
From options, b=4
Option (c) is correct.

Que 17: A train requires 7 seconds to pass a pole while it requires 25 seconds to cross a stationary train which is 378 metres long. Find the speed of the train.
(a) 75.6 km/h
(b) 76.2 km/h
(c) 75.4 km/h
(d) 21 km/h

Solution: Let the length of the train be k
d=s×t
k=s×7 …….(i)
k+378=s×25 ……(ii)
7s+378=25s
s=21 m/s= 21×18/5= 75.6 km/h
Option (a) is correct.

Que 18: Two trains are running on parallel lines in the same direction at speeds of 40 kmph and 20 kmph respectively. The faster train crosses a man in the second train in 36 seconds. The length of the faster train is:
(a) 200 metres
(b) 225 metres
(c) 185 metres
(d) 210 metres

Solution: 36 sec= 36/(60×60)= 1/100 hr
Let the length of the faster train be k
d=s×t
k=(40-20)×1/100= 1/5 km
k= (1/5)×1000= 200 m
Option (a) is correct.

Que 19: Two people, P and Q, start running on a circular track of length 400 m in opposite directions. P’s initial speed is 10 m/s, and Q’s initial speed is 40 m/s. Whenever they meet, P’s speed doubles, and Q’s speed halves. The question is: After what time from the start will they meet for the third time?
(a) 20 sec
(b) 15 sec
(c) 24 sec
(d) 26 sec

Solution: Time= distance/Relative Speed
400/(10+40)+ 400/(20+20)+ 400/(40+10)
=8+10+8
=26 sec
Option (d) is correct.

Que 20: Three persons start running on a circular track of 36 km length at speeds of 7 kmph, 11 kmph, and 15 kmph, respectively, in the same direction. How many points are there where they will meet on the track, including the starting point?
(a) 3
(b) 4
(c) 5
(d) 6

Solution: Time taken for their first meeting at starting point = LCM (36/7, 36/11, 36/15)= 36 hr
Time taken by three persons in their first meeting = LCM of 36/(11-7), 36/(15-7), 36/(15-11)= 9 hr
The number of meeting points= 36/9= 4
Option (b) is correct.

Assignment-2

Que 1: An ant is climbing up a vertical pole that is 63 metres high. The ant climbs up 12 metres and then slips down 5 metres every other hour. How long will it take for the ant to reach the top of the pole?
(a) 18 hours
(b) 17 hours
(c) 16 hours 35 minutes
(d) 16 hours 40 minutes

Solution: 12-5=7 m in 2 hours
63-12=51 m
2 h× 8 alternations=16 hr ⇒7×8= 56 m
Now, the remaining 7 metres will be covered in 7/12 hours
Total time= 16 hr 35 min
Option (c) is correct.

Que 2: A motorcyclist started from point A towards point B. Two hours later, another motorcyclist started from point A to point B as well and arrived at point B at the same time as the first motorcyclist. If both of them had started together from points A and B towards each other, they would have met in 80 minutes. How long did it take the faster motorcyclist to travel from point A to point B?
(a) 6 hours
(b) 3 hours
(c) 2 hours
(d) 4 hours

Solution: Let the time taken by faster bike from A to B = t min
Time taken by slow bike from A to B= t+120 min
Time taken for the first meeting by both the bikes (k)=80 min
Time taken by slow bike from meeting point to B = t+120-80=t+40
Time taken by fast bike from meeting point to A = t-80
80=\sqrt{(t+40)\times (t-80)}.
t=120 min= 2 hr
Option (c) is correct.

Directions for Questions 3 to 5: Read the following and answer the question that follows.
A naughty bird is sitting on top of a car. It sees another car approaching it at a distance of 12 km. The speed of the two cars is 60 km/h each. The bird starts flying from the first Car and moves towards the second car, reaches the second Car and comes back to the first car and so on. If the speed at which the bird flies is 120 km/h then answer the following questions. Assume that the two cars have a crash.

Que 3: The total distance travelled by the bird before the crash is
(a) 6 km
(b) 12 km
(c) 18 km
(d) None of these

Solution: t= 12/(60+60)= 1/10 hr
These cars will crash after 1/10 hour
The total distance travelled by the bird= 120×(1/10)=12 hr
Option (b) is correct.

Que 4: The total distance travelled by the bird before it reaches the second car for the second time is
(a) 10.55 km
(b) 11.55 km
(c) 12.33 km
(d) None of these

Solution: Car A to B⇒ Time (t₁)= 12/(60+120)= 1/15 hr
Remaining distance= 12- 120×1/15= 4 km
Car B to A⇒ t₂= 4/(120+60)= 1/45 hr
Remaining distance= 4- 120×1/45= 4/3 km
Car A to B⇒ t₃=(4/3)/180= 1/135 hr
Total time= 1/15+ 1/45+ 1/135=13/135 hr
The total distance travelled by the bird= 120×(13/135)= 11.55 km
Option (b) is correct.

Que 5: The total number of times that the bird reaches the bonnet of the second car is (theoretically):
(a) 12 times
(b) 18 times
(c) Infinite times
(d) Cannot be determined

Solution: Option (c) is correct.

Que 6: How long will it take for the dog to catch the cat if the dog and cat are currently 25 leaps apart, the dog makes 5 leaps per minute, and the cat makes 6 leaps per minute, with each leap of the dog being equivalent to 2 leaps of the cat?
(a) 12 minutes
(b) 15 minutes
(c) 12.5 minutes
(d) None of these

Solution: Speed of dog= 5 leaps/min
Speed of Cat= 6 leaps/min×(1/2)= 3 leaps/min of dog
Relative speed= 5-3= 2 leaps/min
Time= 25/2= 12.5 min
Option (c) is correct.

Que 7: Ajay and Vijay start running simultaneously from the same point on a circular track in the same direction. If Ajay is running with a uniform speed of ‘a’ km/h and Vijay is running with ‘b’ km/h, they meet for the first time when Ajay is in his 4h round and Vijay is in his 3rd round. Which of the following can be the value of a : b (If a > b)?
(a) 1.4
(b) 1.46
(c) 1.35
Type (1) If only one option is correct.
Type (2) If two options are correct.
Type (3) If all the options are correct.
Type (4) If no option is correct.

Solution: Let the length of track be ‘x’ metres and they meet at ‘y’ metres away from the starting point.
a/b= (3x+y)/(2x+y)=1+x/(2x+y)
(a/b)max=1+1/2=1.5 at y=0
(a/b)min=1+1/3=1.33 at y=x
Correct answer is (3)

Que 8: A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles, the cat runs. If the cat moves to the entrance of tunnel A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat exactly at the exit. Find the ratio of the speed of the train and that of the cat?
(a) 3:1
(b) 4:1
(c) 5:1
(d) None of these

Solution: AB= 8k
If the initial position of the cat is at C then AC=3k and CB=5k
Let the train be initially at a distance ‘d’ from the entrance of the tunnel. The speeds of the train and the cat are x and y respectively.
d/x= 3k/y …….(i)
(d+8k)/x= 5k/y
d/x+ 8k/x= 5k/y
3k/y+ 8k/x= 5k/y
8k/x= 2k/y
x:y= 4:1
Option (b) is correct.

Que 9: At what time did the two trains meet each other if one train started from Delhi at 6:00 a.m. and arrived in Meerut at 10:00 a.m., while the other train started from Meerut at 8:00 a.m. and arrived in Delhi at 11:30 a.m.?
(a) 8:56 AM
(b) 9:20 AM
(c) 9:04 AM
(d) Can not be determined

Solution: Ratio of times= 4: 3.5= 8:7
Ratio of speeds= 7:8
Total distance= 7×4= 28 unit
Distance covered by the first train from 6:00 am to 8:00 am= 7×2=14 unit
Time taken for the first meeting after 8:00 am= 14/(7+8)= 14/15 hr
= (14/15)×60= 56 min
Option (a) is correct.

Que 10: Two candles of the same height are burning simultaneously. The first candle burns completely in 4 hours, while the second candle burns completely in 6 hours. After how much time will the ratio of their heights be 2:3?
(a) 2.4 hr
(b) 2.5 hr
(c) 3.5 hr
(d) 2.3 hr

Solution: LCM(4,6)=12
Let the height of both the candles be 12 units
The rate of burning of these two candles is 12/4 and 12/6 units per hour respectively.
After time ‘t’
12-3t : 12-2t= 2:3
t=2.4 hr
Option (a) is correct.