# Time and Distance Questions and Answers updated daily – Aptitude

Time and Distance Questions: Solved 288 Time and Distance Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Time and Distance Questions

61. The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m and travels towards A at 75 Km/hr. At what time do they meet?

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Correct Ans:11.00 AM

Explanation:

Assume that the train 1 and train 2 meet X hours after 8 a.m.

Then, train 1, starting from A, travels x hours till the trains meet.

Distance travelled by train 1 in x hours = 60x km

Train 2, starting from B, travels (x-1) hours till the trains meet.

Distance travelled by train 2 in (x-1) hours= 75(x-1) km

Total distance travelled = Distance travelled by train 1 + Distance travelled by train 2

330=60x + 75(x-1)

12x+15(x-1)=66

12x+15x-15=66

27x=81

x=81/27

x=3

Hence, the trains meet 3 hours after 8 a.m., i.e. at 11 a.m.

Then, train 1, starting from A, travels x hours till the trains meet.

Distance travelled by train 1 in x hours = 60x km

Train 2, starting from B, travels (x-1) hours till the trains meet.

Distance travelled by train 2 in (x-1) hours= 75(x-1) km

Total distance travelled = Distance travelled by train 1 + Distance travelled by train 2

330=60x + 75(x-1)

12x+15(x-1)=66

12x+15x-15=66

27x=81

x=81/27

x=3

Hence, the trains meet 3 hours after 8 a.m., i.e. at 11 a.m.

Workspace

62. 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?

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Correct Ans:10

Explanation:

It is given that due to stoppages, the bus covers 9 km less because of the stoppages.

Formula for Speed:

Time = Distance/Speed

Difference in the Speed = 54-45 = 9 kmph

Time taken to cover 9 km = (9/54 * 60)min

Time taken to cover 9 km = 10 min

Formula for Speed:

Time = Distance/Speed

Difference in the Speed = 54-45 = 9 kmph

Time taken to cover 9 km = (9/54 * 60)min

Time taken to cover 9 km = 10 min

Workspace

63. The distance between two cities(M and N) is 350 km. A train starts from the city M at 6 am and travels towards city N at the rate of 63 kmph. Another train satrts from city N at 7 am and travels towards the city M at the rate of 77 kmph. At what time will the trains meet?

SHOW ANSWER

Correct Ans:9.03 AM

Explanation:

Let both trains meet after t hours after 6 am

According to the question,

Distance = Speed * Time

(63*t) + (77*(t-1)) = 350

63t + 77t - 77 = 350

140t = 350 + 77 = 427

t = 427/140 hours

= 3(7/140) hours

= 3 hours (7/140)*60 minutes

= 3 hours 3 minutes

Required time = 9 : 03 am

According to the question,

Distance = Speed * Time

(63*t) + (77*(t-1)) = 350

63t + 77t - 77 = 350

140t = 350 + 77 = 427

t = 427/140 hours

= 3(7/140) hours

= 3 hours (7/140)*60 minutes

= 3 hours 3 minutes

Required time = 9 : 03 am

Workspace

64. A bus started its journey from Ramgarh and reached Devgarh in 44 minutes with its average speed of 50 km/hr. If the average speed of the bus is increased by 5 km/hr, how much time will it take to cover the same distance?

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Correct Ans:40 minutes

Explanation:

Distance between Ramgarh and Devgarh = (50*44)/60 = 110/3 km

New speed = 50 + 5 = 55 km/hr

= 55/60 km/min

Therefore, required time = Distance/Speed

= (110/3) * (60/55) = 40 minutes.

New speed = 50 + 5 = 55 km/hr

= 55/60 km/min

Therefore, required time = Distance/Speed

= (110/3) * (60/55) = 40 minutes.

Workspace

65. Car A leaves city C at 5pm and is driven at a speed of 40kmph. 2 hours later another car B leaves city C and is driven in the same direction as car A. In how much time will car B be 9 kms ahead of car A if the speed of car B is 60 kmph?

SHOW ANSWER

Correct Ans:267 min

Explanation:

Relative speed of both the cars = 60 - 40 = 20 kmph

Initial distance travelled by car A when car B was not moving = 40 x 2 = 80 kms

Car B should be 9 km ahead of the A at the required time

Hence it must be 89 km away

Time = 89 / 20 = 4.45 hrs

To convert hours into minutes multiply with 60.

= 4.45 x 60

= 267 mins

Initial distance travelled by car A when car B was not moving = 40 x 2 = 80 kms

Car B should be 9 km ahead of the A at the required time

Hence it must be 89 km away

Time = 89 / 20 = 4.45 hrs

To convert hours into minutes multiply with 60.

= 4.45 x 60

= 267 mins

Workspace

66. Raj while going by bus from home to airport (without any halt) takes 20 minutes less than time taken when the bus halts for some time. The average speed of the bus without any halt is 8 km/hr more than the average speed of the bus when it halts. If the distance from home to airport is 60 km, what is the speed of the bus when it is travelled without stoppage ? (in km/hr)

SHOW ANSWER

Correct Ans:40

Explanation:

Average speed = 2*S

Speed of bus with halt = x km/hr

Therefore, speed of the bus without halt = (x+4) km/hr

According to the question,

Time = Distance/Speed

(60/x) - (60/(x + 4)) = 20/60

60[(x + 4 - x) / x(x + 4)] = 1/3

x(x + 4) = 60*3*8

x

x

x(x - 36) + 40(x - 36) = 0

(x - 36)(x + 40) = 0

x = 36, -40

So, x = 36 km/hr

Hence, the required speed of the bus = 40 km/hr.

_{1}*S_{2}/(S_{1}+ S_{2})Speed of bus with halt = x km/hr

Therefore, speed of the bus without halt = (x+4) km/hr

According to the question,

Time = Distance/Speed

(60/x) - (60/(x + 4)) = 20/60

60[(x + 4 - x) / x(x + 4)] = 1/3

x(x + 4) = 60*3*8

x

^{2}+ 4x - 1440 = 0x

^{2}- 36x + 40x -1440 = 0x(x - 36) + 40(x - 36) = 0

(x - 36)(x + 40) = 0

x = 36, -40

So, x = 36 km/hr

Hence, the required speed of the bus = 40 km/hr.

Workspace

67. A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 50 metres apart, at what speed is the train travelling?

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Correct Ans:60 km/hr

Explanation:

The man can count 21 telephone posts in one minute. Number of gaps between 21 posts is 20 and adjacent posts are 50 metres apart.

It means 20 Ã— 50 = 1000 metres are covered in 1 minute.

distance = 1000 m =1 km

time = 1 min = 1/60 hr

speed = 1/(1/60) = 60 km/hr

It means 20 Ã— 50 = 1000 metres are covered in 1 minute.

distance = 1000 m =1 km

time = 1 min = 1/60 hr

speed = 1/(1/60) = 60 km/hr

Workspace

68. Balu is travelling on his cycle and has calculated to reach point A at 2 pm if he travels at 10 kmph. He will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 pm?

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Correct Ans:12 kmph

Explanation:

Let the distance be x km

Travelling at 10 kmph, Balu will reach point A at 2 pm.

Travelling at 15kmph, Balu will reach point A 12 noon.

Therefore, time taken when travelling at 10 kmph - time taken when travelling at 15 kmph = 2 hours

Time = Distance/speed

â‡’ x/10 - x/15 = 2

â‡’3x - 2x = 2*30

x = 60

Time needed if travelled at 10 kmph = 60/10 = 6 hours

Therefore, to reach at 1pm, his travelling time must be (6 âˆ’ 1) = 5 hours.

Hence, required speed = 60/5 = 12 kmph

Travelling at 10 kmph, Balu will reach point A at 2 pm.

Travelling at 15kmph, Balu will reach point A 12 noon.

Therefore, time taken when travelling at 10 kmph - time taken when travelling at 15 kmph = 2 hours

Time = Distance/speed

â‡’ x/10 - x/15 = 2

â‡’3x - 2x = 2*30

x = 60

Time needed if travelled at 10 kmph = 60/10 = 6 hours

Therefore, to reach at 1pm, his travelling time must be (6 âˆ’ 1) = 5 hours.

Hence, required speed = 60/5 = 12 kmph

Workspace

69. A train-A passes a stationary train B and a pole in 18 sec and 6 sec respectively. If the speed of train A is 54 kmph what will be the length of train B?

SHOW ANSWER

Correct Ans:180 m

Explanation:

Given, speed of train A = 54 kmph

Converting into m/sec

= 3 * 5

=

Given, Train-A passes a pole in 6 sec

--->

= 15 * 6

=

Given, Train-A passes a stationary train B in 18 sec

So,

---> 90 + Length of Train B = 15 * 18

----> Length of Train B = 270 - 90

---->

Converting into m/sec

**Speed of train A**= 54 * (5/18) m/sec= 3 * 5

=

**15 m/sec**Given, Train-A passes a pole in 6 sec

--->

**Length of Train A = Speed of Train A * Time taken to cross a pole**= 15 * 6

=

**90 meter**Given, Train-A passes a stationary train B in 18 sec

So,

**Length of Train A + Length of Train B = Speed of Train A * Time taken to cross a stationary train B**---> 90 + Length of Train B = 15 * 18

----> Length of Train B = 270 - 90

---->

**Length of Train B = 180 meter**
Workspace

70. A man swimming in a stream which flows 1.5 kmph finds that in a given time he can swim twice as far with the sream as he can against it. At what rate does he swim?

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Correct Ans:4.5 kmph

Explanation:

Given that, speed of stream = 1.5 kmph

let Speed of man = x kmph

Then,

= x + 1.5

Upstream speed = Speed of man - speed of stream

= x - 1.5

Given that,

---> x + 1.5 = 2 * (x - 1.5)

---> x + 1.5 = 2x - 3

---> x = 4.5 kmph

let Speed of man = x kmph

Then,

**Downstream speed = Speed of man + speed of stream**= x + 1.5

Upstream speed = Speed of man - speed of stream

= x - 1.5

Given that,

**Downstream speed = 2 * Upstream speed**---> x + 1.5 = 2 * (x - 1.5)

---> x + 1.5 = 2x - 3

---> x = 4.5 kmph

**Speed of man, i.e,, Rate of Man swimming in a stream = 4.5 kmph**
Workspace

71. A Jackal takes 4 leaps for every 5 leaps of goat but 3 leaps of a Jackal are equal to 4 leaps of the goat. compare their speeds

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Correct Ans:16:15

Explanation:

Let the distance covered in 1 leap of the jackal be x and that covered in 1 leap of the goat be y.

Then, 3x = 4y ⇒ x = 4/3 y

For 4x = 16/3 y.

So, Ratio speed of jackal and goat = Ratio of distances covered by them in the same time,

⇒ 4x : 5y = 16y/3 : 5y

⇒ 16/3 : 5

= 16 : 15.

Hence, option D is correct.

Then, 3x = 4y ⇒ x = 4/3 y

For 4x = 16/3 y.

So, Ratio speed of jackal and goat = Ratio of distances covered by them in the same time,

⇒ 4x : 5y = 16y/3 : 5y

⇒ 16/3 : 5

= 16 : 15.

Hence, option D is correct.

Workspace

72. Walking at 3 km/hr . Chintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Chintu"™s from his house is

SHOW ANSWER

Correct Ans:2km

Explanation:

To solve this question, we can apply a short trick approach;

(Product of two speeds/ Difference of two speeds) Ã— Difference between arrival times

Given,

Speed1 = 3 km/hr, Speed2 = 4 km/hr

Time1 = 5 mins late, Time2 = 5 min early

Reqd. Distance

= (3 Ã— 4)/(4 â€“ 3) Ã— (5 + 5)/60 = 12/1 Ã— 10/60 = 2 km

Hence, option B is correct.

(Product of two speeds/ Difference of two speeds) Ã— Difference between arrival times

Given,

Speed1 = 3 km/hr, Speed2 = 4 km/hr

Time1 = 5 mins late, Time2 = 5 min early

Reqd. Distance

= (3 Ã— 4)/(4 â€“ 3) Ã— (5 + 5)/60 = 12/1 Ã— 10/60 = 2 km

Hence, option B is correct.

Workspace

73. The ratio of the speeds of the train and the man is 6 : 1. The length of the train is 650m and crosses a pole in 1 minute 5 seconds. In how much time will the man cross the 240m long platform?

SHOW ANSWER

Correct Ans:2 minutes 24 seconds

Explanation:

Speed of the train = 6x m/s

Speed of the man = x m/s

Length of the train = 650m

Time taken to cross a pole = 1 minute 5 seconds = 65 seconds

Speed = Distance/Time

6x = 650/65

x = 10/6 = 5/3

Speed of the man = 5/3 m/s

Distance/Speed = Time

Man can cross the 240m platform = 240/(5/3) = 144 seconds = 2 minutes 24 seconds

Hence, option D is correct.

Speed of the man = x m/s

Length of the train = 650m

Time taken to cross a pole = 1 minute 5 seconds = 65 seconds

Speed = Distance/Time

6x = 650/65

x = 10/6 = 5/3

Speed of the man = 5/3 m/s

Distance/Speed = Time

Man can cross the 240m platform = 240/(5/3) = 144 seconds = 2 minutes 24 seconds

Hence, option D is correct.

Workspace

74. Shalini was travelling on one side of the Nellai express way with a constant speed of 120 kmph in his car. Hema was travelling with a constant speed of 80 kmph in the opposite direction. When they crossed each other, Shalini decided to take a U-turn and meet her. But before taking a U turn, Shalini had to travel for another 3 minutes. How long will it take for Shalini to meet Hema? [Assume time taken by Shalini to take U turn is negligible]

SHOW ANSWER

Correct Ans:None of these

Explanation:

Distance Travelled by Shalini in 3 minutes = (3/60)*120 = 6 km

Distance travelled by Hema in 3 minutes = (3/60)*80 = 4 km

Hence Shalini after taking U turn will have to travel a distance of 10 km to catch Hema.

Relative speed of Shalini to that of Hema after taking U turn = 120 - 80 = 40 kmph

:. Time taken by Shalini after taking U turn to meet Hema = 10/40 = 15 minutes

Before taking U turn Shalini had to travel for 3 minutes

So, total time taken by Shalini to meet Hema = (3 + 15) minutes = 18 minutes

Hence, option E is correct.

Distance travelled by Hema in 3 minutes = (3/60)*80 = 4 km

Hence Shalini after taking U turn will have to travel a distance of 10 km to catch Hema.

Relative speed of Shalini to that of Hema after taking U turn = 120 - 80 = 40 kmph

:. Time taken by Shalini after taking U turn to meet Hema = 10/40 = 15 minutes

Before taking U turn Shalini had to travel for 3 minutes

So, total time taken by Shalini to meet Hema = (3 + 15) minutes = 18 minutes

Hence, option E is correct.

Workspace

75. A 150 m long train crosses a 450 m long platform in 30 sec. A person is running in same direction and train crosses him in 15 sec then what is speed of person?

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Correct Ans:36 km/hr

Explanation:

Given:

Length of train = 150 m

Length of platform = 450 m

Time taken to cross the platform = 30 sec

Total distance = Length of platform + Length of train

= 150 + 450 = 600 m

WKT,

Therefore, Speed of train = 600/30 = 20 m/sec

Converting mps to kmph,

Speed of train = 20 x (18/5) = 72 km/hr

Let the speed of person be

150 = (72 - X) *15

(150/15) x (18/5) = 72 - X

36 = 72 - X

Length of train = 150 m

Length of platform = 450 m

Time taken to cross the platform = 30 sec

Total distance = Length of platform + Length of train

= 150 + 450 = 600 m

WKT,

**Speed = Distance/Time**Therefore, Speed of train = 600/30 = 20 m/sec

Converting mps to kmph,

Speed of train = 20 x (18/5) = 72 km/hr

Let the speed of person be

**'X' km/hr.****Distance of train = Speed of (train - person) x time**150 = (72 - X) *15

(150/15) x (18/5) = 72 - X

36 = 72 - X

**X = 36 km/hr.**
Workspace

76. Three cars leave A for B in equal time intervals. They reach B simultaneously and then leave for Point C which is 240 km away from B. The first car arrives at C an hour after the second car. The third car, having reached C, immediately turns back and heads towards B. The first and the third car meet a point that is 80 km away from C. What is the difference between the speed of the first and the third car?

SHOW ANSWER

Correct Ans:60 kmph

Explanation:

Given, the cars leave in equal intervals of time and arrive Point A at the same time.

Thus the difference in the time taken between cars 1 and 2 should be equal to the time taken between cars 2 and 3.

**Time = Distance / Speed**

Let S1, S2 and S3 be the speeds of the cars.

(AB/S1) - (AB/S2) = (AB/S2) - (AB/S3)

Since the distance AB is same

----> (1/S1) - (1/S2) = (1/S2) - (1/S3) ---> eqn (1)

From Point B,second car arrived at C an hour earlier than the first,

Time at which First car reaches C = Time at which second car reaches C + 1

---> (240/S1) - (240/S2) = 1

----> (1/S1) - (1/S2) = 1/240 ----> eqn (2)

Given that the first and the third car meet a point that is 80 km away from C.

The 3rd car covered 240 + 80 kms when the first one covered 240 â€“ 80 kms.

Therefore, (240 + 80)/S3 = (240 â€“ 80)/S1

---> 320/S3 = 160/S1

---> (320/ 160) S1 = S3

---->

**2S1 = S3**

Substituting this in eqn (1), we get

(1/S1) - (1/S2) = (1/S2) - (1/2S1)

----> (S2 - S1)/ S1S2 = (2S1 - S2)/ 2S1S2

----> (S2 - S1) = (2S1 - S2)/2

----> 2S2 - 2S1 = 2S1 - S2

-----> 3S2 = 4S1

-----> S2 = 4S1/3

Substituting this in eqn (2), we get

----> (1/S1) - (3/4S1) = 1/240

----> (4 - 3)/4S1 = 1/240

----> 1/4S1 = 1/240

-----> 4S1 = 240

---->

**S1 = 60 km/hr ---> Speed of First car**

**Speed of Third car = S3**= 2S1

= 2 * 60

=

**120 km/hr**

**Difference between the speed of the first and the third car = 120 - 60 = 60 km/hr**

Workspace

77. If the ratio of speed of boat in downstream and speed of stream is 9 : 1, speed of current is 3 km per hr, What would be the distance travelled in upstream by the boat in 5 hours?

SHOW ANSWER

Correct Ans:105 km

Explanation:

Given:

Let the speed of the stream be

Ratio of speed of boat in downstream and speed of stream = 9 : 1

Speed of current = 3 km/hr

Speed of the boat in downstream = 9 x 3 = 27 km/hr

WKT,

= 27 - 3 = 24 km/hr

Speed of boat in upstream = (

= 24 - 3 = 21 km/hr

Distance travelled in upstream by the boat in 5 hours =

= 21 x 5

=

Let the speed of the stream be

**'v' km/hr**and speed of the boat in still water be**'u' km/hr.**Ratio of speed of boat in downstream and speed of stream = 9 : 1

Speed of current = 3 km/hr

Speed of the boat in downstream = 9 x 3 = 27 km/hr

WKT,

**Speed of boat in downstream = (u + v) km/hr****Speed of boat in still water(u) = Speed of boat in downstream - Speed of current(v)**= 27 - 3 = 24 km/hr

Speed of boat in upstream = (

**u - v) km/hr**= 24 - 3 = 21 km/hr

Distance travelled in upstream by the boat in 5 hours =

**Speed of boat in upstream x Time**= 21 x 5

=

**105 km.**
Workspace

78. A 320 metre long train moving with an average speed of 120 km/hr crosses a platform in 24 seconds. A man crosses the same platform in 4 minutes. What is the speed of man in metre/second?

SHOW ANSWER

Correct Ans:2

Explanation:

Let length of the platform be

Speed of the train = 120 kmph

Converting kmph to mps,

Speed of train = 120 x (5/18) = 100/3 m/s

WKT,

320 + X = (100/3) x 24

320 + X = 800

X = 800 - 320

Therefore,

Speed of man = 480/(4 x 60) =

**'X' meters.**Speed of the train = 120 kmph

Converting kmph to mps,

Speed of train = 120 x (5/18) = 100/3 m/s

WKT,

**Distance = Speed x Time****(Length of train + Length of platform) = Speed x Time**320 + X = (100/3) x 24

320 + X = 800

X = 800 - 320

**X = 480 m**Therefore,

**Speed of man = Distance/ Time**Speed of man = 480/(4 x 60) =

**2 m/s.**
Workspace

79. A thief steals a car at 2.30 pm and drives it at 60 kmph. The theft is observed at 3 pm and the owner sets off in another car from same place at 75 kmph. When will he catch the thief?

SHOW ANSWER

Correct Ans:5:00 PM

Explanation:

From the question, we can observe that both the theif and owner are in motion at 3 pm.

So, the distance between them at 3 pm = 30 minutes

WKT,

Distance covered by theif in 30 minutes = 60 x (1/2) = 30 km

Therefore, distance to be covered by theif is

Relative speed between theif and owner = (75 – 60) = 15 kmph

Hence, time taken by owner =

Therefore,

So, the distance between them at 3 pm = 30 minutes

WKT,

**Distance = Speed x Time**Distance covered by theif in 30 minutes = 60 x (1/2) = 30 km

Therefore, distance to be covered by theif is

**30 km**.Relative speed between theif and owner = (75 – 60) = 15 kmph

Hence, time taken by owner =

**Speed/Time**= 30/15 = 2 hoursTherefore,

**owner will catch the thief at 5 pm.**
Workspace

80. A bike starts running with some initial speed and its speed increases every hour by 3 km/kr. It It takes 17 hours to cover a distance of 595 km, then what was the initial speed (in km/hr) of the bike?

SHOW ANSWER

Correct Ans:11

Explanation:

Lets take, the bike covers

WKT,

Given, Speed increases every hour by 3 km/hr.

So, the distance increases by every hour = 3 x 1 = 3 km

In the 1st hour, bike covers = X km

In the 2nd hour, it covers = (X + 3) km

and so on,..

X + (X+3) + ...... = 595

[Since,

(n/2)[2a + (n - 1)d] = 595

Here, n = 17; a = X; d = 3

--> (17/2)[2X + (17 -1)3] = 595

--> [2X + 48] = (595 * 2)/17

--> 2X = 70 - 48

-->

Therefore, in the

Hence, the

**'X' km**in 1 hour.WKT,

**Distance = Speed x Time**Given, Speed increases every hour by 3 km/hr.

So, the distance increases by every hour = 3 x 1 = 3 km

In the 1st hour, bike covers = X km

In the 2nd hour, it covers = (X + 3) km

and so on,..

X + (X+3) + ...... = 595

[Since,

**S**]]_{n}=(n/2)[2a + (n - 1)d(n/2)[2a + (n - 1)d] = 595

Here, n = 17; a = X; d = 3

--> (17/2)[2X + (17 -1)3] = 595

--> [2X + 48] = (595 * 2)/17

--> 2X = 70 - 48

-->

**X = 11 km**.Therefore, in the

**1st hour bike covers 11 km.**Hence, the

**initial speed**=**Distance/Time**= 11/1 =**11 km/hr.**
Workspace

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