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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.
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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
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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
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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.
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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?




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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
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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*S1*S2/(S1 + S2)
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
x2 + 4x - 1440 = 0
x2 - 36x + 40x -1440 = 0
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.
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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 × 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
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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
⇒ x/10 - x/15 = 2
⇒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 − 1) = 5 hours.
Hence, required speed = 60/5 = 12 kmph
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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? 




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Correct Ans:180 m
Explanation:
Given, speed of train A = 54 kmph
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
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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, 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
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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.
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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




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Correct Ans:2km
Explanation:
To solve this question, we can apply a short trick approach;
(Product of two speeds/ Difference of two speeds) × Difference between arrival times
Given,
Speed1 = 3 km/hr, Speed2 = 4 km/hr
Time1 = 5 mins late, Time2 = 5 min early
Reqd. Distance
= (3 × 4)/(4 – 3) × (5 + 5)/60 = 12/1 × 10/60 = 2 km
Hence, option B is correct.
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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? 




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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.
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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] 




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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.
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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, 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.
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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?




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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 – 80 kms.
Therefore, (240 + 80)/S3 = (240 – 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
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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? 




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Correct Ans:105 km
Explanation:
Given:
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.
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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? 




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Correct Ans:2
Explanation:
Let length of the platform be '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.
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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 = 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 hours
Therefore, owner will catch the thief at 5 pm.
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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? 




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Correct Ans:11
Explanation:
Lets take, the bike covers '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, Sn=(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.
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