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Arithmetic: Time Speed Distance Test-1

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Question 1
An express train travels 299 km between two cities.  During the first 111 km of the trip, the train travelled through mountainous terrain. The train travelled 10 km/h slower through mountainous terrain than through level terrain. If the total time to travel between two cities was 7 h, what is the speed of the train on level terrain?
56 km/h
55 km/h
47 km/h
88 km/h
Question 1 Explanation: 
Total distance = 299 km
Distance travelled through mountains 111 km
Therefore the rest of the distance travelled by the train = 188 km
Let the speed of the train = s km/h
So the speed in the mountains will be = (s-10) km/h
According to the question
{188/s + 111/(s-10)} = 7
s = 47 km/h
Question 2
Jugni express which goes from Hyderabad to Chennai leaves Hyderabad at 5:30 am and travels at a constant speed of 50 km/h towards Nalgonda which is 100 km away. At 6:00 am, Khema express leaves from Nalgonda for Hyderabad at a constant speed of 40 km/h. At 6:30 am Mr Shah, the Control Officer realizes that both the trains are on the same track. How much time does Mr Shah have to avert the accident?
20 min
30 min
25 min
15 min
Question 2 Explanation: 
Total distance = 100 km
Distance travelled by Jugni in 1 hour = 50 km
Distance travelled by Khema in ½ h = 20 km
Speed of Jugni = 50km/h
Speed of Khema = 30km/hr
So at 6.30, the distance between Jugni and Khema is 30 km only
Time taken to cover a distance of 30 km (speeds will get added as the trains travel in opposite direction)
= 30 /(50 + 40) = 30/90 = 1/3 hrs
This is 20 minutes. So the right answer for this is option (a)
Question 3
Maninder covers a certain distance on a toy train. If the train moved 4 km/h faster, it would take 30 min less to cover the same distance. If it moved 2 km/h slower, it would have taken 20 min more to cover the same distance. Find the distance.
60 km
45 km
30 km
20 km
Question 3 Explanation: 
Distance covered = d km
Maninder’s speed = s km/h
Therefore time taken would be equal to = d/s
We know that:
t – ½ = d /(s+4)
t + 1/3 = d /(s-2)
Solving these two equations,
d= 60 km
s = 20 km/h
And the time taken will be = 3h
Question 4
The average speed of a train is 20% less on the return journey than during the forward journey. The train halts for half an hour at the destination station before starting on the return journey. If the total time taken for complete(forward and back) journey is 23 h, covering a distance of 1000 km, the speed of the train on the return journey is ___?
60 km/h
40 km/h
50 km/h
55 km/h
Question 4 Explanation: 
Let the speed of the train on onward journey = s km/h
The speed of the train is less than by 20% on the return journey
So the speed will be = 0.8s km/h
Therefore according to the question
500/s + ½ + 500/0.8s = total time taken
1125/s + ½ = 23
s = 1125 x 2/45 = 50 km/h therefore the required speed = 0.80 x 50 = 40km/h
Question 5
Two trains move from station Ladii and station Pamma towards each other at the speed of 50 km/h and 60 km/h respectively. At the meeting point, the driver of the second train felt that the train has covered 120 km more than the first train. What is the distance between Ladii and Pamma?
1320 km
1100 km
1200 km
960 km
Question 5 Explanation: 
At the meeting point, the driver of the second train felt that the train has covered 120 km more.
So if the distance travelled by the train that started at Laddi = d km
Then the distance travelled by the train that started at Pamma = (d+120) km
Since the travel time for both the trains is the same,  we can say that
d/50 = (d+120)/60
Solving this we will get the distance (d) = 600
So the total distance covered by the train = d + (d+120)= 1320 km
Difference between speed = 10 km /h
Meeting time after start = 120/10 = 12h
Distance between stations = (50+60)X12 = 1320 KM
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