on a 120km track a train travels the first 40km at a uniform speed of 40km/h how fast must the train travel the next 80km so as to average 60km/h for the entire trip
Answers
Distance d1 = 30 km, v1 = 30 kmph
=> time taken = t1 = d1/v1 = 1 hour
So distance remaining = d2 = 120 - d1 = 90 km
let speed to cover distance d2 = v2 kmph
=> time taken = d2/v2 = 90/v2 hours
average speed = total distance / total duration of time
60 kmph = 120 km / [1 hour + 90 / v2 hours]
=> 1 = 2 / [ 1 + 90 / v2 ]
=> v2 = 90 kmph
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another way:
as the total distance = 120 km
average speed = 60 kmph
=> total time duration = 120 / 60 = 2 hours
time taken for the first 30 km = 30 km /30 kmph = 1 hours
Hence, time remaining to cover the remaining 90 km = 2 hrs - 1 hrs = 1 hour
=> speed during 90 km = 90 kmph
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Explanation:
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Question
On a 120 km long track, a train travels the first 30 km with a uniform speed of 30 km/h. How fast must the train travel for the rest part of the journey so as to average 60 km/h for the entire trip?
A
90 km/h
B
80 km/h
C
70 km/h
D
60 km/h
Solution
The correct option is (A) 90 km/h
Step 1, Given data
Total distance = 120 km
30 km with a speed 30 km/h
Average speed = 60 km/h
Step 2, Finding the speed
Average speed for entire journey = 60 km/h
So, total time to be taken for the entire journey, T
=
120
k
m
60
k
m
h
−
1
=
2
h
Now, time taken during first half of journey,
t
=
60
k
m
60
k
m
h
−
1
=
1
h
Time taken for second half of journey =
T - t = 2 h - 1 h = 1 h
Distance covered in second part of journey = Total distance - Distance covered in first part of journey
= 120 km - 30 km = 90 km
Hence, average speed for second half =
90
k
m
1
h
=
90
k
m
h
−
1