Line Diagram 38

Directions (1-4) Study the following line graph to answer these questions

Railway Time Schedule of an Express Train X Running Between City A and City H
a – Arrival of train,      
b – Departure of train
A, B, C, D, E, F, G and H – Cities through which the train runs. 
a-d – Indicates stoppage/halting of the train at the city station.
Q1.The average speed of the train maintained between two successive stations was maximum between
(1) E – F
(2) F – G
(3) G – H
(4) both G – H and F – G
(5) None of these
Q2. Between how many pairs of consecutive stations does the speed run below the overall average speed of the entire trip?
(1)4
(2)1
(3) 3
(4)2
(5) None of these
Q3. If the train stops at each city for 30% more time that what it is at the moment, then at what time will it reach the city H after departing from City A as per schedule?
(1) Data insufficient
(2)17 : 03
(3)16 : 41
(4)16 : 58
(5) None of these
Q4.The train begins its onward journey from City A and it is extended to beyond City H to a City M due to some unavoidable reason. The train starts its return journey immediately after it reached City M. The train returns with a speed of 90 km/h without any stoppages in between and reaches City A at 2 : 25 am. Find the distance between City H and City M.
(1)40 km
(2)90 km
(3)70 km
(4)59 km
(5) None of these

Answer & Explanation

1. (4) 
the average speed maintained by the train between different stations E and F 106 km in 5/4 h   =((106×4)  )/5 = 84.8 kmph
Similarly, average speed between F and G = 176 km
In 2h = 88 km/h
Between G and H = (110*4)/5  = 88 km/h

2.(4) 
Over all average speed of the entire trip : 860 km in 11 hr 25 min
i.e. (685/137) h   =(860*12)/137   = 75.3 km/h
Average speed maintained by the train between station
A and B 140 km in 12/3h =  (140*3)/5 = 84 km/h
B and C 91 km in  4/3 h=91*(3/4) km/h=68.25 km/h
C and D 149 km in  103/60=(149*60)103  =86.8 km/h
D and E 88 km in 4/3 h=(88*3)/4=66 km/h
E and F = 84.8 km, FG = 88km/h, GH = 88km/h
Hence, for two stations it runs below the overall average speed

3.(3) 
The train stops at station B for 3 min more
The train stops at station C for 36 sec more
The train stops at station D for 1.5 min more
The train stops at station E for 3 min more
The train stops at station F for 4.5 min more
In all the trains stops 15 min and 36 s more
The train will reach city H after departing from city A at 16h, 40 min, 36 sec i.e., 16h , 41 min approx

4.(4)
Let the distance between city H and city M be x km.
Time taken by the trains from H and A (return)
860/90 = 86/9h
Given, the trains reaches  city A (return) at 2.25 am
i.e, at 26.25(hours)
The train reaches at H (return)
At   =    (26+25/60)-86/9= (16+31/36)h
Time taken from H to M and back
(607/36) – (192/12)= 4/3h
Since the train runs from G to H 88km/h, therefore the train runs 88km/h from city H to city M also. Also speed between city M and city H = 90kmph
x/88+x/90=4/3
x = 59 km

Leave a Comment

Your email address will not be published.