**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

**Q2.**Between how many pairs of consecutive stations does the speed run below the overall average speed of the entire trip?

**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?

**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.

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