# A train runs first 40 km at an average speed 48 km.p.h., second 80 km at an averagespeed 64 km.p.h., then due to the maintenance of the track, travels for 15 minutes at anaverage speed 20 km.p.h....

A train runs first 40 km at an average speed 48 km.p.h., second 80 km at an average

speed 64 km.p.h., then due to the maintenance of the track, travels for 15 minutes at an

average speed 20 km.p.h. and finally covers the remaining distance at an average speed

25 km.p.h. If the average speed of the train for the entire journey is 45 km.p.h., find the

total distance that the train runs.

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### 3 Answers

Let us say the total distance is Pkm.

Time = distance/velocity

For section 1

`t_1 = 40/48 = 5/6`

Similarly;

`t_2 = 80/64 = 1.25`

distance travel at section 3 `= 20xx15/60 = 5km`

Time for section 4 assuming that distance is dkm;

`t_4 = d/25`

Average speed = (total distance)/(total time)

`P = 40+80+5+d = 125+d`

`45 = (125+d)/(5/6+1.25+0.25+d/25)`

Solving the above will give you;

`d = 25`

*So the total distance run by train is 125+25 = 150km*

Total journey `J` :

`J= 40 +80+20xx 1/4 +25t=125+25 xxt`

Total time `T` :

`T= 40/48+ 80/64+1/4+t=7/3+t` `=(7+3t)/3`

So: Average value speed: `S_a=45=J/T`

`45=(75(5+t))/(7+3t)`

`3=(5(5+t))/(7+3t)`

`3(7+3t)=25+5t`

`21+9t=25+5t`

`4t=4` `t=1`

So the last part of journei is of `25xx1= 25 ` Kms

So: `J=125+25=150` Kms

A train runs first 40 km at an average speed 48 km.p.h.,Thus train took time to cover 40 Km

`t_1=40/48=10/12 hr`

Train cover 80 km at an average speed 64 km.p.h., Thus

`t_2=80/64=5/4 hr`

then due to the maintenance of the track, travels for 15 minutes at an

`t_3=15 min= (1/4)hr`average speed 20 km.p.h. .

finally covers x km remaining distance at an average speed

25 km.p.h. Thus it took time

`t_4=x/25 hr`

Total time for journey`T=t_1+t_2+t_3+t_4`

`=10/12+5/4+1/4+x/25`

`=(10+18)/12+x/25`

`=7/3+x/25`

Total distance `D=40+80+(1/4)xx20+x`

`=120+5+x`

`=(125+x)Km`

Thus average speed = D/T

`45=(125+x)/(7/3+x/25)`

`45(7/3+x/25)=125+x`

`105+(9x)/5=125+x`

`(9x)/5-x=125-105`

`(4x)/5=20`

`x=(20xx5)/4=25`

**Thus totla distance for journey= 125+25=150 Km.**