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

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roshan-rox | Valedictorian

Posted May 14, 2013 at 8:01 AM via web

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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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pramodpandey | College Teacher | Valedictorian

Posted May 14, 2013 at 8:42 AM (Answer #1)

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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.

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oldnick | Valedictorian

Posted May 15, 2013 at 3:35 AM (Answer #2)

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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

 

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted May 14, 2013 at 8:20 AM (Answer #1)

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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

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