Question

A goods train leaves a station at 6 pm,followed by express train,which leaves at 8 pm and travels 20km/h faster than the goods train.
The express train arrives at a station,1040 km away,36 mins before the goods train.assuming that the speeds of both the train remain constant between the two stations,calculate their speeds

Answer 1

Goods train speed= v km/ hr

express train speed= 39+ v km / hr

distance travelled by both= 180 km

Time taken by goods train= t hr

Time taken by express train= t- 2 hours- (36/60) hours=t- 13/5

Now

v= 180/ t km /hr ...... (1)

v+39=180/(t-13/5) km / hr ..... (2)

put value of v from (1) into (2) we get:
180/t+39=180/t-13/5

so, t = 5 and t = -12/5

we take t = 5 hours

∴ speed of goods train= v = 180/5 = 36 km /h

and speed of express train = 36+ 39 = 75 km / hr



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Answer 2

Hello..

Let the speed of the goods 'x' km/hr

speed of express train = (x+20) km/hr

leaves a station at 6pm

express train leaves the station at 8pm

express train reaches its destination 36min before goods train.


Time taken = distance covered /speed

Total distance=1040km

Time taken by goods train to cover
1040 km=1040/x

Time taken by express train to cover 1040 km=1040/(x+20)

As per given condition we have,
1040/x=1040/(x+20)+2(36/60)

1040/x-1040/(x+20)=2(3/5)

(1040x+20800-1040x)/(x²+20x)=13/5

13x²+260x=104000

x²+20x-8000=0

x²+100x-80x-8000=0

x(x+100)-80(x+100)=0

(x-80)(x+100)=0

x=80 (or) x=-100
Here x>0
Therefore x=80

speed of goods train=80 km/hr

speed of express train=100 km/hr.