Math, asked by shraddhamishra211, 10 months ago

A train, an hour after starting, meet with an accident which delays it for half an hour and after this, it proceeds at 3/4th of its former rate and arrives 3.5hrs late. Had the accident happened 90 km farther along the line, it would hav arrived only 3 hour late. The length of trip in km was..?​

Answers

Answered by Anonymous
5

Answer:

Normal time be x for remaining distance, but time taken = 4x/3

Difference in time = (1+1/2+4x/3)−(1+x)=72

x=9

2nd case of going further 90 km which lessens the difference by 1/2

If it loses 90 km in 1/2 hr.

Total difference was 3 + 1/2 out of which 1/2 was for repair

Difference in travelling is 3+1/2-1/2=3 hrs

It will lose 3 hrs in 90 x 6=540 km

Now it travelled for 10 hr hours, which is 1 hr + 540 km

So 540 in 9hr

So 60 km per hour

Total distance = 540+60=600km

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