Train Speed

A train ran at a certain speed for one hour and then due to a technical problem, it ran at 3/5th of its original speed and reached the destination 2 hrs late. The driver said, had the problem occurred 50 miles later, we would have gained 40 minutes. What is the distance between the two cities that the train
traveled?
This question is related to Infosys Interview

Questions by sudarshan2u

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amanbaua

  • Sep 10th, 2010
 

This question can be solved in 2 parts
1) Find the original speed 2) Find the original time

1)
The difference of 40 min is in this 50 miles
Time taken when accident took at original place=50/3s/5

Time taken when acc took 50 miles later=50/s
We can say 50*5/3s -50/s=40/60

Which we get as s=50

2) Can we say speed decreases by 40%
Hence time increases by by 66.66%
Let us take orig time as x
2/3*x=2 hence x=3

Hence total distance=50*3=150km

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hsirah

  • Oct 1st, 2010
 

Total distance is 200 km.

Case 1: t1 + t2 = T +120
Case 2: t1 + t2_ + t3_ = T+80

t2 -(t2_+t3_) = 40

(t2-t3_)-t2 = 40

t2 = x2/(3/5*x1/60)
t3_ = x3_/(3/5*x1/60)

t2 = x2_/x1/60

Also x2 - x3_=50

you can solve for x1 with these eqns: x1 =50

also in case 1 : x2 = A-50
and T = A/x1/60

with this you can solve for A, Final answeer A =200 

In this question, we have to solve for original time and original speed and then we will calculate the total distance.
Let x & y be the speed & time respectively.
Therefore, total distance=xy mile

FIRST CASE:
(1 hour + time taken to travel after problem) = (original time + 2 hours)
=> 1+ (xy-x)/(3/5x) = y+2
=> (5y-5)/3 = y+1
=> 5y-5 = 3y+3
=> 2y = 8
=> y=4 hour

SECOND CASE:
distance travelled in 1 hour = x mile
i.e. total time taken to travel the journey of xy mile = ( (x+50)/x + (xy-(x+50))/3/5x)
according to question if problem would have occurred after 50 mile then driver would have saved 40 min
=> (y+2) - ( (x+50)/x + (xy-(x+50))/3/5x) = 40/60
=> 6 - (18x-100)/3x = 2/3
=> 18x-18x+100 = 2x
=> 2x = 100
=> x=50 mile/hour

Thus, we have time=4 hour and speed=50mile/hour
Hence total distance= 4*50 mile = 200 mile

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