A train travels at a particular speed for a duration of one hour, after which one of its engine malfunctions reducing its speed to 3/5th of the actual speed before the occurrence of fault in engine. It travels at this speed for 2 hours to reach at its destination. If the fault had occurred 50 miles later on, the train would have reached its destination 45 minutes early. Find the distance traveled by the train.
Let the actual speed of train be X miles/hour. after fault its speed will be reduced to 3X/5 miles/hour. If we could find the speed of the train, we will get the solution!!! by seeing one can find that time difference( 45 minutes) occur between two cases, in the 50 miles distance traveled by the train at two different speeds(X and 3X/5). => If the time taken to cover 50 miles at speed 3X/5 is 't' hours then time taken to travel 50 miles at speed X is t-(45/60) = (t-0.75) hours. We can write ...=> (3X/5)*t = X*[t-0.75] = 50 miles -->(i) solving we get... t=15/8 hours. then substituting value of 't' in first term of eq. (i) => (3X/5)*15/8=50 solving we get... X= 400/9 miles/hour. Answer : So, solving the problem with first case we get the distance traveled by the train as 97.77 miles.
Case first : let us assume speed to train be x miles/hours Total distance travelled in case 1 = x*1 + 3x/5*2 = 11x/5 milesCase 2 : Total distance travelled in case 2 = x+50 + 3x/5*(2-45/60) = 50 +7x/4 comparing these two equations we have x= 1000/9Hence total distance =11x/5 = 11*1000/9 *1/5 = 244.44 miles
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RE: A train travels at a particular speed for a durati...
Case first : let us assume speed to train be x miles/hours Total distance travelled in case 1 x*1 + 3x/5*2 11x/5 milesCase 2 : Total distance travelled in case 2 x+50 + 3x/5*(2-45/60) 50 +7x/4 comparing these two equations we have x 1000/9Hence total distance 11x/5 11*1000/9 *1/5 244.44 miles
RE: A train travels at a particular speed for a durati...
I get 366.67 miles. Let v train full velocity Let d distance Case 1 - train goes (1 hour * v) + (2 hours * 3 v / 5) d Therefore d v * 2.2 hours Case 2 - train covers 50 miles 45 minutes faster at v than at v * 3/5 ( 50 miles / (.75 hours) ) v * 2 / 5 0.4 v v 166 miles / hourplug v into answer from 1 and you get d 166 mph * 2.2 hours 366.67 miles.
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