A train which travels at a uniform speed due to mechanical fault after traveling for an hour goes at 3/5 th of the original speed and reaches the destination 2 hours late. If the fault occured after traveling another 50 miles the train would have reached 40 minutes earlier. What is the distance between the two stations ?


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Question Number 94820 by jdmath last updated on 21/May/20

$A t r a i n w h i c h t r a v e l s a t a u n i f o r m s p e e d d u e t o m e c h a n i c a l f a u l t a f t e r$ $t r a v e l i n g f o r a n h o u r g o e s a t 3 / 5 t h o f t h e o r i g i n a l s p e e d a n d r e a c h e s t h e$ $d e s t i n a t i o n 2 h o u r s l a t e . I f t h e f a u l t o c c u r e d a f t e r t r a v e l i n g a n o t h e r 50$ $m i l e s t h e t r a i n w o u l d h a v e r e a c h e d 40 m i n u t e s e a r l i e r . W h a t i s t h e$ $d i s t a n c e b e t w e e n t h e t w o s t a t i o n s ?$
	Answered by prakash jain last updated on 21/May/20

	$Assume distance = s km$ $speed = v (km / hr)$ $expected time = \frac{s}{v} = t (I)$ $case when fault occured after 1 hour .$ $distace traveled in 1 hour = v$ $remaining distancd = s - v$ $time taken to cover remaining = \frac{s - v}{\frac{3}{5} v}$ $total time = 1 + \frac{5 (s - v)}{3 v} = t + 2 = \frac{s}{v} + 2$ $given that train is late by 2 hrs wrt (I)$ $1 + \frac{5 (s - v)}{3 v} = t + 2 = \frac{s}{v} + 2 (II)$ $case when fault occured after$ $after addition 50 kms$ $distace traveled in 1 hour = v$ $time taken to cover extra 50 km = \frac{50}{v}$ $remaining distancd = s - v - 50$ $time taken to cover remaining = \frac{s - v - 50}{\frac{3}{5} v}$ $total time = 1 + \frac{50}{v} + \frac{5 (s - v - 50)}{3 v}$ $given that train is late by 40 \min wrt (I)$ $40 minutez = \frac{2}{3} hrs$ $1 + \frac{50}{v} + \frac{5 (s - v - 50)}{3 v} = t + \frac{2}{3} = \frac{s}{v} + \frac{2}{3} (III)$ $solve (I 1) & (III) to calculate$ $required value .$
	Commented by necxxx last updated on 21/May/20

	$m r . P r a k a s h J a i n w e l c o m e b a c k . I$ $s i n c e r e l y m i s s y o u o n t h i s f o r u m .$
	Commented by jdmath last updated on 21/May/20

	$t h a n k s r l y s i r$
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