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 47302 by JDlix last updated on 08/Nov/18

$A t r a i n w h i c h t r a v e l s a t a u n i f o r m s p e e d d u e t o m e c h a n i c a l$ $f a u l t a f t e r t r a v e l i n g f o r a n h o u r g o e s a t 3 / 5 t h o f t h e o r i g i n a l$ $s p e e d a n d r e a c h e s t h e d e s t i n a t i o n 2 h o u r s l a t e . I f t h e f a u l t$ $o c c u r e d a f t e r t r a v e l i n g a n o t h e r 50 m i l e s t h e t r a i n w o u l d h a v e$ $r e a c h e d 40 m i n u t e s e a r l i e r . W h a t i s t h e d i s t a n c e b e t w e e n t h e$ $t w o s t a t i o n s ?$
	Answered by math1967 last updated on 08/Nov/18

	$l e t d i s t a n c e b e t w e e n t w o s t a t i o n y m i l e s$ $o r i g i n a l s p e e d o f t r a i n = x m i l e s / h r$ $f o r 1 h r t r a i n c o v e r s x m i l e s$ $r e m a i n i n g d i s t . = (y - x) m i l e s$ $∴ 1 + \frac{y - x}{\frac{3 x}{5}} = \frac{y}{x} + 2$ $\Rightarrow \frac{5 y - 5 x}{3 x} - \frac{y}{x} = 2 - 1$ $\Rightarrow \frac{2 y - 5 x}{3 x} = 1 \Rightarrow 2 y = 8 x \Rightarrow y = 4 x$ $f r o m 2 n d c o n d i t i o n$ $\frac{50}{\frac{3 x}{5}} - \frac{50}{x} = \frac{40}{60} \Rightarrow \frac{250 - 150}{3 x} = \frac{2}{3}$ $∴ x = 50 ∴ y = 4 \times 50 = 200$ $d i s t . b e t w e e n t w o s t n . = 200 m i l e s A n s .$
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