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-

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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