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 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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