After travelling for 30 minutes a train meets an accident- due to which it has to stop for 45 minutes- Due to the accident its speed is also reduced to 23 of its former value and the train reaches its destination 1 hour 30 minutes late- Had the accident occurred 60 km later- the train would have reached 30 minutes earlier- The length of journey is

The correct option is D 120 km #160;Let the distance be d and speed be v. Therefore the normal running time is d v . ...

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After travell...

Question

After travelling for 30 minutes a train meets an accident, due to which it has to stop for 45 minutes. Due to the accident its speed is also reduced to $\frac{2}{3}$ of its former value and the train reaches its destination 1 hour 30 minutes late. Had the accident occurred 60 km later, the train would have reached 30 minutes earlier. The length of journey is

90 km

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120 km

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150 km

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180 km

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Solution

The correct option is D 120 km
$L e t t h e d i s t a n c e b e d a n d s p e e d b e v . T h e r e f o r e t h e n o r m a l r u n n i n g t i m e i s \frac{d}{v} . I n t h e f i r s t c a s e, r u n n i n g t i m e i s (d + 1 \frac{1}{2}) h r a n d i n t h e s e c o n d c a s e, t h e r u n n i n g t i m e i s (\frac{d}{v} + 1) h r . N o w i n e a c h c a s e c a l c u l a t e t h e r u n n i n g t i m e s, e q u a t e t h e m w i t h t h e r e s p e c t i v e t i m e s a s s h o w n a b o v e, s o l v e s i m u l t a n e o u s e q u a t i o n s i n d a n d v .$

$L e t t h e d i s t a n c e b e d k m a n d s p e e d o f t h e t r a i n b e v k m / h r . T h e n n o r m a l t i m e t o r e a c h t h e d e s t i n a t i o n i s \frac{d}{v} h r . I n t h e f i r s t c a s e d u e t o l a t e r u n n i n g b y 1 \frac{1}{2} h r t h e t o t a l t i m e t_{1} = (\frac{d}{v} + \frac{3}{2}) h r \Rightarrow t_{1} = \frac{2 d + 3 v}{2 v} h r - - - - (1) S i m i l a r l y i n t h e s e c o n d c a s e, t h e t r a i n i s l a t e b y o n e h o u r . ∴ t_{2} = \frac{d}{v} + 1 = \frac{d + v}{v} h r - - - - (2) F i r s t c a s e - N o w t h e t r a i n t r a v e l s 30 k m w i t h a s p e e d v, s t o p s f o r 45 m i n . = \frac{3}{4} h r f o r a c c i d e n t . ∴ T o t a l t i m e t a k e n u p t o t h e p o i n t o f a c c i d e n t i s (\frac{30}{v} + \frac{3}{4}) h - - - - (a) R e s t o f t h e d i s t a n c e i s (d - 30) k m a n d t h e t r a i n r u n s w i t h a s p e e d o f \frac{2}{3} v . ∴ t i m e t a k e n t o r e a c h t h e d e s t i n a t i o n i s (d - 30) \times \frac{3}{2 v} h r = \frac{3 d - 90}{2 v} h r - - - - (b) ∴ T o t a l t i m e t a k e n b y t h e t r a i n i s (a) + (b) = t_{1} = (\frac{30}{v} + \frac{3}{4} + \frac{3 d - 90}{v}) h r . t a k i n g t_{1} f r o m e q u a t i o n (1) w e h a v e . \frac{30}{v} + \frac{3}{4} + \frac{3 d - 90}{2 v} = \frac{2 d + 3 v}{2 v} \Rightarrow 2 d - 3 v = 60 - - - - (3) S e c o n d c a s e - T h e a c c i d e n t h a p p e n s 60 k m l a t e r . T h a t i s i t r u n s (60 + 30) k m = 90 k m w i t h a s p e e d o f v k m / h r . ∴ T i m e t a k e n i s \frac{90}{v} h r .$
$T h e n i t s t o p s f o r \frac{3}{4} h r ∴ T o t a l t i m e t a k e n u p t o t h i s p o i n t i s (\frac{90}{v} + \frac{3}{4}) h r = \frac{360 + 3 v}{4 v} h r - - - - (a) A f t e r t h e a c c i d e n t t h e d i s t a n c e l e f t = (d - 90) k m a n d t h e s p e e d = \frac{2 v}{3} k m / h r ∴ T h e t o t a l t i m e t a k e n t o c o v e r (d - 90) k m i s (d - 90) \times \frac{3}{2 v} h r = \frac{3 d - 270}{2 v} h r - - - - (b) ∴ T h e t o t a l t i m e t a k e n t o r e a c h t h e d e s t i n a t i o n i s (a) + (b) = t_{2} = [\frac{360 + 3 v}{4 v} + \frac{3 d - 270}{2 v}] h r . r e p l a c i n g t_{2} f r o m e q u a t i o n 2. \frac{360 + 3 v}{4 v} + \frac{3 d - 270}{2 v} = \frac{d + v}{v} \Rightarrow 2 d - v = 180 - - - - (4) M u l t i p l y i n g e q u a t i o n (4) w i t h 3 a n d s u b t r a c t i n g f r o m e q u a t i o n (3) w e g e t 4 d = 480 \Rightarrow d = 120 k m (A n s)$

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