A-train-after-travelling-70-km-from-a-station-A-towards-a-station-B-develops-a-fault-in-the-engine-at-C-and-covers-the-remaining-journey-to-B-at-3-4-of-its-earlier-speed-and-arrives-at-B-1-hour-a

Question Number 17610 by Tinkutara last updated on 08/Jul/17

A train, after travelling 70 km from a

station A towards a station B, develops

a fault in the engine at C, and covers

the remaining journey to B at \frac{3}{4} of its

earlier speed and arrives at B 1 hour

and 20 minutes late . If the fault had

developed 35 km further on at D, it

would have arrived 20 minutes sooner .

Find the speed of the train and the

distance from A to B .

Commented by Tinkutara last updated on 08/Jul/17

Answered by alex041103 last updated on 08/Jul/17

L e t S b e t h e d i s t a n c e f r o m A t o B

a n d v - t h e s p e e d o f t h e t r a i n

T h e n

\frac{S}{v} + \frac{4}{3} = \frac{S - 70}{\frac{3}{4} v} + \frac{70}{v} ∣ - \frac{70}{v}

\frac{S - 70}{v} + \frac{4}{3} = \frac{4}{3} \frac{S - 70}{v}

\Rightarrow \frac{S - 70}{v} = 4 o r S = 4 v + 70

A l s o t h e f o l l o w i n g i s t r u e

\frac{S}{v} + 1 = \frac{S - 105}{\frac{3}{4} v} + \frac{105}{v} ∣ - \frac{105}{v}

\frac{S - 105}{v} + 1 = \frac{4}{3} \frac{S - 105}{v} ∣ - \frac{S - 105}{v}

1 = \frac{1}{3} \frac{S - 105}{v} \Rightarrow S = 3 v + 105

\Rightarrow 3 v + 105 = 4 v + 70

\Rightarrow v = 35 k m / h a n d S = 210 k m

N o t e :

20 m i n u t e s s o o n e r m e a n s 20 m i n s

s o o n e r t h a n 1 h o u r a n d 20 m i n u t e s

i . e . 1 h o u r l a t e r t h a n p l a n n e d

Commented by Tinkutara last updated on 09/Jul/17

Thanks Sir!