Question

Two trains, Calcutta Mail and Bombay Mail, start at the same time from stations Kolkata and Mumbai respectively towards each other. After passing each other, they take 12 hours and 3 hours to reach Mumbai and Kolkata respectively. If the Calcutta Mail is moving at the speed of 48 km/h, the speed of the Bombay Mail is

Answer 1

Let the two Trains pass after x hours,

then for Calcutta mail, we have

Distance travelled in x hours = 48x kms

Distance travelled after passing Bombay mail = $48\times 12 = 576 Km$

Now, For Bombay Mail, we have

Distance travelled in x hours = 576Km

Speed = $\frac{576}{x} Km/hr$

Distance travelled after passing Calcutta Mail = 48x Km

$Time = \frac{Distance}{Speed}$

$3 = \frac{48x}{\frac{576}{x} }$

$3 = \frac{48x^{2} }{576}$

$48x^{2} = 1728$

$x^{2} = \frac{1728}{48}$

$x^{2}  = 36$

$x = 6$

Therefore, speed of Bombay Mail $= \frac{576}{6}$

                                                         $= 96 Km/hr$

Answer 2

Answer:

96km/hr∠

Step-by-step explanation:

t1=12 , t2=3

t1∠t2

a=48km/hr

speed of b=a√(t1/t2)

                 =48 √(12/3)

                 =48 √4

                  =48×2

                  =96 km/hr