Question

two trains crosses each other in 14 sec when they are moving in opposite direction, and when they are moving in same direction they crosses each other in 3 minute2 sec. find the speed of the faster train by what percent more than the speed of the slower train?

Answer 1

Let the speed of the slower train be = V
Let the speed of the faster train be =  a * V,    where a is the number of times faster train moves compared to the other.

When the trains are moving in opposite directions, the relative speed of the two trains wrt each other = V + a V = (1+ a) V.

Total distance to cross with this relative velocity = L (say). This is the sum of the lengths of the two trains.

Time taken to cross each other = L / [ (a+1) V] = 14 sec   --- (1)

When the trains are moving in the same direction, their relative speed 
    = a V - V = (a - 1) V

Time duration for cross each other = L / [ (a-1) V ] = 182 sec  ------ (2)

Divide equation (2) by equation (1):

(a+1) / (a - 1)  =  182/14 = 13

Solving this we get:  12 a = 14
                                 a = 7/6

% required = [ a V - V ] / V  * 100 = (a - 1) 100 = 100/6 = 16.67 %

Answer 2

Answer:

suppose trains x and y..

Explanation:

x+y=14(seconds).......(1)

x-y=182(seconds).....(2)

...

after solving two equations

x=98 & y=84

there for ...x is faster train ...more than y by 14

..hence

14/84*100= 16.67%