plzz solve these aptitude

[sweet_madhu]

1>A train blows a siren one hour after starting from the station. After that it travels at 3/5th of its speed it reaches the next station 2 hours behind schedule. If it had a problem 50 miles farther from the previous case,it would have reached 40 minutes sooner. Find the distance between the two stations.
2> An army 50 miles long marches at a constant rate. A courier standing at the rear moves forward and delivers the message to the first person and then turns back and reaches the rear of the army as the army completes 50 miles. Find the distance travelled by the courier.


plzz help me to solve these aptitude.....tell me the approach to solve these aptitude

[RisingSun]

Sol 1 : X-------------------M----------N--------------------Y

start ----> end

M is 1 hour away from start X after which the train starts travelling at 3/5th of speed s = 3s/5

N is point 50 km ahead of M.

Case 1 . X-M-Y stopppage causes delay by 2 hours.

Case 2 . X-N-Y stoppage causes delay by 1 hour 20 mins ( 40 mins sooner).

Clearly difference in case timings is due to MN being travelled at different speeds i .e.

3s/5 rather than at s itself causing a delay of 40 mins.

Therefore, 50/(3s/5) - 50/s = 40 / 60 or 50 *(5/3s - 1/s) = 2/3 or s =50 mph

And distance can be found from the fact that travelling MY at 3s/5 causes delay by 2 hours

i.e. x/30 - x/50 = 2 or x =150 miles.

Hence total distance = 50(1hour travel) + 150 = 200 miles.

[RisingSun]

D = distance by courier
w = distance by army
v = length of army

D = 2k + 2v - w
where:
k = v(j - v) / (v + w - j)
j = sqrt(v^2 + w^2)This can be amazingly simplified to:
D = v + sqrt(v^2 + w^2)
So if v=60 and w=45:
D = 60 + sqrt(60^2 + 45^2) = 60 + 75 = 135