Aptitude - Problems on Trains - Discussion

Solution for 220
__________ =6 is
(250+5x/18)

220*18
_______ =6 => 220*18=6(250+5x)
250+5x
=> 220*3=(250+5x)
=>660=250+5x
=> 660-250=5x
=>410=5x
=>x=82

Here the ans is in km/hr. so frm starting onwards we sould calculate
the problem without converting into m/s.

Let speed of d second train= xkm/hr

Distance covered=(108+112)=220m=0.22km

Time=6sec= 6/3600hr =1/600hr

So
1/600hr= 0.22km/(50+x)
x=82km/hr

Formula : T=(x+y)/(u+v).

Where,
x,y -- Distance.
u,v -- Speed.

6 = (108+112)/((50*5/8)+V).

6v+250/3 = 220.

Therefore, v = 205/9 m/sec.

v = 205/9*(18/5) = 82km/hr.

Let me explain in details.

First train length, a = 108m.
Its speed, u = 50 km/hr = 50*5/18 m/sec= 125/9 m/sec.

Second train length,b = 112m.
Let, its speed, v = x m/sec.
Each other crosses in 6 sec.

So, we have, t= ((a+b)/(u+v)).
6= ((108+112)/((125/9)+x)).

(125/9)+x = 220/6.
x = (220/6)-(125/9)= 205/9 m/sec.

Converting in km/hr.

(205/9)*(18/5) = 82 km/hr.

Well I have a doubt, how did you get 5/18? Sorry if it is a stupid question!

I would like to bring this error to your notice that it is 82 m/s and not 82 km/hr.

Hai I have another option.

D = 220.
S = 50 kmph.
T = 6.
Other speed = x.

So speed = (x+50)*5/18.

T = D/S. Find T subtract 50 that's answer = 82 kmph.

Let me explain in detail.

Let the speed of second train be xkm/hr.

Therefore, relative speed = (50+x)km/hr (since it is opposite direction the speeds should be added). Then no need to confuse yourself by converting the speed to m/s.

Take it as it is, so as to get our answer in km/hr (as the given options are in km/hr). Now just convert the distances 108m & 112m into km/hr.

So add the distances & then convert it.

D = 108+112 = 220m = (220 x 18/5) = 792km/hr.

So now distance is in km/hr.

T = D/S.
6 = 792/(x+50).
6x+300 = 792.

Therefore x = 82km/hr.

(108+112)/6 = 50 + x.
(220/6)*(18/5) = 50 + x.

132 = 50 + x.
x = 82 Km/hr.

How does 660 comes in last step?