Aptitude - Problems on Trains - Discussion

A = 27 - 23sec = 4sec.
B = 23 - 17sec = 6sec.
A:B = 4:6 = 2:3 in Sec.
A:B = 3:2 in Speed.

The explanation are excellent, but we asked the the speed, so why we calculated the time taken?

If we think let x and y be the length of the trains then, x/27,by/17 will be the speed of the trains respectively and the equation becomes x+y/23b=bx/27+y/17.

Solving we get x/y=81/34.

x+y = total distance.
23 = total time.
Speed = x/27+y/27.
Speed = distance/time.

Therefore x+y/23 = x/27+y/17 is the equation,
Can you tell me why this equation is wrong?

Hello there if anyone did't understood the concept behind it let me explain in my way.
So, everyone know when two bodies are moving towards each other or having an velocity in opposite direction the velocity is add up (e.g you are going on car at 20 m/s and from front the other car is coming at 10 m/s so you will cross each other at 30 m/s) V1+V2.

Now same concept applies here,

S1+S2 = V1T1+V2T2 (S=VT).
S1+S2 / V1+V2 = T1+T2.
AND T1+T2 is 23.
So S1+S2/V1+V2 = 23.

NOW for S1= 27V1 and S2 = 17V2.
Put values in eq and you will get the ratio,
27V1+17V2/V1+V2 = 23.
V1/V2 = 3/2.

Thanks.

s=vt.
t= s/v.
s= distance, v=speed.

27+3 = 30.
17+3 = 20.
30/20 = 3:2.

I can't understand the answer clearly. please explain how we get 3 : 2?

Train A length=x;time A=27;speed A=x/27.
Train B length=Y;time B=17;speed B=Y/17.

Now relative speed between train A and B is=(x+Y)/23.

Now we equating these equation through relative speed;
(x+Y)/23=x/27+Y/17.

=> (x+Y)/23=(17x+27Y)/459,
=>459x+459Y=391x+621Y,
=>68x=162Y,
=>x/Y=162/68,
=>(x/27)/(Y/17)=(162/27)/(68/17),
=>speed A/speed B=3/2,
=>speed A : speed B = 3:2.

How possible 3:2?

Because the first train takes more time than the second one. But the speed ratio is 3:2. Actually the second train moves fast than first.

Very nice explanation, Thanks @Poornima.