Aptitude - Problems on Trains - Discussion

T1=27-23=4,T2=23-17=6.
Speed=distance/time.

"Speed" inversely propotional to "time"
(S1/S2)=(T2/T1)=6/4=3/2,
S1:S2=3:2.

@Priyesh Verma.

How can be a train, which taking time than the another, is consider as faster?

In the above example, the time taken by 1 train is 23sec and the other one is 17 sec but the ratio shows that 1train is faster than second.just imagine that length of the train is 54m.

Then,
54/27::54/17~=2:3.

Speed=(distance/time).

It rewritten as (distance/speed)=time.
Same case:
(relative distance/relative speed)=time
We understand the only thing how to calculate relative speed and relative distance,
Consider car A travel at x km/hr and car B travels at y km/hr
Both A and B travel in same direction means
Their relative speed=subtraction of two speed (x+y)km/hr.
Relative distance= (speed1*time1)-(s2+t2).
=27x-17y.

Here 27 and 17 are time1 and time2.
X and y are speed1 and speed2.

Case2:
CarA and carB travels opposite direction.
Relative speed= x+y(addition of speeds)
Relative distance=(speed1*time1)+(s2*t2)
So we get. = 27x+17y.
Finally (relative distance/relative speed)=time,
Therefore (27x+17y)/(x+y)=23,

Maybe my solutions help for you

Very clear explanations. Thanks.

860m, It consists of two cases.

Two trains are moving in opposite direction with the speed of 36 k.m. /h and 54 k.m. /h cross each other in 12 seconds. The length of the 2nd train is half of the 1st train. The 1st train crosses a platform in 1. 30 minutes. Then what is the length of the platform? Please solve this.

@Sudhan.

Isn't 27-17 = 10? How do you say it's 6?

Make it simple.

27-17 =6 :27-23 =4
6:4 or 3:2.

(x+y) I am not understanding that where this comes here this question, please give answer.

Relative time 1st train= 27-23=4,
second train= 23-17=6,
1st:2nd= 6:4=3:2,
Ans:3:2.