ok, there is no distance let it be x.
the speed of both will be different.
time is same.
so. D=s1xt X=s1xt

D=S2xt
S2/S1= ...................................
..........

 

ok, there is no distance let it be x.
the speed of both will be different.
time is same.
so. D=s1xt X=s1xt

D=S2xt
S2/S1= ...................................
..........

Hint: Ask yourself where train 1 is when the front of train 2 is 'flush' with the observer

 

I am not getting how to solve further without watching answer.
please help me again.

 

Hint: As yourself where train 1 is when the front of train 2 is 'flush' with the observer

What this mean?

 

What this mean?

Another typo - please re-read post #142

 

@RRITESH KAKKAR

I's 7:45 AM here and I'd best get my um... beauty rest... My best advice at this point is that you rephrase the problem in your own words -- if all else fails go ahead and solve it via standard algebraic procedure... Talk to you later!

Best regards
HP

 

Hint: Ask yourself where train 1 is when the front of train 2 is 'flush' with the observer

Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
ok, the first was still crossing the man 10sec were left on 1st.

 

I's 7:45 AM here and I'd best get my um... beauty rest... My best advice at this point is that you rephrase the problem in your own words -- if all else fails go ahead and solve it via standard algebraic procedure... Talk to you later!

Best regards
HP

Here is 7:16pm you were talking with me in early morning great .
what is your job.

 

Here is 7:16pm you were talking with me in early morning great .
what is your job.

See post #101 -- Gotta go!

 

X-Rays

Doctor?

 

X-Rays

Doctor?

Actually, I was merely saying that I am essentially a 'layabout' enamored of my hobbies -- FWIW: EHT (also known as 'high voltage') and radiography make for very interesting hobbies! - of course, as with all things useful, they are dangerous in unskilled or careless 'hands'...

So... How's the math problem going?

Best regards
HP

PS -- It occurred to me, based upon your earlier post -- That the time-of-day at your location (at the time of this writing) is 12:50 AM? --- Best to get your rest!

 

@RRITESH KAKKAR

I've just discovered, corrected and annotated another major typo in post #130 --- This is becoming embarrassing!

 

Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
ok, the first was still crossing the man 10sec were left on 1st.

I am still on this problem.

 

What is the time on forum Running ?

 

Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:

The speed in mtr/sec of first is 12.7m/sec, secound is 10m/sec time is 36sec
how to solve forward?

 

Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:

The speed in mtr/sec of first is 12.7m/sec, secound is 10m/sec time is 36sec
how to solve forward?

Hint: The difference in the trains' velocities might be regarded as 'rate' in this exercise

 

What is the time on forum Running ?

That depends upon your 'time zone' setting in 'preferences'...

For your location the correct setting is "UTC +5:30"

 

Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:

The speed in mtr/sec of first is 12.7m/sec, secound is 10m/sec time is 36sec
how to solve forward?
Hint: The difference in the trains' velocities might be regarded as 'rate' in this exercise

Then how to do this?

 

Re: The velocity ratio exercise...

I am still on this problem.

Ok... I'll assume you're weary of approaching this from the 'trick question perspective' and wish to solve the exercise in 'standard fashion'?

Hint: The conventional (and highly convenient) method of solving problems of this sort is to take a 'simultaneous approach' (CIP - write an equation for each train wherein the respective train's velocity is expressed by a unique variable -- Then solve the equations simultaneously (via addition or substitution)...

Hint 2: You're gonna kick yourself -- when you 'spot' the 'trick'

Best regards
HP

 

Then how to do this?

Look again at post #156 --- It's merely a Distance, Rate, Time problem