So, there are two trains on parallel tracks facing opposite directions. The starting and ending points of the trains form a rectangle 5m x 150m. Two men standing at the tail end of each train start moving towards each other along the line of the tâ€¦

So, there are two trains on parallel tracks facing opposite directions. The starting and ending points of the trains form a rectangle 5m x 150m. Two men standing at the tail end of each train start moving towards each other along the line of the train at the same time when the trains start moving such that each train is travelling in the direction of the man present in the train. The speeds of the men are 18 Kmph and 36 Kmph respectively with respect to the corresponding trains in which they are moving. The speed of trains are 3 times the speed of the corresponding men. After how much time, does the distance between the two men become minimum, given that the tracks are separated by 5m?

Bharathi.

P.S: Please post spoilers and answers in the thread itself but only after 1 PM July 16, 2003. Any ambiguity in the question can be posted anytime

So, there are two trains on parallel tracks. The starting and ending points form a rectangle 5 x 150. Two men standing at the tail of each train start moving towards each other in the train at the same time when the trains start moving such that each train is travelling in the direction of the man present in the train. The speeds of the men are 18 Kmph and 36 Kmph respectively with respect to the corresponding trains in which they are moving. The speed of trains are 3 times the speed of the corresponding men. After how much time, does the distance between the two men become minimum?

Bharathi.

P.S: Please post spoilers and answers in the thread itself but only after 1 PM July 16, 2003. Any ambiguity in the question can be posted anytime

Hey, is this really supposed to be a CAT question?

In my opinion, a math question should not leave anything for assumptions or common sense. The problem statement should be clear, precise and unambiguous -UNLESS what you are posing is some sort of a riddle or something.

I see too much ambiguity in this:

- So, there are two trains on parallel tracks ::

**heading the same way or the opposite way?**

- The starting and ending points form a rectangle 5 x 150. ::

**starting and ending points of WHAT? the trains or the tracks?.... AND....what are the units...kilometers? meters? ......AND .....what does 150 correspond to ...the length of the train/track or the separation between them?**

- Two men standing at the tail of each train ... ::

**You mean two men standing one at the tail of each train? .... Though the rest of the problem makes it obvious(?) sentence construction should also be accurate - simply because one can't expect an accurate answer after posing a vague question.**

- ...start moving towards each other in the train ... ::

**eh? "Towards" ... you mean along the imaginary straight line that joins them? .. err...anyway?**

Waiting ...

...

...

Still Waiting ...

bhars18 SaysThe speeds of the men are 18 Kmph and 36 Kmph respectively with respect to the corresponding trains in which they are moving. The speed of trains are 3 times the speed of the corresponding men. After how much time, does the distance between the two men become minimum?

Do we know the length of the train...i am comin up with some crazy solutions....and i need the lenght of train to fig the stuff out!!!

man am so screwed...fill us in on this one!

raghuveer_v SaysHey, is this really supposed to be a CAT question?

The question is. But the phrasing was mine :D.

In my opinion, a math question should not leave anything for assumptions or common sense. The problem statement should be clear, precise and unambiguous -UNLESS what you are posing is some sort of a riddle or something.

Agreed....

I see too much ambiguity in this:

Question edited now. Hope u'll forgive a mere mortal with poor english... er... question phrasing capabilities....

Waiting ...

...

...

Still Waiting ...

Sorry I kept u waiting :).

If there are still more ambiguities, please point out.

Thanks.

Bharathi.

After how much time, does the distance between the two men become minimum

I got the answer as 5 Seconds...though the Qn..sounded wierd...i made some

Assumptions :)

T1

----------------------------

>------------------------

T2

Length of T1=Length of T2 = 150 m

Speed of M1(person in the train T1) = 18Kmph = 5m/s

Speed of T1=15m/s (thrice M1)

Relative speed 1 = 5 + 15 = 20m/s

Speed of M2(person in the train T2) = 36Kmph = 10m/s

Speed of T2=30m/s (thrice M2)

Relative speed 2 = 10 + 30 = 40m/s

Total distance to be covered = 150m + 150m = 300m (as M1 and M2 will be closest when they have a distance of 5 mts(breadth of the track)

So the time they will be closest will be = 300 (Rel. Speed 1 + Rel Speed 2)

= 300/(20+40) = 5 seconds.

P.S. Bharathi thot i solved it in 5 secs..though the answer i got itself is 5 secs anyways am posting it....took some trouble to make those look like trains!!!

Considering that the farthest distance between the trains is 150 m as shown in the fig,

I would proceed as follows.

In 1 sec, M1 would have gone 20 m towards the other end and M2 would have proceeded 40 metres.

So M1 would have gone *150 = 50 m and M2 would have gone 150-50 =100 m when they meet(i.e distance between them is 5 m).

Time taken = 50/20 = 100/40 = 2.5 secs

right?

Vedu

I am sorry Bharti, if I sounded harsh in my previous post. Thanks for the clarifications.

My solution:

Actual Speed of M1 = kmph

Actual Speed of M2 = kmph

(Relative) Speed of approach of the men towards each other = sum of the above two = 216 kmph = 216000 mph

Distance to be covered = 150m

Time (in hours) = distance/speed = 150m / 216000 mph

Time (in seconds) = 150*3600/216000 = **2.5 sec**

I got the answer as 5 Seconds...though the Qn..sounded wierd...i made some

Assumptions :)

T1

----------------------------

>------------------------

T2

Khufu,

We were told that the starting and ending points OF THE TRAINS (initially) make a RECTANGLE (err...so it's not redundant, afterall)....in your configuration, you have a parallelogram instead!

Picture this:

... :> ...

... <: ...="">

... ...

HTH

"I got a better deal...

what if I give you a **bump!**...

...and you give me the right answer"

the relative speed of the man(m1) with respect to his own train is 18kmph.

so 3x - x = 18, x = 9kmph = 2.5 m/s,,,, speed of his train is 7.5m/s

for m2, similarly, his actual speed is 18 kmph = 5m/s,,, and speed of his train is 15m/s

now, speed of m1 with respect to the approaching man m2 on the other train will be the sum of speed of his own train and his speed.

so relative speed of m1 is 2.5+ 7.5 = 10m/s

relative speed of m2 with respect to approaching man m1 is similarly,, 5+15 = 20 m/s

so time taken is 150/30 s = 5s

Bhars?

we sure would like to know the answer.....i didnt really understand the question properly!!!!!

Bhars?

we sure would like to know the answer.....i didnt really understand the question properly!!!!!

I thought you'd see Raghu's solution as well

5/2 is the answer as he gets it.

Bharathi.

Average speed of a vehicle is 60kmph when stoppages r nt considered . When stoppages r considered the average speeed becomes 48 kmph . On an average how many minutes per hour are the stoppages ?????????? plz give me a detailed explanation ğŸ˜ƒ ğŸ˜ƒ chao for now

rani_das SaysAverage speed of a vehicle is 60kmph when stoppages r nt considered . When stoppages r considered the average speeed becomes 48 kmph . On an average how many minutes per hour are the stoppages ?????????? plz give me a detailed explanation ğŸ˜ƒ ğŸ˜ƒ chao for now

Assuming distance to be 60 KM .....it takes 1 hr without stoppage ....if he takes stoppage he takes

60/48 hrs i.e (60/4*60 minutes .......75 minutes ................this implies he stopped for 15 minutes ...............

75 minutes 15 minutes stops

so in 60 minutes

60/75*15

12 minutes /hr ..........

rani_das SaysAverage speed of a vehicle is 60kmph when stoppages r nt considered . When stoppages r considered the average speeed becomes 48 kmph . On an average how many minutes per hour are the stoppages ?????????? plz give me a detailed explanation ğŸ˜ƒ ğŸ˜ƒ chao for now

Another method...

x = normal time for reaching the destination.

y = total stoppage time.

60/48 = (x+y)/x

=> x = 4y.

Average time of stoppage = y/(x+y) = 1/5 .

I am sorry Bharti, if I sounded harsh in my previous post. Thanks for the clarifications.

My solution:

Actual Speed of M1 = kmph

Actual Speed of M2 = kmph

(Relative) Speed of approach of the men towards each other = sum of the above two = 216 kmph = 216000 mph

Distance to be covered = 150m

Time (in hours) = distance/speed = 150m / 216000 mph

Time (in seconds) = 150*3600/216000 =2.5 sec

See how u got the speed of train as 72 kmph and 144 kmph. The actual speed of M1 shd be 2.5 mps and actual speed of M2 shd be 5mps, so that speed of T1 is 7.5mps and speed of T2 is 15mps. This will suffice the given condition that relative speed of M1 wrt T1 is 5mps(18kmph) and relative speed of M2 wrt T2 is 10mps(36kmph).

Anyways these all calculations are not necessary as we are straightaway given the relative speeds of T1 n T2.

So we can assume that both the trains are stationary n only the two men r moving towards each other.

Now

total distance to be covered by both of them before they meet is 150 m and the rel spped of M1 wrt M2 is 15mps.

so time taken =150/15=10 swhich shd be he right answer.

What say guys:)

Ravi n Vikram complete one round of 900 in 9s n 15s respectively

thn their speeds r 100m/s n 60m/s wth rlative speed of 40 m/s

thn guys can u plz explain me how cme ravi overtakes vikram in every 20 seconds????:whatthat:

how ths overtaking time is calculated?????????????:whatthat:

Ravi n Vikram complete one round of 900 in 9s n 15s respectively

thn their speeds r 100m/s n 60m/s wth rlative speed of 40 m/s

thn guys can u plz explain me how cme ravi overtakes vikram in every 20 seconds????:whatthat:

how ths overtaking time is calculated?????????????:whatthat:

its not 20 s but 22.5 seconds.

ravi and vikram start off at the same time.....

lets assume that ravi first overtakes vikram after 't' seconds.

now in this 't' seconds ravi would have covered 100t metres.....and vikram 60t metres.......but since ravi is overtaking vikram he should have covered one round more than vikram i.e. 900 metres.

so 100t - 60t = 900

or 40t = 900 and t = 22.5 seconds

now similarly for the next time ravi overtakes vikram we can assume safely that they are starting off afresh from the point where ravi first overtook vikram....

hope this answers ur question

Ravi n Vikram complete one round of 900 in 9s n 15s respectively

thn their speeds r 100m/s n 60m/s wth rlative speed of 40 m/s

thn guys can u plz explain me how cme ravi overtakes vikram in every 20 seconds????:whatthat:

how ths overtaking time is calculated?????????????:whatthat:

i guess for these type of problems u can directly use the following formula

time taken to overtake= /

i.e t=900/40 =22.5