### Discussion :: Problems on Trains - General Questions (Q.No.2)

Swati said: (Jul 7, 2010) | |

Why is it divided by 18/5? |

Jothi said: (Jul 12, 2010) | |

Please explain to sutract -5 but ans is 50. |

M.Saravanan said: (Jul 14, 2010) | |

Please explain 18/5. |

Shwetha A V said: (Jul 16, 2010) | |

60*60/1000=18/5 |

Saji said: (Jul 19, 2010) | |

18/5 is to convert m/s to km/hr |

Ankita said: (Jul 20, 2010) | |

Why its substracted? |

Gowtham said: (Jul 22, 2010) | |

Hey why we findind relative speed??? They asking only speed na???? |

Vijayagopal said: (Jul 26, 2010) | |

Why 125/10 the 10 is divided 125 please explain it. |

Rahul said: (Jul 28, 2010) | |

@gowtham It would be only speed if the man(REFERENCE POINT) was stationary and only the train was moving. Here as both trian and REFERENCE POINT(man) are moving relative motion comes into picture. |

Rahul said: (Jul 28, 2010) | |

@ankita when a man moves in the direction of train, the train speed is decreased by his own speed. If he moves in the opposite directon he finds the train speed to be much greater. So when is moving in the same direction his speed must be subtracted from that of train speed. If he moves in the opposite direction to that of train, then his speed must be added to the train speed. moving in same directon: SUBTRACT moving in opposite directon: ADD |

Rahul said: (Jul 28, 2010) | |

@ vijayagopal here both man and train are moving in the same direction and the train crosses the man in 10 seconds. speed = distance(or length)/time. As both man and train are moving, relative motion comes into picture. therefore: relative speed= (length of the train)/(time taken by the train to cross the man who is in motion) relative speed= 125/10 |

N Ramesh Chandra said: (Aug 6, 2010) | |

Hai Madhu N Swathi, ****When two bodies say x km/hr and y km/hr are moving in the same direction then their relative speed is (x-y) km/hr ****When we want to convert km/hr in to m/s we multiply with 5/18. *** Let the speed of the train be x kmph Speed of the train relative to man(x-5)kmph (x-5)*5/18 now as we know that Distance/speed=time now,speed is (x-5)*5/18; distance is 125m; time is 10 sec; now u vl get the answer *********** |

Thilaga said: (Aug 13, 2010) | |

Nice explanation.Thanks. |

Umair Chudhary said: (Aug 21, 2010) | |

If someone feel difficult to understand 5/18 i.e; m/s To convert kmph to mps 1000/60*60 m/s It is same thing. |

Sreenivas Reddy said: (Sep 3, 2010) | |

Yeah superb. |

Kamalakannan said: (Sep 5, 2010) | |

Answer = 45 km/hr is relative speed of the train. Now, you calculate (u-v). The Speed of the train 50 km/hr answer is correct. |

Antony said: (Sep 8, 2010) | |

Please explain 18/5. |

Dattatray said: (Sep 8, 2010) | |

18/5 = 1 km/hr = 1000/3600 m/s = 10/36 m/s = 5/18 m/s. |

Sheela said: (Sep 8, 2010) | |

1km=1000m. 1hour=3600sec. So to convert from km/hr to m/sec, we need to divide 3600/1000 i.e,. 1km/1hr=3600s/1000m which is equal to 18/5. If it was asked to convert m/s to km/hr then it would be 1000m/3600s= 5/18 |

Saranya said: (Sep 9, 2010) | |

I could not understand this problem is there any way to understand this more easily. |

Abhishek said: (Sep 10, 2010) | |

Distance = time * speed 125 = 10 * speed distance/time = speed 125/10 = (25/2)m/s = speed ---------------------------------------------------------- Answer should be in km/hr so convert m/s km = 1000m and hr = 3600s m = km/1000 and s = hr/3600 m/s = (3600/1000)km/hr m/s = (18/5)km/hr ---------------------------------------------------------- The condition was speed = (25/2)m/s = (25/2)(18/5)km/hr = 45 km/hr The speed was relative to man so we have taken time in consideration of man -------------------------------------------------------- See further in in view answer. ---------------------------------------------------------- |

Nagapriya said: (Sep 17, 2010) | |

Now I understand that,. Distance = Speed*Time. Time = Distance/Speed. Speed= Distance/Time. OK guys, very thanks to your teaching. |

Amit said: (Sep 21, 2010) | |

Can any one tell me why we have mulipled 25/2 to 18/5..why it should not be multiplied by 5/18 and also how ? x - 5 = 45 ....as I understand it should be 40 km/hr x = 50 km/hr. Because x=45. 45-5=40....hence anyone pls clarify. |

Srikanth Mirdati said: (Sep 25, 2010) | |

Could anyone please tell me how we get to 25/2 by dividing 125/10. I want to know how the division of 125/10 is done to get 25/2 an answer would be appreciated greatly. |

Shabana said: (Sep 25, 2010) | |

Hi, Really the explanation given by Dattatray & Rahul was superb. Thanks guys. |

Manoj Chauhan said: (Sep 28, 2010) | |

125/10 = 12.5 3600/1000 = 36/10 (you should delete the last two zeros) 36/10 = 3.6 x = 12.5 * 3.6 So that x = 45 per hour right (this is relative speed of train) x = 45 (you add 5 km per hour for train speed because 45 is relative speed of train). Therefore x = 50 (the required answer). |

Nitesh Nandwana said: (Sep 28, 2010) | |

Hai Rahul, well explanation about relative speed and same and opposite directions. Good job keep it up. |

K.C.Yogannal said: (Oct 5, 2010) | |

@Vijayagopal. As speed =length/time they have done 125/10. Where 125 is the length and 10 is the time. Hope you understood. |

Atique said: (Oct 11, 2010) | |

Nice Explanation Guys. Thanks a Lot. |

Anitha said: (Oct 27, 2010) | |

Why 18/5 please tell me? |

Anitha Alavala said: (Oct 27, 2010) | |

See the answer for 18/5 is how? 1 km=1000m & 1hr=3600sec by this we get 1km/1hr = 5/18 m/sec.But the answer must be in km/hr....But 5/18 is in m/sec. we have to convert it to km/hr. 1 km = 1000m => m = 1 km/1000 hr/3600) = (1km/1000)*(3600/1hr) = 3600km/1000hr => 18/5 km/hr friends...its a perfect answer...have a look. |

San said: (Oct 27, 2010) | |

I cant understand relative speed formula ie) x-5 is this standard formula ? |

Sathish said: (Nov 3, 2010) | |

(x-5) is not a standard formula x is taken as speed of train and 5 is man's speed, to calculate train's speed from relative speed we subtract (x-5). |

Chakit said: (Nov 9, 2010) | |

x - 5 = 45 x= 45+5 x = 50 km/hr. |

Sanya said: (Nov 10, 2010) | |

relative speed when two movable things whose speeds x,y are moving in same direction is =X-Y, relative speed when two movable things whose speeds x,y are moving in OPPOSITE direction is =X+Y Its an standard formula |

Onam said: (Nov 19, 2010) | |

5/18 is conversion of km/hr to m/s and 18/5 is conversion digit m/s to km/s |

Vijay Kumar Kuppani said: (Nov 21, 2010) | |

DONT WORRY FRIENDS ITS VERY SIMPLE instead of going for a new formula.we can simply get the answer by using the standard formula.i,e t=(a+b)/(u-v) u can see this formula from the general formulas of this exercise. where t-time taken by the train to cross the man=10 sec; a-length of the train=125meters; b=0(because the train is supposed to cross a person); u=speed of the train(you are supposed to find out); v=speed of the man who is running=5km/hr=5*(5/18)=25/18m/sec; by substituting the values in the above equation you can get the value.. |

Nas said: (Dec 4, 2010) | |

Nice explanation friends. ! |

Priyavel said: (Dec 8, 2010) | |

Both (train and man) moves in same direction so we subtract. Both moves in opposite direction means we add it. We convert m into km, sec into hour so we multiply 18/5. |

Jelli said: (Dec 11, 2010) | |

Thanks to all. I got many idea's to solve this problem. |

Kannips said: (Dec 12, 2010) | |

speed=distance/time.so speed=125/10.....25/2*18/5=45... the question is same direction=x-5=45 x=50 |

Kumar said: (Dec 12, 2010) | |

Why 125 is divided by 10? |

Prithika said: (Dec 19, 2010) | |

@kumar cause speed=distance/time |

Raghu said: (Dec 21, 2010) | |

How it is 3600/1000. It will come as 1000/3600 only. Please explain. |

Saranya said: (Dec 21, 2010) | |

hai raghu, in the question man's speed is given in km/hr so we gonna convert train's speed to km/hr which is actually in m/s 1 m/s = 3600 km/1000 hr 125/10 m/s= (125/10)*(3600/1000) |

Tahira said: (Dec 21, 2010) | |

Hai fnds. Howz the 12.5 is converted into 25. Please explain. |

Mani said: (Dec 23, 2010) | |

Hi Tahira, You just cancel 125/10 using multiples of 5. 125/10=25/2=12.5. |

Rekha said: (Dec 23, 2010) | |

Ramesh chandran. CLEAR explanaTION. THANKS. |

Shanmuga Priya.Tex said: (Dec 30, 2010) | |

NICE EXPLANATION ABHISHEK. |

Ravi U S said: (Jan 1, 2011) | |

Hi friends,the problem is very simple.Think in this way.Assume that the man is not moving(standing like a pole),now findout the speed of the train using the data. LENGTH OF THE TRAIN=125m. TIME TAKEN BY THE TRAIN TO CROSS THE MAN(assume standing)=10sec w k t velocity=distance(length of the train)/time. v=(125/10)m/s v=(25/2)m/s convert m/s into km/hr v=(25/2)* (3600/1000) or (25/2)*(18/5) v=45 km/hr 45 km/hr is the train speed if the person is standig(not moving). But in the problem person is moving and that to he is moving in the same direction in which train going,so that's y add the speed of the man to the speed of the train then we will got the correct answer. speed of the train+speed of the man=45+5=50 km/hr. |

Murali said: (Jan 8, 2011) | |

hai!frinds,. in this above problem they given speed of the one object. we know that if the two objectives are move in the same direction then the formula is"(A-B)"(assume A is one objective and B is another objective).so,nw we know one objective speed(taken as B speed we know),that is subtutited in above. then s=(A-5).(here it is the speed of the train) 1.in the problem train(objective) speed is given in the form of seconds,so convert the seconds into the km/hr.because of the one objective speed is give in the form of km/hr and also answer is given in the form of km/hr. 2.while the conversion we have to follow the following, i)if we covert m/sec inti km/hr then multiplis with"8/15" ii)if we convert km/hr into the m/sec then multiplies with "15/8" 3.we know the formula for speed is "speed=Distance/Time"(s=D/T) s=125/10 m/sec [here 125 is the distance,and 10 is the speed of the objective] the conversion s=125/10*(18/5)km/hr =45. 4.in starting we initialse the speed(s)=A-5 then subtuted that speed in the above, A-5=45 A=50. therefore the speed of the train(another objective) is "A=50". |

Amol said: (Jan 10, 2011) | |

Hi Friend's Thanks for teaching it really works dude , 1km=1000m and 1hr =36000 then 5/18 m/sec it's simple. Thanks for that .........! |

Amit said: (Jan 22, 2011) | |

Hi friend's Hey ravi nice explanation. |

Manishkumar said: (Jan 23, 2011) | |

@ all apply logic don't simply go for routine speed of man is 5km/hr and in mt/sec is 1.39. so total distance covered by main in 10 sec is 13.9 m so total distance covered by train in 10 sec is 125+13.9=138.9 so total distance covered by train in 1 sec is 13.89m and if we convert this in km/hr i.e. 13.89*3600/1000= 50km/hr |

Ravi U S said: (Jan 25, 2011) | |

Thanks Amit |

Rajesh Motana said: (Feb 12, 2011) | |

Train moves with speed 125m/10 sec,.. So need to convert it as KM/ Hr ,..To convert 125m to KM we have to divide it with 1000 so it is 125/1000 and to convert 10 sec to hrs we have to divide it with 3600 so it is 10/3600. Then it becomes (125/1000)Km/(10/3600)hr. So we can change it to (125/1000)*(3600/10).. It is (5/40)*(360) = (360*5)/40 = 180/4 = 45 Km/Hr. .. Now we have got the speed in 1 Hour ie 45 Km.. As the man is travelling at 5 Km/Hr.. In that 1 Hr he already travelled 5 Km's.. So the train speed should be 45+5 Km's to cross that person in that 1 Hr.. Now answer is 50 Km/Hr.. I hope it is clear,... Take a paper a write exactly what i have given above,.. I am sure u guys will like it.. |

Abhishek said: (Feb 13, 2011) | |

Can anyone find another way of solving this? Apart from relative speed? |

Jeevanantham said: (Feb 18, 2011) | |

How to relative 45 with man running speed. Please explain it any one. |

Vimal said: (Mar 3, 2011) | |

This is good Question i understand that, some friend disturb for 18/5. its simple its unit for km/hr. let see that1km=1000m & 1hr = 3600 second so 1000/3600 = 5/18 . And see that If we want to small unit convert in to big unit we have to divide & if we want to big unit convert in to small we have to multiply so here they dividing and write 18/5 ok so encourage your self my friend go ahead. VIMAL SONI & DUSHYANT SONI |

Vasuedvareddy said: (Mar 7, 2011) | |

@Vasudeva reddy 1km = 1000m =100,000cm 1 inch = 2.54cm 1 ft=12 inches = (12*2.54)cm 1km = 0.621371192 mile 1.609344 km = 1 mile |

Sarah said: (Mar 17, 2011) | |

Hi I can't understand friends. What is this x-5 ? Please anyone explain clearly. |

Suganthan.M said: (Mar 28, 2011) | |

We need the speed of the train we have only one clue about the train is 125m and 10s speed=distance/time (i.e)125/10= 12.5 m/s but we need the answer in km/hr so convert it 12.5*(18/5)=12.5*3.6 =45 km/hr almost we got it then about the direction, it is in same direction Now the speed of the train as x, because of same direction subtract it from x x-5=45 x=45+5 x=50km/hr |

Chetana said: (Apr 1, 2011) | |

Thanks frnd. I understand properly. |

Gowthami said: (Apr 21, 2011) | |

Nice explanation from you frnds but ravi said that we had add that relative speed, I can't understand from his point of view. |

Jitendra Prajapati said: (Apr 27, 2011) | |

Very good explanation by Suganthan. |

Meera said: (May 13, 2011) | |

Thanks mr. Suganthan. Its really very nice to understand thanks:) ) ). |

Shashank said: (May 25, 2011) | |

Hi. Please tell me why x-5 ? |

Binod Verma said: (May 29, 2011) | |

When two things are running in the same direction or opposite direction then we say that motion is in relative between them. So at first we find the relative motion and then find the speed of train. If someone want to find out speed of man that are given in the question. |

Dhandapani said: (May 30, 2011) | |

Simple you can multiply 5*10. 5-a man running 5km/hr and the train reached in 10 seconds. 5*10=50km/hr. |

Washeru said: (Jun 5, 2011) | |

Good expl. frm Rahuul |

Ragavan said: (Jun 10, 2011) | |

As from the problem. two contrain have to remember *If the man moving towards train. so that train pass the man faster then speed so add the distance. *If not move towards train. The train pass the man late.so minus the distance |

Arul said: (Jun 30, 2011) | |

Hey the quesion is about the speed of the train. jjust bcoz aman runs... will the speed of the train increase...? a little help here pls... |

Salahuddin said: (Jul 4, 2011) | |

To avoid confusion Simple convert the trains speed for 1 hr - metres/seconds, 125m/10 sec. So 125 * 6 * 60 / 10 * 6 * 60 = 45000 mtrs/ 3600 . So to convert it, divide by 1000, its 45 kms/hr. As the man running in the same direction at a speed of 5kms/hr add 45 + 5, if the person moves in opposite direction then subtract. Just logic guys |

Kaushik Rathod said: (Jul 5, 2011) | |

Answer in Another way :- Speed = 5*18/5 = 18 km/hr Time = 10 sec. X + 125 / 10 = 18 X + 125 = 180 X = 180-125 X=55 Now, Speed = X-5 So , Speed = 55 - 5 Speed = 50 km/hr |

Ajith said: (Jul 11, 2011) | |

So in the above problem let us consider that the speed of the person is 5 km\hr . Leave the 5 as it is and then convert the km\hr into m\s that means 1 km=1000m and 60min=3600sec Now when actually converted 3600\1000 that would give u the two zero gets cut and remains is 36\10 that would give u after dividing each by 2 i.e 18\5which means (18m\5s)1.FRACTION Now the major calculation, Two object over here is Train and the Man consider them as A and B where A is the Train as we have to find the speed of the train then B is the speed of the Man is 5 so we get A-5 We had 125m the length of train time 10 seconds already these are in meters and in seconds no need to convert them so 125m\10s u get (25m\2sec) 2.FRACTION Now, finally get the first fraction and the second fraction together i.e (18\5)*(25\2)u get 45 as answer when u calculate then A-5=45 ,when u move the negative -5 to this side of the equals too sign it converts to positive giving u A=45+5 i.e 50 so A=50 and A is the speed of train!!hope u'll understood!!take care!! |

Monisha said: (Jul 15, 2011) | |

Great job guys!!! I understood well than my faculty teaches... |

Raja said: (Aug 2, 2011) | |

Thank you friends I understood well never before. Every one has given their own ideology with respect to problem. |

Umesh said: (Aug 9, 2011) | |

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is: ====> dist. travelled by man in 10 sec. = (5*5/18) *10=250/18 A train will have to travel its own length + dist.travelled by man. (when train will pass the man he would have travelled 125/9 m in same dir.) i.e. 125+125/9 meters = 1250/9 speed of train = (1250/9)/10 = 125/9 m/s=125/9 *18/5 = 50 km/hr |

Tarun Dumra said: (Aug 26, 2011) | |

Nice expalnation by ankita. |

Vandna said: (Sep 4, 2011) | |

Can anyone solve this? A train moving at 36Kmph takes 12 seconds to cross a platform. It takes 6 seconds to cross a man running at the rate of 9Kmph in the opposite direction. Find the length of the train. |

Shiwam said: (Sep 9, 2011) | |

@vandna It is option checking question so option must be given. Since it has 3 variables plateform length,train length & train speed. it might be done by option checking. |

Vijaya said: (Sep 9, 2011) | |

Hi amit and Sarah Here x - 5 = 45 x= 45 + 5 (when minus (-) simble comes from right of = to left of = it will become +. |

G.Humayun said: (Sep 23, 2011) | |

Thanks friends its very useful. |

Preetha said: (Sep 23, 2011) | |

Hi friends, I can't understand this please anyone help me in some other method. |

Gnana said: (Sep 30, 2011) | |

Very nice vandha. |

Sagar said: (Oct 4, 2011) | |

Finally we need to find speed me train. But in problem train speed is compared with moving object i.e man. So (total speed of train-speed of man=relative speed). |

Pradeep said: (Nov 7, 2011) | |

Can you please explain me what is relative speed here and why it is used here ? |

Nagaraj said: (Nov 20, 2011) | |

It should be taken as 1000/60*60 beacuse distance covered/ time taken. |

M.Kanna said: (Nov 23, 2011) | |

18/5 explanation is super k/m h/r put 1000/3600=5/18 answer is very nice apserve then nice. |

Mahendra said: (Nov 26, 2011) | |

Nice explanation sheela thank you all who commented. |

Pradeep said: (Dec 9, 2011) | |

From The Question 5Km/hr So please explain it 1000/3600. give me reason. |

Kirubha said: (Dec 21, 2011) | |

FRNDS is it correct or not tell me plz.... MAN'S SPEED = 5 KM/HR Train length =125 m. let assume train speed=x. then we get relative speed (rs) of train and man is become rs=x-5. To conver km/hr into m/s. = (x-5) / (5/18) = 5x-25 / 18 speed = distance/time 125/10 = 5x-25 / 18 2250 = 50x-250 x = 2000 / 50 x = 40 Is it correct? |

Swapnil....... said: (Dec 24, 2011) | |

Why is it addition of 45km/hr & 5km/hr? Because I thing train & man is going to same direction that time sutract but in this case addition of 45km/Hr & 5km/hr please explain it why is it addition. |

Challu said: (Dec 29, 2011) | |

Think you are in a train, another one moving side by you in parallel, what will you experience? the train is in less speed of actualone. To find original speed you have to add your train speed. |

Preethi said: (Jan 4, 2012) | |

You are right kirubha. Nice explanation challu. |

Preethi said: (Jan 4, 2012) | |

Thanks Abhishek now I clear about relative speed. |

Hari said: (Jan 6, 2012) | |

Excuse me? Why we divided 125 with 10?. Please explain me. |

Ahmed said: (Jan 7, 2012) | |

Explanation given by ramesh chandra is very good and short, Please follow his process without wasting your time. We need such a good explanation like that. |

Uma said: (Jan 13, 2012) | |

Manoj explanation s superb. Thank you. |

Vinod Salunke said: (Jan 18, 2012) | |

Nice explanation by ramesh chandra. |

Akhil said: (Jan 19, 2012) | |

18/5 = 1 km/hr = 1000/3600 m/s = 10/36 m/s = 5/18 m/s. |

Nirmal said: (Jan 22, 2012) | |

Vijay Kumar Kuppani you did solve problem very nicely again thanks. |

Shobi said: (Feb 11, 2012) | |

Hi friends, I ave confuesion in relative speed. Can I have some more explanation about relative speed? |

Shro said: (Feb 15, 2012) | |

Thanku Rahul. Your explanation was easy to understand and cleared my doubts too :). |

Jrr said: (Feb 17, 2012) | |

Abhishek's answer explaines this clearly...Thanks Abhishek!!! |

Ranjit R Pawar said: (Feb 27, 2012) | |

Thanks rahul. |

Anish said: (Mar 27, 2012) | |

Hi Manishkumar,Vijaykumarkuppani,Umesh U hav cleard the points exactly how itz done. Thanks a lot... |

Parvez Khan said: (Apr 4, 2012) | |

How was the relative speed calculated ? |

Ambikun said: (Apr 4, 2012) | |

There is no need of calculating relative speed. It can be solved easily by calculating the distance traveled by the man in 10 sec and the total distance covered by the train in the same time. This is very simple. |

Sanghaniakshay said: (Apr 12, 2012) | |

Thanks abhishek. |

Greeshma Reddy said: (Apr 18, 2012) | |

How did this equation x-5=45 form? |

Pradyumna Kumar Das said: (May 11, 2012) | |

If the man and train in opposite direction then it relative speed is x+5. And speed is x+5=45. =>x=40. Am I right? |

Sada said: (May 24, 2012) | |

Any basic formula is there for this problem kindly explain. |

Bhavesh Sharma/ Gadgad786 said: (May 27, 2012) | |

Why we are multiplying the 18/5 ? or we can ask that why we are divide ? The answer is simple. We are converting the 'kilo-meters/hour' value into 'meters/second'. Let me explain about converting km/hr into m/s in detail. 1 km = 1000 meters. 1 hr = 3600 seconds. 1 km/hr = 1000/3600 m/s = 10/36 m/s = 5/18 m/s. Therefore 1 km/hr = 5/18 m/sec. Similarly 60 km/hr = 60 x 5/18 m/sec. You may ask me, why should I convert it from km/hr to m/sec? |

Nil said: (Jun 3, 2012) | |

If we convert km/hr into m/sec i.e. 1000/3600 it's come 5/18. So why it is calculation is 125/10*18/5 instead of 5/18? |

Ravi said: (Jun 7, 2012) | |

It is given that Speed of man=5km/hr Distance travelled by man in 10s=125/9m Distance travelled by train in 10s= {125+(125/9)}m Hence Speed of train = [{125+(125/9)/10]*3600 = 50 km/hr |

Shailesh said: (Jul 7, 2012) | |

Speed of train=(length of train/time) is a formula hence, 125/10 m/sec now (125/10)*(18/5)=45.. multiply by 18/5 to become km/h now 45 is relative speed sr=s1-s2 45=s1-5 s1=50 |

Chandu said: (Jul 20, 2012) | |

Why it is subtracted ? please explain. |

Vikram said: (Jul 26, 2012) | |

A Train 125m long passes a man,running at 5km/hr. It means that train speed =5000m/3600sec.------(1hr=60*60=3600sec & 1km=1000meters). & Second Train running on same Direction & Passed in 10sec's. It means Second train cross distance within 10sec's. Speed of Train for meters/10sec = 5000m/3600sec*10sec=1.388888888889m So,that train speed for km/hr=speed of train for 10 sec * remaining 3590 Seconds(3600-10). =1.3888888889*3590=49.8611111111111111=50km/hr. Therefore, A speed of that train=50km/hr. |

Bulldozzer said: (Jul 29, 2012) | |

This is simple. Length of the train is given and the seconds is given. So we have m/sec there. To convert it to km/hr, 1km= 1000 m and 1 hr = 3600 sec. So km/hr= 1000/3600. So we have 18/5. Now we got our km/hr. Well that guy is running at 5 km speed. So we take the train speed as x and subtracting the 5 km speed from that. We have x-5 as the train speed. Now x-5= 45. So x= 45+5=50. :-). |

Kishore Kumar said: (Aug 3, 2012) | |

Your running at a speed of 5km/hr (1km=1000 meter and 1hr=3600seconds,which will be 1000/3600=10/36=5/18) Note :for 1 km/hr Therefore 5(given km)*5/18=1.39 ------------------------------- Meanwhile 125m long train crosses you in 10 seconds along with you in same diretion. Then!! 125/10 i.e its in the form of (125)m/(10)s. ------------------------------- So add those two values in the form of m/s,which is 13.89=a ------------------------------- Now question to find speed of the train,since we found in the form of m/s,but the options provided are in the form of KM/HR ------------------------------- So to convert M/S to KM/HR : (a*18/5)=(13.89*3.6) Gives answer as 50 km/hr |

Sunil said: (Aug 18, 2012) | |

How/ why Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. Please any one explain me why let speed =(x-5)km hr. |

Harish said: (Aug 24, 2012) | |

It is brain storming and conceptual question .Thank you for giving more explanation on this question.Thank you very much....! |

Karthik said: (Aug 24, 2012) | |

125/10 = 25/2 25/2*18/5 = 450/10 45 friends, i don't know how to solve this problems after this so plz anyone help me out |

Anbu said: (Sep 4, 2012) | |

Just try this 1 sec 5 km/hr means 10 sec is 50 km/hr thats it.. becoz the man running at same time train is moving .. |

Preethi said: (Sep 16, 2012) | |

Abisheks answer is realy nice. Thank you abi. |

Mamatha said: (Sep 18, 2012) | |

To convert km/h into m/s we multiply that value with 5/18.why is this because 1km=1000m 1h=60min*60sec=3600sec now 1km/h=1000/3600=5/18 if we want to convert m/s into km/h we have to multiply with 18/5. hope you can understand. |

Susmitha said: (Sep 18, 2012) | |

x-5 is nothing but here x refers to the speed of train and as both train and boy are moving in the same direction relative velocity comes into picture and here relative velocity is the difference of two velocities. |

Mani Sharma said: (Sep 20, 2012) | |

If train moving same direction = SUBTRACTION OCCUR. If train moving opposite direction = ADDITION OCCUR. |

Sundeep said: (Sep 27, 2012) | |

Let the speed of the train is x km/hr Then the relative speed is (x-5)km/hr Therefore, Time = 10sec Distance = 125m Speed = (x-5)km/hr As we are calculating in m/sec so we have to convert km/hr to m/sec. (x-5)*1000/3600 m/sec or (x-5)*5/18 {by cancelling by 200. 200*5=1000,200*18=3600} Therefore , Speed = distance/ time (x-5)5/18 = 125/10 x-5 = 125*18/10*5 x-5 = 2250/50 x-5 = 45 x = 45+5 x = 50 Therefore x is 50km/hr (speed of the train) |

Vinothini said: (Oct 4, 2012) | |

Hai friends I want to give small tips. If there train passes in same direction just subtract it. If train passes in different or opposite direction direction just add it. If you need to convert km/pr use this formula (a*5%18) for speed and m/sec the formula is (a*18%5). |

Bibin said: (Oct 28, 2012) | |

Why you made it 18/5, earlier it was 5/18, please explain me. |

Jayan said: (Nov 20, 2012) | |

Here x=45 is speed of per 1km. question is 5km/hr . So x-5=45 x=45+5 x=50 Thats all. |

Anvesh said: (Nov 29, 2012) | |

l=125m; s=5*5/18; t=10s; l=t*s; s=t/l; s=125/10; s=12.5*18/5; the man was also going same direction so, s=45+5; s=50; |

Charles @ Redhills said: (Dec 5, 2012) | |

Let the speed of the train is 'x' Length=125m Speed=10s --> as per "Length of the train = speed X time" --> 125m = 10s.x --> x=(125m/10s) X (360/360)=(360 X 125m)/3600s --> =45000m/3600s = 45km/1hr As the above value is relative with the speed of man whose speed of running is 5km/1hr also to be added for the exact speed of the train. There fore.. -->exact speed of the train is = (45km/1hr)+(5km/1hr)=50km/1hr I believe its simple and right answer. |

Jagadeesh said: (Dec 31, 2012) | |

Let relative velocity of train with respect to man be vtm,actual velocity of train Be vt and velocity of man be vm. We know that vtm=vt-vm, But from problem vtm is =length of train/time taken =(125*18/5)/10 =45 km/hr. From above actual speed of train is,vt=vtm+vm =45+5=50 km/hr. |

Preethi said: (Jan 30, 2013) | |

A train of length 110mtr travelling at a speed of 80km/hr. Find the time taken by that train to overtake a train having length 130mtr travelling at a speed of 62km/hr? |

Malik said: (Feb 25, 2013) | |

@Preethi. Relative speed u-v = 80-62= 18km/hr=(18* 5/18)= 5 m/s. Total length a+b = 110+130= 240 m. Time to overtake = (a+b)/(u-v)[see formula 9 from formula section] = 240/5 = 48 sec. Am I correct? |

Diksha said: (May 24, 2013) | |

How we come to know that the speed 5km\hr is the speed of the man because it is not given in the question? |

Aruntracer said: (Jun 4, 2013) | |

@Diksha. A train 125 m long passes a man, running at 5 km/hr in the same direction means the , in between man and running is for confusing you it should be taken as 125 m long train passes a man running at 5 km/hr. |

Kavya Reddy said: (Jun 12, 2013) | |

Why we are using 18/5 instead of 5/8 is as follows, 1km = 1000m. 1hr = 3600s. So 1km/hr = (1000/3600)m/s. Hence 1km/hr = (5/8)m/s. Then 5/8 becomes reverse as 8/5 and it will becomes as, (8/5)km/hr = m/s. Again you get a doubt that why it converts into m/s because check our problem has 125 m and also they gave 10 seconds. So (125/10)m/s. |

Raja Simhan said: (Jun 14, 2013) | |

Here the train is crossing a moving object rather than a stationary object like a pole. Then how it is possible to say its relative speed to man is (125/10) m/sec ? |

Md Irfan Azam said: (Jun 19, 2013) | |

This solution easy to understand: Convert man speed from km/hr into meter/sec. 5km/hr = 5*1000m/3600sec or 25m/18sec. Since 18sec = 25meter. 10sec = ? meter. So it would be 25*10/18 = 125/9 meter. So man complete 125/9 meter in 10 sec. Train complete distance in 10 sec = 125 meter (train length) + 125/9 meter of man. So it would be 10 = 125+125/9 or 10 sec = (1125+125) /9 meter or 10*9 sec=1125+125 meter or 90 sec = 1250 meter or 125 meter/9sec or convert it into km/hr so it be 125m/9s * 18/5 = 50 km/hr. So train speed is 50 km/hr. To convert km/hr into m/s it is multiplied by 5/18 or 1000/3600. And from m/s into km/hr multiplied by 18/5 k done. |

Mathivanan said: (Jul 8, 2013) | |

Can anyone explain how to find length of train when the train speed and person speed given? |

Chandan said: (Jul 13, 2013) | |

60min -> 5km. 1min->? =>(1*5)/60 = 1/12km. For 1 second -> (1/12)/60 = 1/720km. For 10 second ->(1/720)*10 = 1/72km. It means train covered 1/72km in 10second, then calculate how much distance that train can cover in 60 min(1 hour). Because they've given the choices in km/hr. =>(1/72)*(60*60)=50km/hr. Note: Length of the train is not required in the calculation. |

Ashutosh Maurya said: (Aug 1, 2013) | |

Distance covered by man = 5*5/18*10. = 250/18 Meter. Total Distance covered by Train = 125+250/18 = 2500/18 Meter. So Train speed = (2500/18)/10 Meter/Second. or = 2500/180*18/5 Km/Hr. = 50 Km/Hr. |

Ravichandra said: (Aug 2, 2013) | |

You're running at a speed of 5km/hr. (1 km = 1000 meter and 1hr = 3600 seconds, which will be 1000/3600 = 10/36 = 5/18). Note : for 1 km/hr. Therefore 5(given km)*5/18 = 1.39. ------------------------------- Meanwhile 125m long train crosses you in 10 seconds along with you in same direction. Then! 125/10 i.e its in the form of (125)m/(10)s. ------------------------------- So add those two values in the form of m/s, which is 13.89 = a. ------------------------------- Now question to find speed of the train,since we found in the form of m/s, but the options provided are in the form of KM/HR. ------------------------------- So to convert M/S to KM/HR : (a*18/5)=(13.89*3.6). Gives answer as 50 km/hr. |

Mishti said: (Aug 13, 2013) | |

I solved this question in a different way: When we solve the questions in which train passes a pole(or a standing man), in that case the distance traveled is equal to the length of the train. in this case, 1. The distance travelled would be = The distance traveled by the train assuming the man was standing+The distance travelled by the moving man in those 10 seconds. = 0.125km+5*(10/3600). = 0.139km. Where, 0.125 is the length of the train in km. 5 is the value of speed of the man i.e. km/h. 10/3600 is the time given but converted in hr. 2. Now the speed. As we know speed = distance/time. Above calculated value divided by 10/3600 hr. = 0.139/(10/3600). = 0.139/0.00278. = 50 km/hr. |

Nirmal K Sahoo said: (Aug 14, 2013) | |

Man speed 5 km/hr, So man's speed per second (5*5/18) m/s. After 10 sec, train passes the man, so, man cover distance in 10 sec is : 5*5*10/18 = 13.89m. So, train covered total distance in 10 sec is : 125 m+13.89 m = 138.89 m. So, speed of train is :138.89/10 m/s = 13.889*3600/1000 Km/hr = 50 Km/hr. |

Praveen said: (Aug 23, 2013) | |

Friends its a simple logic that if a men travel along the same direction of the train then the train speed is subtracted if it travel in opposite direction then train speed is added. Hope you understand take this logic in your mind and solve any problem. |

Gayu said: (Aug 27, 2013) | |

Why its divided by 15/8? need some brief explanation. |

Akansha said: (Aug 28, 2013) | |

18/5 is used to change meter per second to kilometer per hour. And I have no idea why it says -5 as it should be +5. |

Veena said: (Aug 30, 2013) | |

Why train speed has to be subtracted with man speed? |

Navaneedhar said: (Aug 31, 2013) | |

Please explain the subtract value of (x-5)? |

Arjun said: (Sep 20, 2013) | |

Speed of man*time = 5*10 = 50. |

Iyyanar said: (Sep 28, 2013) | |

Please give a clear cut definition for relative velocity. |

Aditya said: (Oct 5, 2013) | |

The train and man are moving in the same direction so. If we want to take out the speed of any one of them, then we have to take one object as stationery. |

Roopa said: (Oct 8, 2013) | |

Hi friends, Here length of train = speed*time. Likewise speed of train = length/time. -> 125/10 = (25/2)m/sec. Now given 5km/hr. So speed is (x-5)km/hr or (x-5)(5/18). Here we goes -> (25/2)m/sec = (x-5)(5/18). -> (25/2)*(18/5) = (x-5). -> 45 = (x-5). -> x = 45+5. -> x = 50 km/hr. |

Shanker said: (Oct 12, 2013) | |

We know that 1 km = 1000 meters, 1 hr = 60 minutes, 1 minute = 60 seconds. And 1 hr = 60*60 seconds (i.e.360 seconds). Then 1 km/hr = 1000 meters/360 seconds. This becomes 5 meters/18 seconds((5/18) m/s). So 1km/hr=5/18 m/s //line1. But in this question it is given in m/s. Then from the above line 1 we can write 1 m/s=18/5 km/hr. |

Shafi said: (Oct 21, 2013) | |

Simple trick to handle: S= D/T Where S-speed, D-distance, T-time Now: Train data. D = 125m, T = 10sec. find speed S = 125m/10sec = 25/2mps. Note object running in same direction of train will be positive and in opposite direction will be negative. SPEED OF MEN 5 km/hr+(25/2x5/18). ANSWER 50 KM/HR. |

Eric said: (Oct 24, 2013) | |

@Shafi. This method is quite simple and explain the whole concept, thanks for the nice explanation. |

Bikramjit said: (Nov 9, 2013) | |

1 km/hr -------------> 5/18 m/sec. 5/18 m/sec ------------> 1 km/hr. 1 m/sec -------------> 1/(5/18)km/hr = 18/5 km/hr. 25/2 m/sec -----------> (18/5)*(25/2) = 45 km/hr. If train speed x km/hr, then (x-5)km/hr. x-5 = 45. x = 45+5. x = 50 km/hr(answer). |

Eswaran.P said: (Dec 18, 2013) | |

Here using x-5=45. Please explain? |

Vaishnapriya said: (Dec 27, 2013) | |

Consider x as the train speed(which is to be determined). We know a relative speed formula in same direction, (u-v). Using the given data find out the relative speed i.e 45. Now, Relative speed = (train speed-5). |

Selvakum said: (Jan 9, 2014) | |

(u-v). Using the given data find out the relative speed i.e. 45. |

Sureshbabu Kasturi said: (Jan 20, 2014) | |

If two things are at same speed in same direction. We subtract that, So right 125-5=120. Wrong is 125+5=130. |

Srinivas Rao said: (Jan 22, 2014) | |

If both things are in the same direction, then we have to do subtraction right but why we have to do it. |

Sankar said: (Jan 25, 2014) | |

Man has traveled 13.8 m in 10 sec. i.e 5km/hr = 1.38m/s. 1.38*10s=13.8m he has traveled in 10 sec. So 125m train has passed 13.8m in 10 sec. So speed of the train is 13.8m/s i.e 49.68km/hr = 50 km/hr. |

Priyanka Sharma said: (Feb 25, 2014) | |

Hello friends, If you want to convert m/s into km/hr then multiply with 18/5. For example: 5 m/s = 5m/s*18/5 = 18km/hr. If you want to convert km/hr into m/s then multiply with 5/18. For example: 108 km/hr = 108km/hr*5/18 = 30m/s. |

Satish said: (Mar 5, 2014) | |

What is the need of taking 15/10 and explain how it came? |

Akshay said: (Mar 6, 2014) | |

Why should multiply 18/5? |

Deepak Meena said: (Mar 7, 2014) | |

It's simple way. We know that the speed of train is 125m/10s. Then 125/10 = 25/2. This is the speed of train in 1 second. Now speed in 1 hr = 25/2*3600 = 45. And men is going in 5km/1hr so speed of train is 50km/hr. |

Krish said: (May 17, 2014) | |

How to be solve this 25/2*18/5 = 45? |

Aakash said: (Jun 4, 2014) | |

125m/10s*18/5=45. The man running in opposite direction at a speed of 5 km so 45+5=50. |

Ramkumar said: (Jul 5, 2014) | |

Why use 60*60/1000(18/5) I do not understand that? Please explain. |

Givid said: (Jul 8, 2014) | |

I have a doubt on question. Whose speed is 5km/hr? either the man's or train's? please explain. |

Muhammed Areeb said: (Jul 25, 2014) | |

Why use 60*60/1000(18/5) I do not understand that? |

Gwenbiz said: (Aug 4, 2014) | |

@Jothi, x-5 = 45. We add 5 to both sides so that we stay with x. x-5+5 = 45+5. Therefore, So x = 50. |

Navi said: (Aug 5, 2014) | |

Hi @Swati look here let me explain: L = Speed*time. And, Speed = length/time. Length = 25/2. Time = 5/18. And, 25/2/5/18 = 25/2*18/5. 9*5 = 45km/hr. x-5 = 45. 45+5 = x. x = 50km/hr. |

Shibin said: (Aug 6, 2014) | |

Relative speed = x-5 (x is the speed of train). (x-5)*5/18 = 125/10. x-5 = 125*18/10*5. x-5 = 125*9/25. x-5 = 5*9. x-5 = 45. x = 50. |

Jabi Mir said: (Aug 13, 2014) | |

The only thing to remember is that when objects are moving in the same direction,then relative speed is equal to subtraction of their respective speeds and when the objects are moving in opposite direction,then relative speed is equal to addition of their respective speeds. Therefore, here v1 = 5 km/h, v2=?, t=10sec => 10/3600 hr., Distance = 125 mt => 125/1000 kms. Now we know, Speed=distance/time. or we can say, Relative speed = distance/time. Since here relative speed= v2-v1 (objects moving in same direction). Therefore, v2-5=125/1000 whole divided by 10/3600. or, v2-5=125/1000 * 3600/10, on solving we get, v2-5= 45 => v2 = 45+5 => V2=50 Km/hr. |

Hemavathy said: (Aug 13, 2014) | |

How the will be 50? Speed = 25/2(m/s). Converting(m/s)to(km/hr). 25/2*18/5 = 45. |

Murugan Mj said: (Aug 30, 2014) | |

@Hemavathy. Since man is moving in same direction with train the relative speed is s-5 so, s-5 = 45. s = 50 km/hr. Suppose if man move in opposite direction of train then, s+5 = 45. s = 40 km/hr. |

Pranil T. Mhatre said: (Sep 14, 2014) | |

Length of the train = 125 m. Speed of the man = 5 km/hr. Speed of the man in meters = (5*1000)/(60*60)=(50/36)=(25/18). =1.388 m/sec. Since train has covered the distance in 10 secs. Let us calculate the distance covered by the man in 10 secs. distance covered by the man in 10 secs = 13.88 m. Therefore the displacement(d) = 125 m+13.88 m = 138.88 m. Speed of the train = (138.88/10) = 13.88 m/s. Therefore The Speed of the train in Hours = 13.88 * 60 * 60 = 50000 m/sec = 50 Km/hr. |

Mahinia said: (Oct 14, 2014) | |

Train is same dxn and man is also some dxn it means it means speed is minus x-5 = 125/10*18/5 = 50. |

Sravani said: (Oct 28, 2014) | |

Why relative speed? They asked only speed? |

Prammeela.T said: (Oct 29, 2014) | |

Why we relate the speed of man and train length? |

Mohit Kumar said: (Nov 12, 2014) | |

1 km = 1000 m/1 hr = 3600 sec = 5/18 but we want to convert just opposite. So numerator and denominator will be interchange. |

Nikhil said: (Nov 22, 2014) | |

We have added the 5 km/hr to the final answer because the speed that comes out is with respect to man and to calculate the actual speed of the train we have added 5 km/hr to the final answer. |

Ravindra said: (Dec 10, 2014) | |

X-5 = 45 is right because train and man same side. In same side it is correct. |

Varma said: (Dec 16, 2014) | |

How he equated 45 to x-5? |

Vashu said: (Dec 17, 2014) | |

What is 5/18 and where its come in? |

Raja said: (Dec 19, 2014) | |

18/5 is to convert m/s to km/hr. 5/18 is to convert km/hr to m/sec. 1 km = 1000 m 1 hr = 3600 s. M = 1 km/1000 s =1 hr/3600. M/s = (3600/1000) km/hr. = (5/18) km/hr. ------------------------------------------------. Km/hr = (1000/3600) m/s. = (5/18) m/s. |

Sushma said: (Jan 4, 2015) | |

Why should we use x-5 here please explain once again. |

Hunt said: (Jan 14, 2015) | |

Relative velocity of train with respect to a fixed point (Let assume a pole). v = u-5 [u is the velocity of train & 5 is of man & in same direction so negative sign]. Now by simple equation. Length = Speed x Time. 125 = (u-5*5/18)10. So u = 250/18 m/s = 50 km/h. |

Sadhana said: (Jan 21, 2015) | |

Please explain the km/s. We should get the km/s is 5/18, but why we would get the 18/5. How is possible it is? Please explain this friends. |

Rishi said: (Jan 30, 2015) | |

Answer must be 55, Because train has to cover 125m+distance covered by man in 10 sec which is 5*10/3600. |

Vishal Patel said: (Feb 2, 2015) | |

L = Length of a train. S1 = Speed of train. S2 = Speed of man. T = Time taken by train. *************** Here is simple formula ********************* T = L/(S1 - S2) * (18/5). ************************************************ 10 = 125/(S1 - 5) * (18/5). => (S1 - 5) = (125/10)*18/5. => (S1 - 5) = (25/2)*18/5. => (S1 - 5) = 45. => S1 = 45 + 5. Ans : S1(train's speed) = 50 km/h. |

Jasper said: (Feb 2, 2015) | |

How it is saying that relative speed of the train is 45? Its only the speed of the train, can anyone tell me? |

Munna Raj said: (Feb 4, 2015) | |

Let me explain about converting km/hr into m/s in detail. 1 km = 1000 meters. 1 hr = 3600 seconds. 1 km/hr = 1000/3600 m/s. = 10/36 m/s. = 5/18 m/s. Therefore 1 km/hr = 5/18 m/sec. |

Revathi said: (Feb 6, 2015) | |

Nice explanations are there but still I have a doubt. Can anyone tell me which speed has to subtract in relative motion? |

Nanda said: (Feb 9, 2015) | |

Relative speed = (x-5) km/hr and x-5 = 45. What is the inner meaning for that can any one say? |

Nanda said: (Feb 9, 2015) | |

@ Revathi. What ever the speed we got first 45 km/hr that is not train speed, that is relative speed because it will come by the intervention of the man speed. Now we have to cal calculate train speed by follows. Let the speed of the train be x km/hr. Relative speed(45kms) = x-5(man speed). Now x = relative speed(45kms) + 5(right side 5 will come to left side). Finally x(speed of train) = 45+5 = 50. |

Gogul Prasath said: (Feb 9, 2015) | |

Relative speed is difference between speed of train and speed of man. x: Speed of train. 5: Speed of man. Difference is (x-5) = 45. 45 is relative speed. |

Nish said: (Feb 11, 2015) | |

When we convert km/h to m/sec in that situation we multiple 5/18 and same opposition 18/5. |

Jansi said: (Feb 14, 2015) | |

Could you explain how to calculate downstream in still water with example? |

Suman said: (Feb 28, 2015) | |

Here given that (x-5) km/hr. What is 5? Please explain. |

Sohail said: (Mar 7, 2015) | |

Why did you put 18/5 instead 5/18? Where 5 denotes the meter and 18 denotes the hours (1000/3600) this should be as (25/2 x 5/18)? I am a little unclear. |

Velu said: (Mar 14, 2015) | |

Here they are asking the train speed. But why we are finding the relative speed. |

Deepak said: (Apr 12, 2015) | |

How we where we divide and multiply by 5 and 18? |

Abid said: (Apr 19, 2015) | |

How to calculate what Km/Hr is train running if its length is 130 m and cross it own start point in 15 seconds? |

Pprakshit said: (Apr 27, 2015) | |

Speed of the train relative to man = 125/10. How they wrote it? |

Mammuchowdary said: (Apr 27, 2015) | |

Moving in same direction : SUBTRACT. Moving in opposite direction : ADD. |

Shyam said: (Apr 29, 2015) | |

Can anyone please explain me why it is divided by 18/5 in place of 5/18? Reason is 1 km/hr = 1000 m/3600 sec = 10/36 = 5/18. So why it is used 18/5 in place of 5/18? Please help me understand? |

Jaimik Chaudhary said: (Jun 3, 2015) | |

Distance travel by man in 10s = (5*1000*10)/(3600) = 13.89 m. Now this much more distance the train has to travel due to mobile reference which is man. Total distance travel by train = (125+13.89)m = 138.89 m. Speed = (138.89*3600)/(10000) = 50 km/hr. Comparison: If I split the 138.89 into 125+13.89. = 125/10+13.89/10 = 45+5 = 50 km/hr. |

Sanal said: (Jun 17, 2015) | |

A different way to calculate this: Length of the train: 125 m. Distance traveled by man in 10 sec: 5*(5/18) m/s = 25/18 m for 1 sec. i.e 4(25/18)*10 = 125/9 m for 10 sec. Total distance traveled by train: 125 + 125/9 = 1250/9 m in ten seconds. Speed for traveling 1250/9 m in 1 sec = (1250/9)/10 = 13.888. In km = 13.888*(18/5) = 50 km/hr. |

Bharatj said: (Jun 20, 2015) | |

Relative speed of any two object say x km/h and why km/h moving in the same direction would always be (x-y) km/h. Here the speed of the train is related to the man. Hence relative speed = 125/10 m/s = 25/2 m/s. The speed of the man is 5 km/h. When you convert it for m/s the speed will be (5x5/18) m/s = 25/18 m/s. We know that relative speed, R.S = (x-y) m/s or km/h. Here x (unknown) , y= 25/18 m/s and R.S = 25/2 m/s. Substituting we get, X = (25/2)+(25/18). X = 250/18 m/s (or) (250/18) x (18/5) km/h. So X = 50 km/h. Therefore the speed of the train will be 50 km/h. Hope you got it!. |

Rohit said: (Jul 4, 2015) | |

Why is the distance covered by man in 10 sec not considered? I believe the total distance would be the length of train + distance covered by man in 10 sec. Considering this the relative speed of train comes out to 55 km/hr. Please suggest. |

Plabon said: (Jul 5, 2015) | |

Let the speed of train= x km/hr. So relativee speed = distance/time. Relative speed of train = (x- 5) km/hr. x-5 = 0.125/10/3600. So x- 5 = 0.125x3600/10 = 45. x = 45+5 = 50 km/hr answer. |

Aayushi said: (Jul 11, 2015) | |

In 10 sec man will cover 125/9 m. In the relative motion distance should be added in either cases (opposite direction or same direction). So relative speed should be (125+(125/9))/10. Please explain. |

Deepak said: (Jul 23, 2015) | |

Nice trick. |

Samcena said: (Jul 24, 2015) | |

I made a relation like this (125m/v1-5*5/18) =10 sec. But the answer is not coming ? I converted 5 km/hr to m/s. Then I put it in above equation. I put all the terms in m and sec (not in km and hour). My answer is coming to be 30. How is it? Help me out |

Sai said: (Jul 28, 2015) | |

This concept is relative to distance not to time so you cannot take a relation like you have mentioned. |

Stark said: (Jul 28, 2015) | |

Correct your equation to 125/v1-(10*5)/(5*18) = 10 you will get. v1 = 13.88 m/s. ==> v1 = 50 kmph. |

Deepak said: (Aug 20, 2015) | |

In first problem you mentioned that meters per sec is equal to 5/18 but then you are taking like 18/5 why so please tell me. |

Swamy Chukkala said: (Aug 21, 2015) | |

Well, 10 sec ----125 mt. 36 sec----- = (36/10)*125 = 450 mt = 450/10 = 45 + 5 = 50 km/hr. |

Geeth said: (Sep 2, 2015) | |

Thank you all guys I got many ideas. |

Manav said: (Sep 7, 2015) | |

Everything is ok and easy explanation. But one thing is questionable. If the train is crossing a man who is running in the same direction with a speed of 5km/hr, than than how the distance travelled in 10 secs is equal to the length of the train. Don't you think dear friends the distance travelled by the man should also be added to the length of the train. |

Shikhar Singh said: (Sep 9, 2015) | |

Thanks for all related answer its easier for me too solve query like this. |

Mayurkapse said: (Sep 10, 2015) | |

18/5 divided because the conversion of m/sec to km/hr. |

Dhamodharan.R said: (Sep 14, 2015) | |

What is the meaning of relative speed and also meaning of x-45=5? |

Mandla Npvks said: (Sep 14, 2015) | |

Solution: The train length is 125 m. One man speed is 5 km/h. Time is 10 s. The velocity of man with respect to train is v=distance/time. 125 m/10 s = 12.5 m/s. We convert into km/h we multiplied with 18/5. Then s = 12.5*18/5 = 45 km/h. The train speed is 45+5 = 50 km/h. |

Neeraja said: (Sep 18, 2015) | |

Why it is subtracted by 5? |

Khushwanth said: (Sep 22, 2015) | |

HEY NEERAJA. Distance = Time*Speed 125 = 10*Speed. Distance/Time = Speed 125/10 = (25/2) m/s = Speed. ----------------------------------------------------------. Answer should be in km/hr so convert m/s. Km = 1000 m and hr = 3600 s. M = km/1000 and s = hr/3600. M/s = (3600/1000) km/hr. M/s = (18/5) km/hr. ----------------------------------------------------------. The condition was speed = (25/2) m/s. = (25/2)(18/5) km/hr. = 45 km/hr. The speed was relative to man so we have taken time in consideration of man. --------------------------------------------------------. See further in in view answer. |

Kaushik said: (Oct 12, 2015) | |

I have a doubt were in some case we consider 18/5 and in some case we consider 5/18? Can anyone please let me know in which of the cases we consider what? |

Priya said: (Oct 14, 2015) | |

To convert kilo meter/hour to meter/sec we consider 5/18. To convert meter/sec to kilometer/hour we consider 18/5. |

Baji said: (Dec 20, 2015) | |

Here the time 10 seconds is the time taken by the train to cross it's one length and the person running distance in 10 seconds. So the travelled distance of the person in 10 seconds is = 5 km/hr*5/18*10 = 13.8888 mts. And the train distance is 125 mts. The total distance is = 138.88888. So the finally the velocity of train is 138.8888/10. = 13.8888888 mts/sec. = 13.888888*18/5. = 50 km/hr. |

Nagamani said: (Jan 23, 2016) | |

In the given question 5 km\hr. So why you are calculated only km per hour? |

Kamal said: (Jan 26, 2016) | |

Options only in km/hr. |

Seena said: (Jan 27, 2016) | |

Can someone please tell me? When we make the use of relative speed? |

Vaitheesh Kumar said: (Feb 18, 2016) | |

Guys I have one doubt for finding out the relative speed we should need to subtract but here it's divided please explain me how? |

Pattinson said: (Feb 21, 2016) | |

18/5 is to convert seconds to hours. |

Priyan Singh said: (Mar 7, 2016) | |

How can be its 50 ? It's 45 km/hr. If we put answer 45 km as speed of train. Time = 125/45*5/18 = 10 m/s. But if we 50 we can't get its time. 10 m/s. Explain someone ? |

Ankur said: (Mar 16, 2016) | |

A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is: Kindly solve it with full description. |

Sahil said: (Mar 27, 2016) | |

Nice answer with explanation. It is very easy to understand. |

Sourav said: (Mar 29, 2016) | |

Why did we not converted 125 m to km? |

Rajeev said: (Apr 2, 2016) | |

Nice explanation. |

Panasa Harish Kumar said: (Apr 3, 2016) | |

Dear Friends, Understand the question: Train speed relative to man (X-Y) = ? X is Train speed. Y is Man speed (given as 5 km/hr). We have length(distance) train passes = 125m. Train passed in time = 10 seconds. Hence, speed = distance/time. speed = 125/10 meters/seconds. Speed = 25/2 m/s. Let's convert m/s to km/hr. ****Remember While conversion of km/hr to m/s, multiply with 5/18 or 0.277. Similarly, conversion m/s to km/hr, multiply with 18/5 or 3.6. (these are not blind values, we can get if we calculate)**** = 25/2 X 18/5 km/hr. = 45 km/hr. Let's see, Train speed relative to man. X-Y = ? X-Y = 45 km/hr. as we know Y = 5 km/hr. X-5 = 45. X = 45+5 = 50 km/hr. |

Nishant said: (Apr 4, 2016) | |

Guys, we easily get 45 km/hr then we subtract from 45-5 then we get an answer 40 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45. x = 50 km/hr. |

Suma said: (Apr 11, 2016) | |

Can someone please tell me? When we make the use of relative speed? |

Krishna said: (Apr 12, 2016) | |

Is there any shortcut method to learn the relative motion? |

Shashi said: (Apr 13, 2016) | |

@Suma Relative speed comes when we measure speed between any objects moving at a time. Here man and train are moving at a time. |

Shashi said: (Apr 13, 2016) | |

@Krishna. The thought of shortcut methods comes when we are not able to understand the problem. My word is understanding a problem clearly is itself a shortcut. |

Shikha said: (Apr 25, 2016) | |

Why did we take relative speed twice ? Once with the main i.e 125/10 * 18/5 then again, x- 5. Please clarify my doubt. |

John said: (Apr 26, 2016) | |

@Swati To convert km/h into m/sec. |

Pankaj Barwal said: (May 24, 2016) | |

@John. To convert km/h into m/sec we multiply speed by 5/18. Because 1km = 1000m &1hour = 3600 sec. Therefore 1000/3600 = 5/18. And also for convert m/sec into km/h we multiply speed by 18/5. 3600/1000 = 18/5. |

Vishal Motwani said: (Jun 9, 2016) | |

Important points to remember: The Relative speed when two movable things whose speeds x, y are moving in the same direction is = x - y. The Relative speed when two movable things whose speeds x, y are moving in OPPOSITE direction is =x + y. |

Saggy said: (Jun 22, 2016) | |

Man is moving in same direction of train if man stop them train will less time to cross him but man is walking in same direction therefor man speed will be add in train speed if man walking in opposite direction of train then it minus from that. |

Yrs said: (Jun 30, 2016) | |

Both train and man move same direction then relative speed is the sum of train speed and man speed. |

Nani said: (Jul 6, 2016) | |

Thank you friends, now I get clarity. I am thinking speed is 45kmph and boy running speed is 5kmph so same direction we need to subtract the speed so I got 40kmph. But now I clarified we got 45kmph is already subtracted with 5kmph so now we need to find train speed so we must add 5kmph to 45kmph 50kmph is ams. |

Ayushi said: (Jul 13, 2016) | |

Why x - 5? Not as 5- x? |

Amrit Ashu said: (Jul 16, 2016) | |

A different approach: We have been given the running speed of the man i.e 5km/hr converting this into m/Sec we get 1.38 m/sec. In 10 seconds this man will cover 13.8 m. So now the total distance the train has to cover in 10 seconds is 125 meters(its own length) and 13.8 m which equals 138.8 m. Using speed, time and distance formula we get S= 138.8/10 =13.8m/sec. Convert this into Km/h we get 49.8 which is approx. 50 km/h. |

Shweta said: (Jul 20, 2016) | |

A solution to this problem is so simple. Speed of train = s1; Speed of men = s2; S2 = 5km/hr. The length of train =125m. Time =10 sec. So, convert the time and sec i.e. m/s in km/hr. 125/10 * 18/5; //convert the m/s to km/hr formula is a * 18/5. Formula is: s1- s2 = (L1/T). We got the equation. S1 - 5 = (125/10) * (18/5). S1- 5 = 45km/hr;. S1 = 45 + 5. S1 = 50km/hr. |

Pramod.M said: (Jul 24, 2016) | |

Length(Distance) of the train 125m, time is 10 sec. The speed of man is 5 km/hr. Speed = D/T i.e. 125/10 = 25/2m/s. Therefore speed in km = 25/2 * 18/5 we got 45km/hr. Moving in the same direction we must subtract i.e. 45 - 5 = 40 km/hr. If the man moves against the train we must add 5km/hr to 45 km/hr. |

Mubarak said: (Jul 26, 2016) | |

Thank you guys for your best explanations really easily understandable, for the non maths learners. You people are doing good, good things never stop. Please keep on posting. |

Jaydev Ramnath said: (Aug 4, 2016) | |

Why it is 18/5? Please explain. |

Er.Deepak said: (Aug 9, 2016) | |

Man speed is 5km/hr. So, km = 1000metrs. Per hour = 60mintes = 3600second. 3600/1000 = 36/10 = 18/5. Trains cross the man in 10 seconds. Length of the train is =125m. Relative man trains speed = 125/10 = 25/2. So 25/2 x 18/5 = 45km/hr. We let train speed is = x km/hr. Relative to man turn speed = x - 5 = 45. x = 45 + 5 = 50km/hr. |

Sathiyaraj said: (Sep 5, 2016) | |

Since train and man run in the same direction, So, in 10 seconds, train crosees 125 mts (train length) + 125/9 mts, (5 km/hr => 5000/3600, 10 seconds => 125/9), So train speed , 36(125 + 125/9) (10 seconds x 36 => 3600 seconds => 1 hr), Answer : 50, 000 mts = 3600 secs => 50 km/hr. |

David said: (Sep 16, 2016) | |

@Jaydev Ramnath To convert km/hr into m/s we should multiply with 18/5. |

Peter said: (Sep 24, 2016) | |

Train Is 125 M. Man Speed 5 KM/H. Train Passed Man in 10 Seconds. HR/KM = 3600 seconds/1000 MR = 36/10 divided both on 2 = 18/5. Train Passed in 10 seconds with his tall 125 M so 125/10 divided on 5 = 25/2. (18/5 * 25/2) = 45 KM/H speed + man speed 5 KM/H = 50 :). |

Nandhini said: (Sep 27, 2016) | |

The given speed is 5 km/hr. How it comes 18/5? |

Prem Kumar said: (Sep 29, 2016) | |

What are the formulas used to find the time? |

Eldhose said: (Sep 29, 2016) | |

Time(t) = Distance(s)/Speed. S = ut + .5at^2 v = u + at 2as = (v^2 - u^2) Where, a = acceleration, v = final vel, u = ini.vel, s = dist, t = time. In these train equation, v and a are taken as zero. |

Sophia said: (Oct 5, 2016) | |

The best formula is: Step1: distance = speed x time. Then, speed = distance /time (125/10)m/sec. => (125/10)m/sec = (25/2)m/sec. Step 2: convert m/sec to km/hr. Note: 1000m - 1km. 3600sec - 60min - 1hr. 25m - 0.025km. 2sec - 0.0005hr. Now (25/2)m/sec = (0.025/0.0005)km/hr. (0.025/0.0005)km/hr = 50km/hr. |

Mounish said: (Oct 12, 2016) | |

Distance = speed * time. Speed = distance/time. Speed = (x - 5)km/hr == (x - 5) * 5/18 m/sec. Distance = 125 m. Time = 10 sec. (x - 5) * 5/18 = 125/10. x - 5 = 125/10 * 18/5. x - 5 = 45. |

Pabitra Mohan Das said: (Oct 19, 2016) | |

Why it is divided by 18/5? |

Rishabh Suthar said: (Oct 20, 2016) | |

Same direction then we do subtract and opposite direction then we do addition. Why? |

Hasan Mahmud said: (Dec 4, 2016) | |

Good job. |

Laxman said: (Dec 8, 2016) | |

Nice & excellent explanations, Thank you all. |

Srikanth said: (Dec 12, 2016) | |

Another method friends but bit lengthy. D = S X T. 125 = (A - 25/18)10, 25 = (18A - 25)/9, 18A - 25 = 225, 18A = 250, A = 250/18 X 18/5 = 50 KMPH. |

Vignesh.Rv said: (Dec 22, 2016) | |

Anyone tell me why its divided by 18/5? |

Ramya said: (Jan 10, 2017) | |

Please give the shortcut to solve the problem. |

Ashish Dagar said: (Jan 14, 2017) | |

Here, one thing I got to know is that, you all have missed a point and that was making a sense and hence considerable. We can assume a stationary man or a pole of negligible length but this man was moving with a speed of 5km/hr and hence covered almost (10*(5*5/18)) meters in 10 seconds and so the length should be considered. Thus the answer is 55Km/hr instead of 50Km/hr. |

Chinnu said: (Jan 24, 2017) | |

Here, one thing I got to know is that, you all have missed a point and that was making a sense and hence considerable. We can assume a stationary man or a pole of negligible length but this man was moving with a speed of 5km/hr and hence covered almost (10 * (5 * 5/18)) meters in 10 seconds and so the length should be considered. Thus the answer is 55Km/hr instead of 50Km/hr. |

Jeevan said: (Jan 26, 2017) | |

Shortcut 125/10 = trains speed -mans speed. Lowest form = trains speed - 5. 25/2 into km/h = 25/2 * 18/5. = 5/1 * 9/1. = 45km/h. 45km/h = trains speed - mans speed. 45km/h = trains speed - 5. THEREFORE 5 + 45 = TRAINS SPEED (transposing method). = 50km/h. |

Vinutha said: (Mar 2, 2017) | |

Can anyone explain, how we will get to know that, the given length and time is for relative speed, and why can't it be related to speed of train? |

Sushanth Ratan said: (Mar 30, 2017) | |

@Jothi. -5 if it goes on the other side it becomes plus (+). |

Ajay said: (Apr 13, 2017) | |

As per my calculation, 50 km/hr is the right answer. |

Nishant said: (Apr 26, 2017) | |

But it should have been given that we have to find the relative velocity of the man. So, 45km/hr should be the right answer. |

Priyanka said: (May 12, 2017) | |

Why we consider v=5 to subtract u - v? |

Venkatesh said: (May 21, 2017) | |

Thank you @Nagapriya. |

Ganga said: (May 24, 2017) | |

Why we subtract (x-5)? |

Shailesh Hiremath said: (May 25, 2017) | |

So, kilometre & minutes will be calculated as 1 km=1000 mtrs and 60 minutes =3600 seconds. If you divide 3600/1000 = 5/18. |

Sunil said: (May 25, 2017) | |

Thank you all for the given explanations. |

Basava G said: (May 26, 2017) | |

First, we find total length to across moving man=125+((5*10^3)÷(3600))*10. = 125+13.888. = 138.88m. The speed of train = total/time. = 138/10, = 13.88m/s. This one convert into km/hrs. = 13.88 * 3600, = 50km/hrs. |

Manjunath said: (Jun 11, 2017) | |

Hi, anyone please explain to me why we should take 5/18? |

Rachit said: (Jun 20, 2017) | |

RELATIVE SPEED OF TRAIN WITH RESPECT TO MAN RUNNING= (x-5)km/hr. TIME=DISTANCE/RELATIVE SPEED. Therefore we know that- to convert km/hr to m/s we have to use, (x-5) * 5/18)m/sec. TIME=DISTANCE/RELATIVE SPEED 10=125/(x-5)*5/18. 10=TIME,125=DISTANCE,(x-5)*5/18=RELATIVE SPEED CALCULATION- 10=125/(x-5)*5/18, 10=125*18/5x-25, 10=2250/5x-25, 10(5x-25)=2250, 50x-250=2250, 50x=2500, x=50km/hr. |

Ananya said: (Jun 24, 2017) | |

How can the answer be 50 that's the point x-5 would be 40 kmph? Then how can b be the answer? |

Gaurav said: (Jun 27, 2017) | |

Hey someone tell me, please. Why are you using 18/5 then 5/18? Because the unit is m/s = 5/18 and you are using 18/5 in this problem. |

Saikat Dey said: (Jul 1, 2017) | |

Thanks to all for such nice explanations. |

Vishu said: (Jul 2, 2017) | |

Hey guys for the same question. How to find the time taken by the train to overtake the man. Instead of finding the speed? |

Amit Thakur said: (Jul 12, 2017) | |

Thanks for these explanations. |

Paras said: (Jul 19, 2017) | |

Hers, x is supposed as speed of train. 5 is substracted because it is relative. Both are going in same direction. If the direction was supposed to be opposite then 5 should have been added. |

Dinesh said: (Jul 19, 2017) | |

Can you explain relative speed (x-5)? |

Akshit Mahajan said: (Jul 19, 2017) | |

Could anyone tell. Why we had done x-5? |

Arbaz said: (Jul 25, 2017) | |

Well said @Rahul. |

Chinmay said: (Jul 31, 2017) | |

Thank you @Rahul. |

Sathish Chandran said: (Aug 2, 2017) | |

May I know why we multiply the speed by 18/5? |

Vishal said: (Aug 2, 2017) | |

@Rachit. The answer should be in m/s according to it not in km/hr. |

Sivalaxman said: (Aug 3, 2017) | |

They are two formulas for this, 1. X mts/sec = X*18/5 km/hr, 2. X km/hr = X*5/18mts/sec, When you see the question the options are given in km/hr then we 1. Formula for this solution. |

Siva Laxman said: (Aug 3, 2017) | |

The relatie speed of the train = if opp direction = s1+s2. Same direction s1~s2. |

Rony Shaha said: (Aug 10, 2017) | |

Speed*time=distance or pole's length, How I can calculate length of train? |

Chandrasekar said: (Aug 12, 2017) | |

Why using 18/5? Please explain. |

Bala said: (Aug 26, 2017) | |

Why are you divided 125/10 at the 1st step? Please explain. |

Ashwini said: (Aug 26, 2017) | |

Thank to all the given explanation. |

Jeevitha said: (Aug 29, 2017) | |

Here 18/5 used. Because to convert 125m to km/hr from km/hr to m/sec multiply by 5/18. |

Jeevitha said: (Aug 29, 2017) | |

Somebody asking that why x-5 here. Note that the man running in the same direction. So we have to subtract. If it is opp we have to add. Because man beats train spd. So man power should be subtracted. Then only we can get an exact speed of the train. |

Shreya Parmar said: (Aug 31, 2017) | |

Using [a+b] / [u-v] = t. a = size of train = 125 b = size of man = 0 u = speed of train = ? v = speed of man = 5km/hr =5*5/18 = 25/18 m/s t = time by which train crosses the man =10 sec substitute the values in the formula we have, [125-0]/[u- 25/18] = 10 further solving we get 250/18 =u; which is in m/s now convert it into km/h by multiplying LHS with 18/5 we have u =50km/h which is the right answer. |

Apple said: (Aug 31, 2017) | |

What is 5/18? please explain. |

Amar G J said: (Sep 4, 2017) | |

It is the conversion of km/hr to m/s. km/hr = 1000/3600 = 5/18m/s. |

Angel said: (Sep 12, 2017) | |

@Bala. Speed=(Distance/Time). Here distance is the length of the train which is 125 m and the time is 10 s speed=125/10. |

Ashis said: (Sep 16, 2017) | |

5 km/h convert into m/s =5*1000/3600 or 5*5/18 =1.38, Distance traveled by man in 10 sec = 1.38*10=13.8, Total distance traveling by train =125m +13.8m =138.8, Final speed of train in 10 sec = 13.88m/s=49.968km/h. Some error because of 1.38 is approximate. |

Chithra said: (Nov 7, 2017) | |

1 km = 1000 meters. 1 hr = 3600 seconds. 1 km/hr = 1000/3600 m/s. = 10/36 m/s. = 5/18 m/s. Therefore 1 km/hr = 5/18 m/sec. Similarly, the reciprocal of m/sec is used to calculate km/hr. |

Priya said: (Nov 18, 2017) | |

As both the man and train are moving in the same direction than why the relative is subtracted by 5 i.e, x-5 it should be x+5? |

Yuvraj said: (Nov 27, 2017) | |

Speed=Distance(length of train)/Time; s=125/10; But s is (speed of train and man together i.e man's speed is 5km/hr so convert 5*(5/18)& train and man going in same direction that's why -X (let the speed of train) (X*(5/18))-(5*(5/18))=125/10, x-5=45, x=50. |

Mustufa said: (Dec 29, 2017) | |

As we find train speed pass near to man is 45 km/hr and man also running at speed of 5 km/hr so we add train speed with man speed so we get the exact speed of train. |

Vasu Erothi said: (Jan 2, 2018) | |

a km/hr to m/s conversion. a km/hr =[a*5/18]m/s. |

Arup said: (Jan 4, 2018) | |

What is 18/5? Please explain. |

Nivetha said: (Jan 6, 2018) | |

What is the meaning of (x-5) then, x-5=45? Please explain this line. |

Gilbert said: (Jan 8, 2018) | |

I think, 18/5 is seconds per meter not metres per second. |

Prosenjit said: (Feb 3, 2018) | |

Let the speed of the train x in one second. So their combined speed=x-{5000/(60*60)}=x-1.39 in one second, so, 125/x-1.39=10, x=13.89 (in one second), So 50000m=50 km in hour. |

Imran said: (Feb 21, 2018) | |

@Nivetha. (x-5) is additional speed for overcrossing, the relative speed is 50-5=45. So here answer is 50 because (x-5) is crossing speed. |

Bijayalaxmi Priyadarsini said: (Mar 8, 2018) | |

Hello please solve this question. A 180meter long train crosses a man walking the same direction at a speed of 7. 2km per hour in 10 seconds what is the speed of train? |

Ladura said: (Mar 8, 2018) | |

What is the distance covered by the man in 10 seconds? |

Dk Singh said: (Mar 19, 2018) | |

I think, 18/5 is seconds per meter not metres per second. |

Kumud Singh said: (Mar 31, 2018) | |

1 km = 1000 meters. 1 hr = 3600 seconds. 1 km/hr = 1000/3600 m/s. = 10/36 m/s. = 5/18 m/s. @Gilbert & Dk Singh. Therefore 1 km/hr = 5/18 m/sec. Similarly, the reciprocal of m/sec is used to calculate km/hr. |

Kumud Singh said: (Mar 31, 2018) | |

@Ladura. Distance covered by the man in 10 sec, Speed = 5*5/18 = 25/18 m/sec, Distance = 25/18*10 = 125/19 m. |

Andrews Anbarasu said: (Apr 12, 2018) | |

Your explanation is good @Ravi. Thanks all. |

Koser said: (Apr 12, 2018) | |

Why is it divided by 125/10? |

Shivam Tomar said: (Apr 13, 2018) | |

Speed of the train relative to man = 125/10 m/sec. I am confused in this step, please clarify it. |

Satya said: (Apr 23, 2018) | |

Why is it divided by 18/5 and how it is? Please explain. |

Sarang said: (May 3, 2018) | |

Distance=speed * time. In same direction speed is added, 125=speed*10, 125/10= speed, 12.5 *18/5, 45 relative speed of train, 5 + 45 = 50. |

Apoorva E said: (May 4, 2018) | |

Whenever you convert speed from km/hr to m/sec you should multiply the speed by 5/18. Vice versa when you convert m/sec to km/hr you should multiply the speed by 18/5. |

Bharat Gupta said: (May 6, 2018) | |

Please explain the relative speed = (x - 5) km/hr. |

Vikash Gupta said: (Jun 2, 2018) | |

Relative velocity is 125/10, how? |

Suraj Kumar Swami said: (Jun 8, 2018) | |

x-5 = 45 the result should be 40, then why it is 50? |

Bouvthika R.M said: (Jun 10, 2018) | |

Here please explain for what reason it becomes (X-5). I can't understand. |

Veni Balamurugan said: (Jun 11, 2018) | |

How come 25/2? Please explain. |

Subu D. said: (Jun 15, 2018) | |

125 = speed /10. Speed = 125/10= 25/2. |

Sonu Kumar From Bihar said: (Jun 19, 2018) | |

@Suraj Kumar It is X-5=45, So, X=45+5=50. @Bouvthika. 45km/hr is the relative speed of train w.r.t. the man running speed, but we let the actual speed of the train is X km/hr, so actual speed of train- man running speed = relative speed of the train. |

Pankaj said: (Jun 24, 2018) | |

Please explain 5/18 and 18/5. Why using this? |

Phanikumar said: (Jul 9, 2018) | |

Speed of the train relative to man = 125/10 m/sec (meters per second). But you need to find the speed in KMPh. Ex; 1kmph = 5/18 m/sec, so that 1m/sec = 18/5 kmph. Because of that 25/2 is multiplied with 18/5, then we got the speed in KMPh. Similarly (25/2)*(18/5)=45kmph. But the both man and train is moving in same direction. So, X-5=45. x=50kmph. |

Tanvi said: (Jul 28, 2018) | |

Why using 18/5? Please explain. |

Narendran Sainathan said: (Jul 31, 2018) | |

18/5 is to convert meter per second to kilo meter per hour. Or you can also use 3600 (seconds)/1000 (kilometre). |

Omkar Tanpure said: (Aug 2, 2018) | |

Relative Speed = 45 km/h. Speed of man = 5 km/h. But, we know, here, Relative speed = Speed of train - Speed of the man. Therefore, Speed of train = Relative speed + Speed of man. So, Speed of train = 45 +5 = 50 km/h. |

Jola M said: (Aug 3, 2018) | |

Converting km/hr to m/s is x*5/18 (For eg, 36 km/hr means 36*5/18 =10 m/s). Converting m/s to km/hr is x*18/5 (For eg, 10 m/s means 10*18/5= 36 km/hr). |

Rohit said: (Aug 6, 2018) | |

Whenever we have to convert seconds or minutes to an hour then multiply by 18/5 and if to do vice versa then 5/18. |

Abdus Salam said: (Aug 11, 2018) | |

Thanks a lot for all the given explanation. |

Swati Warang said: (Aug 15, 2018) | |

What is relative speed, can you explain in simple terms? |

Ganesh said: (Aug 22, 2018) | |

They asked the train speed; but why we have to take relative speed? Please tell me. |

Ganesh said: (Aug 22, 2018) | |

Anyone please explain about relative speed. |

Ramu Nayak said: (Aug 22, 2018) | |

Relative velocity is nothing but the velocity of body A to body B. Case 1: A and B moving in the same direction Vr= Va-Vb. (Va>Vb) Vr= Vb-Va (VB>Va) Case 2: A and B moving in opposite directions. Vr=Va+VB. (Velocity not matters). |

Akeer said: (Aug 22, 2018) | |

@Ganesh. The Relative means related to speed. The two trains are in which direction moving based on that we have to take the speed to add or sub. |

Dhivya said: (Aug 24, 2018) | |

L(train) = (speed of train-speed of object) * time. 125m = (St-5)km/hr * 10s, 112.5 * 18/5 = st - 5, St = 50km/hr. |

Farheen said: (Aug 31, 2018) | |

To convert km/hr to m/sec we multiply it with 5/18. If we are converting m/sec to km/hr we multiply it 18/5. |

Vipin said: (Sep 2, 2018) | |

Hi, I am not getting this. Please anyone help me to get it. |

Robi said: (Oct 3, 2018) | |

@Omkar Tanpure. What is the relative speed? And why 45 is called relative speed? Please explain me. |

Ramya said: (Oct 21, 2018) | |

Can anyone Explain this problem to me? I didn't understand this. |

Chandan Azad said: (Nov 1, 2018) | |

T(time required) = Lt(length of train) + Lo(length of object)/ Ut(speed of Train)+,-Uo(speed of object) . ("-" when train and object goes in same direction. "+" when train and object goes in opposite direction.) 10=(125+0)/(Ut-5*5/18). Ut=125/9. Ut=(125/9)*(18/5). Ut=50km/hr. |

Hari Kumar said: (Nov 11, 2018) | |

50 is the right answer. |

Lavanya said: (Dec 5, 2018) | |

Why do we to subtract 45-5? please clarify that. |

Narendra said: (Dec 11, 2018) | |

In same direction, T. S-M. S = 125/10 * 18/5 = 45. M.S = 5KMph. T.S - 5 = 45. T.S = 50sec. |

Abuthahir said: (Dec 27, 2018) | |

X-5 = 45. Here x = 45+5, Then x = 50. |

Anjali said: (Jan 3, 2019) | |

Thanks all for explaining the answer. |

Vinod said: (Jan 4, 2019) | |

Thanks all for explaining the answer. |

Muhammad Zulafqar said: (Jan 21, 2019) | |

The simplest answer is; 10sec/1000Meter =0.01. Then multiply with 5KM i.e 0.01 * 5000meter = 50KM. |

Nithish said: (Jan 28, 2019) | |

Thanks all for the explanations. |

Mozammel Hossain said: (Feb 13, 2019) | |

There have man speed but not train speed. So we known that; speed = distanc/time. = 125/10, = 25/2. Now convert into km/h. 25/2 * 18/5. = 45km/h. Lets train speed is x km/h. So, the relative speed is x-5 km/h. x- 5 = 45, X = 50km/h. |

Kunal Pardeshi said: (Feb 20, 2019) | |

Hi. please solve my query. Since the man is also moving, why the distance travelled by train in 10 seconds is considered equivalent to the length of the train when in actuality the distance covered will be more than that as the man is not stationary. Please, somebody explain this clearly. I want to know the concept and logic behind this. |

P. Sudha said: (Feb 22, 2019) | |

How 18/5 possible? Please explain to me. |

K. Abhishek said: (May 14, 2019) | |

@All. Here, Train took 10 seconds to cross the person who is running at the speed of 5km/hr, then during this 10s the person might have travelled a certain distance, right? so train travelled (125m + distance covered by the man in 10s) in 10s then why the relative speed of train = 125/10? How, will anyone explain it for me? |

Yugapooja said: (May 26, 2019) | |

Thanks for the explanation @Rahul. |

Abhishek said: (Jun 8, 2019) | |

The speed of train with respect to ground=125/10=25/2 (m/sec). Now to covert into km/hr:(25/2)*(3600/1000)=45 Km/hr. Now there is a formula: Velocity of train w.r.t. man(45 km/hr)=velocity of train w.r.t. ground(let it be x) - velocity of man w.r.t. ground(5km/hr). So, 45 = x-5. i.e. x = 50 Km/hr. |

Shashank Shekhar said: (Jun 17, 2019) | |

Train crossing a moving object without length in same direction: Formula is :- Time=length of train / ( speed of train-speed of object). 10 sec = 125 m/ ( x - 5*5/18). So, x=125/9 m/sec = 125/9 * 18/5 km/h = 50 km/h. |

Sai Ganesh said: (Jun 27, 2019) | |

as per my opinion The question is given in m/sec so the answer should be converted into 1 hr = 3600 sec and 1 km = 1000 m. (3600/1000) m/sec as per given question. 125m and 10 sec, (125/10)m/sec=45m/sec. Here the man is moving in the same direction so we should subtract. (x-5)=45. x = 50km/hr. |

Mangaldeep Ghosh said: (Jun 30, 2019) | |

Why ((a+b)/(u-v)) = t does not work? Where a=length of the train (125 meters). b= length of object( 0 meter), u = speed of train ( let, x m/s), t = time taken to cross the object, v = speed of object (5 km/hr). |

Vishwa Telagadi said: (Jul 29, 2019) | |

Man's speed=5 km/hr = 5 * 5/18 = (25/18)m/s. The distance covered by man in 10 sec = (25/18)*10=(250/18)m, Therefore distance covered by train for crossing the man=125+250/18=1250/9m, Therefore speed=(1250/9)/10=125/9 m/s, =(125/9)*3600/1000, =(125/9)*(18/5)=50 km/hr. |

Arun said: (Aug 4, 2019) | |

How we say 45 km/hr is a relative speed? |

Omsingh More said: (Aug 9, 2019) | |

They should mention "relative speed" in the question. Why it is not so? It confuses the examiner! |

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