Discussion :: Problems on Trains - General Questions (Q.No.30)
20x + 25(x - 1) = 110
45x = 135| Shri said: (Aug 11, 2010) | |
| Why (x-1) term is used ? | |
| Amit Kr. Soni said: (Aug 14, 2010) | |
| Hi Shri, We can write here (X-1) b'coz train 'B' starts travelling 1 hr later agains train 'A'. So, At the end, train B will travel 1 hr short, out of full running time to meet each other. i.e 'X' So, we can write total time for train 'B' = (X-1) Thanks |
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| Sathya said: (Aug 26, 2010) | |
| Though the train B starts 1 hour late than train A, The speed of the train varies. | |
| Easwar said: (Nov 14, 2010) | |
| A train leaves a point A at 5 p.m. and reaches another point B a 11 p.m. Another train leaves point B at 7 p.m. and reaches point A at 10 p.m. At what time will the two trains meet? 8 : 15 p.m. 8 : 30 p.m. 8 : 20 p.m. 8 : 00 p.m. can anyone solve this problem |
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| Randhir said: (Nov 17, 2010) | |
| hi Easwar, The speed of one train is twice the other so let speed of 1st train is Y. speed of 2nd train is 2Y. then distance between point A and B is 6Y. Suppose they meet x hours after 5 a.m. Distance covered by 1st train in X hours = YX km. Distance covered by B in (X - 2) hours = 2Y(X - 2) km. Therefore YX + 2Y(X - 2) = 6Y => YX + 2YX - 4Y = 6Y => 3YX = 10Y => X =10 / 3 (in hours) => X =10/3*60 (because converting into minutes) => X =200 minutes 200 minutes means 3 hour 20 minutes So, they meet at 8 : 20 p.m. |
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| Erererer said: (Nov 27, 2010) | |
| X=3 how it convet to 10 am | |
| Easwar said: (Dec 7, 2010) | |
| Hi Erererer, It was assumed that trains r meeting after x hours, so 5 pm(starting time) + 3.20 = 8.20pm Thanks Randhir..... |
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| Faisal said: (Jan 3, 2011) | |
| what if we say they meet after 8AM Distance covered by B in x Hrs =25X Distance covered by A in (x-1)Hrs = 20(x-1) 25x+20(x-1)=110 25x+20x-20=110 25x+20x=130 45x=130 x= 2.8 ( Which in incorrect how come ) |
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| Mahe said: (Feb 3, 2011) | |
| @Faisel, If they meet 8am then Distance covered by b in xhrs is 25x Distance covered by a in(X+1)HRS=20(X+1) (x+1) coz v r considering 1 hr aft train A has started.. So, 25X+20(X+1)=110 45X=90 X=2 Hence 8am + 2 = 10am |
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| Nayana said: (May 17, 2011) | |
| @Randhir In the problem what you have solved above,the distance between point A and B is 6Y. I am not getting how you got 6Y. Can you please explain me that? |
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| Nithi said: (May 23, 2011) | |
| @Randhir I too not getting tht 6y concept..pls explain me... |
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| Teena said: (May 30, 2011) | |
| I dint get how x-1 term is used. | |
| Gai3 said: (Jun 23, 2011) | |
| According to this problem first we have to consider that both trains A & B are start at same time then only we can arrive the answer. In question it is given that, A train starts at 7 a.m. and B train starts at 8 a.m. Let starting time for A train = x and starting time for B train = x-1 . |
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| Sudha said: (Jul 28, 2011) | |
| @Nayana, We know that distance= speed*time So, distance = y*6 = 6y Calculate time between 5pm and 11pm, so totally 6 hours. Thanks. |
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| Gayatri said: (Aug 26, 2011) | |
| @Easwar The speed of one train is twice the other so let speed of 1st train is Y. speed of 2nd train is 2Y. then distance between point A and B is 6Y. Suppose they meet x hours after 5 a.m. Distance covered by 1st train in X hours = YX km. Distance covered by B in (X - 2) hours = 2Y(X - 2) km. Therefore YX + 2Y(X - 2) = 6Y => YX + 2YX - 4Y = 6Y => 3YX = 10Y => X =10 / 3 (in hours) => X =10/3*60 (because converting into minutes) => X =200 minutes 200 minutes means 3 hour 20 minutes So, they meet at 8 : 20 p.m. Hii Randhir How come distance between point A and B is 6Y? Please explain. |
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| Randhir said: (Sep 22, 2011) | |
| @Gayatri, nithi, nayana Speed=distance/time 1st train leaves a point A at 5 p.m. and reaches another point B a 11 p.m.that means time taken by train 6 hours. Time = 6 hour Speed of the 1st train is Y Speed = Y km/h Distance= speed*time (because speed=distance/time) Distance= Y * 6 = 6Y if you are still not getting my point then comes to train 2nd 2nd train leaves point B at 7 p.m. and reaches point A at 10 p.m. that means time taken by train 3 hours. Time = 3 hour Speed of the 1st train is 2Y Speed = 2Y km/h Distance= speed*time (because speed=distance/time) Distance= 2Y * 3 = 6Y |
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| Anusha said: (Sep 26, 2011) | |
| How x=3 is converted into 10, please tell me any one. | |
| Jammy said: (Oct 22, 2011) | |
| 3 = 3 hour.....therefore 7 a.m + 3 hour = 10 a.m | |
| Yogi said: (Nov 22, 2011) | |
| Actually I didn't understand this problem how do you get 10 AM as answer please would anyone get rid of these problem am not getting it. | |
| Yuvaraj said: (Dec 15, 2011) | |
| One more easy method to solve such problems. Given that distance between two stations is 110km. First trains speed is 20kmph and starts at 7 am 2nd trains speeed is 25kmph and starts at 8 am. So at 8 am, the distance covered by first train is 20 kms. Now remaining distance to cover is (110-20 = 90 km). The relative speed of both the trains is 20+45 = 45 kms. So it will take 2 hours to cover 90 km. Hence the trains will meet after 2 hours from 8am i.e. at 10 am. |
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| Knn said: (Dec 30, 2011) | |
| Yuvraj. Cool answer! | |
| Damodar Narayan said: (Jan 27, 2012) | |
| A----------110---------B 7am A---------90--------B 8am(Speed of the 1st train is 20 km/h) A----45----B 9am(relative speed=20+25(45) km/h) AB 10am |
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| Sandy said: (Feb 21, 2012) | |
| Thanks yuvraj. | |
| Ajay said: (Mar 16, 2012) | |
| Its Simple in another way that Total Distance was 110KM.... A start at 7 AM at the speed of 20 KM in 3 Hrs he will cover 60KMS whereas B start at 8 AM @ speed of 25 KM/Per Hour in 2 HRS he will covered 50 KMS that means 7+3=10AM | |
| Arun said: (Apr 18, 2012) | |
| Why we use x-1 not x+1? | |
| Amit Kumar said: (Jun 7, 2012) | |
| Dear Arun, Because train is started 1 hours late, thats why x-1 |
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| Joicy said: (Oct 2, 2012) | |
| Why (X-1) term is used? | |
| Charles @ Redhills said: (Dec 7, 2012) | |
| @Joicy. The train B starts an hour later than the train A. Thus the train time A is assumed as x (started at 7 am) and the train B starts at time 8 am which is later an hour of time x hence its x-1. Hope its clear to you now. |
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| Bhoopesh Choudhary said: (May 30, 2013) | |
| The answer of this question is incorrect if the B train start at 8am and by the speed 25km\hr then it will cover the distance of 50km in next 2 hours but the train A will cove the distance of 60km in next 3 hour by the peed of 20km\hr. So both can be meet at 10 a.m. |
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| Manish Dabhi said: (Sep 10, 2013) | |
| Train A Travel in 1 hour 20 km at 7 am. They both started together 8 am. That's why 110-20 = 90km Remaining, Relative Speed = (20+25) = 45 km/hr. Time = (Distance/Speed). = (90/45) = 2 Hours That's why 8am + 2 hours = 10 am. |
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| Gandhi said: (Sep 25, 2013) | |
| Who ever doesn't understand why they have taken (X-1). Please see the below example. Length of the track between starting position of train A starts at 7A.M and B Starts at 8A.M is 200 km. Let's say A train travels with 20 kmph for 5 hours. Distance covered by train A is = 100 km. Train B Travels with 25 kmph for 4 hrs. Distance covered by train B is =100km. A train => 7+5=12 noon. B train => 8+4=12 noon. If we consider 7 as X then 8 should become (X-1) when we compare with the destination time. |
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| Govinda said: (Nov 10, 2013) | |
| Its very simple. Train A start at 7 AM & B start at 8 AM. A travels 20 km in 1 hr ie, distance between A and B will be. 90 Km at 8 AM. Relative speed = 20+25 = 45Km/hr. Distance = 90km. Time = 90/45 = 2 hr. Total time = 2+1 (that was travelled by train A between 7 to 8 AM. ). Answer will be 7+3 = 10 AM. |
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| Baswaraj said: (Nov 28, 2013) | |
| So x = 3. Then B start 7 a.m so count three numbers like 8,9,10 (after 7 we will get 8) :10 a.m. |
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| Rambabu said: (Jan 13, 2014) | |
| @Easwar. Distance is x and speed first train is = x/6 (because speed=distance/time). And speed of train second train is = x/3. First train start at 5pm. If it travel two hours up to 7pm, the distance covered by this train is ( (x/6) *2) is x/3 and the remaining distance is (x- (x/3) is 2x/3. Now find relative speed (opp direction) is (x/6) + (x/3) =x/2. Remaining distance is 2x/3 and relative speed is x/2 so, ( (2x/3) / (x/2) ) = 4/3hours. Means after 7 pm both trains meet at 4/3 hours later. (4/3) *60=80 minutes so at 8:20 pm both trains meet. Thank you. |
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| Deepak Patgar said: (Aug 21, 2014) | |
| Second train leaves from B towards A. Many of us got confused that it starts from A towards B. And we don't have to make it so complicated. We can generalize this question. first train leaves at 7am. By 8am it covers 20kms. Remaining distance is 90kms (110-20). By 9am first train covers 20kms more and second train covers 25kms. Remaining distance is 45kms (90-(20+25)). At 10am the first train covers another 20kms and second train covers 25kms more. (45-(20+25))=0 kms. Distance between them is zero. Hence they meet at 10am. |
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| Poojitha said: (Nov 5, 2014) | |
| How is it 20x+25(x-1) = 110? | |
| Suresh said: (Jan 11, 2015) | |
| Hai @Poojitha. x is the time taken, 20 is speed. So Speed*Time = Distance. For first train it is 20x. And as the second train started 1 hr later. The time will be x-1. And distance = 25(x-1). Since both the trains are yet to cover 110 km. And travelling in opposite directions to meet each other and we have to calculate the meeting point of time. We have to add them. So, It is 20x+25(x-1) = 110. Hope it is clear now. |
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| Buddhikawd said: (Jan 20, 2015) | |
| A_________________110KM______________________B. 7.00 AM. 8.00 AM. 20 KMH. 25 KMh. 2nd train after 1 hour. So 1st train gone to 20 km. Because totally 110-20 = 90 km. Now 20 kmh+25 kmh = 45 kmh. 90/45 = 2 hours. Now we count 8.00 am+ 2 hours = 10.00 AM. |
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| Gokul said: (Feb 14, 2015) | |
| How 20x + 25(x-1) = 110? | |
| Chai said: (May 6, 2015) | |
| A and B are two stations 390 km a part. A train start from A at 10 AM and travels towards B at 65 kmph. Another train start from B at 11 AM and travels towards A at 35 kmph. At what time do they meet? Solve this problem. | |
| R V said: (May 11, 2015) | |
| One train will cover 20 km in one hour. Now the distance remaining is 90 km. So time take is equal to 90/20+25 = 2 hours means will meet 2 hours after 8 am i.e. 10 am. |
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| Arun said: (May 19, 2015) | |
| @Chai. Answer : 2.15 pm. Is this correct? |
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| Arun said: (May 19, 2015) | |
| @Chai. Sorry I will be confused. I got two answers by solving two different methods. I got 2.15 pm and also I got 4.15 pm. |
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| Mohit said: (Jun 22, 2015) | |
| Let 't' b the time in hrs the two trains meet. Then distance traveled by 1st train is 20t and distance. Traveled by 2nd train is 25 (t-1). Because 2nd train run after 1hr of 1st train. Total distance = 110 km = 20t+25(t-1). Solving we get t = 3hr. 7 a.m.+3hr = 10 a.m. |
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| Sumanth Geras said: (Sep 18, 2015) | |
| Simple solution no formula, no concepts just common sense. Let us consider first train it's speed is 20 km/hr. So it covers 20 kms in 1 hour. Starting from 7 am 1st train second train from 8 am to 9 am. At 8 am it covers 20 km at 9 am it covers 25 km. 9 am it covers 40 km 10 am it covers 50 km. 10 am === 60 km. So at 10 am total distance covered by two trains is 60+50=110 km. Which is the actual distance of separation between two train hence they meet each other so simple. |
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| Anith Kumar said: (Oct 4, 2015) | |
| How we assume x=3 then how we get both meeting time at 10 please explain clearly to me. | |
| Hardik Khatri said: (Feb 21, 2016) | |
| @Chai. Why do we have to take the distance 390 when we add this covered by A after x hours and this covered by B after x-1 hours then why we equal it to 390 km? |
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| Raja said: (Mar 13, 2016) | |
| Here x=3 is answer how to convert into am/pm some other problem. | |
| Anish Mishra said: (Mar 19, 2016) | |
| A train leaves P for Q; at the same time another train leaves Q for P. The trains meet after 6 hours, the train from P and Q having travelled 8 km per hour more than the other. If the distance from P and Q is 528 km, the speed of train from P and Q in km/hour is. Please solve this? |
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| Naresh said: (Apr 23, 2016) | |
| The distance between two stations A and B is 110km. 1st Train starts at 7am at station A at a speed of 20kmph. i.e. 1st train travels 20 kms in one hour. Now The distance is 90kms. 2nd train stats at 8am at and travelling at a speed of 25kmph. So, the Relative speed of two trains is 20 + 25 = 45kmph. => Time = Distance/Speed. Time = 90/45 = 2hrs. Requied answer is 8 A.M + 2hr =10 A.M. |
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| Rahul said: (Apr 27, 2016) | |
| Easy way: d = 110 km train A = starts @ 7.00 with speed = 20 km train B = starts @ 8.00 with speed = 25 km *(5kms difference A & B) So, A+B=45 kms && d = 110 ---> Time = d/s = 110/45 (after cancellation = 22/9) ---> Now add 5 (difference kms between 2 trains speed) to 22 = 27. ---> Now divide 27/9 = 3 ---> Add 3 am to 7 am = 10 am. **Here the difference in time is only 1 hr so added only 5 km's extra. If the time difference was 2 hrs then we 10 and so on. If the answer you get doesn't match with the options then choose the closest options. The simple reason is time has 60 min where as when you are actually calculating it is of 100. Hope you'll are understanding. HOPE THIS WORKS I WAS ACTUALLY TRYING TO MAKE SOME EASY SHIT SO THAT THIS TYPE OF SUMS BECOMES EASY FOR THE US :-) |
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| Keerthu said: (May 19, 2016) | |
| Hi, @Chai. I got 1:25 PM. Is this correct? |
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| Rups said: (Jun 11, 2016) | |
| If the train was late by 3 hours and 12 minutes, then calculate the nearest hours. | |
| Himanshu said: (Jul 16, 2016) | |
| @Yuvaraj, Thanks. | |
| Simran said: (Jul 31, 2016) | |
| @ANISH MISHRA: Let the speed of the train P be xkm/hr and Q be ykm/hr. After 6hrs, P would have travelled 6xkm (distance=speed*time) and similarly Q would have travelled 6ykm. Given, the distance between P and Q is 528, Hence, 6x + 6y = 528 -> (i). Also, given that P having travelled 8km/hr more than Q. So, x = y + 8 -> (ii). Solving equation (i) & (ii), x = 48km/hr (speed of train P). y = 40km/hr (speed of train Q). |
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| Ravi said: (Aug 9, 2016) | |
| Here, the first train is started at 7:00 am at 20 km/h and the second train started at 8 am at the speed of 25 km/ h. Train first started at 7 am @ 20 km/h. 7 am to 8 am covers =20 km 8 am to 9 am covers = 40 km 9 am to 10 am covers =60 km 10 am to 11am covers = 80km 11 am to 12 it covers 100 km Train second started at 8am @ 25 km/h. 8 am to 9 am train covers = 25 km 9 am to 10 am it covers = 50 km 10 am to 11 am it covers = 75 km 11 am to 12 it covers = 100km Now, both it covers 100 km at 12 o'clock. then how these 2 train meets each other at 10 am please help me, sir. |
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| Rabiul said: (Aug 13, 2016) | |
| Thanks @Manish Dabhi. Your explanation is very useful. |
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| Sourav Pal said: (Aug 26, 2016) | |
| Simple formula to be remembered. Total Time Taken = (d = s2 * t)/s1 + s2. Where d = 110km, s2 = 25kmph, s1 = 20kmph, t = (8:00am - 7:00am) = 1hr. From that we get 3hrs. Therefore required Time is (7:00am + 3hrs) = 10:00Am. |
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| Ankita said: (Oct 12, 2016) | |
| Distance between two stations A and B is 100 km, two trains with speed 60 km/h and 40 km/h are running towards each other. What is the distance from station A where both trains will meet? Can anyone find the answer? |
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| Ayush said: (Nov 11, 2016) | |
| 60 km as relative speed of both the trains is 100 km/hr and distance = 100km. Then time taken to meet = distance/speed(relative). T = 100/100 = 1 hr So the trains will meet each other after 1 hour so, Distance covered by first train in 1 hour = 60km/hr * 1hr = 60 km, Distance covered by second train in 1 hour = 40km/hr * 1hr = 40 km, Therefore, the distance from station A where both trains will meet = 60km. |
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| Sonic said: (Jan 3, 2017) | |
| Why it's taken x-1 but not x+1 because train 2 starts 1 hr later? | |
| Sonic said: (Jan 3, 2017) | |
| Why it's taken x-1 but not x+1 because train 2 starts 1 hr later? | |
| Bhat Ashraf said: (Mar 2, 2017) | |
| The difference of speeds of two trains P & Q is 20km/hr. Train P & Q start from points A & B & move towards each other. If the trains meet each other after 2 hr of the journey. What is the speed of the faster train? Please answer it. |
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| Swasti said: (Mar 2, 2017) | |
| How this 3 hr comes? | |
| Mary said: (Mar 29, 2017) | |
| How to solve if they move in the same direction? | |
| Basharat said: (Apr 15, 2017) | |
| This is based on relative speed. If they are in opposite directions just add their speeds if they are in same directions subtract their speeds. |
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| Sachin said: (May 12, 2017) | |
| Here V1 =20 km/hr. V2 = 25km/hr. Time of meet they = 110-20/20 + 25. = 2hr. Then 8+2 = 10am. |
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| Chittaranjan said: (Jul 20, 2017) | |
| A_________________110KM______________________B. 7.00 AM. 8.00 AM. 20 KMH. 25 KMh. 2nd train after 1 hour. So 1st train gone to 20 km. Because totally 110-20 = 90 km. Now 20 kmh+25 kmh = 45 kmh. 90/45 = 2 hours. Now we count 8.00 am+ 2 hours = 10.00 AM. |
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| Nikhil_Vats said: (Oct 26, 2017) | |
| Platform A ____110km____ Platform B. Difference between train A and train B is =1hr. Relative speed of both the trains are = (20kmph+25kmph)= 45kmph. Now think as if you are the boss (what if you can make both the trains depart at 8 AM how much easy it would have become). So, the train A is moving at 20km in 1 hour at 7 AM it means it already has completed 20kms by 8 AM from the given distance of 110 km = so (110-20=90 km) left for train A. And being a boss you now calculate from 8AM for both the trains. Assume, If they both depart at 8 AM Train A now have to go 90 km more Train B have to go 110 km So we will calculate D=S*T ( distance still left = relative speed *time 90=45*time 90/45=2. 2 hours from the time which we shifted like a boss 8 AM so add 2 hours more after 8. Its 10 AM. |
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| Anna said: (Jul 25, 2018) | |
| Thank you very much @Yuvaraj. | |
| Praveen_Nayak said: (Aug 17, 2018) | |
| Thank you @Manish Dabhi. | |
| Manisha said: (Nov 5, 2018) | |
| A train travels in one houe=20 km. Now at 8 o'clock distance between tow trains = 90 km. It is relavent to these two trains speed =20+25km/h = 45km/h. Now time = distance/speed. = 90/45 = 2h. Hence trains meet after 2hours that is 8+2 = 10 a.m. |
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| Viswa said: (Nov 18, 2018) | |
| For equal starting time of two trains (ie 8am, 8am). Train A cover a distance in one hour = 20km. So total distance covered by two trains =110-20. = 90km. The relative speed of trains =20+25. = 45kmph. Two trains are meeting at a time of X hr. The formula is; 90/X = 45, X=2, 8am+2 = 10am. |
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| Raj said: (Jan 22, 2019) | |
| Here, The total distance between them = 110km. (Making equal starting point) By the time train B has started, the Train A has already covered 20km (since speed =20km/hr). Both at 8:00am now. Now the distance between them is; 110-20=90km. Relative speed= 20+25(since opposite direction) =45km/hr. To cover 90 km time taken=90/45=2hrs. 2 hrs from 8:00am=10:00am. (since we have made initial starting point at 8:00am for both the trains). |
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| Nandakumar said: (Jun 1, 2019) | |
| Let x be the meeting distance from A and B. Then A travels (X/20) hrs and B travels (110-X/25) {time=distance/speed}. But A stars 1 hour earlier than B. So; (X/20)-1 = (110-X)/25. solving this; we get X=60kms. Now A travels 60 kms at the speed of 20kmph to meet B in the midway of travel. So now Time is taken to travel to meet B 60 kms from starting point A takes+ 60(distance) / 20(speed) = 3 hrs. From question, 'A' starts to travel at 7 A.M and travels 3 hrs(from final answer) to meet B in the midway of travel. So the time 'A' and 'B' at meet is [7+3]= 10 A.M. |
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| Murli Manohar said: (Jul 20, 2019) | |
| Let they meet X hrs from the 8 a.m. Then, 110 * 1000 = 20 * (5/18) * 3600 + 45(5/18) * X. X=2. So, time= 8+2 = 10 a.m. |
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