Problems on Trains, Boats and Streams

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<ul><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 1</p><p>SREENIVASA INSTITUTE OF TECHNOLOGY ANDMANAGEMENT STUDIES</p><p>Murukambattu Post, Chittoor 517 127 (A.P)AFFILIATED TO JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY </p><p>ANANTAPUR,</p><p>Course materialFor</p><p>Reasoning and Quantitative Aptitude</p><p>Module Name: PROBLEMS ON TRAINS,BOATSAND STREAMS.</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 2</p><p>MODULE OBJECTIVE:</p><p>Decision making for planning, policy and management relies increasingly on thequantitative reasoning, which entails the collection, analysis and interpretation of quantitativedata. This course is designed to introduce principles and techniques to solve trains, boats andstreams related problems.</p><p>Develop logical reasoning in a problem solving framework. One goal is to develop adisciplined logical analysis of word problems. Such reasoning is the foundation for buildingssimple mathematical models of problems-models implicit on trains, boats and streams. Howevera logical mind will serve a person well in any field.</p><p>At the end of these course students:</p><p>1) To be able to understand the types of formulae used to calculate trains, boats and streams.2) To be solving many problems related to trains, boats and streams which can be useful to</p><p>students to face interviews</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 3</p><p>PREREQUISITES:1. a km/hr= (a* 5/18) m/s.</p><p>2. a m / s = (a*18/5) km/hr.</p><p>3 Time taken by a train of length 1 metres to pass a pole or a standing man or a signal post isequal to the time taken by the train to cover 1 metres.</p><p>4. Time taken by a train of length 1 metres to pass a stationary object of length b metres is thetime taken by the train to cover (1 + b) metres.</p><p>5. Suppose two trains or two bodies are moving in the same direction at u m / s and v m/s, whereu &gt; v, then their relatives speed = (u - v) m / s.</p><p>6. Suppose two trains or two bodies are moving in opposite directions at u m / s and v m/s, thentheir relative speed is = (u + v) m/s.</p><p>7. If two trains of length a metres and b metres are moving in opposite directions at u m / s and vm/s, then time taken by the trains to cross each other = (a + b)/(u+v) sec.</p><p>8.If two trains of length a metres and b metres are moving in the same directionat u m / s and vm / s, then the time taken by the faster train to cross the slower train = (a+b)/(u-v) sec.</p><p>9. If two trains (or bodies) start at the same time from points A and B towards each other and aftercrossing they take a and b sec in reaching B and A respectively, then</p><p>(A's speet) : (Bs speed) = (b1/2: a1/2).</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 4</p><p>SOLVED EXAMPLES:</p><p>Ex.I. A train 100 m long is running at the speed of 30 km / hr. Find the time taken byit to pass a man standing near the railway line. (S.S.C. 2001)</p><p>Sol. Speed of the train = (30 x 5/18_) m / sec = (25/3) m/ sec.</p><p>Distance moved in passing the standing man = 100 m.</p><p>Required time taken = 100/(25/3) = (100 *(3/25)) sec = 12 sec</p><p>Ex. 2. A train is moving at a speed of 132 km/br. If the length of the train is110 metres, howlong will it take to cross a railway platform 165 metres long? (Bank P.O.2004)</p><p>Sol. Speed of train = 132 *(5/18) m/sec = 110/3 m/sec.Distance covered in passing the platform = (110 + 165) m = 275 m.Time taken =275 *(3/110) sec =15/2 sec = 7 sec</p><p>Ex. 3. A man is standing on a railway bridge which is 180 m long. He finds that atrain crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of thetrain and its speed?</p><p>(SASKEN 2003)</p><p>Sol. Let the length of the train be x metres,</p><p>Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 secx/8=(x+180)/20 20x = 8 (x + 180) x = 120.Length of the train = 120 m.</p><p>Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 kmEx. 4. A train 150 m long is running with a speed of 68 kmph. In what time will it pass aman who is running at 8 kmph in the same direction in which the train is going?</p><p>(WIPRO 2008)Sol: Speed of the train relative to man = (68 - 8) kmph</p><p>= (60* 5/18) m/sec = (50/3)m/sec</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 5</p><p>Time taken by the train to cross the man I</p><p>= Time taken by It to cover 150 m at 50/3 m / sec = 150 *3/ 50 sec = 9sec</p><p>Ex. 5. A train 220 m long is running with a speed of 59 kmph. In what will</p><p>it pass a man who is running at 7 kmph in the direction opposite to that in which the trainis going?</p><p>sol. Speed of the train relative to man = (59 + 7) kmph= 66 *5/18 m/sec = 55/3 m/sec.</p><p>Time taken by the train to cross the man = Time taken by it to cover 220 m at (55/3) m / sec= (220 *3/55) sec = 12 secEx. 6. Two trains 137 metres and 163 metres in length are running towards each other onparallel lines, one at the rate of 42 kmph and another at 48 kmpb. In what time will theybe clear of each other from the moment they meet?</p><p>Sol. Relative speed of the trains = (42 + 48) kmph = 90 kmph</p><p>=(90*5/18) m / sec = 25 m /sec.</p><p>Time taken by the trains to'pass each other</p><p>= Time taken to cover (137 + 163) m at 25 m /sec =(300/25) sec = 12 secEx. 7. Two trains 100 metres and 120 metres long are running in the same direction withspeeds of 72 km/hr,In howmuch time will the first train cross the second? (TCS 2007)</p><p>Sol: Relative speed of the trains = (72 - 54) km/hr = 18 km/hr</p><p>= (18 * 5/18) m/sec = 5 m/sec.</p><p>Time taken by the trains to cross each other</p><p>= Time taken to cover (100 + 120) m at 5 m /sec = (220/5) sec = 44 sec.Ex. 8. A train 100 metres long takes 6 seconds to cross a man walking at 5 kmph in thedirection opposite to that of the train. Find the speed of the train.?</p><p>Sol:Let the speed of the train be x kmph.</p><p>Speed of the train relative to man = (x + 5) kmph = (x + 5) *5/18 m/sec.</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 6</p><p>Therefore 100/((x+5)*5/18)=6 30 (x + 5) = 1800 x = 55Speed of the train is 55 kmph.</p><p>Ex. 9. A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes.12 secto pass a man walking at 6 kmph in the same direction in which the train is going . Find thelength of the train and the length of the platform. (CAT 2006)Sol:Let the length of train be x metres and length of platform be y metres.</p><p>Speed of the train relative to man = (54 - 6) kmph = 48 kmph= 48*(5/18) m/sec = 40/3 m/sec.In passing a man, the train covers its own length with relative speed.</p><p>Length of train = (Relative speed * Time) = ( 40/3)*12 m = 160 m.</p><p>Also, speed of the train = 54 *(5/18)m / sec = 15 m / sec.(x+y)/15 = 20 x + y = 300 Y = (300 - 160) m = 140 m.</p><p>Ex10. A man sitting in a train which is traveling at 50 kmph observes that a goods train,traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long,find its speed.?(ACCENTURE 2008)</p><p>Sol: Relative speed = 280/9 m / sec = ((280/9)*(18/5)) kmph = 112 kmph.</p><p>Speed of goods train = (112 - 50) kmph = 62 kmph.Ex. 11. Find the time taken by a train 180 m long, running at 72 kmph,in crossing anelectric pole.</p><p>Sol. Speed of the train = (72 x 5/18) m/sec = 20 m/sec.Distance moved in passing the pole = 180 m.Required time taken = (180/20) sec = 9 sec.</p><p>Ex. 12. A train 140 m long is running at 60 kmph. In how much time wiU it pass a platform260 m long?</p><p>Sol. Speed of the train = (60 x 5/18) m/sec =50/3m/sec.Distance covered in passing the platform = (140 + 260) m = 400 m:. Time taken = (400 x 3/50) see = 24 sec.</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 7</p><p>Ex. 13. A man is standing on a railway bridge which is 180 m long. He finds that a traincrosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train andits speed.(INFOSYS 2008)</p><p>Sol. Let the length of the train be x metres.</p><p>Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 seconds.:. X/8 = (x + 180)/20 20x=8(x +180) x=120:. Length of train = 120m.Speed of train = (120/8) m/sec = 15 m/sec = [15 X 18/5 ] Kmph</p><p>= 54 Kmph.</p><p>Ex. 14. A train 150 m long is running with a speed of 68 kmph. In what time will it pass aman who is running at 8 kmph in the same direction in which the train is going ?(BANK PO 2006)</p><p>Sol. Speed of the train relative to man = (68 - 8) kmph= (60 x 5/18)m/see = 50/3 m/sec</p><p>Time taken by the train to cross the man = Time taken by it to cover 150 m at ( 50/3)m/sec= (150 x 3/50 )sec = 9 sec.</p><p>Ex. 15. A train 220 m long is running with a speed of 59 kmph. In what time will it pass aman who is running at 7 kmph in the direction opposite to that in which the train is going?</p><p>(S.S.C 2004)Sol. Speed of the train relative to man = (59 + 7) kmph</p><p>= (66 x 5/18 ) m/sec = ( 55/3) m/sec.Time taken by the train to cross the man</p><p>:. (x + y)/15 = 20 or x + y = 300 or y =(300 160) m = 140m.:. Length of the platform = 140 m.</p><p>Ex. 16. A man sitting in a train which is traveling at 50 kmph observes that a goods train,traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 150 m long,find its speed.(HEXAWARE 2007)</p><p>Sol. Relative speed = (150/9) m/sec =(150/ 9 x 18/5) kmph = 60 kmph.:. Speed of goods train = (60 - 50) kmph = 10 kmph.</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 8</p><p>EXERCISE PROBLEMS:</p><p>1. A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the lengthof the train? (CSC 2008)A. 120 metres B. 180 metres</p><p>C. 324 metres D. 150 metresAnswer &amp; ExplanationAnswer: Option DExplanation:</p><p>Speed= 60 x5</p><p>m/sec=</p><p>50m/sec.18 3</p><p>Length of the train = (Speed x Time) = 50 x 9m = 150 m.3</p><p>2. A train 125 m long passes a man, running at 5 km/hr in the same direction in which thetrain is going, in 10 seconds. The speed of the train is:A. 45 km/hr B. 50 km/hr</p><p>C. 54 km/hr D. 55 km/hrAnswer &amp; ExplanationAnswer: Option BExplanation:</p><p>Speed of the train relative to man =125</p><p>m/sec10</p><p>=</p><p>25m/sec.2</p><p>=</p><p>25x</p><p>18km/hr2 5</p><p>= 45 km/hr.Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.</p><p>x - 5 = 45 x = 50 km/hr.</p><p>3. The length of the bridge, which a train 130 metres long and travelling at 45 km/hr cancross in 30 seconds, is:(CTS 2009)A. 200 m B. 225 m</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 9</p><p>C. 245 m D. 250 mAnswer &amp; ExplanationAnswer: Option CExplanation:</p><p>Speed = 45 x5</p><p>m/sec=</p><p>25m/sec.18 2</p><p>Time = 30 sec.Let the length of bridge be x metres.</p><p>Then,130 + x</p><p>=</p><p>2530 2</p><p>2(130 + x) = 750x = 245 m.</p><p>4. Two trains running in opposite directions cross a man standing on the platform in 27seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratioof their speeds is.(IBM 2008)A. 1 : 3 B. 3 : 2</p><p>C. 3 : 4 D. None of theseAnswer &amp; ExplanationAnswer: Option BExplanation:Let the speeds of the two trains be x m/sec and y m/sec respectively.Then, length of the first train = 27x metres,and length of the second train = 17y metres.</p><p>27x + 17y= 23</p><p>x+ y27x + 17y = 23x + 23y4x = 6yx</p><p>=</p><p>3.</p><p>y 2</p><p>5. A train passes a station platform in 36 seconds and a man standing on the platform in20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?A. 120 m B. 240 m</p><p>C. 300 m D. None of theseAnswer &amp; ExplanationAnswer: Option BExplanation:</p><p>Speed = 54 x5</p><p>m/sec = 15 m/sec.18Length of the train = (15 x 20)m = 300 m.</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 10</p><p>Let the length of the platform be x metres.</p><p>Then,x + 300</p><p>= 1536</p><p>x + 300 = 540x = 240 m.</p><p>6. A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform650 m long?(TCS 2006)A. 65 sec B. 89 sec</p><p>C. 100 sec D. 150 secAnswer &amp; ExplanationAnswer: Option BExplanation:</p><p>Speed =240</p><p>m/sec = 10 m/sec.24</p><p>Required time =240 + 650 sec = 89 sec.</p><p>7. Two trains of equal length are running on parallel lines in the same direction at 46km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The lengthof each train is:A. 50 m B. 72 m</p><p>C. 80 m D. 82 mAnswer &amp; ExplanationAnswer: Option AExplanation:Let the length of each train be x metres.Then, distance covered = 2x metres.Relative speed = (46 - 36) km/hr</p><p>= 10 x5</p><p>m/sec18</p><p>=</p><p>25m/sec9</p><p>2x=</p><p>2536 92x = 100x = 50.</p><p>8. A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge140 m long?(IGATE 2009)</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 11</p><p>A. 40 sec B. 42 sec</p><p>C. 45 sec D. 48 secAnswer &amp; ExplanationAnswer: Option AExplanation:</p><p>Formula for converting from km/hr to m/s: X km/hr = X x5</p><p>m/s.18</p><p>Therefore, Speed = 45 x5</p><p>m/sec=</p><p>25m/sec.</p><p>18 2Total distance to be covered = (360 + 140) m = 500 m.Formula for finding Time =</p><p>DistanceSpeed</p><p>Required time =500 x 2</p><p>sec= 40 sec.</p><p>25</p><p>9. Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengthsare 1.10 km and 0.9 km respectively. The time taken by the slower train to cross thefaster train in seconds is:(PATNI 2008)A. 36 B. 45</p><p>C. 48 D. 49Answer &amp; ExplanationAnswer: Option CExplanation:Relative speed = (60+ 90) km/hr</p><p>= 150 x5</p><p>m/sec18</p><p>=</p><p>125m/sec.3</p><p>Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.Required time = 2000 x</p><p>3sec = 48 sec.125</p><p>10. A jogger running at 9 kmph alongside a railway track in 240 metres ahead of theengine of a 120 metres long train running at 45 kmph in the same direction. In howmuch time will the train pass the jogger?A. 3.6 sec B. 18 sec</p><p>C. 36 sec D. 72 secAnswer &amp; ExplanationAnswer: Option CExplanation:</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 12</p><p>Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.= 36 x</p><p>5m/sec18</p><p>= 10 m/sec.Distance to be covered = (240 + 120) m = 360 m.</p><p>Time taken =360</p><p>sec= 36 sec.</p><p>10</p><p>11. A 270 metres long train running at the speed of 120 kmph crosses another trainrunning in opposite direction at the speed of 80 kmph in 9 seconds. What is the lengthof the other train?(HCL 2009)A. 230 m B. 240 m</p><p>C. 260 m D. 320 m</p><p>E. None of theseAnswer &amp; ExplanationAnswer: Option AExplanation:Relative speed = (120 + 80) km/hr</p><p>= 200 x5</p><p>m/sec18</p><p>=</p><p>500m/sec.9</p><p>Let the length of the other train be x metres.</p><p>Then,x + 270</p><p>=</p><p>5009 9</p><p>x + 270 = 500x = 230.</p><p>12. A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26seconds. What is the length of the goods train?(CTS 2005)A. 230 m B. 240 m</p><p>C. 260 m D. 270 mAnswer &amp; ExplanationAnswer: Option DExplanation:</p><p>Speed = 72 x5</p><p>m/sec= 20 m/sec.</p><p>18Time = 26 sec.Let the length of the train be x metres.Then, x + 250 = 20</p></li><li><p>REASONING AND QUANTITATIVE APTITUDE PROBLEMS ON TRAINS, BOATS AND STREAMS</p><p>SITAMS 13</p><p>26x + 250 = 520x = 270.</p><p>13. Two trains, each 100 m long, moving in opposite directions, cross each other in 8seconds. If one is moving twice as fast the other, then the speed of the faster train is:</p><p>(CAT 2010)A. 30 km/hr B. 45 km/hr</p><p>C. 60 km/hr D. 75 km/hrAnswer &amp; ExplanationAnswer: Option CExplanation:Let the speed of the slower train be x m/sec.Then, speed of the faster train = 2x m/sec.Relative speed = (x + 2x) m/sec = 3...</p></li></ul>