{"id":5993,"date":"2025-06-09T06:55:37","date_gmt":"2025-06-09T06:55:37","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5993"},"modified":"2025-06-09T06:55:37","modified_gmt":"2025-06-09T06:55:37","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-problems-trains-on","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-problems-trains-on\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: PROBLEMS ON TRAINS"},"content":{"rendered":"

PROBLEMS ON TRAINS\u00a0<\/strong><\/h1>\n

IMPORTANT FACTS AND FORMULAE<\/u><\/strong><\/h2>\n

\u00a0<\/u>a km\/hr= (a* <\/em>5\/18) m\/s.<\/u><\/p>\n

    \n
  1. a m \/ s <\/em>= (a*18\/5) km\/hr.<\/u><\/li>\n<\/ol>\n

    \u00a0<\/u>3 Time taken by a train of length 1 <\/em>metres to pass a pole or a standing man or a signal \u00a0post is equal to the time taken by the train to cover 1 <\/em>metres.<\/p>\n

      \n
    1. Time taken by a train of length 1 <\/em>metres to pass a stationary object of length b <\/em>metres is the time taken by the train to cover (1 <\/em>+ b) <\/em>metres.<\/li>\n
    2. Suppose two trains or two bodies are moving in the same direction at u m \/ s and v m\/s, <\/em>where u > v, <\/em>then their relatives speed = (u – v) <\/em>m \/ s.<\/li>\n
    3. Suppose two trains or two bodies are moving in opposite directions at u m \/ s and v m\/s, <\/em>then their relative speed is = (u + v) <\/em>m\/s.<\/li>\n
    4. If two trains of length a <\/em>metres and b <\/em>metres are moving in opposite directions at u m \/ <\/em>s and v <\/em>m\/s, then time taken by the trains to cross each other = (a <\/u><\/em>+ b)\/(u+v) <\/em><\/u>sec.<\/li>\n<\/ol>\n

      8.If two trains of length\u00a0 a metres and b <\/em>metres are moving in the same direction<\/p>\n

      at u m \/ s and v <\/em>m \/ s, then the time taken by the faster\u00a0 train to cross the slower train = (a+b)\/(u-v) sec.<\/p>\n

        \n
      1. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a <\/em>and b <\/em>sec in reaching B and A respectively, then<\/li>\n<\/ol>\n

        (A’s speet) <\/em>: (B\u2019s speed) <\/em>= (b1\/2<\/sup>: <\/em>a1\/2<\/sup>).<\/em><\/p>\n

        \u00a0<\/em>\u00a0SOLVED EXAMPLES <\/u><\/strong><\/p>\n

        \u00a0<\/u><\/strong>Ex.I. A train 100 m long is running at the speed of 30 km \/ hr. Find the time taken \u00a0by<\/strong>\u00a0 it to pass <\/strong>a man standing near the railway line.<\/strong> (S.S.C. 2001)<\/p>\n

        Sol.<\/strong> Speed of the train = (30 x 5\/18_) <\/u>m \/ sec = (25\/3) <\/u>m\/ sec.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Distance moved in passing the standing man = 100 m.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Required time taken = 100\/(25\/3) <\/u><\/em>= (100 *(3\/25)) sec = 1<\/u>2 sec<\/p>\n

        Ex. 2. A train is moving at a speed of 132 km\/br. If the length of the train is<\/strong><\/p>\n

        110 <\/strong>metres, <\/strong>how long will it take to cross a railway platform 165 metres long?<\/strong><\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Section Officers’, 2003)<\/strong><\/p>\n

        Sol.<\/strong> Speed of train = 132 *(5\/18) <\/u>m\/sec = 110\/3 \u00a0m\/sec.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0Distance covered in passing the platform = (110 + 165) m = 275 m.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Time taken =275 *(3\/110) \u00a0sec =15\/2 sec = 7 \u00a0\u00bd sec<\/p>\n

        Ex. 3. A man is standing on a railway bridge which is 180 m long. He finds that \u00a0a train crosses the bridge in 20 seconds but himself in 8 seconds. Find <\/strong>the length of <\/strong>the train and its speed?<\/strong><\/p>\n

        \u00a0<\/strong>Sol.<\/strong> Let the length of the train be x metres,<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 sec<\/p>\n

        \u00a0<\/em>\u00a0\u00a0\u00a0\u00a0 x\/8=(x+180)\/20 <\/em>\u00f3 20x = 8 (x <\/em>+ 180)\u00a0\u00a0\u00a0 <=>\u00a0\u00a0 x <\/em>= 120.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 Length of the train = 120 m.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 Speed of the train = (120\/8) <\/u>m \/ <\/em>sec = m \/ <\/em>sec = (15 *18\/5) <\/u>kmph = 54 km<\/p>\n

        Ex. 4. A train 150 m long is running with a speed of 68 kmph. In <\/strong>what <\/strong>time will it pass a man who is running at 8 kmph in the same direction in which <\/strong>the train is going?<\/strong><\/p>\n

        Sol:<\/strong> Speed of the train relative to man = (68 – 8) kmph<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = (60* 5\/18) m\/sec <\/u>= (50\/3)m\/sec<\/u><\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Time taken by the train to cross the man\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Time taken by It to cover 150 m at 50\/3 \u00a0m \/ <\/em>sec = 150 *3\/ 50 <\/u>\u00a0sec \u00a0= 9sec<\/p>\n

        Ex. 5. A train 220 m long is running with a speed of 59 kmph.. <\/strong>In what <\/strong>will<\/strong><\/p>\n

        it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going?<\/strong><\/p>\n

        sol.<\/strong> Speed of the train relative to man = (59 + 7) kmph<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0= 66 *5\/18 \u00a0<\/u>m\/sec <\/em>= 55\/3 m\/sec.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Time taken by the train to cross the man = Time taken by it to cover 220 m \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0at (55\/3) <\/u>m \/ <\/em>sec = (220 *3\/55) <\/u>sec = 12 sec<\/p>\n

        Ex. 6. Two trains 137 metres and 163 metres in length are running <\/strong>towards each <\/strong>other on parallel lines, one at the rate of 42 kmph and another at 48 kmpb. In what \u00a0time will they be clear of each other from the moment they meet?<\/strong><\/p>\n

        \u00a0<\/strong>Sol.<\/strong> Relative speed of the trains = (42 + 48) kmph = 90 kmph<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =(90*5\/18) m \/ <\/em>sec = 25 m \/<\/em>sec.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Time taken by the \u00a0trains\u00a0 to’pass each other<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Time taken to cover <\/em>(137 + 163) m at 25 m \/sec <\/em>=(300\/25) s<\/u>ec = 12 sec<\/p>\n

        \u00a0<\/u>Ex. 7. Two trains 100\u00a0 metres and 120 metres long\u00a0 are running in the <\/strong>same direction <\/strong>with speeds of \u00a072 km\/hr,In howmuch \u00a0\u00a0time will the first train cross the\u00a0 second?<\/strong><\/p>\n

        Sol:<\/strong> Relative speed of the trains = (72 – 54) km\/hr = 18 km\/hr<\/em><\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = (18 * 5\/18) <\/u>m\/sec <\/em>= 5 m\/sec.<\/em><\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Time taken by the trains to cross each other<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0 = Time taken to cover (100 + 120) m at 5 m \/<\/em>sec = (220\/5) <\/u>sec = 44 sec.<\/p>\n

        Ex<\/strong>. 8. A train \u00a0100 metres long takes 6 seconds to cross a man walking at 5 kmph in the direction <\/strong>\u00a0<\/strong>opposite to that of the train. Find the speed of the train.?<\/strong><\/p>\n

        \u00a0<\/strong>Sol<\/strong>:Let the speed of the train be x kmph.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 \u00a0 Speed of the train relative to man = (x + 5) kmph = (x <\/em>+ 5) *5\/18 m\/sec.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0 Therefore 100\/((x+5)*5\/18)=6 <=> 30 (x <\/em>+ 5) = 1800 <=> x <\/em>= 55<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Speed of the train is 55 kmph.<\/p>\n

        Ex9. A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes.12 sec <\/strong>\u00a0<\/strong>to pass a man walking at 6 kmph in the same direction in which the train is going . Find the length of the train and the length of the platform.<\/strong><\/p>\n

        \u00a0<\/strong>Sol:<\/strong>Let the length of train be x metres and length of platform be y <\/em>metres.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 Speed of the train relative to man = (54 – 6) kmph = 48 kmph<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0 = 48*(5\/18) m\/sec = 40\/3 m\/sec.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 In passing a man, the train covers its own length with relative speed.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 Length of train = (Relative speed * Time) = ( 40\/3)*12<\/u> m = 160 m.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0 Also, speed of the train = 54 *(5\/18)<\/u>m \/ sec = 15 m \/ sec.<\/p>\n

        \u00a0<\/em>\u00a0\u00a0 (x+y)\/15 = 20 <=> x <\/em>+ y <\/em>= 300 <=> Y <\/em>= (300 – 160) m = 140 m.<\/p>\n

        Ex10. A man sitting in a train which is traveling at 50 kmph observes that a goods <\/strong>train, <\/strong>traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.?<\/strong><\/p>\n

        \u00a0<\/strong>Sol<\/strong>: Relative speed = 280\/9 m \/ sec = ((280\/9)*(18\/5)) kmph = 112 kmph.<\/p>\n

        \u00a0\u00a0\u00a0 \u00a0 Speed of goods train = (112 – 50) kmph = 62 kmph.<\/p>\n

        “Quantitative Aptitude” by R.S. Aggarwal is a widely recognized resource for mastering various quantitative topics, including “Problems on Trains.” While accessing the complete book in PDF format for free may infringe upon copyright laws, there are legitimate ways to study this specific chapter:<\/p>\n

          \n
        1. \n

          Online Practice Questions<\/strong>:<\/p>\n

            \n
          • Websites like provide practice questions specifically on “Problems on Trains” from R.S. Aggarwal’s book. This allows you to practice relevant problems directly online.<\/li>\n<\/ul>\n<\/li>\n
          • \n

            Solution Platforms<\/strong>:<\/p>\n

              \n
            • Platforms such as offer solutions and explanations for exercises related to “Problems on Trains.” This can help you understand the methodology behind solving these problems.<\/li>\n<\/ul>\n<\/li>\n
            • \n

              Purchase the Book<\/strong>:<\/p>\n

                \n
              • For comprehensive coverage and practice, consider purchasing the book through authorized sellers. This ensures you have access to all topics and practice exercises.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n

                Please be cautious of unauthorized websites offering free downloads of the complete book, as they may violate copyright laws and could pose security risks. Utilizing authorized resources not only supports the creators but also guarantees the accuracy and quality of the study materials you use in your preparation.<\/p>\n

                RS Aggarwal Quantitative Aptitude PDF Free Download: PROBLEMS ON TRAINS<\/h3>\n

                 <\/p>\n","protected":false},"excerpt":{"rendered":"

                PROBLEMS ON TRAINS\u00a0 IMPORTANT FACTS AND FORMULAE \u00a0a km\/hr= (a* 5\/18) m\/s. a m \/ s = (a*18\/5) km\/hr. \u00a03 Time taken by a train of length 1 metres to pass a pole or a standing man or a signal \u00a0post is equal to the time taken by the train to cover 1 metres. Time […]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[126,127],"tags":[],"class_list":["post-5993","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude","category-rs-aggarwal-quantitative-aptitude-pdf"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5993","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/users\/41"}],"replies":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/comments?post=5993"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5993\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=5993"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=5993"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=5993"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}