{"id":92515,"date":"2025-06-01T11:27:09","date_gmt":"2025-06-01T11:27:09","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=92515"},"modified":"2025-06-01T11:27:09","modified_gmt":"2025-06-01T11:27:09","slug":"while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/","title":{"rendered":"While coming down on an escalator of a station, when a man walks down"},"content":{"rendered":"<p>While coming down on an escalator of a station, when a man walks down twenty-six steps of the staircase, he requires thirty seconds to reach the bottom. However, if he steps down thirty-four steps, it takes eighteen seconds to reach the bottom. How many steps are there in the escalator ?<\/p>\n<p>[amp_mcq option1=&#8221;42&#8243; option2=&#8221;44&#8243; option3=&#8221;46&#8243; option4=&#8221;48&#8243; correct=&#8221;option3&#8243;]<\/p>\n<div class=\"psc-box-pyq-exam-year-detail\">\n<div class=\"pyq-exam\">\n<div class=\"psc-heading\">This question was previously asked in<\/div>\n<div class=\"psc-title line-ellipsis\">UPSC CISF-AC-EXE &#8211; 2018<\/div>\n<\/div>\n<div class=\"pyq-exam-psc-buttons\"><a href=\"\/pyq\/pyq-upsc-cisf-ac-exe-2018.pdf\" target=\"_blank\" class=\"psc-pdf-button\" rel=\"noopener\">Download PDF<\/a><a href=\"\/pyq-upsc-cisf-ac-exe-2018\" target=\"_blank\" class=\"psc-attempt-button\" rel=\"noopener\">Attempt Online<\/a><\/div>\n<\/div>\n<section id=\"pyq-correct-answer\">\nThere are 46 steps in the escalator.<br \/>\n<\/section>\n<section id=\"pyq-key-points\">\nLet E be the total number of steps on the escalator (when stationary).<br \/>\nLet V_m be the speed of the man relative to the escalator (steps\/second).<br \/>\nLet V_e be the speed of the escalator (steps\/second).<br \/>\nThe total distance covered by the man relative to the ground is E steps. This distance is the sum of the steps the man walks relative to the escalator and the steps the escalator moves relative to the ground during the time taken.<br \/>\nE = (Man&#8217;s speed relative to ground) * Time<br \/>\nMan&#8217;s speed relative to ground = V_m + V_e.<br \/>\nE = (V_m + V_e) * T.<\/p>\n<p>Alternatively, consider the total steps E. In time T, the man takes N steps relative to the escalator (so V_m = N\/T). The escalator moves V_e * T steps.<br \/>\nE = N + V_e * T.<\/p>\n<p>Case 1: Man takes N1 = 26 steps, T1 = 30 seconds.<br \/>\nE = 26 + V_e * 30 (Equation 1)<\/p>\n<p>Case 2: Man takes N2 = 34 steps, T2 = 18 seconds.<br \/>\nE = 34 + V_e * 18 (Equation 2)<\/p>\n<p>We have a system of two linear equations with two variables (E and V_e).<br \/>\nEquating the expressions for E:<br \/>\n26 + 30 * V_e = 34 + 18 * V_e<br \/>\n30 * V_e &#8211; 18 * V_e = 34 &#8211; 26<br \/>\n12 * V_e = 8<br \/>\nV_e = 8 \/ 12 = 2\/3 steps\/second.<\/p>\n<p>Substitute the value of V_e into either equation. Using Equation 1:<br \/>\nE = 26 + (2\/3) * 30<br \/>\nE = 26 + 2 * 10<br \/>\nE = 26 + 20<br \/>\nE = 46.<\/p>\n<p>Using Equation 2:<br \/>\nE = 34 + (2\/3) * 18<br \/>\nE = 34 + 2 * 6<br \/>\nE = 34 + 12<br \/>\nE = 46.<br \/>\nBoth equations give the same result.<br \/>\n<\/section>\n<section id=\"pyq-additional-information\">\nThe man&#8217;s speed relative to the escalator is not constant in terms of steps per second, but the number of steps he *takes* is given. The interpretation used is that 26 steps are the steps counted on the treads by the man during his descent in 30 seconds, and 34 steps are counted in 18 seconds. The escalator adds to his progress relative to the ground.<br \/>\n<\/section>\n","protected":false},"excerpt":{"rendered":"<p>While coming down on an escalator of a station, when a man walks down twenty-six steps of the staircase, he requires thirty seconds to reach the bottom. However, if he steps down thirty-four steps, it takes eighteen seconds to reach the bottom. How many steps are there in the escalator ? [amp_mcq option1=&#8221;42&#8243; option2=&#8221;44&#8243; option3=&#8221;46&#8243; &#8230; <\/p>\n<p class=\"read-more-container\"><a title=\"While coming down on an escalator of a station, when a man walks down\" class=\"read-more button\" href=\"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/#more-92515\">Detailed Solution<span class=\"screen-reader-text\">While coming down on an escalator of a station, when a man walks down<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1089],"tags":[1114,1102],"class_list":["post-92515","post","type-post","status-publish","format-standard","hentry","category-upsc-cisf-ac-exe","tag-1114","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>While coming down on an escalator of a station, when a man walks down<\/title>\n<meta name=\"description\" content=\"There are 46 steps in the escalator. Let E be the total number of steps on the escalator (when stationary). Let V_m be the speed of the man relative to the escalator (steps\/second). Let V_e be the speed of the escalator (steps\/second). The total distance covered by the man relative to the ground is E steps. This distance is the sum of the steps the man walks relative to the escalator and the steps the escalator moves relative to the ground during the time taken. E = (Man&#039;s speed relative to ground) * Time Man&#039;s speed relative to ground = V_m + V_e. E = (V_m + V_e) * T. Alternatively, consider the total steps E. In time T, the man takes N steps relative to the escalator (so V_m = N\/T). The escalator moves V_e * T steps. E = N + V_e * T. Case 1: Man takes N1 = 26 steps, T1 = 30 seconds. E = 26 + V_e * 30 (Equation 1) Case 2: Man takes N2 = 34 steps, T2 = 18 seconds. E = 34 + V_e * 18 (Equation 2) We have a system of two linear equations with two variables (E and V_e). Equating the expressions for E: 26 + 30 * V_e = 34 + 18 * V_e 30 * V_e - 18 * V_e = 34 - 26 12 * V_e = 8 V_e = 8 \/ 12 = 2\/3 steps\/second. Substitute the value of V_e into either equation. Using Equation 1: E = 26 + (2\/3) * 30 E = 26 + 2 * 10 E = 26 + 20 E = 46. Using Equation 2: E = 34 + (2\/3) * 18 E = 34 + 2 * 6 E = 34 + 12 E = 46. Both equations give the same result.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"While coming down on an escalator of a station, when a man walks down\" \/>\n<meta property=\"og:description\" content=\"There are 46 steps in the escalator. Let E be the total number of steps on the escalator (when stationary). Let V_m be the speed of the man relative to the escalator (steps\/second). Let V_e be the speed of the escalator (steps\/second). The total distance covered by the man relative to the ground is E steps. This distance is the sum of the steps the man walks relative to the escalator and the steps the escalator moves relative to the ground during the time taken. E = (Man&#039;s speed relative to ground) * Time Man&#039;s speed relative to ground = V_m + V_e. E = (V_m + V_e) * T. Alternatively, consider the total steps E. In time T, the man takes N steps relative to the escalator (so V_m = N\/T). The escalator moves V_e * T steps. E = N + V_e * T. Case 1: Man takes N1 = 26 steps, T1 = 30 seconds. E = 26 + V_e * 30 (Equation 1) Case 2: Man takes N2 = 34 steps, T2 = 18 seconds. E = 34 + V_e * 18 (Equation 2) We have a system of two linear equations with two variables (E and V_e). Equating the expressions for E: 26 + 30 * V_e = 34 + 18 * V_e 30 * V_e - 18 * V_e = 34 - 26 12 * V_e = 8 V_e = 8 \/ 12 = 2\/3 steps\/second. Substitute the value of V_e into either equation. Using Equation 1: E = 26 + (2\/3) * 30 E = 26 + 2 * 10 E = 26 + 20 E = 46. Using Equation 2: E = 34 + (2\/3) * 18 E = 34 + 2 * 6 E = 34 + 12 E = 46. Both equations give the same result.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:27:09+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"While coming down on an escalator of a station, when a man walks down","description":"There are 46 steps in the escalator. Let E be the total number of steps on the escalator (when stationary). Let V_m be the speed of the man relative to the escalator (steps\/second). Let V_e be the speed of the escalator (steps\/second). The total distance covered by the man relative to the ground is E steps. This distance is the sum of the steps the man walks relative to the escalator and the steps the escalator moves relative to the ground during the time taken. E = (Man's speed relative to ground) * Time Man's speed relative to ground = V_m + V_e. E = (V_m + V_e) * T. Alternatively, consider the total steps E. In time T, the man takes N steps relative to the escalator (so V_m = N\/T). The escalator moves V_e * T steps. E = N + V_e * T. Case 1: Man takes N1 = 26 steps, T1 = 30 seconds. E = 26 + V_e * 30 (Equation 1) Case 2: Man takes N2 = 34 steps, T2 = 18 seconds. E = 34 + V_e * 18 (Equation 2) We have a system of two linear equations with two variables (E and V_e). Equating the expressions for E: 26 + 30 * V_e = 34 + 18 * V_e 30 * V_e - 18 * V_e = 34 - 26 12 * V_e = 8 V_e = 8 \/ 12 = 2\/3 steps\/second. Substitute the value of V_e into either equation. Using Equation 1: E = 26 + (2\/3) * 30 E = 26 + 2 * 10 E = 26 + 20 E = 46. Using Equation 2: E = 34 + (2\/3) * 18 E = 34 + 2 * 6 E = 34 + 12 E = 46. Both equations give the same result.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/","og_locale":"en_US","og_type":"article","og_title":"While coming down on an escalator of a station, when a man walks down","og_description":"There are 46 steps in the escalator. Let E be the total number of steps on the escalator (when stationary). Let V_m be the speed of the man relative to the escalator (steps\/second). Let V_e be the speed of the escalator (steps\/second). The total distance covered by the man relative to the ground is E steps. This distance is the sum of the steps the man walks relative to the escalator and the steps the escalator moves relative to the ground during the time taken. E = (Man's speed relative to ground) * Time Man's speed relative to ground = V_m + V_e. E = (V_m + V_e) * T. Alternatively, consider the total steps E. In time T, the man takes N steps relative to the escalator (so V_m = N\/T). The escalator moves V_e * T steps. E = N + V_e * T. Case 1: Man takes N1 = 26 steps, T1 = 30 seconds. E = 26 + V_e * 30 (Equation 1) Case 2: Man takes N2 = 34 steps, T2 = 18 seconds. E = 34 + V_e * 18 (Equation 2) We have a system of two linear equations with two variables (E and V_e). Equating the expressions for E: 26 + 30 * V_e = 34 + 18 * V_e 30 * V_e - 18 * V_e = 34 - 26 12 * V_e = 8 V_e = 8 \/ 12 = 2\/3 steps\/second. Substitute the value of V_e into either equation. Using Equation 1: E = 26 + (2\/3) * 30 E = 26 + 2 * 10 E = 26 + 20 E = 46. Using Equation 2: E = 34 + (2\/3) * 18 E = 34 + 2 * 6 E = 34 + 12 E = 46. Both equations give the same result.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:27:09+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/","url":"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/","name":"While coming down on an escalator of a station, when a man walks down","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:27:09+00:00","dateModified":"2025-06-01T11:27:09+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"There are 46 steps in the escalator. Let E be the total number of steps on the escalator (when stationary). Let V_m be the speed of the man relative to the escalator (steps\/second). Let V_e be the speed of the escalator (steps\/second). The total distance covered by the man relative to the ground is E steps. This distance is the sum of the steps the man walks relative to the escalator and the steps the escalator moves relative to the ground during the time taken. E = (Man's speed relative to ground) * Time Man's speed relative to ground = V_m + V_e. E = (V_m + V_e) * T. Alternatively, consider the total steps E. In time T, the man takes N steps relative to the escalator (so V_m = N\/T). The escalator moves V_e * T steps. E = N + V_e * T. Case 1: Man takes N1 = 26 steps, T1 = 30 seconds. E = 26 + V_e * 30 (Equation 1) Case 2: Man takes N2 = 34 steps, T2 = 18 seconds. E = 34 + V_e * 18 (Equation 2) We have a system of two linear equations with two variables (E and V_e). Equating the expressions for E: 26 + 30 * V_e = 34 + 18 * V_e 30 * V_e - 18 * V_e = 34 - 26 12 * V_e = 8 V_e = 8 \/ 12 = 2\/3 steps\/second. Substitute the value of V_e into either equation. Using Equation 1: E = 26 + (2\/3) * 30 E = 26 + 2 * 10 E = 26 + 20 E = 46. Using Equation 2: E = 34 + (2\/3) * 18 E = 34 + 2 * 6 E = 34 + 12 E = 46. Both equations give the same result.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/while-coming-down-on-an-escalator-of-a-station-when-a-man-walks-down\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CISF-AC-EXE","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cisf-ac-exe\/"},{"@type":"ListItem","position":3,"name":"While coming down on an escalator of a station, when a man walks down"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/92515","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=92515"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/92515\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=92515"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=92515"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=92515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}