{"id":90258,"date":"2025-06-01T10:24:03","date_gmt":"2025-06-01T10:24:03","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90258"},"modified":"2025-06-01T10:24:03","modified_gmt":"2025-06-01T10:24:03","slug":"consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/","title":{"rendered":"Consider an equilateral triangle ABC as given in the following diagram"},"content":{"rendered":"

Consider an equilateral triangle ABC as given in the following diagram :
\nTwo people start at the same time from points A and B with speeds 30 km per hour and 20 km per hour respectively, and move on the sides of the triangle in the clockwise direction. They meet each other for the first time at<\/p>\n

[amp_mcq option1=”point C” option2=”a point between C and A” option3=”a point between A and B” option4=”point A” correct=”option4″]<\/p>\n

\n
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This question was previously asked in<\/div>\n
UPSC CAPF – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
Let the side length of the equilateral triangle be L. The total perimeter is 3L. Person 1 starts at A (speed 30 km\/h) and Person 2 starts at B (speed 20 km\/h), both moving clockwise. Person 1 moves along A->B->C->A… and Person 2 moves along B->C->A->B… The initial distance between them along the clockwise path from A’s perspective to B’s position is L. Their relative speed when moving towards each other along the perimeter is the sum of their speeds: 30 + 20 = 50 km\/h. The time taken for them to cover the initial distance L and meet for the first time is Time = Distance \/ Relative Speed = L \/ 50 hours. In this time, Person 1 travels a distance of 30 * (L\/50) = 3L\/5 km from A. Person 2 travels a distance of 20 * (L\/50) = 2L\/5 km from B. Let’s track their positions: Person 1 starts at A (position 0), moves 3L\/5 along A->B. This point is along AB. Person 2 starts at B (position L from A along A->B->C), moves 2L\/5 along B->C. This point is along BC. This initial distance calculation (L) only considers the segment AB. A better approach for motion on a closed loop is relative speed on the entire loop. The relative speed is 50 km\/h. They will meet when the difference in the total distance covered by them is a multiple of the perimeter (3L), or when their positions modulo 3L are the same. Let P1’s position be d1 and P2’s position be d2, measured from A clockwise. d1 = 30t mod 3L. P2 starts at B (distance L from A), so d2 = (L + 20t) mod 3L. They meet when 30t = L + 20t (mod 3L), which means 10t = L (mod 3L). The smallest positive t occurs when 10t = L, so t = L\/10. At this time, P1 has traveled 30 * (L\/10) = 3L km. Starting from A, traveling 3L km clockwise along the perimeter A->B->C->A means P1 is back at A. At this time, P2 has traveled 20 * (L\/10) = 2L km. Starting from B, traveling 2L km clockwise along B->C->A->B means P2 travels B->C (L) then C->A (L), ending up at A. Thus, they meet for the first time at point A.<\/section>\n
For objects moving on a closed track, they meet when the difference in the distances they have traveled is a multiple of the track length, or by considering their positions modulo the track length. Relative speed can be the sum or difference of speeds depending on direction.<\/section>\n
In this case, although they start at different points, their movement along the perimeter means they will meet when their relative displacement along the perimeter is a multiple of the perimeter length. The equation 10t = L (mod 3L) correctly captures this for their first meeting.<\/section>\n","protected":false},"excerpt":{"rendered":"

Consider an equilateral triangle ABC as given in the following diagram : Two people start at the same time from points A and B with speeds 30 km per hour and 20 km per hour respectively, and move on the sides of the triangle in the clockwise direction. They meet each other for the first … <\/p>\n

Detailed SolutionConsider an equilateral triangle ABC as given in the following diagram<\/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":[1085],"tags":[1114,1102],"class_list":["post-90258","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1114","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nConsider an equilateral triangle ABC as given in the following diagram<\/title>\n<meta name=\"description\" content=\"Let the side length of the equilateral triangle be L. The total perimeter is 3L. Person 1 starts at A (speed 30 km\/h) and Person 2 starts at B (speed 20 km\/h), both moving clockwise. Person 1 moves along A->B->C->A... and Person 2 moves along B->C->A->B... The initial distance between them along the clockwise path from A's perspective to B's position is L. Their relative speed when moving towards each other along the perimeter is the sum of their speeds: 30 + 20 = 50 km\/h. The time taken for them to cover the initial distance L and meet for the first time is Time = Distance \/ Relative Speed = L \/ 50 hours. In this time, Person 1 travels a distance of 30 * (L\/50) = 3L\/5 km from A. Person 2 travels a distance of 20 * (L\/50) = 2L\/5 km from B. Let's track their positions: Person 1 starts at A (position 0), moves 3L\/5 along A->B. This point is along AB. Person 2 starts at B (position L from A along A->B->C), moves 2L\/5 along B->C. This point is along BC. This initial distance calculation (L) only considers the segment AB. A better approach for motion on a closed loop is relative speed on the entire loop. The relative speed is 50 km\/h. They will meet when the difference in the total distance covered by them is a multiple of the perimeter (3L), or when their positions modulo 3L are the same. Let P1's position be d1 and P2's position be d2, measured from A clockwise. d1 = 30t mod 3L. P2 starts at B (distance L from A), so d2 = (L + 20t) mod 3L. They meet when 30t = L + 20t (mod 3L), which means 10t = L (mod 3L). The smallest positive t occurs when 10t = L, so t = L\/10. At this time, P1 has traveled 30 * (L\/10) = 3L km. Starting from A, traveling 3L km clockwise along the perimeter A->B->C->A means P1 is back at A. At this time, P2 has traveled 20 * (L\/10) = 2L km. Starting from B, traveling 2L km clockwise along B->C->A->B means P2 travels B->C (L) then C->A (L), ending up at A. Thus, they meet for the first time at point A. For objects moving on a closed track, they meet when the difference in the distances they have traveled is a multiple of the track length, or by considering their positions modulo the track length. Relative speed can be the sum or difference of speeds depending on direction.\" \/>\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\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Consider an equilateral triangle ABC as given in the following diagram\" \/>\n<meta property=\"og:description\" content=\"Let the side length of the equilateral triangle be L. The total perimeter is 3L. Person 1 starts at A (speed 30 km\/h) and Person 2 starts at B (speed 20 km\/h), both moving clockwise. Person 1 moves along A->B->C->A... and Person 2 moves along B->C->A->B... The initial distance between them along the clockwise path from A's perspective to B's position is L. Their relative speed when moving towards each other along the perimeter is the sum of their speeds: 30 + 20 = 50 km\/h. The time taken for them to cover the initial distance L and meet for the first time is Time = Distance \/ Relative Speed = L \/ 50 hours. In this time, Person 1 travels a distance of 30 * (L\/50) = 3L\/5 km from A. Person 2 travels a distance of 20 * (L\/50) = 2L\/5 km from B. Let's track their positions: Person 1 starts at A (position 0), moves 3L\/5 along A->B. This point is along AB. Person 2 starts at B (position L from A along A->B->C), moves 2L\/5 along B->C. This point is along BC. This initial distance calculation (L) only considers the segment AB. A better approach for motion on a closed loop is relative speed on the entire loop. The relative speed is 50 km\/h. They will meet when the difference in the total distance covered by them is a multiple of the perimeter (3L), or when their positions modulo 3L are the same. Let P1's position be d1 and P2's position be d2, measured from A clockwise. d1 = 30t mod 3L. P2 starts at B (distance L from A), so d2 = (L + 20t) mod 3L. They meet when 30t = L + 20t (mod 3L), which means 10t = L (mod 3L). The smallest positive t occurs when 10t = L, so t = L\/10. At this time, P1 has traveled 30 * (L\/10) = 3L km. Starting from A, traveling 3L km clockwise along the perimeter A->B->C->A means P1 is back at A. At this time, P2 has traveled 20 * (L\/10) = 2L km. Starting from B, traveling 2L km clockwise along B->C->A->B means P2 travels B->C (L) then C->A (L), ending up at A. Thus, they meet for the first time at point A. For objects moving on a closed track, they meet when the difference in the distances they have traveled is a multiple of the track length, or by considering their positions modulo the track length. Relative speed can be the sum or difference of speeds depending on direction.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:24:03+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=\"3 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Consider an equilateral triangle ABC as given in the following diagram","description":"Let the side length of the equilateral triangle be L. The total perimeter is 3L. Person 1 starts at A (speed 30 km\/h) and Person 2 starts at B (speed 20 km\/h), both moving clockwise. Person 1 moves along A->B->C->A... and Person 2 moves along B->C->A->B... The initial distance between them along the clockwise path from A's perspective to B's position is L. Their relative speed when moving towards each other along the perimeter is the sum of their speeds: 30 + 20 = 50 km\/h. The time taken for them to cover the initial distance L and meet for the first time is Time = Distance \/ Relative Speed = L \/ 50 hours. In this time, Person 1 travels a distance of 30 * (L\/50) = 3L\/5 km from A. Person 2 travels a distance of 20 * (L\/50) = 2L\/5 km from B. Let's track their positions: Person 1 starts at A (position 0), moves 3L\/5 along A->B. This point is along AB. Person 2 starts at B (position L from A along A->B->C), moves 2L\/5 along B->C. This point is along BC. This initial distance calculation (L) only considers the segment AB. A better approach for motion on a closed loop is relative speed on the entire loop. The relative speed is 50 km\/h. They will meet when the difference in the total distance covered by them is a multiple of the perimeter (3L), or when their positions modulo 3L are the same. Let P1's position be d1 and P2's position be d2, measured from A clockwise. d1 = 30t mod 3L. P2 starts at B (distance L from A), so d2 = (L + 20t) mod 3L. They meet when 30t = L + 20t (mod 3L), which means 10t = L (mod 3L). The smallest positive t occurs when 10t = L, so t = L\/10. At this time, P1 has traveled 30 * (L\/10) = 3L km. Starting from A, traveling 3L km clockwise along the perimeter A->B->C->A means P1 is back at A. At this time, P2 has traveled 20 * (L\/10) = 2L km. Starting from B, traveling 2L km clockwise along B->C->A->B means P2 travels B->C (L) then C->A (L), ending up at A. Thus, they meet for the first time at point A. For objects moving on a closed track, they meet when the difference in the distances they have traveled is a multiple of the track length, or by considering their positions modulo the track length. Relative speed can be the sum or difference of speeds depending on direction.","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\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/","og_locale":"en_US","og_type":"article","og_title":"Consider an equilateral triangle ABC as given in the following diagram","og_description":"Let the side length of the equilateral triangle be L. The total perimeter is 3L. Person 1 starts at A (speed 30 km\/h) and Person 2 starts at B (speed 20 km\/h), both moving clockwise. Person 1 moves along A->B->C->A... and Person 2 moves along B->C->A->B... The initial distance between them along the clockwise path from A's perspective to B's position is L. Their relative speed when moving towards each other along the perimeter is the sum of their speeds: 30 + 20 = 50 km\/h. The time taken for them to cover the initial distance L and meet for the first time is Time = Distance \/ Relative Speed = L \/ 50 hours. In this time, Person 1 travels a distance of 30 * (L\/50) = 3L\/5 km from A. Person 2 travels a distance of 20 * (L\/50) = 2L\/5 km from B. Let's track their positions: Person 1 starts at A (position 0), moves 3L\/5 along A->B. This point is along AB. Person 2 starts at B (position L from A along A->B->C), moves 2L\/5 along B->C. This point is along BC. This initial distance calculation (L) only considers the segment AB. A better approach for motion on a closed loop is relative speed on the entire loop. The relative speed is 50 km\/h. They will meet when the difference in the total distance covered by them is a multiple of the perimeter (3L), or when their positions modulo 3L are the same. Let P1's position be d1 and P2's position be d2, measured from A clockwise. d1 = 30t mod 3L. P2 starts at B (distance L from A), so d2 = (L + 20t) mod 3L. They meet when 30t = L + 20t (mod 3L), which means 10t = L (mod 3L). The smallest positive t occurs when 10t = L, so t = L\/10. At this time, P1 has traveled 30 * (L\/10) = 3L km. Starting from A, traveling 3L km clockwise along the perimeter A->B->C->A means P1 is back at A. At this time, P2 has traveled 20 * (L\/10) = 2L km. Starting from B, traveling 2L km clockwise along B->C->A->B means P2 travels B->C (L) then C->A (L), ending up at A. Thus, they meet for the first time at point A. For objects moving on a closed track, they meet when the difference in the distances they have traveled is a multiple of the track length, or by considering their positions modulo the track length. Relative speed can be the sum or difference of speeds depending on direction.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:24:03+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"3 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/","url":"https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/","name":"Consider an equilateral triangle ABC as given in the following diagram","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:24:03+00:00","dateModified":"2025-06-01T10:24:03+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Let the side length of the equilateral triangle be L. The total perimeter is 3L. Person 1 starts at A (speed 30 km\/h) and Person 2 starts at B (speed 20 km\/h), both moving clockwise. Person 1 moves along A->B->C->A... and Person 2 moves along B->C->A->B... The initial distance between them along the clockwise path from A's perspective to B's position is L. Their relative speed when moving towards each other along the perimeter is the sum of their speeds: 30 + 20 = 50 km\/h. The time taken for them to cover the initial distance L and meet for the first time is Time = Distance \/ Relative Speed = L \/ 50 hours. In this time, Person 1 travels a distance of 30 * (L\/50) = 3L\/5 km from A. Person 2 travels a distance of 20 * (L\/50) = 2L\/5 km from B. Let's track their positions: Person 1 starts at A (position 0), moves 3L\/5 along A->B. This point is along AB. Person 2 starts at B (position L from A along A->B->C), moves 2L\/5 along B->C. This point is along BC. This initial distance calculation (L) only considers the segment AB. A better approach for motion on a closed loop is relative speed on the entire loop. The relative speed is 50 km\/h. They will meet when the difference in the total distance covered by them is a multiple of the perimeter (3L), or when their positions modulo 3L are the same. Let P1's position be d1 and P2's position be d2, measured from A clockwise. d1 = 30t mod 3L. P2 starts at B (distance L from A), so d2 = (L + 20t) mod 3L. They meet when 30t = L + 20t (mod 3L), which means 10t = L (mod 3L). The smallest positive t occurs when 10t = L, so t = L\/10. At this time, P1 has traveled 30 * (L\/10) = 3L km. Starting from A, traveling 3L km clockwise along the perimeter A->B->C->A means P1 is back at A. At this time, P2 has traveled 20 * (L\/10) = 2L km. Starting from B, traveling 2L km clockwise along B->C->A->B means P2 travels B->C (L) then C->A (L), ending up at A. Thus, they meet for the first time at point A. For objects moving on a closed track, they meet when the difference in the distances they have traveled is a multiple of the track length, or by considering their positions modulo the track length. Relative speed can be the sum or difference of speeds depending on direction.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/consider-an-equilateral-triangle-abc-as-given-in-the-following-diagram\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"Consider an equilateral triangle ABC as given in the following diagram"}]},{"@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\/90258","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=90258"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90258\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90258"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90258"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90258"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}