{"id":90564,"date":"2025-06-01T10:31:06","date_gmt":"2025-06-01T10:31:06","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90564"},"modified":"2025-06-01T10:31:06","modified_gmt":"2025-06-01T10:31:06","slug":"in-a-group-of-100-children-64-children-like-to-play-cricket-53-child","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/","title":{"rendered":"In a group of 100 children, 64 children like to play cricket, 53 child"},"content":{"rendered":"

In a group of 100 children, 64 children like to play cricket, 53 children like to play football and 20 children like to play both cricket and football. How many children do NOT like to play cricket or football ?<\/p>\n

[amp_mcq option1=”3″ option2=”5″ option3=”7″ option4=”9″ correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2020<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThis problem can be solved using the principle of inclusion-exclusion or a Venn diagram.
\nLet C be the set of children who like cricket and F be the set of children who like football.
\nTotal children = 100.
\nNumber of children who like cricket, |C| = 64.
\nNumber of children who like football, |F| = 53.
\nNumber of children who like both cricket and football, |C \u2229 F| = 20.
\nThe number of children who like at least one of the games (either cricket or football or both) is given by the union of the two sets:
\n|C U F| = |C| + |F| – |C \u2229 F|
\n|C U F| = 64 + 53 – 20
\n|C U F| = 117 – 20
\n|C U F| = 97.
\nThis means 97 children like either cricket or football or both.
\nThe number of children who do NOT like to play cricket or football is the total number of children minus the number of children who like at least one game:
\nChildren who like neither = Total children – |C U F|
\nChildren who like neither = 100 – 97
\nChildren who like neither = 3.
\n<\/section>\n
\n– Use the formula for the union of two sets: |A U B| = |A| + |B| – |A \u2229 B|.
\n– Understand that the number of elements in the union represents those who like at least one of the items.
\n– Subtract the number who like at least one from the total number to find those who like neither.
\n<\/section>\n
\nThis type of problem is a classic application of basic set theory. A Venn diagram could also be used: Draw two overlapping circles for Cricket and Football. The overlap is 20. The part of the Cricket circle only is 64 – 20 = 44. The part of the Football circle only is 53 – 20 = 33. The total inside the circles is 44 + 20 + 33 = 97. Those outside the circles are 100 – 97 = 3.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

In a group of 100 children, 64 children like to play cricket, 53 children like to play football and 20 children like to play both cricket and football. How many children do NOT like to play cricket or football ? [amp_mcq option1=”3″ option2=”5″ option3=”7″ option4=”9″ correct=”option1″] This question was previously asked in UPSC CAPF – … <\/p>\n

Detailed SolutionIn a group of 100 children, 64 children like to play cricket, 53 child<\/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":[1288,1102],"class_list":["post-90564","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1288","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nIn a group of 100 children, 64 children like to play cricket, 53 child<\/title>\n<meta name=\"description\" content=\"This problem can be solved using the principle of inclusion-exclusion or a Venn diagram. Let C be the set of children who like cricket and F be the set of children who like football. Total children = 100. Number of children who like cricket, |C| = 64. Number of children who like football, |F| = 53. Number of children who like both cricket and football, |C \u2229 F| = 20. The number of children who like at least one of the games (either cricket or football or both) is given by the union of the two sets: |C U F| = |C| + |F| - |C \u2229 F| |C U F| = 64 + 53 - 20 |C U F| = 117 - 20 |C U F| = 97. This means 97 children like either cricket or football or both. The number of children who do NOT like to play cricket or football is the total number of children minus the number of children who like at least one game: Children who like neither = Total children - |C U F| Children who like neither = 100 - 97 Children who like neither = 3. - Use the formula for the union of two sets: |A U B| = |A| + |B| - |A \u2229 B|. - Understand that the number of elements in the union represents those who like at least one of the items. - Subtract the number who like at least one from the total number to find those who like neither.\" \/>\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\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In a group of 100 children, 64 children like to play cricket, 53 child\" \/>\n<meta property=\"og:description\" content=\"This problem can be solved using the principle of inclusion-exclusion or a Venn diagram. Let C be the set of children who like cricket and F be the set of children who like football. Total children = 100. Number of children who like cricket, |C| = 64. Number of children who like football, |F| = 53. Number of children who like both cricket and football, |C \u2229 F| = 20. The number of children who like at least one of the games (either cricket or football or both) is given by the union of the two sets: |C U F| = |C| + |F| - |C \u2229 F| |C U F| = 64 + 53 - 20 |C U F| = 117 - 20 |C U F| = 97. This means 97 children like either cricket or football or both. The number of children who do NOT like to play cricket or football is the total number of children minus the number of children who like at least one game: Children who like neither = Total children - |C U F| Children who like neither = 100 - 97 Children who like neither = 3. - Use the formula for the union of two sets: |A U B| = |A| + |B| - |A \u2229 B|. - Understand that the number of elements in the union represents those who like at least one of the items. - Subtract the number who like at least one from the total number to find those who like neither.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:31:06+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":"In a group of 100 children, 64 children like to play cricket, 53 child","description":"This problem can be solved using the principle of inclusion-exclusion or a Venn diagram. Let C be the set of children who like cricket and F be the set of children who like football. Total children = 100. Number of children who like cricket, |C| = 64. Number of children who like football, |F| = 53. Number of children who like both cricket and football, |C \u2229 F| = 20. The number of children who like at least one of the games (either cricket or football or both) is given by the union of the two sets: |C U F| = |C| + |F| - |C \u2229 F| |C U F| = 64 + 53 - 20 |C U F| = 117 - 20 |C U F| = 97. This means 97 children like either cricket or football or both. The number of children who do NOT like to play cricket or football is the total number of children minus the number of children who like at least one game: Children who like neither = Total children - |C U F| Children who like neither = 100 - 97 Children who like neither = 3. - Use the formula for the union of two sets: |A U B| = |A| + |B| - |A \u2229 B|. - Understand that the number of elements in the union represents those who like at least one of the items. - Subtract the number who like at least one from the total number to find those who like neither.","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\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/","og_locale":"en_US","og_type":"article","og_title":"In a group of 100 children, 64 children like to play cricket, 53 child","og_description":"This problem can be solved using the principle of inclusion-exclusion or a Venn diagram. Let C be the set of children who like cricket and F be the set of children who like football. Total children = 100. Number of children who like cricket, |C| = 64. Number of children who like football, |F| = 53. Number of children who like both cricket and football, |C \u2229 F| = 20. The number of children who like at least one of the games (either cricket or football or both) is given by the union of the two sets: |C U F| = |C| + |F| - |C \u2229 F| |C U F| = 64 + 53 - 20 |C U F| = 117 - 20 |C U F| = 97. This means 97 children like either cricket or football or both. The number of children who do NOT like to play cricket or football is the total number of children minus the number of children who like at least one game: Children who like neither = Total children - |C U F| Children who like neither = 100 - 97 Children who like neither = 3. - Use the formula for the union of two sets: |A U B| = |A| + |B| - |A \u2229 B|. - Understand that the number of elements in the union represents those who like at least one of the items. - Subtract the number who like at least one from the total number to find those who like neither.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:31:06+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\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/","url":"https:\/\/exam.pscnotes.com\/mcq\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/","name":"In a group of 100 children, 64 children like to play cricket, 53 child","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:31:06+00:00","dateModified":"2025-06-01T10:31:06+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"This problem can be solved using the principle of inclusion-exclusion or a Venn diagram. Let C be the set of children who like cricket and F be the set of children who like football. Total children = 100. Number of children who like cricket, |C| = 64. Number of children who like football, |F| = 53. Number of children who like both cricket and football, |C \u2229 F| = 20. The number of children who like at least one of the games (either cricket or football or both) is given by the union of the two sets: |C U F| = |C| + |F| - |C \u2229 F| |C U F| = 64 + 53 - 20 |C U F| = 117 - 20 |C U F| = 97. This means 97 children like either cricket or football or both. The number of children who do NOT like to play cricket or football is the total number of children minus the number of children who like at least one game: Children who like neither = Total children - |C U F| Children who like neither = 100 - 97 Children who like neither = 3. - Use the formula for the union of two sets: |A U B| = |A| + |B| - |A \u2229 B|. - Understand that the number of elements in the union represents those who like at least one of the items. - Subtract the number who like at least one from the total number to find those who like neither.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/in-a-group-of-100-children-64-children-like-to-play-cricket-53-child\/#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":"In a group of 100 children, 64 children like to play cricket, 53 child"}]},{"@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\/90564","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=90564"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90564\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90564"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90564"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90564"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}