{"id":115606,"date":"2018-04-10T15:55:31","date_gmt":"2018-04-10T18:55:31","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=115606"},"modified":"2018-09-18T16:11:18","modified_gmt":"2018-09-18T19:11:18","slug":"matematica-see-mg-prova-resolvida","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/","title":{"rendered":"Matem\u00e1tica SEE-MG &#8211; prova resolvida"},"content":{"rendered":"<p>Caros alunos,<\/p>\n<p>Vejam a resolu\u00e7\u00e3o das quest\u00f5es de matem\u00e1tica da prova da SEE\/MG, realizada no \u00faltimo domingo.<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Sabe-se que a massa&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Podemos dividir a massa da Terra pela massa da Lua, ficando com:<\/p>\n<p>T \/ L = (5,972 . 10^21) \/ (7,348 . 10^19) = (5,972 \/ 7,348) . 10^2<\/p>\n<p>T \/ L = 0,812 . 100 = 81,2<\/p>\n<p><strong>Resposta: D (82)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Um determinado medicamento \u00e9 vendido&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Veja que \u00e9 poss\u00edvel comprar:<\/p>\n<ul>\n<li>5 cartelas de 4 comprimidos<\/li>\n<li>2 cartelas com 4 e 2 cartelas com 6 comprimidos<\/li>\n<li>1 cartela com 4, 2 com 5 e 1 com 6 comprimidos<\/li>\n<li>4 cartelas de 5 comprimidos<\/li>\n<\/ul>\n<p><strong>Resposta: D (4)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Uma escola realizou&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Somando as quantidades de medalhas de cada um, veja que Carlos ganhou 5 + 6 + 9 = 20 medalhas. Este \u00e9 o maior valor.<\/p>\n<p><strong>Resposta: E (20)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Durante um dia de carnaval&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Das 120 pessoas, vejam que:<\/p>\n<ul>\n<li>12 v\u00e3o e voltam de taxi<\/li>\n<li>4 v\u00e3o e voltam a p\u00e9<\/li>\n<li>10 v\u00e3o e voltam de carona<\/li>\n<li>6 v\u00e3o e voltam de \u00f4nibus<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<p>Ao todo temos 12 + 4 + 10 + 6 = 32 que usaram o mesmo tipo de transporte na ida e na volta. Os demais 120 &#8211; 32 = 88 usaram meios diferentes para ir e voltar.<\/p>\n<p><strong>Resposta: D (88)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0O Tangram Cl\u00e1ssico&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Veja que na figura da letra A temos figuras arredondadas. Isto n\u00e3o \u00e9 poss\u00edvel obter utilizando as pe\u00e7as do Tangram, que n\u00e3o possuem partes arredondadas.<\/p>\n<p><strong>Resposta: A<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Renato est\u00e1 sempre desenhando&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Observe que o tri\u00e2ngulo preenchido vai mudando no sentido hor\u00e1rio, al\u00e9m de passarmos por uma figura com o tri\u00e2ngulo central preenchido e outra sem nenhum preenchido. Na \u00faltima figura do enunciado temos o tri\u00e2ngulo do canto inferior esquerdo preenchido. O pr\u00f3ximo ser\u00e1 o do canto superior, ou seja, alternativa E.<\/p>\n<p><strong>Resposta: E<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Um trem faz uma viagem&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>O trem percorre 54km em 1 hora. Para percorrer 279km:<\/p>\n<p>54km &#8212;&#8212;&#8212; 1 hora<\/p>\n<p>279km &#8212;&#8212;- T horas<\/p>\n<p>&nbsp;<\/p>\n<p>54 x T = 279 x 1<\/p>\n<p>T = 279 \/ 54<\/p>\n<p>T = 270\/54 + 9\/54<\/p>\n<p>T = 5 horas + 1\/6 hora<\/p>\n<p>T = 5 horas + 1\/6 x 60 minutos<\/p>\n<p>T = 5 horas + 10 minutos<\/p>\n<p><strong>Resposta: D (5h e 10min)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Uma foto impressa em&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>A raz\u00e3o seria de (6+6)\/(10+6) = 12\/16 = 3\/4 = 0,75.<\/p>\n<p><strong>Resposta: C (0,75)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Densidade demogr\u00e1fica \u00e9 a&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Podemos fazer c\u00e1lculos aproximados para chegar \u00c0 resposta:<\/p>\n<p>DF = 3 \/ 5,8 = 0,5<\/p>\n<p>ES = 4 \/ 46 = 0,1<\/p>\n<p>MG = 21 \/ 586 ~ 40\/1000 = 0,4<\/p>\n<p>RJ = 16 \/ 43 = 0,3<\/p>\n<p>SP = 45 \/ 248 = 180\/1000 = 0,18<\/p>\n<p>A maior densidade \u00e9 a do distrito federal. Fazendo o c\u00e1lculo correto:<\/p>\n<p>3.039.444 \/ 5779,997 = 525,85\u00a0 hab por km^2<\/p>\n<p><strong>Resposta: E (DF)<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>FUMARC &#8211; SEE\/MG &#8211; 2018)<\/strong>\u00a0Um avi\u00e3o fez uma viagem&#8230;<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Dividindo o custo total pelo tempo de viagem (3,5h), temos o custo m\u00e9dio por hora de viagem:<\/p>\n<p>Custo m\u00e9dio por hora = 596534 \/ 3,5 = 170438 reais por hora<\/p>\n<p>Dividindo esse valor pelo n\u00famero de passageiros, teremos o custo m\u00e9dio por hora para cada passageiro:<\/p>\n<p>Custo m\u00e9dio por hora por passageiro = 170438 \/ 285 = 598 reais<\/p>\n<p><strong>Resposta: C (598)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Caros alunos, Vejam a resolu\u00e7\u00e3o das quest\u00f5es de matem\u00e1tica da prova da SEE\/MG, realizada no \u00faltimo domingo. FUMARC &#8211; SEE\/MG &#8211; 2018)\u00a0Sabe-se que a massa&#8230; RESOLU\u00c7\u00c3O: Podemos dividir a massa da Terra pela massa da Lua, ficando com: T \/ L = (5,972 . 10^21) \/ (7,348 . 10^19) = (5,972 \/ 7,348) . 10^2 [&hellip;]<\/p>\n","protected":false},"author":48,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[1],"tags":[],"tax_estado":[],"class_list":["post-115606","post","type-post","status-publish","format-standard","hentry","category-cursos-e-concursos"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.2 (Yoast SEO v27.2) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Matem\u00e1tica SEE-MG - prova resolvida<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Matem\u00e1tica SEE-MG - prova resolvida\" \/>\n<meta property=\"og:description\" content=\"Caros alunos, Vejam a resolu\u00e7\u00e3o das quest\u00f5es de matem\u00e1tica da prova da SEE\/MG, realizada no \u00faltimo domingo. FUMARC &#8211; SEE\/MG &#8211; 2018)\u00a0Sabe-se que a massa&#8230; RESOLU\u00c7\u00c3O: Podemos dividir a massa da Terra pela massa da Lua, ficando com: T \/ L = (5,972 . 10^21) \/ (7,348 . 10^19) = (5,972 \/ 7,348) . 10^2 [&hellip;]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/\" \/>\n<meta property=\"og:site_name\" content=\"Estrat\u00e9gia Concursos\" \/>\n<meta property=\"article:published_time\" content=\"2018-04-10T18:55:31+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2018-09-18T19:11:18+00:00\" \/>\n<meta name=\"author\" content=\"Coordena\u00e7\u00e3o\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@EstratConcursos\" \/>\n<meta name=\"twitter:site\" content=\"@EstratConcursos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"Coordena\u00e7\u00e3o\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. tempo de leitura\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"NewsArticle\",\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/\"},\"author\":{\"name\":\"Coordena\u00e7\u00e3o\",\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/#\/schema\/person\/cdad40b5818eb228a838461fcd656da7\"},\"headline\":\"Matem\u00e1tica SEE-MG &#8211; 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FUMARC &#8211; SEE\/MG &#8211; 2018)\u00a0Sabe-se que a massa&#8230; RESOLU\u00c7\u00c3O: Podemos dividir a massa da Terra pela massa da Lua, ficando com: T \/ L = (5,972 . 10^21) \/ (7,348 . 10^19) = (5,972 \/ 7,348) . 10^2 [&hellip;]","og_url":"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/","og_site_name":"Estrat\u00e9gia Concursos","article_published_time":"2018-04-10T18:55:31+00:00","article_modified_time":"2018-09-18T19:11:18+00:00","author":"Coordena\u00e7\u00e3o","twitter_card":"summary_large_image","twitter_creator":"@EstratConcursos","twitter_site":"@EstratConcursos","twitter_misc":{"Escrito por":"Coordena\u00e7\u00e3o","Est. tempo de leitura":"2 minutos"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"NewsArticle","@id":"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/#article","isPartOf":{"@id":"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-see-mg-prova-resolvida\/"},"author":{"name":"Coordena\u00e7\u00e3o","@id":"https:\/\/www.estrategiaconcursos.com.br\/blog\/#\/schema\/person\/cdad40b5818eb228a838461fcd656da7"},"headline":"Matem\u00e1tica SEE-MG &#8211; 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