{"id":52596,"date":"2016-10-17T13:39:52","date_gmt":"2016-10-17T16:39:52","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=52596"},"modified":"2018-09-21T17:37:24","modified_gmt":"2018-09-21T20:37:24","slug":"raciocinio-logico-tcm-rj-prova-resolvida-e-gabarito","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/raciocinio-logico-tcm-rj-prova-resolvida-e-gabarito\/","title":{"rendered":"Racioc\u00ednio L\u00f3gico TCM-RJ &#8211; prova resolvida e gabarito"},"content":{"rendered":"<p style=\"text-align: justify;\">Caros alunos, deixo abaixo a resolu\u00e7\u00e3o da prova de Racioc\u00ednio L\u00f3gico do TCM-RJ.<\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Dentre as alternativas, a \u00fanica incorreta \u00e9: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) A soma de dois n\u00fameros irracionais \u00e9 sempre um n\u00famero irracional <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) O conjunto dos reais \u00e9 a uni\u00e3o entre os n\u00fameros racionais e os n\u00fameros irracionais <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) A subtra\u00e7\u00e3o n\u00e3o \u00e9 opera\u00e7\u00e3o no conjunto dos naturais <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) Toda d\u00edzima peri\u00f3dica pertence ao conjunto dos racionais<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">A afirma\u00e7\u00e3o A est\u00e1 errada. Se somarmos um n\u00famero irracional com o seu oposto, o resultado \u00e9 zero. Por exemplo, somando raiz(2) com o n\u00famero -raiz(2), o resultado \u00e9 zero.<\/p>\n<p style=\"text-align: justify;\">Os n\u00fameros reais de fato s\u00e3o compostos pela uni\u00e3o entre racionais e irracionais.<\/p>\n<p style=\"text-align: justify;\">A subtra\u00e7\u00e3o n\u00e3o tem a propriedade do fechamento no conjunto dos naturais, pois se subtraimos 3 &#8211; 7, por exemplo, o resultado \u00e9 -4, n\u00famero que n\u00e3o faz parte dos naturais.<\/p>\n<p style=\"text-align: justify;\">As d\u00edzimas pertencem ao conjunto dos racionais pois podem ser escritas na forma de fra\u00e7\u00e3o.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0O resultado da ra\u00edz c\u00fabica do n\u00famero quatro ao quadrado \u00e9 um n\u00famero entre: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) 1 e 2 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) 3 e 4 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) 2 e 3 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) 1,5 e 2,3<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Queremos a raiz c\u00fabica de 4 ao quadrado, ou melhor, de 16. Para isso, podemos pensar o seguinte:<\/p>\n<p style=\"text-align: justify;\">2 ao cubo \u00e9 igual a 8 (menor que 16)<\/p>\n<p style=\"text-align: justify;\">3 ao cubo \u00e9 igual a 27 (maior que 16)<\/p>\n<p style=\"text-align: justify;\">Logo, a raiz c\u00fabica de 16, n\u00famero que est\u00e1 entre 8 e 27, ser\u00e1 um n\u00famero que est\u00e1 entre 2 e 3.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Sabe-se que a soma dos \u00e2ngulos internos de um pol\u00edgono \u00e9 igual a 1260\u00ba. Se esse pol\u00edgono \u00e9 regular, ent\u00e3o cada \u00e2ngulo externo desse pol\u00edgono \u00e9 igual a: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) 140\u00ba <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) 40\u00ba <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) 126\u00ba <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) 54\u00ba<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">A soma dos \u00e2ngulos internos \u00e9 dada por:<\/p>\n<p style=\"text-align: justify;\">S = (n &#8211; 2) x 180<\/p>\n<p style=\"text-align: justify;\">1260 = (n &#8211; 2) x 180<\/p>\n<p style=\"text-align: justify;\">n &#8211; 2 = 1260 \/ 180<\/p>\n<p style=\"text-align: justify;\">n &#8211; 2 = 7<\/p>\n<p style=\"text-align: justify;\">n = 9<\/p>\n<p style=\"text-align: justify;\">Temos um pol\u00edgono com 9 lados. Este tamb\u00e9m \u00e9 o n\u00famero de v\u00e9rtices e de \u00e2ngulos internos. Cada \u00e2ngulo interno mede, portanto, 1260 \/ 9 = 140 graus. O \u00e2ngulo externo \u00e9 o SUPLEMENTO do \u00e2ngulo interno, ou seja, 180 &#8211; 140 = 40 graus.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Sabe-se que 0,5 horas \u00e9 igual a 30 minutos. Ent\u00e3o 2,4 horas, em minutos, \u00e9 igual a: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) 160 minutos <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) 240 minutos <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) 140 minutos <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d<\/strong><strong>) 144 minutos<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Basta fazermos:<\/p>\n<p style=\"text-align: justify;\">2,4 horas =<\/p>\n<p style=\"text-align: justify;\">2,4 x 60 minutos =<\/p>\n<p style=\"text-align: justify;\">24 x 6 minutos =<\/p>\n<p style=\"text-align: justify;\">144 minutos<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Um comerciante separou suas moedas de dez centavos e vinte e cinco centavos e verificou que haviam 65 moedas e um total de R$ 12,80. Desse modo, o valor total das moedas de vinte e cinco centavos \u00e9: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) R$ 10,50 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) R$ 4,25 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) R$ 2,50 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) R$ 9,50<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Sejam D moedas de 10 centavos e V moedas de 25 centavos. O total de moedas \u00e9 65, portanto:<\/p>\n<p style=\"text-align: justify;\">D + V = 65<\/p>\n<p style=\"text-align: justify;\">D = 65 &#8211; V<\/p>\n<p style=\"text-align: justify;\">A soma dos valores \u00e9 12,80 reais, ou seja,<\/p>\n<p style=\"text-align: justify;\">Dx0,10 + Vx0,25 = 12,80<\/p>\n<p style=\"text-align: justify;\">(65-V)x0,10 + Vx0,25 = 12,80<\/p>\n<p style=\"text-align: justify;\">65&#215;0,10 &#8211; Vx0,10 + Vx0,25 = 12,80<\/p>\n<p style=\"text-align: justify;\">6,5 + Vx0,15 = 12,80<\/p>\n<p style=\"text-align: justify;\">Vx0,15 = 12,80 &#8211; 6,5<\/p>\n<p style=\"text-align: justify;\">Vx0,15 = 6,30<\/p>\n<p style=\"text-align: justify;\">V = 6,30\/0,15<\/p>\n<p style=\"text-align: justify;\">V = 42 moedas de 25 centavos<\/p>\n<p style=\"text-align: justify;\">O valor total, em moedas de 25 centavos, \u00e9 igual a 42 x 0,25 = 10,5 reais.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Dada a fun\u00e7\u00e3o f(x) = 3x -2 e g(x) = (x + 8)\/3 , ent\u00e3o g(f(-1)) \u00e9: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) \u2013 1 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) 0 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) 1 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) 5<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Veja que f(-1) = 3.(-1) &#8211; 2 = -3 &#8211; 2 = -5<\/p>\n<p style=\"text-align: justify;\">g(f(-1)) = g(-5) = (-5 + 8)\/3 = 3\/3 = 1<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Com velocidade m\u00e9dia de 60 km\/h um autom\u00f3vel vai de uma cidade A at\u00e9 uma cidade B em 4 horas. Se a velocidade m\u00e9dia do autom\u00f3vel aumentar em 15 km\/h, ent\u00e3o o tempo para ir da cidade B at\u00e9 a cidade A, pelo mesmo percurso, \u00e9: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) 3 horas <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) 3 horas e 20 minutos <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) 3 horas e 12 minutos <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) 5 horas<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Temos a seguinte situa\u00e7\u00e3o:<\/p>\n<p style=\"text-align: justify;\">Velocidade \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Tempo<\/p>\n<p style=\"text-align: justify;\">60 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a04<\/p>\n<p style=\"text-align: justify;\">75 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0T<\/p>\n<p style=\"text-align: justify;\">Quanto MAIOR a velocidade, MENOR o tempo necess\u00e1rio para percorrer uma dist\u00e2ncia. Vamos inverter a coluna do tempo, dado que as grandezas s\u00e3o inversamente proporcionais.<\/p>\n<p style=\"text-align: justify;\">Velocidade \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Tempo<\/p>\n<p style=\"text-align: justify;\">60 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0T<\/p>\n<p style=\"text-align: justify;\">75 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a04<\/p>\n<p style=\"text-align: justify;\">60&#215;4 = 75xT<\/p>\n<p style=\"text-align: justify;\">240 \/ 75 = T<\/p>\n<p style=\"text-align: justify;\">T = 3,2 horas<\/p>\n<p style=\"text-align: justify;\">T = 3 horas + 0,2 horas<\/p>\n<p style=\"text-align: justify;\">T = 3 horas + 0,2&#215;60 minutos<\/p>\n<p style=\"text-align: justify;\">T = 3 horas + 12 minutos<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Para valorizar seus funcion\u00e1rios, uma empresa ir\u00e1 sortear 3 viagens para a Disney entre seus 10 funcion\u00e1rios, de modo que cada funcion\u00e1rio poder\u00e1 ganhar somente uma viagem. O total de possibilidades distintas de sorteio para esses funcion\u00e1rios \u00e9: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) 120 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) 360 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) 720 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) 420<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Basta combinarmos os 10 funcion\u00e1rios em grupos de 3, afinal a ordem de escolha dos funcion\u00e1rios n\u00e3o importa, e sim o trio que \u00e9 formado. Ficamos com:<\/p>\n<p style=\"text-align: justify;\">C(10,3) = 10x9x8 \/ (3x2x1) = 10x3x4 = 120 formas<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Num envelope foram colocadas todas as 18 letras que formam a palavra CONSTITUCIONALISTA. A probabilidade de retirarmos uma letra desse envelope e ela ser consoante ou a letra O \u00e9:<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) 5\/9<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) 1\/9<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) 11\/18<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) 2\/3<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">O total de letras desta palavra \u00e9 18. Temos 10 consoantes e 2 letras O, portanto o total de letras que nos interessam \u00e9 igual a 12. A probabilidade de tirar uma delas \u00e9:<\/p>\n<p style=\"text-align: justify;\">P = 12 \/ 18 = 6 \/ 9 = 2 \/ 3<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Se as letras da sequ\u00eancia A,C,F,J, &#8230;, est\u00e3o descritas atrav\u00e9s de racioc\u00ednio l\u00f3gico, ent\u00e3o, considerando as 26 letras do alfabeto, a pr\u00f3xima letra da sequ\u00eancia deve ser: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>a) M <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>b) O <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>c) P <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>d) N<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Veja que:<\/p>\n<ul>\n<li style=\"text-align: justify;\">de A para C pulamos 1 letra (B)<\/li>\n<li style=\"text-align: justify;\">de C para F pulamos 2 letras (D, E)<\/li>\n<li style=\"text-align: justify;\">de F para J pulamos 3 letras (G, H, I)<\/li>\n<li style=\"text-align: justify;\">de F para a pr\u00f3xima letra, devemos pular 4 letras, que s\u00e3o: K, L, M, N. A pr\u00f3xima letra ser\u00e1 o O.<\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><strong>Resposta: B<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Caros alunos, deixo abaixo a resolu\u00e7\u00e3o da prova de Racioc\u00ednio L\u00f3gico do TCM-RJ. IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Dentre as alternativas, a \u00fanica incorreta \u00e9: a) A soma de dois n\u00fameros irracionais \u00e9 sempre um n\u00famero irracional b) O conjunto dos reais \u00e9 a uni\u00e3o entre os n\u00fameros racionais e os n\u00fameros irracionais c) A subtra\u00e7\u00e3o [&hellip;]<\/p>\n","protected":false},"author":48,"featured_media":43262,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[],"tags":[1627],"tax_estado":[],"class_list":["post-52596","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","tag-concurso-tcm-rj"],"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>Racioc\u00ednio L\u00f3gico TCM-RJ - prova resolvida e gabarito<\/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\/raciocinio-logico-tcm-rj-prova-resolvida-e-gabarito\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Racioc\u00ednio L\u00f3gico TCM-RJ - prova resolvida e gabarito\" \/>\n<meta property=\"og:description\" content=\"Caros alunos, deixo abaixo a resolu\u00e7\u00e3o da prova de Racioc\u00ednio L\u00f3gico do TCM-RJ. IBFC &#8211; TCM\/RJ &#8211; 2016)\u00a0Dentre as alternativas, a \u00fanica incorreta \u00e9: a) A soma de dois n\u00fameros irracionais \u00e9 sempre um n\u00famero irracional b) O conjunto dos reais \u00e9 a uni\u00e3o entre os n\u00fameros racionais e os n\u00fameros irracionais c) A subtra\u00e7\u00e3o [&hellip;]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.estrategiaconcursos.com.br\/blog\/raciocinio-logico-tcm-rj-prova-resolvida-e-gabarito\/\" \/>\n<meta property=\"og:site_name\" content=\"Estrat\u00e9gia Concursos\" \/>\n<meta property=\"article:published_time\" content=\"2016-10-17T16:39:52+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2018-09-21T20:37:24+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2016\/07\/09202054\/gabarito-e1509444514343.jpeg\" \/>\n\t<meta property=\"og:image:width\" content=\"400\" \/>\n\t<meta property=\"og:image:height\" content=\"267\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\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=\"5 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\/raciocinio-logico-tcm-rj-prova-resolvida-e-gabarito\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/raciocinio-logico-tcm-rj-prova-resolvida-e-gabarito\/\"},\"author\":{\"name\":\"Coordena\u00e7\u00e3o\",\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/#\/schema\/person\/cdad40b5818eb228a838461fcd656da7\"},\"headline\":\"Racioc\u00ednio L\u00f3gico TCM-RJ &#8211; 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