{"id":88485,"date":"2017-09-10T23:15:20","date_gmt":"2017-09-11T02:15:20","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=88485"},"modified":"2018-09-21T15:35:52","modified_gmt":"2018-09-21T18:35:52","slug":"raciocinio-logico-prova-resolvida-colegio-pedro-ii","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/raciocinio-logico-prova-resolvida-colegio-pedro-ii\/","title":{"rendered":"Racioc\u00ednio L\u00f3gico &#8211; prova resolvida &#8211; Col\u00e9gio Pedro II"},"content":{"rendered":"<p style=\"text-align: justify;\">Caros alunos,<\/p>\n<p style=\"text-align: justify;\">Vejam a seguir a resolu\u00e7\u00e3o das quest\u00f5es da prova do Col\u00e9gio Pedro II para o cargo de Assistente em Administra\u00e7\u00e3o, que aconteceu neste domingo.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Considere a proposi\u00e7\u00e3o&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Uma equival\u00eancia manjada da proposi\u00e7\u00e3o condicional \u00e9 obtida invertendo a ordem das duas proposi\u00e7\u00f5es e negando ambas. Ou seja, uma equival\u00eancia de P&#8211;&gt;~Q seria:<\/p>\n<p style=\"text-align: justify;\">~(~Q) &#8211;&gt; ~P<\/p>\n<p style=\"text-align: justify;\">Ou melhor:<\/p>\n<p style=\"text-align: justify;\">Q&#8211;&gt;~P<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Observe a tabela-verdade&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Na primeira linha, como R \u00e9 V, notamos que ~R ser\u00e1 F. Assim, a proposi\u00e7\u00e3o P&#8211;&gt;(~R^Q) fica:<\/p>\n<p style=\"text-align: justify;\">V&#8211;&gt;(F^V)<\/p>\n<p style=\"text-align: justify;\">ou seja,<\/p>\n<p style=\"text-align: justify;\">V&#8211;&gt;F<\/p>\n<p style=\"text-align: justify;\">Esta \u00e9 uma condicional falsa, de modo que x \u00e9 F.<\/p>\n<p style=\"text-align: justify;\">Na segunda linha, o mero fato de P ser F torna toda a proposi\u00e7\u00e3o verdadeira. Isto porque basta o antecedente ser F para que a condicional seja V. Logo, y \u00e9 V.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Dez t\u00e9cnicos administrativos&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Note que 2 t\u00e9cnicos se ausentaram, de modo que no primeiro caso t\u00ednhamos 10 t\u00e9cnicos e, no segundo caso, 8 t\u00e9cnicos. No segundo caso eles v\u00e3o trabalhar 9 horas por dia, ou seja, 1 hora por dia a mais que no primeiro. Podemos esquematizar assim:<\/p>\n<p style=\"text-align: justify;\">T\u00e9cnicos \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Horas\/dia \u00a0 \u00a0 \u00a0 \u00a0 Dias<\/p>\n<p style=\"text-align: justify;\">10 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a08 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a036<\/p>\n<p style=\"text-align: justify;\">8 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a09 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D<\/p>\n<p style=\"text-align: justify;\">Quanto MAIS dias dispon\u00edveis, podemos realizar o trabalho com MENOS t\u00e9cnicos, trabalhando MENOS horas por dia. Logo, devemos inverter as duas primeiras colunas, que representam grandezas inversamente proporcionais ao n\u00famero de dias:<\/p>\n<p style=\"text-align: justify;\">T\u00e9cnicos \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Horas\/dia \u00a0 \u00a0 \u00a0 \u00a0 Dias<\/p>\n<p style=\"text-align: justify;\">8 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a09 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a036<\/p>\n<p style=\"text-align: justify;\">10 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a08 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D<\/p>\n<p style=\"text-align: justify;\">Montando a propor\u00e7\u00e3o:<\/p>\n<p style=\"text-align: justify;\">36\/D = (8\/10).(9\/8)<\/p>\n<p style=\"text-align: justify;\">36\/D = 9\/10<\/p>\n<p style=\"text-align: justify;\">4\/D = 1\/10<\/p>\n<p style=\"text-align: justify;\">D = 40 dias<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Alex, Jo\u00e3o e Pedro&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Para cada um dos rapazes, temos as seguintes possibilidades:<\/p>\n<ul style=\"text-align: justify;\">\n<li>campus: A, B ou C<\/li>\n<li>ensinos: Fundamental, M\u00e9dio ou Ambos<\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Como Pedro vai trabalhar no Campus C, podemos tirar essa possibilidade dos demais. Sobram os campi A e B. Como Alex n\u00e3o vai trabalhar no A, este deve ser de Jo\u00e3o, e Alex fica com o B. Temos:<\/p>\n<ul style=\"text-align: justify;\">\n<li>Alex: campus B<\/li>\n<li>Jo\u00e3o: campus A<\/li>\n<li>Pedro: campus C<\/li>\n<\/ul>\n<p style=\"text-align: justify;\">A afirma\u00e7\u00e3o &#8220;Alex n\u00e3o vai trabalhar s\u00f3 com Ensino Fundamental&#8221; sugere que ele vai trabalhar com Ambos os ensinos. Como Pedro vai ficar com o campus que s\u00f3 tem ensino m\u00e9dio, sobra para Jo\u00e3o o campus com apenas ensino fundamental. Temos:<\/p>\n<ul style=\"text-align: justify;\">\n<li>Alex: campus B, ambos os ensinos<\/li>\n<li>Jo\u00e3o: campus A, ensino fundamental<\/li>\n<li>Pedro: campus C, ensino m\u00e9dio<\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Assim, os campi A, B e C oferecem, respectivamente:<\/p>\n<ul style=\"text-align: justify;\">\n<li>ensino fundamental \/ ensino fundamental e m\u00e9dio \/ ensino m\u00e9dio<\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><strong>Resposta: D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Considere as premissas&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Se um poliglota N\u00c3O \u00e9 inteligente, ele N\u00c3O pode ser aluno do Col\u00e9gio Pedro II, pois todos os alunos deste col\u00e9gio s\u00e3o inteligentes, como diz a primeira premissa.<\/p>\n<p style=\"text-align: justify;\">Logo, podemos afirmar que se um poliglota N\u00c3O \u00e9 inteligente, ele N\u00c3O \u00e9 aluno do Col\u00e9gio Pedro II.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Em 2002, o Euro&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Veja que 1 DEM = 102,72 PTE. Como 1 Euro = 200,48 PTE, podemos dizer que:<\/p>\n<p style=\"text-align: justify;\">1 PTE = 1\/200,48 Euro<\/p>\n<p style=\"text-align: justify;\">Substituindo 1 PTE por 1\/200,48 Euro na primeira equa\u00e7\u00e3o que escrevi, temos:<\/p>\n<p style=\"text-align: justify;\">1 DEM = 102,72 x (1\/200,48) Euro<\/p>\n<p style=\"text-align: justify;\">Como 1 Euro = 0,89 d\u00f3lar, podemos escrever:<\/p>\n<p style=\"text-align: justify;\">1 DEM = 102,72 x (1\/200,48) x 0,89 d\u00f3lar<\/p>\n<p style=\"text-align: justify;\">Efetuando os c\u00e1lculos, temos:<\/p>\n<p style=\"text-align: justify;\">1 DEM = 0,456 d\u00f3lar<\/p>\n<p style=\"text-align: justify;\">A resposta mais pr\u00f3xima \u00e9 a da alternativa C.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Considere as seguintes&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Aqueles &#8220;alguns alunos&#8221; que s\u00e3o inteligentes s\u00e3o tamb\u00e9m dedicados, pois sabemos que todos os alunos s\u00e3o dedicados. Assim, podemos afirmar que existem pessoas que s\u00e3o inteligentes e tamb\u00e9m educadas.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Andy, Betty&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Vamos chamar as amigas pelas iniciais A, B, C, D e E. Como A e D tem menos que C, podemos esquematizar assim:<\/p>\n<p style=\"text-align: justify;\">C &gt; A<\/p>\n<p style=\"text-align: justify;\">C &gt; D<\/p>\n<p style=\"text-align: justify;\">Como A e E tem mais que B:<\/p>\n<p style=\"text-align: justify;\">A &gt; B<\/p>\n<p style=\"text-align: justify;\">E &gt; B<\/p>\n<p style=\"text-align: justify;\">Veja que podemos juntar com a informa\u00e7\u00e3o acima, ficando com:<\/p>\n<p style=\"text-align: justify;\">C &gt; A &gt; B<\/p>\n<p style=\"text-align: justify;\">E &gt; B<\/p>\n<p style=\"text-align: justify;\">C &gt; D<\/p>\n<p style=\"text-align: justify;\">Como D tem mais que E e menos que A:<\/p>\n<p style=\"text-align: justify;\">A &gt; D &gt; E<\/p>\n<p style=\"text-align: justify;\">Veja que todas as pessoas tem pelo menos uma outra que possui MENOS dinheiro que ela:<\/p>\n<p style=\"text-align: justify;\">A tem mais que B; E tem mais que B, C tem mais que A, D tem mais que E.<\/p>\n<p style=\"text-align: justify;\">Somente B n\u00e3o tem mais dinheiro que ningu\u00e9m. Esta \u00e9 a pessoa com menos dinheiro.<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Abr\u00e3o, Ar\u00e3o, &#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Repare que, para formar senhas, devemos permutar as letras de AACDEIII. Trata-se da permuta\u00e7\u00e3o de 8 letras, com repeti\u00e7\u00e3o de 2 letras A e 3 letras I. Calculamos o total assim:<\/p>\n<p style=\"text-align: justify;\">P(8; 2 e 3) = 8! \/ (2! x 3!)<\/p>\n<p style=\"text-align: justify;\">P(8; 2 e 3) = 8x7x6x5x4x3! \/ (2! x 3!)<\/p>\n<p style=\"text-align: justify;\">P(8; 2 e 3) = 8x7x6x5x4 \/ (2!)<\/p>\n<p style=\"text-align: justify;\">P(8; 2 e 3) = 8x7x6x5x4 \/ 2<\/p>\n<p style=\"text-align: justify;\">P(8; 2 e 3) = 8x7x6x5x2<\/p>\n<p style=\"text-align: justify;\">P(8; 2 e 3) = 3360<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEDRO II &#8211; 2017)<\/strong>\u00a0Uma pessoa, com uma calculadora&#8230;<\/p>\n<p style=\"text-align: justify;\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p style=\"text-align: justify;\">Veja que y \u00e9 a raiz quarta de x, ou seja,<\/p>\n<p style=\"text-align: justify;\">y = x<sup>1\/4<\/sup><\/p>\n<p style=\"text-align: justify;\">Veja ainda que 10 \u00e9 a raiz quadrada de y, de modo que y \u00e9 igual a 10 ao quadrado, ou seja,<\/p>\n<p style=\"text-align: justify;\">y = 100<\/p>\n<p style=\"text-align: justify;\">Assim,<\/p>\n<p style=\"text-align: justify;\">100 = x<sup>1\/4<\/sup><\/p>\n<p style=\"text-align: justify;\">Elevando ambos os lados \u00e0 quarta pot\u00eancia:<\/p>\n<p style=\"text-align: justify;\">100<sup>4<\/sup> = x<sup>(1\/4).4<\/sup><\/p>\n<p style=\"text-align: justify;\">100<sup>4<\/sup> = x<sup>1<\/sup><\/p>\n<p style=\"text-align: justify;\">100.000.000\u00a0= x<\/p>\n<p style=\"text-align: justify;\"><strong>Resposta: C<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Caros alunos, Vejam a seguir a resolu\u00e7\u00e3o das quest\u00f5es da prova do Col\u00e9gio Pedro II para o cargo de Assistente em Administra\u00e7\u00e3o, que aconteceu neste domingo. &nbsp; PEDRO II &#8211; 2017)\u00a0Considere a proposi\u00e7\u00e3o&#8230; RESOLU\u00c7\u00c3O: Uma equival\u00eancia manjada da proposi\u00e7\u00e3o condicional \u00e9 obtida invertendo a ordem das duas proposi\u00e7\u00f5es e negando ambas. Ou seja, uma equival\u00eancia [&hellip;]<\/p>\n","protected":false},"author":48,"featured_media":40099,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[3766],"tags":[],"tax_estado":[],"class_list":["post-88485","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-educacional"],"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 - prova resolvida - Col\u00e9gio Pedro II<\/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-prova-resolvida-colegio-pedro-ii\/\" \/>\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 - prova resolvida - Col\u00e9gio Pedro II\" \/>\n<meta property=\"og:description\" content=\"Caros alunos, Vejam a seguir a resolu\u00e7\u00e3o das quest\u00f5es da prova do Col\u00e9gio Pedro II para o cargo de Assistente em Administra\u00e7\u00e3o, que aconteceu neste domingo. &nbsp; PEDRO II &#8211; 2017)\u00a0Considere a proposi\u00e7\u00e3o&#8230; RESOLU\u00c7\u00c3O: Uma equival\u00eancia manjada da proposi\u00e7\u00e3o condicional \u00e9 obtida invertendo a ordem das duas proposi\u00e7\u00f5es e negando ambas. 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