{"id":118816,"date":"2018-04-30T14:03:09","date_gmt":"2018-04-30T17:03:09","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=118816"},"modified":"2018-09-17T18:21:40","modified_gmt":"2018-09-17T21:21:40","slug":"matematica-basa-prova-resolvida-e-gabarito","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/matematica-basa-prova-resolvida-e-gabarito\/","title":{"rendered":"Matem\u00e1tica BASA &#8211; prova resolvida e gabarito"},"content":{"rendered":"<p>Caros alunos,<\/p>\n<p>Vejam a seguir a prova de Matem\u00e1tica de T\u00e9cnico Banc\u00e1rio do Banco da Amaz\u00f4nia (BASA) resolvida.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>CESGRANRIO \u2013 BASA \u2013 2018)<\/strong> Um valor inicial C<sub>0<\/sub> \u00a0foi capitalizado por meio da incid\u00eancia de juros compostos mensais constantes iguais a 6,09%. Ao final de 6 meses, isto \u00e9, ap\u00f3s 6 incid\u00eancias dos juros, gerou-se o montante M. A partir do valor inicial C<sub>0<\/sub>, seria alcan\u00e7ado o mesmo montante M ao final de 12 meses (12 incid\u00eancias), se os juros compostos mensais constantes tivessem sido iguais a<\/p>\n<p>(A) 3,045%<\/p>\n<p>(B) 3,000%<\/p>\n<p>(C) 1,045%<\/p>\n<p>(D) 1,450%<\/p>\n<p>(E) 3,450%<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>No primeiro caso, temos prazo de t = 6 meses, e taxa de j = 6,09% ao m\u00eas. Assim,<\/p>\n<p>M = C x (1+j)<sup>t<\/sup><\/p>\n<p>M = C<sub>0<\/sub> x (1+6,09%)<sup>6<\/sup><\/p>\n<p>M = C<sub>0<\/sub> x (1,0609)<sup>6<\/sup><\/p>\n<p><sup>\u00a0<\/sup><\/p>\n<p>No segundo caso, temos prazo de t = 12 meses. Neste caso, para termos o mesmo capital e mesmo montante:<\/p>\n<p>M = C<sub>0<\/sub> x (1 + j)<sup>12<\/sup><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p>Igualando as duas express\u00f5es do montante:<\/p>\n<p>C<sub>0<\/sub> x (1,0609)<sup>6<\/sup> = C<sub>0<\/sub> x (1 + j)<sup>12<\/sup><\/p>\n<p>(1,0609)<sup>6<\/sup> = (1 + j)<sup>12<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p>Tirando a raiz de 6\u00ba grau dos dois lados, temos:<\/p>\n<p>1,0609 = (1+j)<sup>2<\/sup><\/p>\n<p>1,03<sup>2<\/sup> = (1+j)<sup>2<\/sup><\/p>\n<p>1,03 = 1 + j<\/p>\n<p>j = 0,03 = 3% am<\/p>\n<p><strong>Resposta: B<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>CESGRANRIO \u2013 BASA \u2013 2018)<\/strong> O comprimento de um grande fio corresponde \u00e0 soma dos comprimentos de 24 fios menores. S\u00e3o eles:<\/p>\n<ul>\n<li>12 fios, cada um dos quais com comprimento que mede 14,7 cm;<\/li>\n<li>4 fios, cada um dos quais com comprimento que mede 0,3765 km;<\/li>\n<li>8 fios, cada um dos quais com comprimento que mede 13,125 dam.<\/li>\n<\/ul>\n<p>Esse grande fio foi dividido em 3 fios de igual comprimento, chamados de unidade modelo.<\/p>\n<p>Qual \u00e9 a medida, em metros, do comprimento de uma unidade modelo?<\/p>\n<p>(A) 852,588<\/p>\n<p>(B) 31,434<\/p>\n<p>(C) 6385,500<\/p>\n<p>(D) 2557,764<\/p>\n<p>(E) 94,302<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Podemos colocar todos os comprimentos dos fios na unidade metro, ficando com:<\/p>\n<p>14,7cm = 1,47dm = 0,147m<\/p>\n<p>0,3765km = 3,765hm = 37,65dam = 376,5m<\/p>\n<p>13,125dam = 131,25m<\/p>\n<p>&nbsp;<\/p>\n<p>O comprimento total \u00e9:<\/p>\n<p>Total = 12&#215;0,147 + 4&#215;376,5 + 8&#215;131,25<\/p>\n<p>Total = 2557,764m<\/p>\n<p>&nbsp;<\/p>\n<p>Dividindo em 3 segmentos de mesma medida, cada segmento ter\u00e1 2557,764 \/ 3 = 852,588m.<\/p>\n<p><strong>Resposta: A<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>CESGRANRIO \u2013 BASA \u2013 2018)<\/strong> Considere a sequ\u00eancia num\u00e9rica cujo termo geral \u00e9 dado por \u00a0\u00a0\u00a0a<sub>n<\/sub>=2<sup>1-3n<\/sup>, para n \u2265 1. Essa sequ\u00eancia num\u00e9rica \u00e9 uma progress\u00e3o<\/p>\n<p>(A) geom\u00e9trica, cuja raz\u00e3o \u00e9 -3.<\/p>\n<p>(B) aritm\u00e9tica, cuja raz\u00e3o \u00e9 1\/8.<\/p>\n<p>(C) geom\u00e9trica, cuja raz\u00e3o \u00e9 1\/8.<\/p>\n<p>(D) geom\u00e9trica, cuja raz\u00e3o \u00e9 -6.<\/p>\n<p>(E) aritm\u00e9tica, cuja raz\u00e3o \u00e9 -3.<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Podemos calcular os primeiros termos dessa progress\u00e3o:<\/p>\n<p>a<sub>1<\/sub> = 2<sup>1-3.1<\/sup> = 2<sup>-2<\/sup> = 1\/4<\/p>\n<p>a<sub>2<\/sub> = 2<sup>1-3.2<\/sup> = 2<sup>-5<\/sup> = 1\/32<\/p>\n<p>a<sub>3<\/sub> = 2<sup>1-3.3<\/sup> = 2<sup>-8<\/sup> = 1\/256<\/p>\n<p>&nbsp;<\/p>\n<p>Observe que a divis\u00e3o entre 2 termos consecutivos \u00e9 sempre o mesmo valor:<\/p>\n<p>(1\/256) \/ (1\/32) = (1\/256) x 32 = 32\/256 = 1\/8<\/p>\n<p>&nbsp;<\/p>\n<p>(1\/32) \/ (1\/4) = (1\/32) x 4 = 4\/32 = 1\/8<\/p>\n<p>&nbsp;<\/p>\n<p>Portanto, temos uma PG com raz\u00e3o igual a 1\/8.<\/p>\n<p><strong>Resposta: C<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>CESGRANRIO \u2013 BASA \u2013 2018)<\/strong> No conjunto dos n\u00fameros reais, considere as seguintes duas inequa\u00e7\u00f5es:<\/p>\n<p>Inequa\u00e7\u00e3o 1: 5x \u2013 7 &gt; x<sup>2<\/sup> &#8211; x + 1<\/p>\n<p>Inequa\u00e7\u00e3o 2: x + 6 &gt; -x + 10<\/p>\n<p>Um n\u00famero real x, que \u00e9 solu\u00e7\u00e3o da inequa\u00e7\u00e3o 2, tamb\u00e9m ser\u00e1 solu\u00e7\u00e3o da inequa\u00e7\u00e3o 1, se, e somente se, for solu\u00e7\u00e3o da inequa\u00e7\u00e3o<\/p>\n<p>(A) x<sup>2<\/sup> &#8211; 16 &gt; 0<\/p>\n<p>(B) 1\/x &lt; 1\/4<\/p>\n<p>(C) -x &lt; -4<\/p>\n<p>(D) 4x \u2013 16 &lt; 0<\/p>\n<p>(E) x + 1 &gt; x + 9<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Trabalhando com a primeira inequa\u00e7\u00e3o, temos:<\/p>\n<p>5x \u2013 7 &gt; x<sup>2<\/sup> &#8211; x + 1<\/p>\n<p>0 &gt; x<sup>2<\/sup> &#8211; x + 1 \u2013 5x + 7<\/p>\n<p>x<sup>2<\/sup> \u2013 6x + 8 &lt; 0<\/p>\n<p>Temos:<\/p>\n<p>Delta = (-6)<sup>2<\/sup> \u2013 4.1.8 = 36 \u2013 32 = 4<\/p>\n<p>&nbsp;<\/p>\n<p>A raiz de delta \u00e9 igual a 2. Logo,<\/p>\n<p>x = [-(-6) +\/- 2] \/ (2.1)<\/p>\n<p>x = [6 +\/- 2]\/2<\/p>\n<p>x = 3 +\/- 1<\/p>\n<p>x = 4\u00a0 ou \u00a0x = 2<\/p>\n<p>&nbsp;<\/p>\n<p>A fun\u00e7\u00e3o x<sup>2<\/sup> \u2013 6x + 8 \u00e9 uma par\u00e1bola com concavidade para cima. Ela ser\u00e1 negativa somente entre as duas ra\u00edzes, ou seja, 2 &lt; x &lt; 4 \u00e9 o conjunto solu\u00e7\u00e3o.<\/p>\n<p>&nbsp;<\/p>\n<p>Na segunda inequa\u00e7\u00e3o, temos:<\/p>\n<p>x + 6 &gt; -x + 10<\/p>\n<p>2x \u2013 4 &gt; 0<\/p>\n<p>x &gt; 2<\/p>\n<p>&nbsp;<\/p>\n<p>Os valores de x que atendem a 2\u00aa inequa\u00e7\u00e3o v\u00e3 atender tamb\u00e9m a 1\u00aa inequa\u00e7\u00e3o se forem menores do que 4, ou seja,<\/p>\n<p>x &lt; 4<\/p>\n<p>&nbsp;<\/p>\n<p>Olhando as op\u00e7\u00f5es de resposta, veja que a solu\u00e7\u00e3o da inequa\u00e7\u00e3o 4x \u2013 16 &lt; 0 \u00e9 exatamente x &lt; 4. Logo, este \u00e9 nosso gabarito.<\/p>\n<p><strong>Resposta: D<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>CESGRANRIO \u2013 BASA \u2013 2018)<\/strong> Sabe-se que 30% dos clientes de um banco s\u00e3o do sexo masculino e os 70% restantes s\u00e3o do sexo feminino. Entre os clientes do sexo masculino, a m\u00e9dia do tempo de v\u00ednculo com o banco \u00e9 igual a 4 anos e, entre os clientes do sexo feminino, \u00e9 igual a 6 anos. Considerando-se todos os clientes, de ambos os sexos, qual \u00e9 a m\u00e9dia do tempo de v\u00ednculo de cada um com o banco?<\/p>\n<p>(A) 6 anos<\/p>\n<p>(B) 5,7 anos<\/p>\n<p>(C) 5 anos<\/p>\n<p>(D) 5,3 anos<\/p>\n<p>(E) 5,4 anos<\/p>\n<p><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n<p>Podemos calcular a m\u00e9dia rapidamente assim:<\/p>\n<p>M\u00e9dia = 30%.4 + 70%.6<\/p>\n<p>M\u00e9dia = 0,30&#215;4 + 0,70&#215;6<\/p>\n<p>M\u00e9dia = 1,2 + 4,2<\/p>\n<p>M\u00e9dia = 5,4 anos<\/p>\n<p><strong>Resposta: E<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Caros alunos, Vejam a seguir a prova de Matem\u00e1tica de T\u00e9cnico Banc\u00e1rio do Banco da Amaz\u00f4nia (BASA) resolvida. &nbsp; CESGRANRIO \u2013 BASA \u2013 2018) Um valor inicial C0 \u00a0foi capitalizado por meio da incid\u00eancia de juros compostos mensais constantes iguais a 6,09%. Ao final de 6 meses, isto \u00e9, ap\u00f3s 6 incid\u00eancias dos juros, gerou-se [&hellip;]<\/p>\n","protected":false},"author":48,"featured_media":43262,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[1],"tags":[1019],"tax_estado":[],"class_list":["post-118816","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cursos-e-concursos","tag-basa"],"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 BASA - 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\/matematica-basa-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=\"Matem\u00e1tica BASA - prova resolvida e gabarito\" \/>\n<meta property=\"og:description\" content=\"Caros alunos, Vejam a seguir a prova de Matem\u00e1tica de T\u00e9cnico Banc\u00e1rio do Banco da Amaz\u00f4nia (BASA) resolvida. &nbsp; CESGRANRIO \u2013 BASA \u2013 2018) Um valor inicial C0 \u00a0foi capitalizado por meio da incid\u00eancia de juros compostos mensais constantes iguais a 6,09%. 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