Autor: Professor Allan Gomes | Publicado: 2013-05-11T07:12:00-07:00



AULA CURSINHO INSTITUTO – DIA 11/Maio/2013.

Prof. Msc. Allan Gomes dos Santos

Meu Blog: professorallangomes.blogspot.com
(VAMOS VISITAR!!!!!!)

1. (Prova Brasil de Matemática: espaço e forma)

&nbsp;O desenho abaixo
representa um sólido.

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Uma possível planificação desse sólido é

 

  
  
(A)

  
  
(B)

  
  
(C)

  
  
(D)

 

  
  
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Resolução: Uma sugestão para decidir qual planificação se
relaciona ao sólido proposto é observar o número de faces e seus posicionamentos
no plano estão relacionados.





2. (GUARDA CIVIL-FCC)

A escala de um mapa rodoviário é de 1 para 2 500 000. Se a
distância entre dois pontos nesse mapa é 25 mm, a distância real entre esses
pontos, em km, é igual a:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 



(A) 100 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

(B) 10 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

(C) 62,5 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

(D) 6,25 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 

(E) 1

Resolução: Verificamos que as medidas de comprimento que se
relacionam neste exemplo de mapa é mm (milímetro) e km (quilometro). Portanto, sabendo
que 1mm no mapa representa 2.500.000mm na realidade vamos transformar 2.500.000mm
em km que dá 2,5km (a vírgula volta 6 casas). Ou seja, 1mm no mapa representa
2,5km na realidade. Como a distancia no mapa é 25mm podemos multiplicar 25 x 2,5km
para achar a realidade. Assim, teremos 62,5km. 





3. (Concurso – COPEVE) Hoje
tenho x anos e daqui a 20 anos minha idade será maior que duas vezes a que
tenho hoje. Uma inequação que expressa esta situação é



A) x + 20
&gt; 2x

B) x + 20 &lt; 2x

C) x &lt; 20 − 2x

D) x &gt; 20 − 2x

Resolução: É uma questão de leitura do que tá acontecendo e, também,
conhecer a simbologia de maior (&gt;) (indica para frente) e menor (&lt;) (indica para trás). Então, x + 20 &gt; 2x





4. A idade
de João é inferior em 20% à de Luiz e a de José é superior em 20% à de Luiz. Em
quantos por cento a idade de José é superior à de João?

a) 50% &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 

b) 48% &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 

c) 45%&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 

d) 42%&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 

e) 40%

Resolução: Atribua uma idade a Luiz, por exemplo,
Luiz tem 100 anos, então José tem 120 anos (são 20% a mais) e João 80 (são 20%
a menos). Assim calculando a razão da idade de José para a de João são 

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5.
(Concurso – COPEVE) Uma pessoa
deseja dividir R$ 30.000,00 em duas partes, de modo que a primeira, aplicada a
4% ao ano, produza o mesmo que a segunda, a 6%, durante o mesmo tempo. O valor
da maior dessas partes será:

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 

A) R$
10.000,00.&nbsp; 

B) R$
20.000,00.&nbsp; 

C) R$
12.000,00.&nbsp; 

D) R$
16.000,00.&nbsp; 

E) R$ 18.000,00.

Resolução: Este problema
caracteriza uma relação divisão de partes diretamente proporcional. Assim,
podemos modelar o problema: A/4%(=0,04) = B/6%(=0,06) = A + B(=30.000)/
4%+6%(=0,04+0,06=0,1). Dividindo 30.000 por 0,1 = 300.000. Assim, podemos achar
cada parte separada, pois temos uma igualdade. B/0,06 = 300.000 implica que B =
18.000

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
A/0,04 = 300.000 implica que A = 12.000. 

A maior parte é B = 18.000.







6. (Concurso – COPEVE) Observe
o sistema abaixo.

O gráfico que melhor representa esse sistema é

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A)

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" strokecolor="white"/><![endif]--><!--[if !vml]-->


 

  
  
  
  
 

  
  
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D)

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C)

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<!--[if gte vml 1]><v:shape
 id="Imagem_x0020_19" o:spid="_x0000_i1029" type="#_x0000_t75" style='width:152.25pt;
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<![endif]--><!--[if !vml]--><!--[endif]-->&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <!--[if gte vml 1]><v:shape id="Imagem_x0020_18" o:spid="_x0000_i1028"
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<![endif]--><!--[if !vml]--><!--[endif]-->



Resolução: Podemos definir que o termo independente
de uma função representa o ponto que o gráfico de uma função corta o eixo da
ordenada, ou seja, o eixo do y. Portanto, x + y = 3 implica que y = - x + 3. Também,
2x – y = 1 implica que - y = - 2x + 1 (-1) ficando y = 2x – 1. Então, temos os
pontos 3 e -1. Resposta letra A.



7. (ANGLO) A = (50%)² , B = <!--[if gte vml 1]><v:shape
 id="_x0000_i1025" type="#_x0000_t75" style='width:40.5pt;height:18pt' o:ole="">
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  o:title=""/>
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 <o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1025"
  DrawAspect="Content" ObjectID="_1429775939">
 
<![endif]-->&nbsp;&nbsp;e&nbsp; C = <!--[if gte vml 1]><v:shape id="_x0000_i1026"
 type="#_x0000_t75" style='width:34.5pt;height:30.75pt' o:ole="">
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  DrawAspect="Content" ObjectID="_1429775940">
 
<![endif]-->, pode-se afirmar que :



a)
A&lt;B&nbsp;&nbsp; e A &lt; C

b) A &lt; B e A = C

c)
A &gt; B e A &gt; C

d)
A = 25 , B = 0,1&nbsp; e C = 1 / 4

e)
A = 25,&nbsp; B = 1 e C = 25%



Resolução: São várias formas para resolver. Mas, podemos definir que
é o certo, que 100% = 1. Então, 50% = 1/2 , 100% = 1 e 400% = 4. Fazendo as
substituições: A = (1/2)2 = 1/4 = 0,25; B = raiz de 1 = 1 e que C = 1/4
= 0,25. Portanto, A é menor que B e A é igual a C.



8. (TÉC.JUD.-TRT-FCC) Uma
empresa gerou um lucro de R$ 420 000,00, que foi dividido entre seus três
sócios, da seguinte maneira: a parte recebida pelo primeiro está para a do
segundo assim como 2 está para 3; a parte do segundo está para a do terceiro
assim como 4 está para 5. Nessa divisão, a menor das partes é igual a: 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;


(A) R$ 80 000,00

(B) R$124 000,00

(C) R$ 96 000,00

(D)
R$144 000,00

(E) R$120 000,00&nbsp;&nbsp;&nbsp; 

Resolução: É uma questão que podemos verificar uma
relação de razão. Então, podemos chamar às partes de x, y e z, respectivamente,
as partes que dividem o valor do lucro de 420.000,00. Portanto, pelo que é dado
na situação problema, x/y = 2/3 implica (resolvendo a proporção meios pelos
extremos) que 3x = 2y; na outra situação y/z = 4/5 implica (de novo a proporção)
que 5y = 4z. Assim, temos uma equação básica que x + y + z = 420.000. Observando
as relações, podemos substituir e dando valores para “y”, onde ficamos: 2y/3 +
y + 5y/4 = 420.000. Assim, o valor de y = 144.000. Resolvendo as proporções z =
180.000 e x = 96.000. Então a menor parte é a primeira x.





9. (Concurso – COPEVE) Dentro de um armário estão guardados cinco
livros, a saber, um de Matemática, um de Física, um de Química, um de Português
e outro de História. Sabe-se que: 

I. O livro de Física está entre o livro de História e o de Português.

II. O livro de Matemática não é o primeiro e o livro de Português
não é o último.

III. O livro de Química está separado do livro de Português por
dois livros.

Qual a posição do livro de Matemática?

A) O livro de Matemática é o primeiro.

B) O livro de Matemática é o segundo.

C) O livro de Matemática é o terceiro.

D) O livro de Matemática é o quarto.

E) O livro de Matemática é o quinto.

Resolução: Podemos ver que primeira
situação física não pode ser 1º ou 5ª e, também, o 4º por causa da segunda
situação. Pela a terceira situação química só pode tá na 1ª, assim, português na
4º. Física na 3ª e história na 2º. Enfim, matemática na 5ª.





10. (Concurso – COPEVE) Admitindo a
validade da sentença:

&nbsp;“Todo
historiador viaja muito. Ricardo nunca viajou”. Podemos afirmar que:

A) Ricardo
é historiador.

B) nenhum
historiador conhece Ricardo.

C) existe
um historiador que nunca viajou.

D) Ricardo ainda não é um historiador.

E) Ricardo nunca será um historiador.