How to print answers only in exam?












1















With printanswers I can print all the answers following each of the questions. Now I want to print the answers only without the questions. I need this because I want to upload answers only, not the questions on website.



documentclass[11pt,oneside,A4paper,final,leqno]{exam}
usepackage{tikz}
usepackage{enumerate}
usepackage{amsfonts}
usepackage{amsmath}
usepackage{longtable}
usepackage{xcolor}
renewcommand{sectionmark}[1]{ {markright{rm smallthesection. #1}}}
usepackage[paper=a4paper,margin=1.086571in]{geometry}%
usepackage{etex}

usepackage[mathscr]{eucal}
usepackage{graphics, graphpap}
usepackage{multicol,color}
usepackage{indentfirst}

renewcommand{thefootnote}{}



usepackage{pgf,tikz}
usepackage{mathrsfs}
usetikzlibrary{arrows}

usepackage[absolute,overlay]{textpos}
usepackage{pgf,tikz}
usepackage{mathrsfs}
usetikzlibrary{arrows}
usepackage{amsthm}
defbi{textbf{i}}
defbj{textbf{j}}
defbk{textbf{k}}


usepackage[all]{xy} SelectTips{eu}{}

renewcommandquestionlabel{thequestion.} %replace period with parenthesis
renewcommand{thequestion}{bfseries arabic{question}}
printanswers
begin{document}


section*{Maths Test # 1901 rule{5.92cm}{.4pt}
}




begin{questions}



question What is the probability of picking an ace in two consecutive attempts in a 52 card deck?
begin{multicols}{4}
begin{choices}
choice $displaystylefrac{13}{221}$
choice $displaystylefrac{23}{221}$
choice $displaystylefrac{33}{221}$
choice $displaystylefrac{43}{221}$
end{choices}
end{multicols}
begin{solution} $(C)$.
In the first attempt, the chance not to get an ace is $48/52$. In the second one, the chance is $47/51$. Therefore, the chance not to get an ace in two consecutive attempts is $48*47/51*52=188/221$. Hence, the result must be $1-188/221=33/221$.
end{solution}




question Given three points $A(1,5), B(4,1)$ and $C(5,8)$. What is the angle $widehat{ACB}$?
begin{multicols}{4}
begin{choices}
choice $90$
choice $45$ %b
choice $60$
choice $120$
end{choices}
end{multicols}
begin{solution} $(B)$
We have $overrightarrow{AB}=(3,-4)$ and $overrightarrow{AC}=(4,3)$. Therefore, $ABC$ is an isosceles triangle with $widehat{BAC}=90$. Thus, the answer is $widehat{ACB}=45$.
end{solution}

question Suppose the ball moves freely inside the square domain with constant speed and the reflection off the boundary elastic and subject to a familiar law: the angle of incidence equals the angle of reflection. Put the ball at the center of the domain. At which angle $alpha$ does the ball have to start so that it will hit one of the four corners of the domain?
begin{center}

begin{tikzpicture}
clip(-4.5,1.5) rectangle (2.5,8.5);
%fill[line width=.4pt,fill=white] (2.,2.) -- (2.,8.) -- (-4.,8.) -- (-4.,2.) -- cycle;
draw[line width=.4pt,color=black] (-0.6928530532823249,2.) -- (-0.6928530532823249,2.3071469467176757) -- (-1.,2.3071469467176757) -- (-1.,2.) -- cycle;
draw [shift={(-1.,5.)},line width=.6pt,color=black] (0,0) -- (270.:1) arc (270.:303.6900675259798:1) -- cycle;
draw [line width=.6pt] (2.,2.)-- (2.,8.);
draw [line width=.66pt] (2.,8.)-- (-4.,8.);
draw [line width=.6pt] (-4.,8.)-- (-4.,2.);
draw [line width=.6pt] (-4.,2.)-- (2.,2.);
draw [line width=.6pt] (-1.,5.)-- (-1.,2.);
draw [line width=.6pt] (-1.,5.)-- (1.,2.);
begin{scriptsize}
draw (-0.1,4.6081570530868605) node {$alpha$};
end{scriptsize}
end{tikzpicture}
end{center}


begin{choices}
choice $displaystyletan{alpha}=frac{1}{2017}$
choice $displaystyletan{alpha}=frac{1}{2018}$ %b
choice $displaystyletan{alpha}=frac{2}{2019}$
choice $displaystyletan{alpha}=2020$
end{choices}


end{questions}

end{document}









share|improve this question



























    1















    With printanswers I can print all the answers following each of the questions. Now I want to print the answers only without the questions. I need this because I want to upload answers only, not the questions on website.



    documentclass[11pt,oneside,A4paper,final,leqno]{exam}
    usepackage{tikz}
    usepackage{enumerate}
    usepackage{amsfonts}
    usepackage{amsmath}
    usepackage{longtable}
    usepackage{xcolor}
    renewcommand{sectionmark}[1]{ {markright{rm smallthesection. #1}}}
    usepackage[paper=a4paper,margin=1.086571in]{geometry}%
    usepackage{etex}

    usepackage[mathscr]{eucal}
    usepackage{graphics, graphpap}
    usepackage{multicol,color}
    usepackage{indentfirst}

    renewcommand{thefootnote}{}



    usepackage{pgf,tikz}
    usepackage{mathrsfs}
    usetikzlibrary{arrows}

    usepackage[absolute,overlay]{textpos}
    usepackage{pgf,tikz}
    usepackage{mathrsfs}
    usetikzlibrary{arrows}
    usepackage{amsthm}
    defbi{textbf{i}}
    defbj{textbf{j}}
    defbk{textbf{k}}


    usepackage[all]{xy} SelectTips{eu}{}

    renewcommandquestionlabel{thequestion.} %replace period with parenthesis
    renewcommand{thequestion}{bfseries arabic{question}}
    printanswers
    begin{document}


    section*{Maths Test # 1901 rule{5.92cm}{.4pt}
    }




    begin{questions}



    question What is the probability of picking an ace in two consecutive attempts in a 52 card deck?
    begin{multicols}{4}
    begin{choices}
    choice $displaystylefrac{13}{221}$
    choice $displaystylefrac{23}{221}$
    choice $displaystylefrac{33}{221}$
    choice $displaystylefrac{43}{221}$
    end{choices}
    end{multicols}
    begin{solution} $(C)$.
    In the first attempt, the chance not to get an ace is $48/52$. In the second one, the chance is $47/51$. Therefore, the chance not to get an ace in two consecutive attempts is $48*47/51*52=188/221$. Hence, the result must be $1-188/221=33/221$.
    end{solution}




    question Given three points $A(1,5), B(4,1)$ and $C(5,8)$. What is the angle $widehat{ACB}$?
    begin{multicols}{4}
    begin{choices}
    choice $90$
    choice $45$ %b
    choice $60$
    choice $120$
    end{choices}
    end{multicols}
    begin{solution} $(B)$
    We have $overrightarrow{AB}=(3,-4)$ and $overrightarrow{AC}=(4,3)$. Therefore, $ABC$ is an isosceles triangle with $widehat{BAC}=90$. Thus, the answer is $widehat{ACB}=45$.
    end{solution}

    question Suppose the ball moves freely inside the square domain with constant speed and the reflection off the boundary elastic and subject to a familiar law: the angle of incidence equals the angle of reflection. Put the ball at the center of the domain. At which angle $alpha$ does the ball have to start so that it will hit one of the four corners of the domain?
    begin{center}

    begin{tikzpicture}
    clip(-4.5,1.5) rectangle (2.5,8.5);
    %fill[line width=.4pt,fill=white] (2.,2.) -- (2.,8.) -- (-4.,8.) -- (-4.,2.) -- cycle;
    draw[line width=.4pt,color=black] (-0.6928530532823249,2.) -- (-0.6928530532823249,2.3071469467176757) -- (-1.,2.3071469467176757) -- (-1.,2.) -- cycle;
    draw [shift={(-1.,5.)},line width=.6pt,color=black] (0,0) -- (270.:1) arc (270.:303.6900675259798:1) -- cycle;
    draw [line width=.6pt] (2.,2.)-- (2.,8.);
    draw [line width=.66pt] (2.,8.)-- (-4.,8.);
    draw [line width=.6pt] (-4.,8.)-- (-4.,2.);
    draw [line width=.6pt] (-4.,2.)-- (2.,2.);
    draw [line width=.6pt] (-1.,5.)-- (-1.,2.);
    draw [line width=.6pt] (-1.,5.)-- (1.,2.);
    begin{scriptsize}
    draw (-0.1,4.6081570530868605) node {$alpha$};
    end{scriptsize}
    end{tikzpicture}
    end{center}


    begin{choices}
    choice $displaystyletan{alpha}=frac{1}{2017}$
    choice $displaystyletan{alpha}=frac{1}{2018}$ %b
    choice $displaystyletan{alpha}=frac{2}{2019}$
    choice $displaystyletan{alpha}=2020$
    end{choices}


    end{questions}

    end{document}









    share|improve this question

























      1












      1








      1








      With printanswers I can print all the answers following each of the questions. Now I want to print the answers only without the questions. I need this because I want to upload answers only, not the questions on website.



      documentclass[11pt,oneside,A4paper,final,leqno]{exam}
      usepackage{tikz}
      usepackage{enumerate}
      usepackage{amsfonts}
      usepackage{amsmath}
      usepackage{longtable}
      usepackage{xcolor}
      renewcommand{sectionmark}[1]{ {markright{rm smallthesection. #1}}}
      usepackage[paper=a4paper,margin=1.086571in]{geometry}%
      usepackage{etex}

      usepackage[mathscr]{eucal}
      usepackage{graphics, graphpap}
      usepackage{multicol,color}
      usepackage{indentfirst}

      renewcommand{thefootnote}{}



      usepackage{pgf,tikz}
      usepackage{mathrsfs}
      usetikzlibrary{arrows}

      usepackage[absolute,overlay]{textpos}
      usepackage{pgf,tikz}
      usepackage{mathrsfs}
      usetikzlibrary{arrows}
      usepackage{amsthm}
      defbi{textbf{i}}
      defbj{textbf{j}}
      defbk{textbf{k}}


      usepackage[all]{xy} SelectTips{eu}{}

      renewcommandquestionlabel{thequestion.} %replace period with parenthesis
      renewcommand{thequestion}{bfseries arabic{question}}
      printanswers
      begin{document}


      section*{Maths Test # 1901 rule{5.92cm}{.4pt}
      }




      begin{questions}



      question What is the probability of picking an ace in two consecutive attempts in a 52 card deck?
      begin{multicols}{4}
      begin{choices}
      choice $displaystylefrac{13}{221}$
      choice $displaystylefrac{23}{221}$
      choice $displaystylefrac{33}{221}$
      choice $displaystylefrac{43}{221}$
      end{choices}
      end{multicols}
      begin{solution} $(C)$.
      In the first attempt, the chance not to get an ace is $48/52$. In the second one, the chance is $47/51$. Therefore, the chance not to get an ace in two consecutive attempts is $48*47/51*52=188/221$. Hence, the result must be $1-188/221=33/221$.
      end{solution}




      question Given three points $A(1,5), B(4,1)$ and $C(5,8)$. What is the angle $widehat{ACB}$?
      begin{multicols}{4}
      begin{choices}
      choice $90$
      choice $45$ %b
      choice $60$
      choice $120$
      end{choices}
      end{multicols}
      begin{solution} $(B)$
      We have $overrightarrow{AB}=(3,-4)$ and $overrightarrow{AC}=(4,3)$. Therefore, $ABC$ is an isosceles triangle with $widehat{BAC}=90$. Thus, the answer is $widehat{ACB}=45$.
      end{solution}

      question Suppose the ball moves freely inside the square domain with constant speed and the reflection off the boundary elastic and subject to a familiar law: the angle of incidence equals the angle of reflection. Put the ball at the center of the domain. At which angle $alpha$ does the ball have to start so that it will hit one of the four corners of the domain?
      begin{center}

      begin{tikzpicture}
      clip(-4.5,1.5) rectangle (2.5,8.5);
      %fill[line width=.4pt,fill=white] (2.,2.) -- (2.,8.) -- (-4.,8.) -- (-4.,2.) -- cycle;
      draw[line width=.4pt,color=black] (-0.6928530532823249,2.) -- (-0.6928530532823249,2.3071469467176757) -- (-1.,2.3071469467176757) -- (-1.,2.) -- cycle;
      draw [shift={(-1.,5.)},line width=.6pt,color=black] (0,0) -- (270.:1) arc (270.:303.6900675259798:1) -- cycle;
      draw [line width=.6pt] (2.,2.)-- (2.,8.);
      draw [line width=.66pt] (2.,8.)-- (-4.,8.);
      draw [line width=.6pt] (-4.,8.)-- (-4.,2.);
      draw [line width=.6pt] (-4.,2.)-- (2.,2.);
      draw [line width=.6pt] (-1.,5.)-- (-1.,2.);
      draw [line width=.6pt] (-1.,5.)-- (1.,2.);
      begin{scriptsize}
      draw (-0.1,4.6081570530868605) node {$alpha$};
      end{scriptsize}
      end{tikzpicture}
      end{center}


      begin{choices}
      choice $displaystyletan{alpha}=frac{1}{2017}$
      choice $displaystyletan{alpha}=frac{1}{2018}$ %b
      choice $displaystyletan{alpha}=frac{2}{2019}$
      choice $displaystyletan{alpha}=2020$
      end{choices}


      end{questions}

      end{document}









      share|improve this question














      With printanswers I can print all the answers following each of the questions. Now I want to print the answers only without the questions. I need this because I want to upload answers only, not the questions on website.



      documentclass[11pt,oneside,A4paper,final,leqno]{exam}
      usepackage{tikz}
      usepackage{enumerate}
      usepackage{amsfonts}
      usepackage{amsmath}
      usepackage{longtable}
      usepackage{xcolor}
      renewcommand{sectionmark}[1]{ {markright{rm smallthesection. #1}}}
      usepackage[paper=a4paper,margin=1.086571in]{geometry}%
      usepackage{etex}

      usepackage[mathscr]{eucal}
      usepackage{graphics, graphpap}
      usepackage{multicol,color}
      usepackage{indentfirst}

      renewcommand{thefootnote}{}



      usepackage{pgf,tikz}
      usepackage{mathrsfs}
      usetikzlibrary{arrows}

      usepackage[absolute,overlay]{textpos}
      usepackage{pgf,tikz}
      usepackage{mathrsfs}
      usetikzlibrary{arrows}
      usepackage{amsthm}
      defbi{textbf{i}}
      defbj{textbf{j}}
      defbk{textbf{k}}


      usepackage[all]{xy} SelectTips{eu}{}

      renewcommandquestionlabel{thequestion.} %replace period with parenthesis
      renewcommand{thequestion}{bfseries arabic{question}}
      printanswers
      begin{document}


      section*{Maths Test # 1901 rule{5.92cm}{.4pt}
      }




      begin{questions}



      question What is the probability of picking an ace in two consecutive attempts in a 52 card deck?
      begin{multicols}{4}
      begin{choices}
      choice $displaystylefrac{13}{221}$
      choice $displaystylefrac{23}{221}$
      choice $displaystylefrac{33}{221}$
      choice $displaystylefrac{43}{221}$
      end{choices}
      end{multicols}
      begin{solution} $(C)$.
      In the first attempt, the chance not to get an ace is $48/52$. In the second one, the chance is $47/51$. Therefore, the chance not to get an ace in two consecutive attempts is $48*47/51*52=188/221$. Hence, the result must be $1-188/221=33/221$.
      end{solution}




      question Given three points $A(1,5), B(4,1)$ and $C(5,8)$. What is the angle $widehat{ACB}$?
      begin{multicols}{4}
      begin{choices}
      choice $90$
      choice $45$ %b
      choice $60$
      choice $120$
      end{choices}
      end{multicols}
      begin{solution} $(B)$
      We have $overrightarrow{AB}=(3,-4)$ and $overrightarrow{AC}=(4,3)$. Therefore, $ABC$ is an isosceles triangle with $widehat{BAC}=90$. Thus, the answer is $widehat{ACB}=45$.
      end{solution}

      question Suppose the ball moves freely inside the square domain with constant speed and the reflection off the boundary elastic and subject to a familiar law: the angle of incidence equals the angle of reflection. Put the ball at the center of the domain. At which angle $alpha$ does the ball have to start so that it will hit one of the four corners of the domain?
      begin{center}

      begin{tikzpicture}
      clip(-4.5,1.5) rectangle (2.5,8.5);
      %fill[line width=.4pt,fill=white] (2.,2.) -- (2.,8.) -- (-4.,8.) -- (-4.,2.) -- cycle;
      draw[line width=.4pt,color=black] (-0.6928530532823249,2.) -- (-0.6928530532823249,2.3071469467176757) -- (-1.,2.3071469467176757) -- (-1.,2.) -- cycle;
      draw [shift={(-1.,5.)},line width=.6pt,color=black] (0,0) -- (270.:1) arc (270.:303.6900675259798:1) -- cycle;
      draw [line width=.6pt] (2.,2.)-- (2.,8.);
      draw [line width=.66pt] (2.,8.)-- (-4.,8.);
      draw [line width=.6pt] (-4.,8.)-- (-4.,2.);
      draw [line width=.6pt] (-4.,2.)-- (2.,2.);
      draw [line width=.6pt] (-1.,5.)-- (-1.,2.);
      draw [line width=.6pt] (-1.,5.)-- (1.,2.);
      begin{scriptsize}
      draw (-0.1,4.6081570530868605) node {$alpha$};
      end{scriptsize}
      end{tikzpicture}
      end{center}


      begin{choices}
      choice $displaystyletan{alpha}=frac{1}{2017}$
      choice $displaystyletan{alpha}=frac{1}{2018}$ %b
      choice $displaystyletan{alpha}=frac{2}{2019}$
      choice $displaystyletan{alpha}=2020$
      end{choices}


      end{questions}

      end{document}






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      asked 1 hour ago









      ThumboltThumbolt

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