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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}