Rounding a percentage to the nearest multiple of $frac{1}{n}$












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If I take a percentage like $60%$ I can easily round it to a multiple of $frac{1}{n}$ where $n=2$ like this...



$$60%doteq50%$$
$$50%=frac{1}{2}$$



...or where $n=3$ like this.



$$60%doteq 66%$$
$$66%doteq frac{2}{3}$$



But what if $n$ was a not-so-friendly number, like 43? How do I round $60%$ to the nearest multiple of a fraction like $frac{1}{43}$ without doing so much guessing and checking?



Is there a consistent method for rounding $k%$ to the nearest multiple of $frac{1}{n}$ with minimal use of the guess and check method?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    If I take a percentage like $60%$ I can easily round it to a multiple of $frac{1}{n}$ where $n=2$ like this...



    $$60%doteq50%$$
    $$50%=frac{1}{2}$$



    ...or where $n=3$ like this.



    $$60%doteq 66%$$
    $$66%doteq frac{2}{3}$$



    But what if $n$ was a not-so-friendly number, like 43? How do I round $60%$ to the nearest multiple of a fraction like $frac{1}{43}$ without doing so much guessing and checking?



    Is there a consistent method for rounding $k%$ to the nearest multiple of $frac{1}{n}$ with minimal use of the guess and check method?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      If I take a percentage like $60%$ I can easily round it to a multiple of $frac{1}{n}$ where $n=2$ like this...



      $$60%doteq50%$$
      $$50%=frac{1}{2}$$



      ...or where $n=3$ like this.



      $$60%doteq 66%$$
      $$66%doteq frac{2}{3}$$



      But what if $n$ was a not-so-friendly number, like 43? How do I round $60%$ to the nearest multiple of a fraction like $frac{1}{43}$ without doing so much guessing and checking?



      Is there a consistent method for rounding $k%$ to the nearest multiple of $frac{1}{n}$ with minimal use of the guess and check method?










      share|cite|improve this question









      $endgroup$




      If I take a percentage like $60%$ I can easily round it to a multiple of $frac{1}{n}$ where $n=2$ like this...



      $$60%doteq50%$$
      $$50%=frac{1}{2}$$



      ...or where $n=3$ like this.



      $$60%doteq 66%$$
      $$66%doteq frac{2}{3}$$



      But what if $n$ was a not-so-friendly number, like 43? How do I round $60%$ to the nearest multiple of a fraction like $frac{1}{43}$ without doing so much guessing and checking?



      Is there a consistent method for rounding $k%$ to the nearest multiple of $frac{1}{n}$ with minimal use of the guess and check method?







      fractions percentages






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      asked Jan 7 at 3:04









      Diriector_DocDiriector_Doc

      1207




      1207






















          3 Answers
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          1












          $begingroup$

          Let $m = text{round}left(dfrac{k cdot n}{100}right)$, i.e. compute $dfrac{k cdot n}{100}$ and round it to the nearest integer.



          Then, the nearest multiple of $dfrac{1}{n}$ to $k%$ is $dfrac{m}{n}$.



          This works since the following statements are equivalent:



          $dfrac{m}{n}$ is the nearest multiple of $dfrac{1}{n}$ to $k%$



          $dfrac{m-tfrac{1}{2}}{n} < dfrac{k}{100} le dfrac{m+tfrac{1}{2}}{n}$



          $m-dfrac{1}{2} < dfrac{kn}{100} le m+dfrac{1}{2}$



          $m$ is the nearest integer to $dfrac{kn}{100}$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Round $kn%$ to the nearest integer and that's your numerator!



            eg $60%*43=25.8$ so $60%≈frac{26}{43}$.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Find the nearest integer to $k%$ of $n$. In your example, we can say that 1 is the nearest integer to 60% of 2 i.e. 1.2 and 2 is the nearest integer to 60% of 3 i.e. 1.8.



              Hope it helps:)






              share|cite|improve this answer









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                Let $m = text{round}left(dfrac{k cdot n}{100}right)$, i.e. compute $dfrac{k cdot n}{100}$ and round it to the nearest integer.



                Then, the nearest multiple of $dfrac{1}{n}$ to $k%$ is $dfrac{m}{n}$.



                This works since the following statements are equivalent:



                $dfrac{m}{n}$ is the nearest multiple of $dfrac{1}{n}$ to $k%$



                $dfrac{m-tfrac{1}{2}}{n} < dfrac{k}{100} le dfrac{m+tfrac{1}{2}}{n}$



                $m-dfrac{1}{2} < dfrac{kn}{100} le m+dfrac{1}{2}$



                $m$ is the nearest integer to $dfrac{kn}{100}$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Let $m = text{round}left(dfrac{k cdot n}{100}right)$, i.e. compute $dfrac{k cdot n}{100}$ and round it to the nearest integer.



                  Then, the nearest multiple of $dfrac{1}{n}$ to $k%$ is $dfrac{m}{n}$.



                  This works since the following statements are equivalent:



                  $dfrac{m}{n}$ is the nearest multiple of $dfrac{1}{n}$ to $k%$



                  $dfrac{m-tfrac{1}{2}}{n} < dfrac{k}{100} le dfrac{m+tfrac{1}{2}}{n}$



                  $m-dfrac{1}{2} < dfrac{kn}{100} le m+dfrac{1}{2}$



                  $m$ is the nearest integer to $dfrac{kn}{100}$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Let $m = text{round}left(dfrac{k cdot n}{100}right)$, i.e. compute $dfrac{k cdot n}{100}$ and round it to the nearest integer.



                    Then, the nearest multiple of $dfrac{1}{n}$ to $k%$ is $dfrac{m}{n}$.



                    This works since the following statements are equivalent:



                    $dfrac{m}{n}$ is the nearest multiple of $dfrac{1}{n}$ to $k%$



                    $dfrac{m-tfrac{1}{2}}{n} < dfrac{k}{100} le dfrac{m+tfrac{1}{2}}{n}$



                    $m-dfrac{1}{2} < dfrac{kn}{100} le m+dfrac{1}{2}$



                    $m$ is the nearest integer to $dfrac{kn}{100}$






                    share|cite|improve this answer









                    $endgroup$



                    Let $m = text{round}left(dfrac{k cdot n}{100}right)$, i.e. compute $dfrac{k cdot n}{100}$ and round it to the nearest integer.



                    Then, the nearest multiple of $dfrac{1}{n}$ to $k%$ is $dfrac{m}{n}$.



                    This works since the following statements are equivalent:



                    $dfrac{m}{n}$ is the nearest multiple of $dfrac{1}{n}$ to $k%$



                    $dfrac{m-tfrac{1}{2}}{n} < dfrac{k}{100} le dfrac{m+tfrac{1}{2}}{n}$



                    $m-dfrac{1}{2} < dfrac{kn}{100} le m+dfrac{1}{2}$



                    $m$ is the nearest integer to $dfrac{kn}{100}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 7 at 3:12









                    JimmyK4542JimmyK4542

                    41.3k245107




                    41.3k245107























                        1












                        $begingroup$

                        Round $kn%$ to the nearest integer and that's your numerator!



                        eg $60%*43=25.8$ so $60%≈frac{26}{43}$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Round $kn%$ to the nearest integer and that's your numerator!



                          eg $60%*43=25.8$ so $60%≈frac{26}{43}$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Round $kn%$ to the nearest integer and that's your numerator!



                            eg $60%*43=25.8$ so $60%≈frac{26}{43}$.






                            share|cite|improve this answer









                            $endgroup$



                            Round $kn%$ to the nearest integer and that's your numerator!



                            eg $60%*43=25.8$ so $60%≈frac{26}{43}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 7 at 3:13









                            timtfjtimtfj

                            2,503420




                            2,503420























                                0












                                $begingroup$

                                Find the nearest integer to $k%$ of $n$. In your example, we can say that 1 is the nearest integer to 60% of 2 i.e. 1.2 and 2 is the nearest integer to 60% of 3 i.e. 1.8.



                                Hope it helps:)






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Find the nearest integer to $k%$ of $n$. In your example, we can say that 1 is the nearest integer to 60% of 2 i.e. 1.2 and 2 is the nearest integer to 60% of 3 i.e. 1.8.



                                  Hope it helps:)






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Find the nearest integer to $k%$ of $n$. In your example, we can say that 1 is the nearest integer to 60% of 2 i.e. 1.2 and 2 is the nearest integer to 60% of 3 i.e. 1.8.



                                    Hope it helps:)






                                    share|cite|improve this answer









                                    $endgroup$



                                    Find the nearest integer to $k%$ of $n$. In your example, we can say that 1 is the nearest integer to 60% of 2 i.e. 1.2 and 2 is the nearest integer to 60% of 3 i.e. 1.8.



                                    Hope it helps:)







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 7 at 3:14









                                    MartundMartund

                                    1,965213




                                    1,965213






























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