How to divide and set an increment within a total?












0












$begingroup$


I have total=90. I want to divide this number between 5 employees.
For each employee from the second, I want to add an increment of 30% but the sum must remain 90.



Variables:

employees=5;

total=90;

increment=30%



E.g.



 Employee     Partial

   1st          10

   2nd          13

   3rd          16.9

   4th          21.7 

   5th          28.4


Is there a formula to achieve this?






fractions







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I have total=90. I want to divide this number between 5 employees.
    For each employee from the second, I want to add an increment of 30% but the sum must remain 90.



    Variables:
    
employees=5;
    
total=90;
    
increment=30%
    

    
E.g.
    

    
 Employee     Partial
    
   1st          10
    
   2nd          13
    
   3rd          16.9
    
   4th          21.7 
    
   5th          28.4


    Is there a formula to achieve this?






    fractions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have total=90. I want to divide this number between 5 employees.
      For each employee from the second, I want to add an increment of 30% but the sum must remain 90.



      Variables:
      
employees=5;
      
total=90;
      
increment=30%
      

      
E.g.
      

      
 Employee     Partial
      
   1st          10
      
   2nd          13
      
   3rd          16.9
      
   4th          21.7 
      
   5th          28.4


      Is there a formula to achieve this?






      fractions







      share|cite|improve this question









      $endgroup$




      I have total=90. I want to divide this number between 5 employees.
      For each employee from the second, I want to add an increment of 30% but the sum must remain 90.



      Variables:
      
employees=5;
      
total=90;
      
increment=30%
      

      
E.g.
      

      
 Employee     Partial
      
   1st          10
      
   2nd          13
      
   3rd          16.9
      
   4th          21.7 
      
   5th          28.4


      Is there a formula to achieve this?






      fractions




      fractions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 23:00









      NineCattoRulesNineCattoRules

      1032




      1032






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          So you want the first employee to get some starting amount $x$.



          Then you want the second employee to get $x + 30%$ - 30% more than the previous person. Equivalently, this is $1.3x$, obviously.



          The third employee gets 30% more than the previous guy - again, $1.3$ times that amount. So they get $1.3 times 1.3x = (1.3)^2 x$.



          You can see how this goes: employee $n$ gets $(1.3)^{n-1}x$.



          In that light, we could just add everything up, set it equal to $90$, combine like terms, and find $x$:



          $$x + (1.3)x + (1.3)^{2}x + (1.3)^{3}x + (1.3)^{4}x = 90$$



          Adding like terms and all isn't difficult here, but there's a more general way to do this for, say, REALLY long series. You see, what we have on the left of the equals sign is known as a geometric series: each term is the previous, times some number $r$:



          $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x$$



          This is what we call a "finite" geometric series, because it ends after ever-so-many terms. (Some geometric series go on infinitely and they have some special nuances to deal with. You're only dealing with a finite one so I'll skip over those details.) It can be shown that the sum of a finite geometric series can be given by



          $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x = left( frac{r^{n+1}-1}{r-1} right) x$$



          (In your case, we take $r = 1.3$.) Solving for $x$ becomes significantly easier this way.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            If you pay the first employee an amount $P$



            Then the total paid to the five employees will be the sum of the geometric series ...



            $$T=P(1+1.3+1.3^2+ 1.3^3+1.3^4) $$



            a formula does exist for this type of sum



            $$T = P bigg(frac{1-1.3^5}{1-1.3}bigg)
            $$
            so if your total is $90 , then the first employee should be paid



            $$P = 90bigg(frac{1-1.3}{1-1.3^5}bigg) $$



            Which works out to be close to $9.95






            share|cite|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026310%2fhow-to-divide-and-set-an-increment-within-a-total%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              So you want the first employee to get some starting amount $x$.



              Then you want the second employee to get $x + 30%$ - 30% more than the previous person. Equivalently, this is $1.3x$, obviously.



              The third employee gets 30% more than the previous guy - again, $1.3$ times that amount. So they get $1.3 times 1.3x = (1.3)^2 x$.



              You can see how this goes: employee $n$ gets $(1.3)^{n-1}x$.



              In that light, we could just add everything up, set it equal to $90$, combine like terms, and find $x$:



              $$x + (1.3)x + (1.3)^{2}x + (1.3)^{3}x + (1.3)^{4}x = 90$$



              Adding like terms and all isn't difficult here, but there's a more general way to do this for, say, REALLY long series. You see, what we have on the left of the equals sign is known as a geometric series: each term is the previous, times some number $r$:



              $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x$$



              This is what we call a "finite" geometric series, because it ends after ever-so-many terms. (Some geometric series go on infinitely and they have some special nuances to deal with. You're only dealing with a finite one so I'll skip over those details.) It can be shown that the sum of a finite geometric series can be given by



              $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x = left( frac{r^{n+1}-1}{r-1} right) x$$



              (In your case, we take $r = 1.3$.) Solving for $x$ becomes significantly easier this way.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                So you want the first employee to get some starting amount $x$.



                Then you want the second employee to get $x + 30%$ - 30% more than the previous person. Equivalently, this is $1.3x$, obviously.



                The third employee gets 30% more than the previous guy - again, $1.3$ times that amount. So they get $1.3 times 1.3x = (1.3)^2 x$.



                You can see how this goes: employee $n$ gets $(1.3)^{n-1}x$.



                In that light, we could just add everything up, set it equal to $90$, combine like terms, and find $x$:



                $$x + (1.3)x + (1.3)^{2}x + (1.3)^{3}x + (1.3)^{4}x = 90$$



                Adding like terms and all isn't difficult here, but there's a more general way to do this for, say, REALLY long series. You see, what we have on the left of the equals sign is known as a geometric series: each term is the previous, times some number $r$:



                $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x$$



                This is what we call a "finite" geometric series, because it ends after ever-so-many terms. (Some geometric series go on infinitely and they have some special nuances to deal with. You're only dealing with a finite one so I'll skip over those details.) It can be shown that the sum of a finite geometric series can be given by



                $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x = left( frac{r^{n+1}-1}{r-1} right) x$$



                (In your case, we take $r = 1.3$.) Solving for $x$ becomes significantly easier this way.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  So you want the first employee to get some starting amount $x$.



                  Then you want the second employee to get $x + 30%$ - 30% more than the previous person. Equivalently, this is $1.3x$, obviously.



                  The third employee gets 30% more than the previous guy - again, $1.3$ times that amount. So they get $1.3 times 1.3x = (1.3)^2 x$.



                  You can see how this goes: employee $n$ gets $(1.3)^{n-1}x$.



                  In that light, we could just add everything up, set it equal to $90$, combine like terms, and find $x$:



                  $$x + (1.3)x + (1.3)^{2}x + (1.3)^{3}x + (1.3)^{4}x = 90$$



                  Adding like terms and all isn't difficult here, but there's a more general way to do this for, say, REALLY long series. You see, what we have on the left of the equals sign is known as a geometric series: each term is the previous, times some number $r$:



                  $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x$$



                  This is what we call a "finite" geometric series, because it ends after ever-so-many terms. (Some geometric series go on infinitely and they have some special nuances to deal with. You're only dealing with a finite one so I'll skip over those details.) It can be shown that the sum of a finite geometric series can be given by



                  $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x = left( frac{r^{n+1}-1}{r-1} right) x$$



                  (In your case, we take $r = 1.3$.) Solving for $x$ becomes significantly easier this way.






                  share|cite|improve this answer









                  $endgroup$



                  So you want the first employee to get some starting amount $x$.



                  Then you want the second employee to get $x + 30%$ - 30% more than the previous person. Equivalently, this is $1.3x$, obviously.



                  The third employee gets 30% more than the previous guy - again, $1.3$ times that amount. So they get $1.3 times 1.3x = (1.3)^2 x$.



                  You can see how this goes: employee $n$ gets $(1.3)^{n-1}x$.



                  In that light, we could just add everything up, set it equal to $90$, combine like terms, and find $x$:



                  $$x + (1.3)x + (1.3)^{2}x + (1.3)^{3}x + (1.3)^{4}x = 90$$



                  Adding like terms and all isn't difficult here, but there's a more general way to do this for, say, REALLY long series. You see, what we have on the left of the equals sign is known as a geometric series: each term is the previous, times some number $r$:



                  $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x$$



                  This is what we call a "finite" geometric series, because it ends after ever-so-many terms. (Some geometric series go on infinitely and they have some special nuances to deal with. You're only dealing with a finite one so I'll skip over those details.) It can be shown that the sum of a finite geometric series can be given by



                  $$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x = left( frac{r^{n+1}-1}{r-1} right) x$$



                  (In your case, we take $r = 1.3$.) Solving for $x$ becomes significantly easier this way.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 23:08









                  Eevee TrainerEevee Trainer

                  7,00811337




                  7,00811337























                      2












                      $begingroup$

                      If you pay the first employee an amount $P$



                      Then the total paid to the five employees will be the sum of the geometric series ...



                      $$T=P(1+1.3+1.3^2+ 1.3^3+1.3^4) $$



                      a formula does exist for this type of sum



                      $$T = P bigg(frac{1-1.3^5}{1-1.3}bigg)
                      $$
                      so if your total is $90 , then the first employee should be paid



                      $$P = 90bigg(frac{1-1.3}{1-1.3^5}bigg) $$



                      Which works out to be close to $9.95






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        If you pay the first employee an amount $P$



                        Then the total paid to the five employees will be the sum of the geometric series ...



                        $$T=P(1+1.3+1.3^2+ 1.3^3+1.3^4) $$



                        a formula does exist for this type of sum



                        $$T = P bigg(frac{1-1.3^5}{1-1.3}bigg)
                        $$
                        so if your total is $90 , then the first employee should be paid



                        $$P = 90bigg(frac{1-1.3}{1-1.3^5}bigg) $$



                        Which works out to be close to $9.95






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          If you pay the first employee an amount $P$



                          Then the total paid to the five employees will be the sum of the geometric series ...



                          $$T=P(1+1.3+1.3^2+ 1.3^3+1.3^4) $$



                          a formula does exist for this type of sum



                          $$T = P bigg(frac{1-1.3^5}{1-1.3}bigg)
                          $$
                          so if your total is $90 , then the first employee should be paid



                          $$P = 90bigg(frac{1-1.3}{1-1.3^5}bigg) $$



                          Which works out to be close to $9.95






                          share|cite|improve this answer









                          $endgroup$



                          If you pay the first employee an amount $P$



                          Then the total paid to the five employees will be the sum of the geometric series ...



                          $$T=P(1+1.3+1.3^2+ 1.3^3+1.3^4) $$



                          a formula does exist for this type of sum



                          $$T = P bigg(frac{1-1.3^5}{1-1.3}bigg)
                          $$
                          so if your total is $90 , then the first employee should be paid



                          $$P = 90bigg(frac{1-1.3}{1-1.3^5}bigg) $$



                          Which works out to be close to $9.95







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 4 '18 at 23:16









                          WW1WW1

                          7,3251712




                          7,3251712






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026310%2fhow-to-divide-and-set-an-increment-within-a-total%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              How to change which sound is reproduced for terminal bell?

                              Image
                              3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                              Read more

                              Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                              Image
                              up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                              Read more

                              Can I use Tabulator js library in my java Spring + Thymeleaf project?

                              Image
                              0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                              Read more