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

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          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









            $endgroup$













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              2 Answers
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              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






























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