How to divide and set an increment within a total?
$begingroup$
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
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
$endgroup$
add a comment |
$begingroup$
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
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
$endgroup$
add a comment |
$begingroup$
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
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
$endgroup$
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
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
asked Dec 4 '18 at 23:00
NineCattoRulesNineCattoRules
1032
1032
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
$endgroup$
add a comment |
$begingroup$
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
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
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.
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
$endgroup$
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.
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
answered Dec 4 '18 at 23:08
Eevee TrainerEevee Trainer
5,3881836
5,3881836
add a comment |
add a comment |
$begingroup$
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
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
$endgroup$
add a comment |
$begingroup$
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
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
$endgroup$
add a comment |
$begingroup$
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
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
$endgroup$
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
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
answered Dec 4 '18 at 23:16
WW1WW1
7,2651712
7,2651712
add a comment |
add a comment |
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