I feel like a fifth grader now for asking this and this on http://math.stackexchange.com and not seeing the obvious solution. To avoid further humiliation on the internet, i used a notepad to find and solve an easier problem. See if you can solve it.
Problem 1:
A marketing company with 3 salesmen A,B,C has total revenue of 10,000 $.
A, B, C have the following revenues:
A = 2000 $
B = 3000 $
C = 5000 $
How will the marketing company increase revenues by 10 %?
Solution 1.1:
By increasing the revenues of A,B,C by 10%
A = 2200 $
B = 3300 $
C = 5500 $
Total revenue = 11000 $ = 10 %
By increasing the revenue of A by 1000 $ or 10 % of total
A = 2000 $ + 1000 $ = 3000 $
Total revenue = 11000 $ = 10 %
Solution 1.3:
Assuming dA, dB, dC denote the increase in revenues for A,B,C, all possible solutions are given by the algebraic equation:
dA + dB + dC = 1000 $
Solution 1.4:
A reasonably achievable solution can be obtained by taking the weighted percentages of Total revenues:
dA = 10 % * 2000 / 10000 = 2 % * 10000= 200 $
dB = 10 % * 3000 / 10000 = 3 % * 10000= 300 $
dC = 10 % * 5000 / 10000 = 5 % * 10000= 500 $
Increase in Revenue = 1000 $ = 10 %
Problem 2:
Revenues have two parts - Income and Expense
For A = 2000 $
Ia = 80% = 1600 $
Ea = 20% = 400 $
For B = 3000 $:
Ib = 70% = 2100 $
Eb = 30% = 900 $
For C = 5000 $:
Ic = 60% = 3000 $
Ec = 40% = 2000 $
Total income = 6700 $ = 67 %
Total expense = 3300 $ = 33 %
How will the marketing company increase only Incomes by 10 %?
Solution 2.1
A simpler solution is to increase Ia, Ib, Ic by 10% each.
Ia = 1600 + 160 = 1760 $
Ib = 2100 + 210 = 2310 $
Ic = 3000 + 300 = 3300 $
Total income = 6700 + 670 = 10 % increase
Solution 2.2
Another simpler solution is to increase only Ia by 670 $ or 10 % of total
Ia = 1600 + 670
Total income = 6700 + 670 = 10 % increase
Solution 2.3:
Assuming dIa, dIb, dIc denote the increase in Income for A,B,C, the solution is given by the algebraic equation with 3 unknowns:
dIa + dIb + dIc = 670
Solution 2.4
A reasonably achievable solution can be obtained by taking the weighted percentages of total income:
dIa = 10 % * 2000 / 10000 = 2 % of 6700= 134 $
dIb = 10 % * 3000 / 10000 = 3 % of 6700= 201 $
dIc = 10 % * 5000 / 10000 = 5 % of 6700= 335 $
Increase in Income = 670 $ = 10 %
Problem 3:
Retrospectively, How could the marketing company have increased incomes by 10 % and decreased expenses by 10 % for the same revenues last year?
Solution 3.1:
Solution 2.3 increases only total income by 10 % keeping the total expense constant.
This results in the proportion of income and expense not being 67 % + 33 % anymore
To increase income by 10 % and decrease expense by 10 % for the same total revenues last year
Total income (old) = 6700 $ = 67 %
Total expense (old) = 3300 $ = 33 %
Total revenue (old) = 6700 + 3300 = 10000 $
Total income (new) = 77 % = 7700 $
Total income (new) = 23 % = 2300 $
Total revenue (new = 7700 + 2300 = 10000 $
The solution is given by the solutions to the 2 equations with 6 unknowns:
dIa + dIb + dIc = New income - Old income = 77% * 10,000 - 6700 = 1000
dEa + dEb + dEc = New expense - Old expense = 23% * 10,000 - 3300 = -1000
Solution 3.2
A reasonably achievable solution can be obtained by taking the weighted percentage of total Increase/Decrease in Income/Expense
Total income increase = 1000 $
Total expense decrease = -1000 $
By using weighted percentages of total increase in income:
dIa = 2 % of +1000 = 200 $
dIb = 3% of +1000 = 300 $
dIc = 5% of +1000 = 500 $
dI = +1000 = +10%
By using weighted percentages of total decrease in expense:
dEa = 2 % of -1000 = -200 $
dEb = 3% of -1000 = -300 $
dEc = 5% of -1000 = -500 $
dE = -1000 = -10%
