over the last three years, sales person A has earned twice as much as Salesperson B, who
in turned has average 50%more than salesperson C. the three combined to make $220.000
over the three years. how much did sales person B earn on average in any year
Equations: A=2B, B=0.5C+C. You can also say A=2(0.5C+C), A=C+2C, A=3C
Now, we have A+B+C=$220
Plugging in the values, we have 3C+0.5C+C+C=220
5C+0.5C=220
5.5C=220
C=40
And, B= 0.5(40)+40=20+40=\(\boxed{60}\).
Or, is that a comma after that 220?
Let the sales of Person C = S
The sales of Person B =1.5S
The sales of Person A =2 x 1.5S =3S
S + 1.5S + 3S =$220,000
5.5S = $220,000
S =$220,000 / 5.5
S =$40,000 - total earning of salesperson C over a 3-year period.
$40,000 x 1.5 =$60,000 - total earning of salesperson B over a 3-year period.
$60,000 / 3 =$20,000 - average earning of salesperson B per year for 3 years.
Edited: Sorry-Took the amount as $200,000 instead of $220,000
Let's assume that 220.000 is supposed to be 220,000
Let salesperson C earn E during this period
Then B has earned 50% more than C = 1.5C = 1.5E
And A has earned twice B = 2B = 2(1.5)C = 3C = 3E
So
A + B + C = 220000
3E + 1.5E + E = 220000
5.5E = 220000 [ divide both sides by 5.5 ]
E = 40000 = amt C earned in the period
And B earned 1.5C = 1.5 (40000) = 60000
So...in any one year, B's average earnings = 60000 / 3 = 20000