To help raise funding for attending a science competition out of town, John and his three friends sell candies in their neighborhood. The average of each person’s sale is $200. John’s sale is 1/3 of Steve’s sale, Steve’s sale is 1/4 of Paul’s sale, and Charlie's’ sale is 2/3 of Paul’s sale. How much is each person’s sale?
John's sale = $X.
Steve's sale = $3X.
Paul's sale = $12X.
Charlie's sale = $8X.
x + 3x + 12x + 8x = 4*$200.
X = ?.
3X = ?.
12X = ?.
8X = ?.
Thanks Henry
To solve this problem, we'll start by assigning some variables to the sales of each person. Let's call John's sale J, Steve's sale S, Paul's sale P, and Charlie's sale C.
We're given that the average of each person's sale is $200, so we can set up the equation:
(J + S + P + C) / 4 = $200
Now let's use the given information to create equations relating the different sales.
1) John’s sale is 1/3 of Steve’s sale:
J = (1/3) * S
2) Steve’s sale is 1/4 of Paul’s sale:
S = (1/4) * P
3) Charlie's’ sale is 2/3 of Paul’s sale:
C = (2/3) * P
Now we can substitute these equations into the first equation to solve for the sales.
(J + (1/3)S + (1/4)P + (2/3)C) / 4 = $200
Substituting the relationships between the sales, we have:
(J + (1/3)S + (1/4)P + (2/3)C) / 4 = $200
(J + (1/3) * (1/4) * P + (1/4)P + (2/3) * (2/3) * P) / 4 = $200
(J + (1/12)P + (1/4)P + (4/9)P) / 4 = $200
(J + (3/36)P + (9/36)P + (16/36)P) / 4 = $200
(J + (28/36)P) / 4 = $200
(J + (7/9)P) / 4 = $200
J + (7/9)P = 4 * $200
J + (7/9)P = $800
Now we have two equations with two variables, J and P. Let's solve this system.
J = (1/3)S (Equation from above)
J + (7/9)P = $800 (Equation from above)
We can substitute the first equation into the second equation:
(1/3)S + (7/9)P = $800
Since we don't know the value of S, we can go no further. The information provided in the problem is not sufficient to determine the exact values of each person's sale.