A hair product company sells three types of hair products for $30, $20, and $10 per unit. In one year, the total revenue for the three products was $800,000, which corresponded to the sale of 40,000 units. The company sold half as many units of the $30 product as units of the $20 product. How many units of each
product were sold?
30x+20y+10z = 800000
x+y+z = 40000
x = y/2
x=10,000
y=20,000
z=10,000
Let's represent the number of units sold for the $30 product as x.
According to the information given, the number of units sold for the $20 product would be double the number of units sold for the $30 product. Therefore, the number of units sold for the $20 product would be 2x.
The number of units sold for the $10 product can be calculated by subtracting the total number of units sold for the $30 and $20 products from the total number of units sold, which is 40,000. Therefore, the number of units sold for the $10 product is 40,000 - (x + 2x).
To calculate the total revenue, we multiply the number of units sold for each product by their respective prices and add them together. The revenue equation can be written as:
30x + 20(2x) + 10(40,000 - 3x) = 800,000
Now, we can solve this equation to find the value of x, which represents the number of units sold for the $30 product.
30x + 40x + 400,000 - 30x = 800,000
40x = 800,000 - 400,000
40x = 400,000
x = 10,000
Therefore, the number of units sold for the $30 product is 10,000.
The number of units sold for the $20 product would be double the number of units sold for the $30 product, so it would be 2 * 10,000 = 20,000.
The number of units sold for the $10 product can be calculated by subtracting the total number of units sold for the $30 and $20 products from the total number of units sold: 40,000 - (10,000 + 20,000) = 10,000.
In conclusion, the company sold 10,000 units of the $30 product, 20,000 units of the $20 product, and 10,000 units of the $10 product.
To solve this question, let's assign variables to the number of units sold for each product.
Let's say:
- x represents the number of units sold for the $30 product
- y represents the number of units sold for the $20 product
- z represents the number of units sold for the $10 product
We are given that the total revenue for all three products in one year is $800,000. We can set up an equation based on this information:
30x + 20y + 10z = 800,000
We also know that the total number of units sold is 40,000 units. We can set up another equation based on this information:
x + y + z = 40,000
Furthermore, we are given that the company sold half as many units of the $30 product as units of the $20 product. This translates to: x = 0.5y.
Now we have a system of three equations with three variables:
30x + 20y + 10z = 800,000
x + y + z = 40,000
x = 0.5y
To solve this system, we can use substitution or elimination method. Let's solve it using substitution:
Using the third equation, substitute x in the first and second equations:
30(0.5y) + 20y + 10z = 800,000
0.5y + y + z = 40,000
Simplifying these equations, we get:
15y + 20y + 10z = 800,000
1.5y + y + z = 40,000
Combining like terms, we get:
35y + 10z = 800,000
2.5y + z = 40,000
We can multiply the second equation by 10 to eliminate decimals:
25y + 10z = 400,000
Now we have a system of two equations with two variables:
35y + 10z = 800,000
25y + 10z = 400,000
By subtracting the second equation from the first equation, we can eliminate z:
(35y + 10z) - (25y + 10z) = 800,000 - 400,000
10y = 400,000
y = 40,000 / 10
y = 40,000
Now we can substitute the value of y into one of the original equations to find x:
x = 0.5y
x = 0.5 * 40,000
x = 20,000
Finally, we can substitute the values of x and y into the equation x + y + z = 40,000 to find z:
20,000 + 40,000 + z = 40,000
z = 0
Therefore, the solution is:
x = 20,000 units of the $30 product
y = 40,000 units of the $20 product
z = 0 units of the $10 product
So, the company sold 20,000 units of the $30 product, 40,000 units of the $20 product, and 0 units of the $10 product.