A car manufacturing company (the Company) can produce a car for $3000.00 and a truck for $5000.00. Suppose "C" cars and "T" trucks are sold to the dealer at mark-ups, respectively, of 20 and 30 percent. If the Company made a profit of $27 million on sales of $137 million in one particular year, how many cars and trucks were sold by the company in that year?
I have to get 2 equations from this and use the substitution or elimination method
I tried to solve it using those equations but my answer for T is a negative number which I know it can't be
The first equation would be
3000 C + 5000 T = 137,000,000 (total sales)
The second equation would be
3000*.20*C + 5000*.30*T = 27,000,000 (profit)
This can be rewritten
600 C + 1500 T = 27,000,000
Now solve those two equations for C and T, the number of cars and trucks sold, using either method.
I get the same result as you, and checked my work. I have no explanation
Sometimes one winds up with an "erroneous" solution.
To solve this problem, let's set up two equations based on the given information.
Equation 1: The total cost of manufacturing and selling all cars and trucks is equal to the sum of the cost of cars and the cost of trucks.
Equation 2: The total revenue generated from selling all cars and trucks is equal to the sum of the revenue from cars and the revenue from trucks.
Let's represent the number of cars sold as "C" and the number of trucks sold as "T".
Equation 1:
Cost of cars + Cost of trucks = Total cost
(3000 * C) + (5000 * T) = Total cost
Equation 2:
Revenue from cars + Revenue from trucks = Total revenue
[(3000 * C) * 1.2] + [(5000 * T) * 1.3] = Total revenue
Now, let's substitute the given information and solve the equations.
Given:
Total profit = $27 million
Total revenue = $137 million
We can rewrite the equation for profit using the equation for revenue:
Total revenue - Total cost = Total profit
[(3000 * C) * 1.2] + [(5000 * T) * 1.3] - [(3000 * C) + (5000 * T)] = 27 million
Now simplify and rearrange the equation:
3600C + 6500T - 3000C - 5000T = 27 million
600C + 1500T = 27 million
We also know:
[(3000 * C) * 1.2] + [(5000 * T) * 1.3] = 137 million
Now, we have a system of equations:
Equation 1: 600C + 1500T = 27 million
Equation 2: 3600C + 6500T = 137 million
We can solve this system of equations using the substitution or elimination method to find the values of C (cars sold) and T (trucks sold).