Let’s call the number of cars sold by dealership A as C and the number of trucks as T.
From the problem, we know that C + T = 164. (Equation 1)
We also know that dealership B sells twice as many cars as dealership A, so 2C. It also sells half as many trucks as dealership A, so T/2. Therefore, 2C + T/2 = 229. (Equation 2)
To find the number of trucks dealership B sold, we need to find the value of T.
First, let's solve Equation 1 for C.
C = 164 - T.
Now, substitute C in Equation 2 with (164-T).
2(164-T) + T/2 = 229.
Distribute 2 to the terms in the parentheses.
328 - 2T + T/2 = 229.
To eliminate the fraction, multiply every term by 2.
656 - 4T + T = 458.
Combine like terms.
656 - 3T = 458.
Subtract 656 from both sides.
-3T = 458 - 656.
Simplify the right side of the equation.
-3T = -198.
Divide both sides by -3 to solve for T.
T = -198/-3 = 66.
Therefore, dealership B sold 66 trucks. Answer: \boxed{66}.
A company owns two dealerships, both of which sell cars and trucks. Dealership A sells a total of 164 cars and trucks. Dealership B sells twice as many cars and half as many trucks as Dealership A, and sells a total of 229 cars and trucks. (4 pointsd) Considering that Dealership B sold half as many trucks as Dealership A, how many trucks did Dealership B sell?
1 answer