Bunuel
The 50 participants of a management training seminar ate dinner at a certain restaurant. They had 3 choices for their meal: vegetarian lasagna for $12, blackened catfish for $15, or stuffed pork chops for $18. Each participant ordered exactly 1 meal and the total cost of the meals ordered by the participants was $810. How many participants of the management training seminar ordered blackened catfish?
(1) Six more people ordered catfish than lasagna.
(2) Twice as many pork chop meals were ordered as catfish meals.
- Number of participants who ordered vegetarian lasagna = \(x\)
- Number of participants who ordered blackened catfish = \(y\)
- Number of participants who stuffed pork chops = \(z\)
\(x + y + z = 50\) ---- (1)
\(12*x + 15*y + 18*z = 810\)
\(4x + 5y + 6z = 270\) ---- (2)
Statement 1(1) Six more people ordered catfish than lasagna.\(y = x + 6\)
Using Equation 1
\(x + x + 6 + z = 50\)
\(2x + z = 43\)
Using Equation 2
\(4x + 5(x+6) + 6z = 270\)
\(9x + 30 + 6z = 270\)
\(9x + 6z = 240\)
We have two equations and two variables, hence we can find a unique value of x, y, and z. The statement alone is sufficient. We can eliminate B, C, and E.
Statement 2(2) Twice as many pork chop meals were ordered as catfish meals.\(z = 2y\)
Using Equation 1
\(x + y + 2y = 50\)
\(x + 3y = 50\)
Using Equation 2
\(4x + 5y + 6(2y) = 270\)
\(4x + 17y = 270\)
We have two equations and two variables, hence we can find a unique value of x, y, and z. The statement alone is sufficient.
Option D