A cafeteria serves three choices for lunch: turkey, chicken, or veggie. The turkey plate costs $4.80, the chicken plate costs $2.40, and the vegetable plate costs $1.20. If the lunches are mixed making 6 orders for $3.40 each, with twice as many chicken plates as veggie plates, find the number of each lunch plate mixed. Provide the answer along with an explanation of how you arrived at the answer.

Question

anonymous · Answer

Let x be the number of veggie plates. From the problem statement, 2x has to be the number of chicken plates. Hence 6-x is the number of turkey plates.
Solve the following expression for x:$$\frac{2x* \frac{240}{100}+\frac{120}{100}x+(6-3x)\frac{480}{100}}{6}=\frac{340}{100}$$Move the RHS to the LHS and simplify.$$-\frac{7}{5} (-1+x)=0$$$$x=1 $$$$x=1,\text{ 2}x=2 \text{ and } (6-3x)=3 $$Verify the answer.$$1.20+2*2.40+3*4.80=6*3.40\text{  }?$$$$20.40=20.40\text{  }\text{yes}$$

hero · Answer

x = turkey
y = chicken
z = veggie

turkey plate       = 4.80x
chicken plate     = 2.40y
vegetable plate = 1.20z


6(3.40) =20.40
2z= y

x + 2z + z = 6
x + 3z =6

4.80x + 2.40y + 1.20z = 20.40
4.80x + 2.40(2z) + 1.20z = 20.40
4.80x + 4.80z + 1.20z = 20.40
4.80x + 6z = 20.40

x + 3z = 6
4.80x + 6z = 20.40

z = (6-x)/3
z = (20.40 - 4.80x)/6

3(20.40 - 4.80x) = 6(6-x)
61.20 - 14.40x = 36-6x
25.20 = 8.40x

x = 3
y = 2
z = 1