Answer :
Answer:
Total dinners served at the banquet = 42
Step-by-step explanation:
chicken:beef:fish = 7:5:2
Chicken = [tex]\frac{7}{2} Fish[/tex] = [tex]\frac{7}{5} Beef[/tex]
Beef = [tex]\frac{5}{2} Fish[/tex] = [tex]\frac{5}{7} Chicken[/tex]
Fish = [tex]\frac{2}{7} Chicken[/tex] = [tex]\frac{2}{5} Beef[/tex]
Fish > 5...........(i)
Beef + Fish < 30
[tex]\frac{5}{2} Fish[/tex] + Fish < 30
Fish < 8.57........(ii)
Beef + [tex]\frac{2}{5} Beef[/tex] < 30
Beef < 21.43.........(iii)
[tex]\frac{5}{7} Chicken[/tex] + [tex]\frac{2}{7} Chicken[/tex] < 30
Chicken < 30.........(iv)
Chicken < 25........(v)
[tex]\frac{7}{2} Fish[/tex] < 25
Fish < 7.143........(vi)
[tex]\frac{7}{5} Beef[/tex] < 25
Beef < 17.857........(vii)
From (i), (ii) and (vi)
Fish > 5; Fish < 8.57; Fish <7.143;
Therefore: 5 < Fish < 7.143
So fish is either 6 or 7
From (iii) and (vii)
Beef < 21.43; Beef < 17.857;
Therefore Beef < 17.857
If Fish = 6, Beef = 15;
If Fish = 7, Beef = 17.5
From (iv) and (v)
Chicken < 30; Chicken <25;
Therefore Chicken < 25
If Fish = 6, Chicken = 21;
If Fish = 7, Beef = 24.5
Since meals must be whole numbers, Fish = 6, Beef = 15, Chicken = 21
Total = chicken + beef + fish = 21 + 15 + 6 = 42
Answer:
42 dinners
Step-by-step explanation:
The ratio chicken:beef:fish is 7k:5k:2k
More than 5 fish dinners were served. The smallest whole number greater than 5 is 6. At least 6 fish dinners were served.
The ratio of the dinners is 7k:5k:2k
2k >= 6
k >= 3
Using k = 3, we get:
21:15:6, and 21 + 15 + 6 = 42 total meals
Then beef + fish = 21 + 6 = 27 which is less than 30.
15 is less than 25
Both conditions are met.
Using k = 4, we get:
28:20:8, and 28 + 20 + 8 = 56 total meals
Then beef + fish = 28 + 8 = 36 which is greater than 30.
The first condition is not met, so this cannot be the answer.
Answer: 42 dinners