Question

Each of the dinners served at a banquet was either chicken or beef or fish. The ratio of the number of chicken dinners to the number of beef dinners to the number of fish dinners served at the banquet was 7:5:2, respectively. If there were more than 5 fish dinners served at the banquet, what was the total number of dinners served at the banquet?

Answer 1

Answer:

Min 42 dinners

Step-by-step explanation:

The ratio is the following:

C : 7

B : 5

F : 2

That means that every 2 fish dinners, they served 5 beefs and 7 chickens.

If they served more than 5 fish dinners, that means at least 6 fish dinners.

And by proportion,

$\frac{F}{B} = \frac{2}{5} =\frac{6}{X}$

X= (6*5)/2= 15 BEEF

$\frac{B}{C} = \frac{5}{7} =\frac{15}{X}$

X= (15*7)/5= 21 CHICKEN

ADDING: 21+15+6= 42

So the minimum amount of dinners served was 42.