A problem that is similar to the examples talked about are St Marks Community babraque served 250 dinners. A childs plate cost $3.50 and an adults plate cost $7.00. A total of $1347.50 was collected . How many of each plate was served?
x = number of child's plates
y = number of adult
x + y = 250
3.50x + 7.00y = 1347.50
solve these two equations together and solve for x and y
2.50
To solve this problem, we need to set up equations based on the given information.
Let's assume the number of child's plates served is "C" and the number of adult's plates served is "A."
From the problem, we know that a child's plate costs $3.50, and an adult's plate costs $7.00. We also know that a total of 250 dinners were served, and the total amount collected is $1347.50.
So, we can set up two equations based on this information:
1. The number of child's plates served plus the number of adult's plates served equals the total number of dinners served:
C + A = 250
2. The total amount collected from child's plates plus the total amount collected from adult's plates equals the total amount collected:
3.50C + 7.00A = 1347.50
Now we have a system of equations. We can solve this system of equations using substitution or elimination method.
Let's solve this system of equations using substitution method:
From equation 1, we can express the value of C in terms of A:
C = 250 - A
Substituting this value of C into equation 2:
3.50(250 - A) + 7.00A = 1347.50
Simplifying the equation:
875 - 3.50A + 7.00A = 1347.50
3.50A + 7.00A = 1347.50 - 875
10.50A = 472.50
Dividing both sides by 10.50:
A = 472.50 / 10.50
A = 45
Now, substitute this value of A back into equation 1 to find the value of C:
C + 45 = 250
C = 250 - 45
C = 205
Therefore, 205 child's plates and 45 adult's plates were served.