There were 610 people at a play. The admission price was $3.00 for adults and $1.00 for children. The admission receipts were $1,250.00. How many adults and children attended?
a + c = 610, therefore c = 610 - a
3a + c = 1250
Substitute 610-a for c in second equation and solve for a. Insert that value into the first equation and solve for c. Check by inserting both values into the second equation.
290 children---- 290X 1.00= 290
320 adults ----- 320X 3.oo= 960
therefore 290+ 960= 1250
To solve this problem, we can use a system of equations.
Let's assume that the number of adults is represented by 'A' and the number of children is represented by 'C'.
From the given information, we know two things:
1. The total number of people who attended the play is 610. So, we have the equation: A + C = 610.
2. The total admission receipts were $1,250.00. The price for each adult ticket is $3.00, and the price for each child ticket is $1.00. So, we have the equation: 3A + 1C = 1250.
Now we can solve this system of equations using substitution or elimination.
Let's solve by substitution:
From equation 1, we have A = 610 - C.
Substitute this value for A in equation 2 to get:
3(610 - C) + C = 1250.
Simplify the equation:
1830 - 3C + C = 1250.
Combine like terms:
1830 - 2C = 1250.
Subtract 1830 from both sides:
-2C = -580.
Divide both sides by -2:
C = 290.
Now substitute this value for C in equation 1 to find the number of adults:
A + 290 = 610.
A = 610 - 290.
A = 320.
Therefore, there were 320 adults and 290 children who attended the play.