Adult tickets for a play cost $16 and child tickets cost $6. If there were 25 people at a performance and the theater collected $260 from ticket sales, how many adults and how many children attended the play?
By system of equations:
C=number of child tickets sold
A=number of adult tickets sold
C+A = 25
6C+16A = 260
Solve for A and C (by substitution).
Solving the problem in the head without pencils or paper or calculator:
Adult tickets cost $10 more than child tickets. If all 25 were child tickets, they would have gathered 6*25=$150.
To get $260, there must have been (260-150)/10=11 adults.
So there were 11 adults and 14 children.
To solve this problem, we can set up a system of equations. Let's define "A" as the number of adult tickets sold and "C" as the number of child tickets sold.
From the given information, we have the following two equations:
1) A + C = 25 (equation 1), since the total number of people at the performance is 25.
2) 16A + 6C = 260 (equation 2), since the total revenue from adult tickets ($16 each) and child tickets ($6 each) is $260.
To solve this system of equations, we can use substitution or elimination method. Let's use the substitution method.
From equation 1, we have A = 25 - C. Now we can substitute this value for A in equation 2:
16(25 - C) + 6C = 260
Simplifying the equation:
400 - 16C + 6C = 260
-10C = 260 - 400
-10C = -140
Dividing both sides by -10:
C = -140 / -10
C = 14
Now substitute the value of C = 14 into equation 1:
A + 14 = 25
A = 25 - 14
A = 11
Therefore, there were 11 adults and 14 children who attended the play.