The Bishop's Peak 4-H club is having its annual fundraising dinner. Adults pay $35 apiece and children pay $20 apiece. If the number of adult tickets sold is twice the number of children's tickets sold, and the total income for the dinner is $1,890, how many of each kind of ticket did the 4-H club sell?
to find the # of children's tickets, you just solve
35(2c) + 20c = 1890
then double that for the adults.
kids tickets ---- x
adult tix's ----- 2x
solve for x:
20x + 35(2x) = 1890
To solve this problem, we can start by assigning variables to represent the unknown quantities. Let's assume the number of children's tickets sold is "x". Therefore, the number of adult tickets sold would be twice that, which is "2x".
Next, we can determine the revenue from the children's tickets. Since each child ticket costs $20 and "x" children's tickets were sold, the revenue from the children's tickets would be 20x.
Similarly, the revenue from the adult tickets can be calculated. Each adult ticket costs $35, and "2x" adult tickets were sold, so the revenue from the adult tickets would be 35 * 2x, which simplifies to 70x.
According to the problem, the total income for the dinner is $1,890. So we can write the equation:
20x + 70x = 1,890
Combining like terms gives us:
90x = 1,890
Next, we want to isolate "x", so we divide both sides of the equation by 90:
x = 1,890 / 90
Simplifying the right side gives us:
x = 21
Therefore, the number of children's tickets sold is 21. Since the number of adult tickets sold is twice that, the number of adult tickets sold would be:
2x = 2 * 21 = 42
So, the 4-H club sold 21 children's tickets and 42 adult tickets.