a cashier collected $312 from the sale of 50 tickets. An adult ticket costs $8, a child ticket costs $4. How many tickets of each type did the cashier sell?
whew :)
I never claimed to make sense, but can do arithmetic.
8 a + 4 c = 312
1 a + 1 c = 50
==================
2 a + 1 c = 78
1 a + 1 c = 50
--------------------- subtract
a + 0 = 28
etc
wait, it doesn't make sense, can you explain?
actually, solved it
I dont understand
To find the number of adult tickets and child tickets sold, we can set up a system of equations based on the information given.
Let's say the number of adult tickets sold is x, and the number of child tickets sold is y.
From the problem, we know that the total number of tickets sold is 50:
x + y = 50
We also know that the total amount collected from the sales is $312:
8x + 4y = 312
To solve this system of equations, we can use the method of substitution or elimination.
Method 1: Substitution
1. Solve one equation for one variable in terms of the other. Let's solve the first equation for x:
x = 50 - y
2. Substitute this expression for x in the second equation:
8(50 - y) + 4y = 312
3. Simplify and solve for y:
400 - 8y + 4y = 312
-4y = -88
y = -88/-4
y = 22
4. Substitute this value of y back into the first equation to find x:
x + 22 = 50
x = 50 - 22
x = 28
Therefore, the cashier sold 28 adult tickets and 22 child tickets.
Method 2: Elimination
1. Multiply the first equation by 4 to make the coefficients of y in both equations the same:
4x + 4y = 200
2. Subtract this equation from the second equation to eliminate y:
8x + 4y - (4x + 4y) = 312 - 200
8x + 4y - 4x - 4y = 112
4x = 112
x = 112/4
x = 28
3. Substitute this value of x back into the first equation to find y:
28 + y = 50
y = 50 - 28
y = 22
Once again, we find that the cashier sold 28 adult tickets and 22 child tickets.