A group of nine people, some adults and some children, went to the movie theater. Adult admission was $8 and child admission was $5. The amount charge was $55. Could this amount be correct?
To determine if the given amount of $55 is correct for a group of nine people, we can set up an equation based on the number of adults and children in the group.
Let's assume that there are 'x' adults in the group and 'y' children. Since there are a total of nine people, we can write the equation:
x + y = 9 ---(1)
The adult admission cost is $8, so the total cost for adults would be 8x. The child admission cost is $5, so the total cost for children would be 5y. According to the given information, the total charge was $55, so we can write the equation:
8x + 5y = 55 ---(2)
Now we have a system of two equations (1) and (2). We can solve this system to determine if the amount is correct.
Multiplying equation (1) by -5, we get:
-5x - 5y = -45 ---(3)
Adding equations (2) and (3), we eliminate the 'y' variable:
8x + 5y + (-5x - 5y) = 55 + (-45)
3x = 10
x = 10/3
Since 'x' represents the number of adults and it is not a whole number, the given amount of $55 cannot be correct for a group of nine people.
To determine if the given amount of $55 could be correct for a group of nine people with different ticket prices for adults and children, we can try solving it using algebra.
Let's assume the number of adults in the group is "x" and the number of children is "y."
The cost for adults would be $8x, and the cost for children would be $5y.
Given that the total charge was $55, we can set up the following equation:
8x + 5y = 55
Now, we can try different combinations of values for "x" and "y" to see if the equation is satisfied.
Here are a few potential combinations for "x" and "y" that add up to nine people:
1. x = 5 (5 adults) and y = 4 (4 children)
2. x = 4 (4 adults) and y = 5 (5 children)
3. x = 6 (6 adults) and y = 3 (3 children)
4. x = 3 (3 adults) and y = 6 (6 children)
By substituting these values into the equation, we can check if the total charge equals $55:
1. 8(5) + 5(4) = 40 + 20 = 60 (not equal to $55)
2. 8(4) + 5(5) = 32 + 25 = 57 (not equal to $55)
3. 8(6) + 5(3) = 48 + 15 = 63 (not equal to $55)
4. 8(3) + 5(6) = 24 + 30 = 54 (equal to $55)
Therefore, only when there are 3 adults and 6 children in the group would the total charge be $55.
a + c = 9
8 a + 5 c = 55
solve the system of equations for (a) and (c)
if the solutions are not integer values (whole numbers)
... then something is amiss with the $55 charge