. A total of 1000 passengers rode a bus during the morning rush hour. A child’s fare is $0.25
and the adult fare is $0.75. The total revenue from the fares during the morning rush hour
was $650. How many children and how many adults rode the bus?
Let x be the number of children and y be the number of adults.
We can set up a system of equations:
x + y = 1000 (since there were 1000 passengers in total)
0.25x + 0.75y = 650 (since the total revenue from the fares was $650)
To solve for x and y, we can use substitution or elimination.
Using substitution, we can solve for x in the first equation and substitute into the second equation:
x = 1000 - y
0.25(1000 - y) + 0.75y = 650
250 - 0.25y + 0.75y = 650
0.5y = 400
y = 800
Then, we can substitute y = 800 into the first equation to solve for x:
x + 800 = 1000
x = 200
Therefore, there were 200 children and 800 adults who rode the bus during the morning rush hour.
Let's assume the number of children riding the bus is "c" and the number of adults riding the bus is "a".
We can start by setting up a system of equations to represent the given information:
Equation 1: c + a = 1000 (total number of passengers)
Equation 2: 0.25c + 0.75a = 650 (total revenue from the fares)
To solve this system of equations, we can use the method of substitution.
From Equation 1, we can rewrite it as c = 1000 - a.
Substituting this value into Equation 2, we have:
0.25(1000 - a) + 0.75a = 650
Simplifying the equation, we get:
250 - 0.25a + 0.75a = 650
0.5a = 400
a = 400 / 0.5
a = 800
Now, we can substitute this value back into Equation 1 to find the value of c:
c + 800 = 1000
c = 1000 - 800
c = 200
Therefore, there were 200 children and 800 adults who rode the bus during the morning rush hour.