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's assume that the number of children that rode the bus during the morning rush hour is x, and the number of adults is y.
So we have two equations based on the given information:
1) x + y = 1000 (since the total number of passengers is 1000)
2) 0.25x + 0.75y = 650 (since the total revenue from fares is $650)
We can solve these equations simultaneously to find the values of x and y.
First, let's solve the first equation for x in terms of y:
x = 1000 - y
Substituting this value of x into the second equation:
0.25(1000 - y) + 0.75y = 650
250 - 0.25y + 0.75y = 650
0.5y = 400
y = 400 / 0.5
y = 800
Now substituting this value of y back into the first equation:
x + 800 = 1000
x = 1000 - 800
x = 200
Therefore, there were 200 children and 800 adults who rode the bus during the morning rush hour.
To solve this problem, we can use a system of equations.
Let's assume the number of children riding the bus is 'C' and the number of adults riding the bus is 'A'.
From the given information, we have two equations:
Equation 1: C + A = 1000 (The total number of passengers is 1000)
Equation 2: 0.25C + 0.75A = 650 (The total revenue from fares is $650)
To solve the system of equations, we can use substitution or elimination method. In this case, let's use the substitution method.
From Equation 1, we can express C = 1000 - A.
Substitute this value of C into Equation 2:
0.25(1000 - A) + 0.75A = 650
Simplify the equation:
250 - 0.25A + 0.75A = 650
0.5A = 400
A = 400/0.5
A = 800
Substitute the value of A back into Equation 1 to find 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.
c + a = 1000
.25 c + .75 a = 650
solve the system