Mrs. Rodrigues was selling two kinds of raffle tickets for $1.50 and $3.50. The number of $1.50 tickets sold was 5 more than three times the number of $3.50 tickets sold. She earned a total of $1,607.50. How many $1.50 tickets were sold?
expensive tickets --- x
cheap tickets ------ 3x+5
3.5x + 1.5(3x+5) = 1607.5
solve for x
To solve this problem, we can set up a system of equations.
Let's say x represents the number of $1.50 tickets sold and y represents the number of $3.50 tickets sold.
From the problem, we know that the number of $1.50 tickets sold was 5 more than three times the number of $3.50 tickets sold. This can be written as:
x = 3y + 5 (equation 1)
We also know that Mrs. Rodrigues earned a total of $1,607.50 from selling the tickets. The total amount earned can be calculated by multiplying the price of each ticket by the number of tickets sold and adding them together:
1.50x + 3.50y = 1607.50 (equation 2)
Now, we have a system of two equations:
x = 3y + 5
1.50x + 3.50y = 1607.50
We can solve this system of equations using substitution or elimination.
Let's use substitution method:
From equation 1, we can express x in terms of y:
x = 3y + 5
Substitute this value of x into equation 2:
1.50(3y + 5) + 3.50y = 1607.50
Distribute the 1.50:
4.50y + 7.50 + 3.50y = 1607.50
Combine like terms:
8y + 7.50 = 1607.50
Subtract 7.50 from both sides:
8y = 1600
Divide both sides by 8:
y = 200
Now, we have found the value of y, which represents the number of $3.50 tickets sold.
To find the number of $1.50 tickets sold (x), substitute the value of y into equation 1:
x = 3(200) + 5
x = 600 + 5
x = 605
Therefore, Mrs. Rodrigues sold 605 $1.50 tickets.