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?
To solve this problem, let's first define some variables.
Let:
x = the number of $1.50 tickets sold
y = the number of $3.50 tickets sold
Now let's set up the given information as equations.
1) The number of $1.50 tickets sold was 5 more than three times the number of $3.50 tickets sold:
x = 3y + 5
2) The total amount earned from selling these tickets is $1,607.50:
1.50x + 3.50y = 1,607.50
Now we have a system of two equations with two variables. We can solve this system using substitution or elimination method.
Let's use the substitution method to solve it.
Step 1: Solve equation 1 for x in terms of y:
x = 3y + 5
Step 2: Substitute this value of x into equation 2:
1.50(3y+5) + 3.50y = 1,607.50
Simplify and solve for y:
4.50y + 7.50 + 3.50y = 1,607.50
8y + 7.50 = 1,607.50
8y = 1,607.50 - 7.50
8y = 1,600
y = 1,600 / 8
y = 200
Step 3: Substitute the value of y back into equation 1 to find x:
x = 3(200) + 5
x = 600 + 5
x = 605
Therefore, Mrs. Rodrigues sold 605 $1.50 tickets.
Fun this question
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Let x = # of 1.50 and y = # of 3.50
x = 3y + 5
1.50x + 3.50y = 1607.50
Substitute 3y+5 for x in the second equation and solve for y. Insert that value into the first equation and solve for x. Check by putting both values into the second equation.