Question

For the school play, the advance tickets cost $3, while tickets at the door cost $5. Thirty more tickets were sold at door in advance, and $2630 was collected. How many of each kind of ticket was sold?

Answer 1

Answer:

310 tickets were sold in advance before school pay and 340 door tickets were sold in advance.

Step-by-step explanation:

Given:

Let number of tickets sold in advance be 'x'.

Cost of advance ticket = $3

Cost of door tickets = $5

Also given:

Thirty more tickets were sold at door in advance

Hence number of door tickets sold = $x+30$

Total Money Collected = $2630

Now we can say that Total Money Collected is equal to sum of number of tickets sold in advance multiplied by Cost of advance ticket and number of door tickets sold multiplied by Cost of door tickets.

Framing in equation form we get;

$3x+(x+30)5=2630$

Solving the equation to find the value of x we get;

$3x+5x+150=2630$

Combining the like terms we get;

$3x+5x=2630-150\\\\8x= 2480$

Now Dividing 8 on both side using division property we get;

$\frac{8x}{8} =\frac{2480}{8}\\ \\x=310$

Substituting the value of x to find number of door tickets been sold.

number of door tickets sold = $x+30=310+30 =340$

Hence, 310 tickets were sold in advance before school pay and 340 door tickets were sold in advance.