Question

At the city museum, child admission is $6.20 and adult admission is $9.50. On Friday, three times as many adult tickets as child tickets were sold, for a total sales of $1179.80. How many child tickets were sold that day? Number of chad tickets:

Answer 1

34 child tickets

Explanation:

let c = # child tickets

let a = # adult tickets

System of equations:

6.20c+9.50a=1179.80

a=3c

substitute 3c for a:

6.20c +9.50(3c) = 1179.80

34.7c=1179.80

c=34

Answer 2

34 child tickets were sold that day.

Explanation:

Let's use C to represent the number of child tickets sold and A to represent the number of adult tickets sold.

From the problem, we know that:

• Child admission = $6.20
• Adult admission = $9.50
• A = 3C (since three times as many adult tickets as child tickets were sold)
• Total sales = $1179.80

We can set up an equation based on the information given:

`\qquad\large\rm{6.2C + 9.5A = 1179.8}`

Substitute A with 3C:

`\qquad\large\rm{6.2C + 9.5(3C) = 1179.8}`

Simplify:

`\qquad\large\rm{6.2C + 28.5C = 1179.8}`

`\qquad\qquad\large\rm{34.7C = 1179.8}`

`\qquad\qquad\quad\large\boxed{\boxed{\bold{\:\:C \approx 34\:\:}}}`

`\therefore` 34 child tickets were sold that day.