Read and Understand: We want to find the number of child and adult tickets.  We know the total number of tickets sold, the total revenue and the cost of a child and adult ticket.

Define and Translate: Let c = the number of children and a = the number of adults in attendance.  Revenue comes from number of tickets sold multiplied by the price of the ticket.  We will get revenue for adults by multiplying [latex]$50.00[/latex] times a.  [latex]$25.00[/latex] times c will give the revenue from the number of child tickets sold.

Write and Solve: We can use a table as we did in the mixture problems section to organize the information we have.  Although a table is not necessary, it can help you get started.  For this problem, we labeled columns as amount, value, and total revenue because that is the information we are given.

The total number of people is [latex]2,000[/latex].

The total revenue is [latex]$70,000[/latex]. We can use this and the revenue from child and adult tickets to write an equation for the revenue.[latex]25c+50a=70,000[/latex]

The number of people at the game that day is the total number of child tickets sold plus the total number of adult tickets, [latex]c+a=2,000[/latex]

We now have a system of linear equations in two variables.[latex]\begin{array}{r}c+a=2,000\,\,\,\\ 25c+50a=70,000\end{array}[/latex].

We can use any method of solving systems of equations to solve this system for a and c.  Substitution looks easiest because we can  solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].

[latex]\begin{array}{c}c+a=2,000\\ a=2,000-c\end{array}[/latex]

Substitute the expression [latex]2,000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].

[latex]\begin{array}{r} 25c+50\left(2,000-c\right)=70,000\,\,\,\, \\ 25c+100,000 - 50c=70,000\,\,\,\, \\ -25c=-30,000 \\ c=1,200\,\,\,\,\,\,\, \end{array}[/latex]

Substitute [latex]c=1,200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{array}{r}1,200+a=2,000 \\ a=800\,\,\,\,\,\, \end{array}[/latex]

Answer

We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the game that day. The marketing group may want to focus their advertising toward attracting young people.

Read and Understand: We want to find the number of quarters and the number of dimes in the jar.  We know that dimes are [latex]$0.10[/latex] and quarters are [latex]$0.25[/latex], and the total number of coins is [latex]11[/latex].

Define and Translate: We will call the number of quarters q and the number of dimes d. The part of the total [latex]$1.85[/latex] that comes from quarters will be determined by how many quarters and the fact that each one is worth [latex]$0.25[/latex], so [latex]$0.25q[/latex] represents the amount of [latex]$1.85[/latex] that is quarters.  The same idea can be used for dimes, so [latex]$0[/latex].10d represents the amount of [latex]$1.85[/latex] that is dimes.

Write and Solve: We can label a new table with the information we are given.

We can write our two equations, remember that we need two to solve for two unknowns.

 [latex]\begin{array}{r}q+d=11\,\,\,\\0.25q+0.10d=1.85\end{array}[/latex]

Substitution looks like the easiest path to a solution, solve for q.

[latex]\begin{array}{c}q+d=11\\ q=11-d\end{array}[/latex]

Substitute this into the other equation, and solve for d.

[latex]\begin{array}{r}0.25\left(11-d\right)+0.10d=1.85\,\,\,\, \\2.75-0.25d+0.10d=1.85\,\,\,\,\\ 2.75-0.15d=1.85\\-0.15d=-0.9\\\,\,\,\,\,\,\, d=6\end{array}[/latex]

Substitute [latex]d=6[/latex] into the first equation to solve for [latex]q[/latex].

[latex]\begin{array}{r}q+6=11 \\q=5\,\,\,\,\,\, \end{array}[/latex]

Answer

We have [latex]6[/latex] dimes and [latex]5[/latex] quarters.