Question

In the following exercises, solve each ticket or stamp word problem.

A church luncheon made $\$842$.

Adult ticket cost $\$10$ each and children's ticket cost $\$6$ each.

The number of children is $12$ more than twice the number of adults.

How many of each ticket were sold?

Answer 1

Step-1: Find $x$:

Let $x$ be the number of adult tickets.

The number of children is $12$ more than twice the number of adults.

Multiply $2$ by $x$ and add $12$ to obtain the number of child tickets.

The number of child tickets are $2x+12$.

Adult ticket cost $\$10$ each and children's ticket cost $\$6$ each.

Multiply $10$ by $x$ and $6$ by $2x+12$ and equate it to $842$.

$$\begin{gathered} 10x+6\left(2x+12\right)=842 \\ \Rightarrow 10x+12x+72=842 \\ \Rightarrow 22x=770 \\ \Rightarrow x=\frac{770}{22}\left(\mathrm{Divide}770\mathrm{by}22\right) \\ \therefore x=35 \end{gathered}$$

Step-2: Find ticket.

Substitute $x=35$ in the expression $2x+12$, to find children's ticket number.

$\begin{array}{rcl}2x+12& =& 2\left(35\right)+12\\ & =& 70+12\\ & =& 82\end{array}$

Therefore, the number of adult's ticket is $35$ and the number of children's ticket is $82.$