# The corporate team-building event will cost $36 if it has 18 attendees. How many attendees can there be, at most, if the budget for the corporate team-building event is$78?

For $78 the count of attendees is 39 #### Explanation: $\textcolor{b l u e}{\text{The cheating method - Not really a cheat!}}$Consider the attendance for$36 to be an attendance of I set
The count of sets for $78 is $\frac{78}{36}$So the count of attendees is $\frac{78}{36} \times 18 = 39$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{The long way round}}$Using ratio but in fraction format (this is NOT a fraction) Let the unknown count be $x$Initial condition:->18/($36)
Target condition:->x/($78) The ratio of the numbers in each case is the same. So we can write: $\left(\text{count")/("cost}\right) \to \frac{18}{36} \equiv \frac{x}{78}$Multiply both sides by 78 $\textcolor{g r e e n}{\frac{18}{36} \textcolor{red}{\times 78} \text{ "=" } \frac{x}{78} \textcolor{red}{\times 78}}$What follows is the process that is behind cancelling out $\textcolor{g r e e n}{39 \text{ "=" } x \times \frac{\textcolor{red}{78}}{78}}$But $\frac{78}{78} = 1$and $1 \times x = x$$\text{count } = x = 39\$