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?

1 Answer
May 21, 2017

Answer:

Using a sort of cheat method!

For $78 the count of attendees is 39

Explanation:

#color(blue)("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 #78/36#

So the count of attendees is #78/36xx18=39#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("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:

#("count")/("cost")->18/(36)-=x/78#

Multiply both sides by 78

#color(green)(18/36color(red)(xx78)" "=" "x/78color(red)(xx78))#

What follows is the process that is behind cancelling out

#color(green)(39" "=" "x xx(color(red)(78))/78)#

But #78/78=1# and #1xx x=x#

#"count "=x=39#