Question

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

Answer 1

Hint: Now first we will use unitary method to find the cost if the event has 1 attendees. To do so we will equate the ratio in both cases. Now we will again use the ratio to find the number of attendees if the cost is $\$78$. . Hence we get the solution of the equation.

Complete step by step solution:
Now we are given that the cost of the event is $\$36$ if it has 18 attendees.
We want to find the maximum number of attendees if the budget is $\$78$.
Now let us first calculate the cost of the event for 1 attendee.
Let the cost of the event for 1 attendee be x. Now we are given that the cost of the event for 18 attendees is $\$36.$ Let us equate the ratio of the cost and number of attendees in both cases. Hence we get,
$\Rightarrow \dfrac{36}{18}=\dfrac{x}{1}$
Now we know that $36=18\times 2$ Hence using this we get,
$\Rightarrow \dfrac{18\times 2}{18}=\dfrac{x}{1}$
Now cancelling 18 from numerator and denominator we get,
$\Rightarrow x=2$
Hence the cost of 1 attendee is $\$2$
Now let us say there the maximum number of people that attend the event be y. We know that the budget of the event is $=\$78$.
Hence again equating the ratio we get,
$\Rightarrow \dfrac{2}{1}=\dfrac{78}{y}$
Now cross-multiplying the equation we get,
$\Rightarrow 2y=78$
Now dividing the whole equation by 3 we get,
$\Rightarrow y=39$

Hence we have the maximum number of attendees will be 39.

Note: Now note that while taking ratio we should have units in numerator or denominator of both fractions. Let us say we have written the number of attendees in numerator and the cost in denominator, then while equating it to the other fraction we will again write number of attendees in numerator and cost in denominator.