Question

The admission fee at an amusement park is $1.75 for children and $6.20 for adults. On a certainday, 303 people entered the park, and the admission fees collected totaled $1220. How manychildren and how many adults were admitted?number of children equalsnumber of adults equals

Answer 1

T solve the question, we will make a system of equations

Let there are x children and y adults in that day

Since there are 303 people on that day, then

Add x and y, then equate the sum by 303

`x+y=303\rightarrow(1)`

Since the admission fee for children is $1.75

Since the admission fee for adults is $6.20

Since the park collected $1220 from admissions, then

Multiply x by 1.75 and y by 6.20, then add the products and equate the sum by 1220

`1.75x+6.20y=1220\rightarrow(2)`

Now, we have a system of equations to solve it

Multiply equation (1) by -6.20 to make the coefficients of y equal in values and different in signs

`\begin{gathered} (-6.20)(x)+(-6.20)(y)=(-6.20)(303) \\ -6.20x-6.20y=-1878.60\rightarrow(3) \end{gathered}`

Add equations (2) and (3) to eliminate y

`\begin{gathered} (1.75x-6.20x)+(6.20y-6.20y)=(1220-1878.60) \\ -4.45x=-658.6 \end{gathered}`

Divide both sides by -4.45

`\begin{gathered} (-4.45x)/(-4.45)=(-658.6)/(-4.45) \\ x=148 \end{gathered}`

Substitute x in equation (1) by 148 to find y

`148+y=303`

Subtract 148 from both sides

`\begin{gathered} 148-148+y=303-148 \\ y=155 \end{gathered}`

There were 148 children and 155 adults