The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended? Need help asap and broken down so I'll understand. Much appreciated

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27-10-2014
Mathematics

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The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended? Need help asap and broken down so I'll understand. Much appreciated

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MonaDee MonaDee · Answer 1 · 2014-10-27T01:01:58-04:00

To solve this problem, you'll need to create a system of equations and then solve for the variables. For my two equations, x = # of children and y = # of adults

We know from the problem that the total amount of people who attended the fair is 2200, so that tells us that the number of children (x) plus the number of adults (y) will give us 2200. So, that's our first equation.

x + y = 2200

We also know that the total amount of money collected was $5050. This tells us that the number of children's tickets sold (1.50 * x) plus the number of adult's tickets sold (4 * y) will give us $5050. So, that's our second equation.

1.50x + 4y = 5050

Now, you take both equations and solve for the variables.
x + y = 2200
1.50x + 4y = 5050

x + y = 2200
x = -y + 2200

1.50(-y + 2200) + 4y = 5050
-1.50y + 3300 +4y = 5050
3300 + 2.5y = 5050
2.5y = 1750
y = 700

Now that you know that y =700, you plug that information into either of the two equations and solve for x. I'm going to use the first equation because it's easier.
x + y = 2200
x + 700 = 2200
x = 1500

So, x = 1500 and y = 700

sunshinet1983 sunshinet1983 · Answer 2 · 2022-05-24T18:51:43-04:00

sunshinet1983 sunshinet1983

24-05-2022

Answer:

c + a = 2200

1.50c + 4a = 5050

Step-by-step explanation:

6.12 Quiz: Applications - Systems of Equations

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