ticket for a concert cost $5.00 for adults, $3.00 for children and $2.00 for senior citizens. Revenue was $500.00 for 180 tickets. 10 times more adults than children attended. How many senior tickets were sold?
a + c + s = 180
5a + 3c + 2s = 500
a = 10c
replace 'a' with '10c' in the first two equations.
This will give you two equations in two variables. You can then use the elimination/addition method to solve.
Do you need more help?
To solve this problem, we need to set up a system of equations based on the given information.
Let's say the number of adult tickets is A, the number of child tickets is C, and the number of senior tickets is S.
From the given information, we can set up the following equations:
1. The total number of tickets sold is 180:
A + C + S = 180
2. The price of an adult ticket is $5, the price of a child ticket is $3, and the price of a senior ticket is $2. The total revenue is $500:
5A + 3C + 2S = 500
3. It is given that there were 10 times more adults than children attending the concert:
A = 10C
Now, we can solve these equations simultaneously to find the values of A, C, and S.
Substitute equation (3) into equation (1) to eliminate A:
10C + C + S = 180
11C + S = 180
Now, we can express S in terms of C by rearranging the equation:
S = 180 - 11C
Substitute equation (3) into equation (2) to eliminate A:
5(10C) + 3C + 2S = 500
50C + 3C + 2S = 500
Substitute the expression for S from equation (1) into equation (2):
50C + 3C + 2(180 - 11C) = 500
50C + 3C + 360 - 22C = 500
31C = 140
C = 140 / 31 ≈ 4.5161
Since we can't have a fraction of a ticket, we need to determine the number of senior tickets sold (S) rounded down to the nearest whole number. Let's substitute the value of C into equation (1) to find S:
11C + S = 180
11(4.5161) + S = 180
49.6771 + S = 180
S = 180 - 49.6771
S ≈ 130.3229
Therefore, the number of senior tickets sold is approximately 130.
Note: In this case, we are rounding down the number of senior tickets to the nearest whole number, as it is not possible to sell a fraction of a ticket.