Question 1119157: A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3357 tickets overall. It has sold 157 more $20 tickets than $10 tickets. The total sales are $64 comma 64,640. How many tickets of each kind have been sold?
# $10 tickets sold
# $20 tickets sold
# $30 tickets sold
Found 3 solutions by ankor@dixie-net.com, ikleyn, josgarithmetic:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! let a = no. of $10 ticket
let b = no. of $20 tickets
let c = no. $30 tickets
:
Write an equation for each statement
:
A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30.
The total sales are 64,640.
10a + 20b + 30c = 64640
simplify divide by 10
a + 2b + 3c = 6464
:
The team has sold 3357 tickets overall.
a + b + c = 3357
:
It has sold 157 more $20 tickets than $10 tickets.
b = a + 157
-a + b = 157
:
How many tickets of each kind have been sold?
:
We solve this as a matrix using that feature on a ti83 or similar
a + 2b + 3c = 6464
a + b + c = 3357
-a + b + 0 = 157
enter
1, 2, 3, 6464
1, 1, 1, 3357
-1,1 0 = 157
Resulting in
a = 1150
b = 1307
c = 900
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
This problem is for ONE unknown.
Let x = #of $10 tickets.
Then the number of $20 tickets is (x+157).
Then the number of $30 tickets is the rest 3357 - x - (x+57) = 3300-2x.
The money equation is
10x + 20*(x+57) + 30*(3300-2x) = 64640.
Simplify and solve for x.
The lesson to learn from this solution:
This problem is for ONE unknown - not for three.
This problem is for ONE unknown - not for three.
This problem is for ONE unknown - not for three.
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To see many other similar solved problems, look into the lesson
- Advanced word problems to solve using a single linear equation
in this site.
Answer by josgarithmetic(39617)  (Show Source):
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