SOLUTION: Prom tickets beforehand were $31 and at the door they were $40 a total of 200 tickets were sold. Total money collected for both was 6,641.00 - How many tickets were sold at the d

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Question 1110285: Prom tickets beforehand were $31 and at the door they were $40 a total of 200 tickets were sold. Total money collected for both was 6,641.00
- How many tickets were sold at the door?
Found 3 solutions by josgarithmetic, mananth, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Standard two-variable linear system application exercise

x, before hand
y, at the door
system%28x%2By=200%2C31x%2B40y=6641%29

You can choose either substitution method or elimination method.

x=200-y
-
31%28200-y%29%2B40y=6641-------solve this for y, answers the question.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Prom tickets beforehand were $31 and at the door they were $40 a total of 200 tickets were sold. Total money collected for both was 6,641.00
- How many tickets were sold at the door?
Let beforehand tickets sold be x
door tickets sold be y
x+y = 200................1
Value equation
31x+40y = 6641...........2
Multiply (1) by 31
31x+31y=6200
31x+40y = 6641
subtract
-9y = -441
y =49
you can calculate x by plugging y

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a quick mental math way to solve problems like this....

If all 200 tickets were $40 tickets, the total would have been $8000; that is $8000-$6641 = $1359 more than the actual total.

The difference between the two ticket prices is $9, so the number of $31 tickets was 1359/9 = 151.

151 tickets were sold beforehand; 49 were sold at the door.