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melomori [17]
3 years ago
12

a high school play costs an up front $30,000 to produce, plus an additional cost of $1000 for each performance. every sold out p

erformance of the play brings in $2500 of revenue to the school. how many sold out performances are needed so that the school breaks even on the production of the play
Mathematics
1 answer:
beks73 [17]3 years ago
8 0
So based on the information here, it costs $3,000 up front but an additional $1,000 for each play and each sold out performance brings in $2,500

For the sake of simplicity, when solving it, I won't be putting the dollar sign

So the equation is the following:

3,000+1,000x=2,500x 

So you need to get "x" by itself

First, minus 1,000x from each side

So it is now 3,000=1,500x

Now you divide 3,000 by 1,500 which is 2

<span>x=2

</span>So that means two performances need to sell out in order to break even with the cost

-------------------
<span>
To check:
</span>
$3,000+$1,000(2)=$2,500(2)

$1,000×2=$2,000 and $3,000+$2,000=$5,000

$5,000=$2,5000(2)

$2,500×2=$5,000

$5,000=$5,000
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First, let's write what we know.

We can represent the number of students in the play from each class as L, G, and C. We know that L = 7, and if Gardener has 4 more students than Cho, then G = C + 4.

Then, taking the third line, we can write an inequality:

C < L < G

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If C is greater than 3 and less than 7 and is an integer, than means C is 4, 5, or 6.

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C + L + G

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So we have our expression for the number of students in the play. Then, we need to find the total number of students. We know that 2C + 11 will be 30% of the total, so if T is the total, we can find T.

0.3T = 2C + 11

(Divide by 3/10 or multiply by 10/3 on both sides)

T = 20/3 C + 110/3

We know from before that C is 4, 5 or 6. We can plug these into our equation here to find which one produces a whole number.

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T = 210/3

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T = 20/3 * 6 + 110/3

T = 230/3

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