The manager of a theater wants to know whether the majority of its patrons are adults or children. One day; $5400$ tickets were sold and the receipts totaled $$ 41, 640$ . The adult admission is $$ 9.50$ , and the children's admission is $$ 6.50$ . How many adult patrons were there?

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Solution

Step-1: Formation of equation from given question:
Let the number of adult be $x$ and the children be $y$
$\Rightarrow x + y = 5400$ ..(i)
The adult admission is $$ 9.50$ , and the children's admission is $$ 6.50$
$\Rightarrow 9.50 x + 6.50 y = 41640$ …(ii)
Step-2: Solving both the equation:
Multiply $6.50$ in equation (i) and then subtract with equation (ii), we get
$\Rightarrow 9.50 x + 6.50 y = 41640 \Rightarrow 6.50 x + 6.50 y = 35100 (-) (-) (-)______________________3 x = 6540 x = \frac{6540}{3} = 2180$
Hence, the number of adult patrons were $2180$ .

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The manager of a theater wants to know whether the majority of its patrons are adults or children. One day; 5400 tickets were sold and the receipts totaled $41,640. The adult admission is $9.50, and the children's admission is $6.50. How many adult patrons were there?

The manager of a theater wants to know whether the majority of its patrons are adults or children. One day; $5400$ tickets were sold and the receipts totaled $$ 41, 640$ . The adult admission is $$ 9.50$ , and the children's admission is $$ 6.50$ . How many adult patrons were there?