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3 years ago

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Tickets for a certain show cost $17, $21, or, for VIP seats, $40. If ten times as many $17 tickets were sold as VIP tic

kets, and the number of $17 tickets sold was 57 more than the sum of the number of $21 tickets and VIP tickets, sales of all three kinds of tickets would total $51 comma 471. How many of each kind of ticket would have been sold?

1 answer:

lawyer [7]3 years ago

8 0

Answer:

Tickets sold:

VIP $=126$

$17 tickets $=1,260$

$21 tickets $=9\cdot 126+57=1,191$

Step-by-step explanation:

Let x be the number of VIP tickets.

If ten times as many $17 tickets were sold as VIP tickets, then the number of $17 tickets is $10x.$

If the number of $17 tickets sold was 57 more than the sum of the number of $21 tickets and VIP tickets, then $10x+57=x+y$ and the number y of $21 tickets is $9x+57.$

Amounts earned:

VIP tickets $=\$40x$

$17 tickets $=\$17\cdot 10x=\$170x$

$21 tickets $=\$21\cdot (9x+57)=\$(189x+1,197)$

Total $=\$(40x+170x+189x+1,197)=\$(399x+1,197)$

The sales of all three kinds of tickets would total $51,471, so

$399x+1,197=51,471\\ \\399x=51,471-1,197\\ \\399x=50,274\\ \\x=126$

Tickets sold:

VIP $=126$

$17 tickets $=1,260$

$21 tickets $=9\cdot 126+57=1,191$

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