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A high school drama club is selling tickets for their annual musical. There is a maximum of 600 tickets for the show. The tickets cost \$5 (if bought before the day of the show) and \$7 (if bought on the day of the show).

Let x represent the number of tickets sold before the day of the show and y represent the number of tickets sold the day of the show. To meet the expenses of the show, the club must sell at least \$3500 worth of tickets.

The club sells 330 tickets before the day of the show. Is it possible to sell enough additional tickets on the day of the show to meet the expenses of the show?

Jan 26, 2018

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330 x \$5 =\$1,650 - money raised before the day of the show

\$3,500 - \$1,650 =\$1,850 - money that must be raised on the day of the show

600 - 330 =270 tickets to be sold on the day of the show

270 x \$7 =\$1,890 money that CAN be raised on the day of the show.

\$1,890 - \$1,850 = \$40 additional money that WILL be raised if ALL the remaining 270 tickets are sold on the day of the show.

So, the answer is "YES", on the condition that you can sell at least \$1,850 / \$7 =~ 265 tickets on the day of the show.

Jan 26, 2018
#2
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From the problem statement, we can make these two inequalities:

x + y  ≤  600            and            5x + 7y  ≥  3500

If  x = 330 , then...

330 + y  ≤  600

y  ≤  270

and

5(330) + 7y  ≥  3500

1650 + 7y  ≥  3500

7y  ≥  1850

y  ≥  264 + 2/7      To sell at least  264 and 2/7  tickets, we have to sell at least 265 tickets.

y  ≥  265

Is there a possible value of  y  such that   y  ≤  270   and   y  ≥  265   ?

yes, for example,  y = 265 , or  y  =  266 , or  y = 267

Jan 26, 2018
edited by hectictar  Jan 26, 2018