<p>for a concert tickets that were purchased in advance of the day of the concert cost 5 dollars each and tickets purchased the day of the concert cost 8 dollars each. the total amount collected in ticket sales was the same as if every ticket purchased had cost 5.5 dollars. If 100 tickets were purchased in advance, what was the total number of tickets purchased.</p>
<p>How do you solve this algebraically?</p>
<p>Let a be the number of advance tickets purchased, and let d be the number of tickets purchased the day of the concert.</p>
<p>5a + 8d = 5.5(a + d) and a = 100.</p>
<p>So, </p>
<p>5(100) + 8d = 5.5(100 + d)
500 + 8d = 550 + 5.5d
2.5d = 50
d = 50/2.5 = 20</p>
<p>So the total number of tickets purchased was 100 + 20 = 120.</p>
<p>Where:
b is the number of tickets purchased before the day of the concert;
d is the number of tickets purchased the day of the concert</p>
<p>(5b + 8d)/(b+d) = 5.5
b = 100 Substitute 100 for b</p>
<p>(5*100 + 8d)/(100 + d) = 5.5
(500 + 8d)/(100 + d) = 5.5 Multiply by (100 + d) on both sides</p>
<p>500 + 8d = 550 + 5.5d
2.5d = 50<br>
d = 20</p>
<p>Therefore b + d equals 100 + 20, which equals 120.</p>
<p>Edit: One minute late lol</p>
<p>Or, with no algebra at all:</p>
<p>The average price was $5.5. Each $8 ticket is $2.50 above the average. Each $5 ticket is .50 below the average. So each $8 ticket provides a surplus that is enough to balance the deficit of 5 separate $5 tickets. Since 100 of those $5 tickets were sold, you must also have sold 100/5 = 20 of the $8 tickets for a total of 120 tickets sold.</p>
<p>Many SAT average problems become easier when you think of surpluses balancing deficits. I like to envision piles of rocks, initially all the same height – the given average – then move the rocks around until it fits the problem.</p>