A jazz concert brought in $207,000 on the sale of 8,100 tickets. If the tickets sold for $20 and $30 each, how many of each type of ticket were sold?
If you can help me set the question up, I can do the math. I just need to see an example similar to this
you need to write two equations. let's call the ticket types C (cheap) and E (expensive). C tickets sell for $20. E tickets sell for $30. we know that C + E = the number of tickets sold we also know that 20*C + 30*E = total ticket revenue That's two equations in two unknowns, and you can solve that pretty easily by elimination or substitution. What do you get?
you could even do this with only one variable. Let C be the number of cheap tickets sold. As we know that 8100 tickets are sold in total, we sell C cheap tickets and 8100-C expensive tickets, right? Then our revenue is 20*C — C cheap tickets @ $20 each 30*(8100-C) — 8100-C expensive tickets @ $30 each ------------ 207,000 or 20C + 30(8100-C) = 207,000 solve that for C. once you have C, you can find the number of expensive tickets with 8100-C.
Thankee-sai! My professor uses MyMathLab and the online text was a wee bit confusing. This helps a lot.
do you have an answer? I'll check it for you
Give me a minute to work this out with pen and pencil
with a little practice you'll be able to do it in your head :-)
That's my goal. To (a) do this without a calculator and (b) to see the answer by looking at the question
as soon as you give me the answer I'll tell you another way you could do it in your head
c=3600
if i did the math correctly
yep! how many expensive tickets?
8100 - 3600 cheap = 4500 expensive
or did i get that backwards with cheap/expensive?
my logic says more cheap than expensive will sell, therefore I need to switch the labels
well, let's check the answer: 3600 * 20 + 4500 * 30 = 7200 + 135000 = 207000 3600 + 4500 = 8100 both equations give correct answers with the proposed solution, so it is correct.
let's try it with them switched: 4500*20 + 3600*30 = 90000 + 108000 = 198000 oops 4500 + 3600 = 8100 notice how one equation still gave a valid answer (the second one here) but the other did not. you have to verify the solution in all equations to be sure that it is truly correct.
now, I said I'd give you a quick mental route to computing this: assume that all of the tickets that sell are the cheap ones. how much money comes in? 20*8100 = 162000. Well, we know that they actually raked in 207000, and we are 207000-162000 = 45000 short. Each expensive ticket makes 30-20 = 10 more than a cheap ticket, so if we trade some of those cheap tickets for expensive tickets, we can boost the revenue up to the 207000. How many do we need to switch? That's easy: we have a 45000 shortfall, and each expensive ticket sold instead of a cheap ticket brings in an extra 10 bucks, so 45000/10 = 4500 expensive tickets. took longer to type than it did to do!
indeed ;)
of course you could assume they were all expensive tickets, 8100*30 = 243000, that's 243000-207000 = 36000 over the revenue quota. each ticket traded down to a cheap ticket cuts the revenue by 10 bucks, 36000/10 = 3600 cheap tickets instead of expensive.
yeah and I always reduce the number of trailing zeroes when I can while working the problem, then reinserting them back in at the end. same result, different path
well, I think this horse, if not completely dead, is at least down on the ground, whimpering in pain :-)
either way, many thanks! i appreciate you helping me think it through rather than just giving me the answer. Give a man a fish, he will eat for a day. Teach a man to fish and he will eat for a lifetime
well, he'll come up with a number of reasons to go fishing, and buy expensive gadgets, at least :-)
RIGHT?!
have a good one, time to go make breakfast!
blessed be
Join our real-time social learning platform and learn together with your friends!