I ran an event yesterday and the people working the door didn’t mark down the number of people who came in. I need that number! But I’m exhausted and can’t math this right now. Help?
I have $1305 cash. Some people paid $25 for half the event, and some paid $45 for the whole event. There are almost certainly more of the whole event people than the half people. Can I solve this? I have credit card sales, but they are broken down by individual transaction, so are easier to identify.
Possibly 19 people paid $45, and 18 paid $25.
If there are exactly the same number of whole- and half-day people, you got about 37 people: 1305/(45+25) * 2.
If you got 2 whole-day per half, you got about 34 people: 1305 / (45 + 45 + 25) * 3
So, without any more specific numbers, I’d say you got about 35 people, plus or minus a few (even if everybody paid full price, you still got 29 people).
(number of people paying full price), (number of people paying half price) is one of below:
4, 45
9, 36
14, 27
19, 18
24, 9
29, 0
To clarify, the reason I say “possibly” is that in the real world, people make mistakes counting money, making change, etc. From a purely theoretical point of view, that combination is the only combination that satisfies all of your requirements, assuming that at least one person paid $25, and there are more $45 payers than $25 payers.
Upon review, I overlooked the “24, 9” possibility that Asymptotically Fat produced. So that is also a possible answer.
You can get a range of answers, all of which fit the constraint, because you have two variables and only one equation.
Obviously, you know more than you’re explicitly letting on, because you can estimate the number of people who came to within an order of magnitude (was it closer to 10 or 100?), and you likely know it more precisely than that, which means you could use a piece of software to set up and solve multiple problems with different trial values for the equation you don’t have, which is full + half = total. I’d personally use Maxima or Octave for this kind of thing.
Six-way simulpost!
24,9 is the only reasonable answer, unless there is money missing.
The next mathematical solutions on ether side are either , approx 50-50 , and 100% -0.
and OP said something like 75%- 25% …
This sounds remarkably like a homework problem.
My first thought when I read it.
It sounds not at all like a homework problem. A homework problem would have a single answer which was sharply well-defined. The problem as presented does not.
Sneaky bastard OP probably knew that.
Which is a joke, but actually is interesting as a marker of high conceptual understanding of any problem in math.
I don’t see that in the OP. All I see is that there are “almost certainly more of the whole event people than the half people.” To me, the hedging suggests that 19 whole, 18 half is a reasonable possibility, too.
E. Cannot be calculated based on the information given
Unless the homework was for Diophantine equations.
At the school where I work the lower math class has assignments where the correct answer to a problem could be “Not enough information to solve.”