# Help with math fail **URL:** \ **Category:** Factual Questions\ **Created:** [January 10, 2016, 10:53pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749 "2016-01-10T22:53:22Z")\ **Posts on this page:** 16\ **Page:** 1
**Author:** ![CastleAnthrax](https://avatars.discourse-cdn.com/v4/letter/c/ecd19e/32.png) [@CastleAnthrax](https://boards.straightdope.com/u/CastleAnthrax)\ **Post date:** [January 10, 2016, 10:53pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/1 "2016-01-10T22:53:22Z")
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. ---
**Author:** ![wellanuff](https://avatars.discourse-cdn.com/v4/letter/w/57b2e6/32.png) [@wellanuff](https://boards.straightdope.com/u/wellanuff)\ **Post date:** [January 10, 2016, 11:06pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/2 "2016-01-10T23:06:00Z")
Possibly 19 people paid $45, and 18 paid $25. ---
**Author:** ![Ethilrist](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ethilrist/32/4604_2.png) [@Ethilrist](https://boards.straightdope.com/u/Ethilrist)\ **Post date:** [January 10, 2016, 11:06pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/3 "2016-01-10T23:06:39Z")
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). ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [January 10, 2016, 11:08pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/4 "2016-01-10T23:08:18Z")
> [@CastleAnthrax](#): > > 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. (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 ---
**Author:** ![wellanuff](https://avatars.discourse-cdn.com/v4/letter/w/57b2e6/32.png) [@wellanuff](https://boards.straightdope.com/u/wellanuff)\ **Post date:** [January 10, 2016, 11:08pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/5 "2016-01-10T23:08:56Z")
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. ---
**Author:** ![Derleth](https://avatars.discourse-cdn.com/v4/letter/d/b9e5f3/32.png) [@Derleth](https://boards.straightdope.com/u/Derleth)\ **Post date:** [January 10, 2016, 11:11pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/6 "2016-01-10T23:11:23Z")
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](http://maxima.sourceforge.net/) or [Octave](https://www.gnu.org/software/octave/) for this kind of thing. ---
**Author:** ![Derleth](https://avatars.discourse-cdn.com/v4/letter/d/b9e5f3/32.png) [@Derleth](https://boards.straightdope.com/u/Derleth)\ **Post date:** [January 10, 2016, 11:12pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/7 "2016-01-10T23:12:29Z")
Six-way simulpost! ---
**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\ **Post date:** [January 11, 2016, 6:21am UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/8 "2016-01-11T06:21:35Z")
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% … ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [January 11, 2016, 6:05pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/9 "2016-01-11T18:05:20Z")
This sounds remarkably like a homework problem. ---
**Author:** ![I.Dunno](https://avatars.discourse-cdn.com/v4/letter/i/9f8e36/32.png) [@I.Dunno](https://boards.straightdope.com/u/I.Dunno)\ **Post date:** [January 11, 2016, 6:06pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/10 "2016-01-11T18:06:52Z")
> [@ftg](#): > > This sounds remarkably like a homework problem. My first thought when I read it. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [January 11, 2016, 7:09pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/11 "2016-01-11T19:09:32Z")
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. ---
**Author:** ![Leo\_Bloom](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/leo_bloom/32/10377_2.png) [@Leo\_Bloom](https://boards.straightdope.com/u/Leo_Bloom)\ **Post date:** [January 11, 2016, 7:43pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/12 "2016-01-11T19:43:35Z")
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. ---
**Author:** ![pulykamell](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/pulykamell/32/3166_2.png) [@pulykamell](https://boards.straightdope.com/u/pulykamell)\ **Post date:** [January 11, 2016, 7:50pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/13 "2016-01-11T19:50:54Z")
> [@Isilder](#): > > 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% … 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. ---
**Author:** ![gigi](https://avatars.discourse-cdn.com/v4/letter/g/a587f6/32.png) [@gigi](https://boards.straightdope.com/u/gigi)\ **Post date:** [January 11, 2016, 10:54pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/14 "2016-01-11T22:54:18Z")
> [@Chronos](#): > > 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. E. Cannot be calculated based on the information given ---
**Author:** ![rowrrbazzle](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rowrrbazzle/32/414_2.png) [@rowrrbazzle](https://boards.straightdope.com/u/rowrrbazzle)\ **Post date:** [January 12, 2016, 3:44am UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/15 "2016-01-12T03:44:01Z")
> [@Chronos](#): > > 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. Unless the homework was for Diophantine equations. ---
**Author:** ![Doug\_K](https://avatars.discourse-cdn.com/v4/letter/d/b5ac83/32.png) [@Doug\_K](https://boards.straightdope.com/u/Doug_K)\ **Post date:** [January 12, 2016, 1:08pm UTC](https://boards.straightdope.com/t/help-with-math-fail/742749/16 "2016-01-12T13:08:40Z")
> [@rowrrbazzle](#): > > 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.”