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A friend asked me this question earlier today, and it made me wonder how to come up with a general solution (where each variable is an integer):

I have a vocabulary test tomorrow at school. On it, there will be $W$ terms listed, and I will have to define $X$ of them. However, before the test, I am given a list of $Y$ possible terms that could be on the list given on the test. How many terms must I be able to define in order to assure that I get at least a mark of $Z$% correct on the test?

I've tinkered with some numbers a tad, and came up with some possibilities; i.e. if $W = 10$, $X = 4$, $Y = 19$, and $Z = 100$, then the individual needs to know $13$ terms. I'm having a bit of trouble, however, coming up with a generalized solution.

Any help is appreciated.

Rodrigo de Azevedo
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Cisplatin
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1 Answers1

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To guarantee a particular minimum grade, simply assume that every term you don't study will be on the test.

For the example given, if every term not studied is on the test, to get 100%, you need to find 4 terms that you've studied, no matter which 10 terms appear. Hence you can only afford to skip 6 terms, and must therefore study $19-6=13$ terms.

More details, as per request:

To achieve grade of $Z$, you must get at least $\frac{Z}{100}X$. Hence you can only afford to skip $W-\lfloor \frac{Z}{100} X\rfloor$ terms. Consequently, you must study at least

$$Y-\left(W-\bigg\lfloor \frac{Z}{100} X\bigg\rfloor\right)=Y-W+\bigg\lfloor \frac{Z}{100} X\bigg\rfloor$$

terms. Note that $\lfloor \cdot \rfloor$ denotes the floor function.

Rodrigo de Azevedo
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vadim123
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    Why the downvote? This isn't pessimism, this is a legitimate answer. – vadim123 May 18 '13 at 04:15
  • Is there any chance you could expand this into a generalized answer? For example, XY + C^D? That's where I'm having a tad of trouble. – Cisplatin May 18 '13 at 04:18
  • @Sim: I think the problem is the distinction between assure and "assure with sufficient probability". If you ask how many times I need to flip a coin to assure at least 20 heads or 20 tails I will say 39. But the probability I will fail at 38 is very small. This is a good answer to the question asked, where you said assure. – Ross Millikan May 18 '13 at 04:58