AIME 2004 I · 第 5 题

Solution

Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he gets right on day 2.

These inequalities follow:

$$\frac{a}{q} < \frac{160}{300} = \frac{8}{15}$$ $$\frac{b}{500-q} < \frac{140}{200} = \frac{7}{10}$$ Solving for a and b and adding the two inequalities:

$$a + b < \frac{8}{15}q + (350 - \frac{7}{10}q)$$ $$a + b < 350 - \frac{1}{6}q$$ From here, we see the largest possible value of $a+b$ is $349$.

Checking our conditions, we know that $a$ must be positive so therefore $q$ must be positive. A quick check shows that for $2\le q \le 5$, $q$ follows all the conditions and results in $a+b=349$.

This makes Beta's success ratio $\frac{349}{500}$. Thus, the answer is $m+n = 349 + 500 = \boxed{849}$.

Simple Solution

We see that we want Beta to have more points where there is a higher Alpha success rate (that way, the score is shifted more towards the higher Alpha score). With that in mind, $.7 > 8/15$. Thus, we might as well put as many points as possible into the second day.

That said, we need Beta to have a positive integer score on either end per requirements. The best way is for Beta to go $\frac{1}{2}$ on the first day. That way, we "waste" the least points- both in the numerator and denominator.

Thereafter, we see that, letting $x$ be the number of points Beta gained on the second day, $\frac{x}{498} < \frac{7}{10}$; thus $\max(x) = 348$.

Aha. $348+1+500= \boxed{849}$.

~minor $\LaTeX$ edit by Yiyj1