2004 AIME I Problems/Problem 5

Problem

Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

Solution

Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number it gets right. Let $b$ be the number it 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 $q=6$ follows all the conditions and results in $a+b=349$ .

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

2004 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2004 AIME I Problems/Problem 5

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