Difference between revisions of "2004 AIME I Problems/Problem 5"
m (→Simple Solution) |
|||
(8 intermediate revisions by 6 users not shown) | |||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | Let <math>q</math> be the number of questions Beta takes on day 1 and <math>a</math> be the number | + | Let <math>q</math> be the number of questions Beta takes on day 1 and <math>a</math> be the number he gets right. Let <math>b</math> be the number he gets right on day 2. |
These inequalities follow: | These inequalities follow: | ||
Line 20: | Line 20: | ||
From here, we see the largest possible value of <math>a+b</math> is <math>349</math>. | From here, we see the largest possible value of <math>a+b</math> is <math>349</math>. | ||
− | Checking our conditions, we know that a must be positive so therefore q must be positive. A quick check shows that <math>q | + | Checking our conditions, we know that <math>a</math> must be positive so therefore <math>q</math> must be positive. A quick check shows that for <math>2\le q \le 5</math>, <math>q</math> follows all the conditions and results in <math>a+b=349</math>. |
This makes Beta's success ratio <math>\frac{349}{500}</math>. Thus, the answer is <math>m+n = 349 + 500 = \boxed{849}</math>. | This makes Beta's success ratio <math>\frac{349}{500}</math>. Thus, the answer is <math>m+n = 349 + 500 = \boxed{849}</math>. | ||
+ | |||
+ | ==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, <math>.7 > 8/15</math>. 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 <math>\frac{1}{2}</math> on the first day. That way, we "waste" the least points- both in the numerator and denominator. | ||
+ | |||
+ | Thereafter, we see that, letting <math>x</math> be the number of points Beta gained on the second day, <math>\frac{x}{498} < \frac{7}{10}</math>; thus <math>\max(x) = 348</math>. | ||
+ | |||
+ | Aha. <math>348+1+500= \framebox{849}</math>. | ||
+ | |||
+ | ~minor <math>\LaTeX</math> edit by Yiyj1 | ||
== See also == | == See also == | ||
{{AIME box|year=2004|n=I|num-b=4|num-a=6}} | {{AIME box|year=2004|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:31, 9 January 2024
Contents
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 where
and
are relatively prime positive integers. What is
?
Solution
Let be the number of questions Beta takes on day 1 and
be the number he gets right. Let
be the number he gets right on day 2.
These inequalities follow:
Solving for a and b and adding the two inequalities:
From here, we see the largest possible value of
is
.
Checking our conditions, we know that must be positive so therefore
must be positive. A quick check shows that for
,
follows all the conditions and results in
.
This makes Beta's success ratio . Thus, the answer is
.
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, . 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 on the first day. That way, we "waste" the least points- both in the numerator and denominator.
Thereafter, we see that, letting be the number of points Beta gained on the second day,
; thus
.
Aha. .
~minor edit by Yiyj1
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.