Difference between revisions of "2001 AIME I Problems/Problem 2"
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Let x be the mean of S. Let a be the number of elements in S. | Let x be the mean of S. Let a be the number of elements in S. | ||
Then, | Then, | ||
− | <cmath>\ | + | <cmath>\frac{ax+1}{a+1}=x-13</cmath> and <cmath>\frac{ax+2001}{a+1}=x+27</cmath> |
− | <cmath>\ | + | <cmath>\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}</cmath> |
+ | <cmath>\frac{2000}{a+1}=40</cmath> so <cmath>2000=40(a+1)</cmath> | ||
+ | <cmath>a=49</cmath> | ||
+ | We plug that into our very first formula, and get: | ||
+ | <cmath>\frac{49x+1}{50}=x-13</cmath> | ||
+ | <cmath>49x+1=50x-650</cmath> | ||
+ | <cmath>x=651</cmath> | ||
== See Also == | == See Also == | ||
{{AIME box|year=2001|n=I|num-b=1|num-a=3}} | {{AIME box|year=2001|n=I|num-b=1|num-a=3}} |
Revision as of 23:03, 21 December 2007
Problem
A finite set of distinct real numbers has the following properties: the mean of is less than the mean of , and the mean of is more than the mean of . Find the mean of .
Solution
Let x be the mean of S. Let a be the number of elements in S. Then, and so We plug that into our very first formula, and get:
See Also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |