Difference between revisions of "2005 AMC 10A Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | Since the average of <math>20</math> numbers is <math>30</math>, their sum is <math>20\cdot30=600</math>. | + | Since the [[arithmetic mean|average]] of the first <math>20</math> numbers is <math>30</math>, their sum is <math>20\cdot30=600</math>. |
Since the average of <math>30</math> other numbers is <math>20</math>, their sum is <math>30\cdot20=600</math>. | Since the average of <math>30</math> other numbers is <math>20</math>, their sum is <math>30\cdot20=600</math>. | ||
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So the sum of all <math>50</math> numbers is <math>600+600=1200</math> | So the sum of all <math>50</math> numbers is <math>600+600=1200</math> | ||
− | Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \ | + | Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math> |
− | ==See | + | ==See also== |
− | + | {{AMC10 box|year=2005|ab=A|num-b=5|num-a=7}} | |
− | + | {{MAA Notice}} | |
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Revision as of 16:29, 8 December 2020
Problem
The average (mean) of numbers is , and the average of other numbers is . What is the average of all numbers?
Solution
Since the average of the first numbers is , their sum is .
Since the average of other numbers is , their sum is .
So the sum of all numbers is
Therefore, the average of all numbers is
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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