Difference between revisions of "2005 AMC 12A Problems/Problem 5"
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== Problem == | == Problem == | ||
| − | The average ( | + | The average (mean) of <math>20</math> numbers is <math>30</math>, and the average of <math>30</math> other numbers is <math>20</math>. What is the average of all <math>50</math> numbers? |
<math> | <math> | ||
(\mathrm {A}) \ 23 \qquad (\mathrm {B}) \ 24 \qquad (\mathrm {C})\ 25 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 27 | (\mathrm {A}) \ 23 \qquad (\mathrm {B}) \ 24 \qquad (\mathrm {C})\ 25 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 27 | ||
</math> | </math> | ||
| + | |||
== Solution == | == Solution == | ||
| − | <math>\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}</math> | + | |
| + | ===Solution 1=== | ||
| + | The sum of the first <math>20</math> numbers is <math>20 \cdot 30</math> and the sum of the other <math>30</math> numbers is <math>30\cdot 20</math>. Hence the overall average is <math>\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}</math>. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | This is just the harmonic mean, so the answer is <math>\frac{2 \cdot 20 \cdot 30}{20+30}=24 \ \mathrm{(B)}</math>. | ||
| + | |||
| + | Solution by franzliszt | ||
== See also == | == See also == | ||
| Line 12: | Line 20: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:47, 1 July 2025
Problem
The average (mean) of 
numbers is 
, and the average of other numbers is . What is the average of all 
numbers?
Solution
Solution 1
The sum of the first numbers is 
and the sum of the other numbers is 
. Hence the overall average is 
.
Solution 2
This is just the harmonic mean, so the answer is 
.
Solution by franzliszt
See also
| 2005 AMC 12A (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
