2005 AMC 12A Problems/Problem 5: Difference between revisions
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== Solution == | == Solution == | ||
<math>\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}</math> | The sum of the first 20 numbers is <math>20 \cdot 30</math> and the sum of the other 30 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>. | ||
== See also == | == See also == | ||
Revision as of 21:26, 25 December 2011
Problem
The average (mean) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?
Solution
The sum of the first 20 numbers is 
and the sum of the other 30 numbers is 
. Hence the overall average is 
.
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 | |
