Difference between revisions of "2005 AMC 10A Problems/Problem 6"
Charles3829 (talk | contribs) m (→Solution) |
Sevenoptimus (talk | contribs) m (Improved formatting) |
||
| Line 2: | Line 2: | ||
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? | 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) } \ 26\qquad \mathrm{(E) } \ 27 </math> |
==Solution== | ==Solution== | ||
| − | Since the | + | Since the 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>. | ||
| − | 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}=\boxed{\ | + | Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=\boxed{\mathrm{(B) } \ 24}</math>. |
==Video Solution== | ==Video Solution== | ||
Revision as of 16:20, 1 July 2025
Contents
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 
.
Video Solution
~Charles3829
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
