Difference between revisions of "2020 AMC 10A Problems/Problem 2"
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The arithmetic mean of the numbers <math>3, 5, 7, a,</math> and <math>b</math> is equal to <math>\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15</math>. Solving for <math>a+b</math>, we get <math>a+b=60</math>. Dividing by <math>2</math> to find the average of the two numbers <math>a</math> and <math>b</math> gives <math>\frac{60}{2}=\boxed{\text{(C) }30}</math>. | The arithmetic mean of the numbers <math>3, 5, 7, a,</math> and <math>b</math> is equal to <math>\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15</math>. Solving for <math>a+b</math>, we get <math>a+b=60</math>. Dividing by <math>2</math> to find the average of the two numbers <math>a</math> and <math>b</math> gives <math>\frac{60}{2}=\boxed{\text{(C) }30}</math>. | ||
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== See Also == | == See Also == | ||
Revision as of 21:04, 31 January 2020
Problem 2
The numbers 
and 
have an average (arithmetic mean) of 
. What is the average of 
and ?
Solution
The arithmetic mean of the numbers and is equal to 
. Solving for 
, we get 
. Dividing by 
to find the average of the two numbers and gives 
.
~aryam
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by num-b=1 |
Followed by Problem 3 | |
| 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. 
