Difference between revisions of "2020 AMC 10A Problems/Problem 2"
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+ | ==Problem 2== | ||
The numbers <math>3, 5, 7, a,</math> and <math>b</math> have an average (arithmetic mean) of <math>15</math>. What is the average of <math>a</math> and <math>b</math>? | The numbers <math>3, 5, 7, a,</math> and <math>b</math> have an average (arithmetic mean) of <math>15</math>. What is the average of <math>a</math> and <math>b</math>? | ||
− | + | <math>\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60</math> | |
− | {{AMC10 box|year=2020|ab=A|num- | + | == Solution == |
+ | |||
+ | 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 $\frac{60}{2}=\boxed{\text{(C) }30}. | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2020|ab=A|before=num-a=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:02, 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 $\frac{60}{2}=\boxed{\text{(C) }30}.
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by num-a=1 |
Followed by Problem 3 | |
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All AMC 10 Problems and Solutions |
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