2020 AMC 10A Problems/Problem 2: Difference between revisions
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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 | |
| 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 | ||
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