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>?
  
==See Also==
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<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-b=1|num-a=3}}
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== Solution ==
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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 $\frac{60}{2}=\boxed{\text{(C) }30}.
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== See Also ==
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{{AMC10 box|year=2020|ab=A|before=num-a=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:02, 31 January 2020

Problem 2

The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$. What is the average of $a$ and $b$?

$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$

Solution

The arithmetic mean of the numbers $3, 5, 7, a,$ and $b$ is equal to $\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15$. Solving for $a+b$, we get $a+b=60$. Dividing by $2$ to find the average of the two numbers $a$ and $b$ gives $\frac{60}{2}=\boxed{\text{(C) }30}.

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
num-a=1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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