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Difference between revisions of "2005 AMC 12A Problems/Problem 5"

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== Problem ==
 
== Problem ==
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The average ([[mean]]) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?
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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>
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== Solution ==
 
== Solution ==
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<math>\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}</math>
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===Solution 1===
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The sum of the first <math>20</math> numbers is <math>20 \cdot 30</math> and the sum of the other <math>30</math> numbers is <math>30\cdot 20</math>. Hence the overall average is <math>\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}</math>.
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===Solution 2===
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This is just the harmonic mean, so the answer is <math>\frac{2 \cdot 20 \cdot 30}{20+30}=24 \ \mathrm{(B)}</math>.
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Solution by franzliszt
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 14:47, 1 July 2025

Problem

The average (mean) of $20$ numbers is $30$, and the average of $30$ other numbers is $20$. What is the average of all $50$ numbers?

$(\mathrm {A}) \ 23 \qquad (\mathrm {B}) \ 24 \qquad (\mathrm {C})\ 25 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 27$

Solution

Solution 1

The sum of the first $20$ numbers is $20 \cdot 30$ and the sum of the other $30$ numbers is $30\cdot 20$. Hence the overall average is $\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}$.

Solution 2

This is just the harmonic mean, so the answer is $\frac{2 \cdot 20 \cdot 30}{20+30}=24 \ \mathrm{(B)}$.

Solution by franzliszt

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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