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

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== Solution ==
 
== Solution ==
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===Solution 1===
 
The sum of the first 20 numbers is <math>20 \cdot 30</math> and the sum of the other 30 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>.
 
The sum of the first 20 numbers is <math>20 \cdot 30</math> and the sum of the other 30 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. Answer is <math>\frac{2*20*30}{20+30}=24 \ \mathrm{(B)}</math>.
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Solution by franzliszt
  
 
== See also ==
 
== See also ==

Latest revision as of 18:15, 11 July 2020

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. Answer is $\frac{2*20*30}{20+30}=24 \ \mathrm{(B)}$.

Solution by franzliszt

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
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

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