2005 AMC 12A Problems/Problem 5

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

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

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2005 AMC 12A Problems/Problem 5

Contents

Problem

Solution

Solution 1

Solution 2

See also