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

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

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