# 2009 AMC 12A Problems/Problem 3

## Problem

What number is one third of the way from $\frac14$ to $\frac34$? $\textbf{(A)}\ \frac {1}{3} \qquad \textbf{(B)}\ \frac {5}{12} \qquad \textbf{(C)}\ \frac {1}{2} \qquad \textbf{(D)}\ \frac {7}{12} \qquad \textbf{(E)}\ \frac {2}{3}$

## Solution

### Solution 1

We can rewrite the two given fractions as $\frac 3{12}$ and $\frac 9{12}$. (We multiplied all numerators and denominators by $3$.)

Now it is obvious that the interval between them is divided into three parts by the fractions $\boxed{\frac 5{12}}$ and $\frac 7{12}$.

### Solution 2

The number we seek can be obtained as a weighted average of the two endpoints, where the closer one has weight $2$ and the further one $1$. We compute: $$\dfrac{ 2\cdot\frac 14 + 1\cdot\frac 34 }3 = \dfrac{ \frac 54 }3 = \boxed{\dfrac 5{12}}$$

## See Also

 2009 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS