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:

$\dfrac{ 2\cdot\frac 14 + 1\cdot\frac 34 }3 = \dfrac{ \frac 54 }3 = \boxed{\dfrac 5{12}}$

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 12 Problems and Solutions

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2009 AMC 12A Problems/Problem 3

Contents

Problem

Solution

Solution 1

Solution 2

See Also