2006 CEMC Fermat Problems/Problem 5

Problem

Three cubes have edges of lengths $4$ , $5$ , and $6$ . The average (mean) of their volumes is

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$\text{ (A) }\ 120 \qquad\text{ (B) }\ 125 \qquad\text{ (C) }\ 1125 \qquad\text{ (D) }\ 261 \qquad\text{ (E) }\ 135$

The average of the volumes of the cubes will be the sum of the volumes of the cubes divided by the number of cubes.

The first cube has an edge of $4$ , so its volume is $4^3 = 64$ .

The second cube has an edge of $5$ , so its volume is $5^3 = 125$ .

The last cube has an edge of $6$ , so its volume is $6^3 = 216$ .

The sum of these is $64 + 125 + 216 = 189 + 216 = 405$ .

To find the average of the volumes, we can then divide by $3$ because there are $3$ cubes.

$\frac{405}{3} = \boxed {\textbf {(E) } 135}$

~anabel.disher