A rectangular box has interior dimensions 6-inches by 5-inches by 10-inches. The box is filled with as many solid 3-inch cubes as possible, with all of the cubes entirely inside the rectangular box. What percent of the volume of the box is taken up by the cubes?

The box would have 99% of its volume taken up.

Step-by-step explanation: The box has dimensions as follows;

Length = 6 inches

Width = 5 inches

Height = 10 inches

Therefore the volume of the box shall become

Volume = L x W x H

Volume = 6 x 5 x 10

Volume = 300 cubic inches

Also a 3 inch cube would have its volume given as follows (

Volume = 3 x 3 x 3 (All sides of a cube has equal lengths)

Volume = 27 cubic inches

To find out how many of 3-inch cubes can fit in, divide 300 by 27 and that equals 11.11.

Hence you can have at most 11 cubes in the box. The total volume of 11 cubes is given as 11 x 27 which equals 297. Therefore, the percentage of the box taken up completely by the cubes is given as;

Percentage = (Volume of cubes/Volume of box) x 100

Percentage = (297/300) x 100

Percentage = 99

Therefore the box would have 99% of its volume taken up by the cubes.

based on context clues you gave i'm assuming the answer would be c.