What is the maximum number of rectangular blocks measuring 3 inches by 2 inches by 1 inch that can be packed into a cube-shaped box whose interior measures 6inches on an edge?
A. 24
B. 8
C. 30
D. 36

Hint: In order to solve this problem we will find the volume of both the blocks and assume that n number of boxes gets fitted into the big box then we will equate the volume of n number of boxes with the bigger volume to get the value of n. Doing this will solve your problem and will give you the right answer.

Complete step-by-step solution:
It is given that,
Dimensions of the Rectangular block are (l, b, h) = (3, 2, 1).
And that is the length of an edge of the cube-shaped box = 6.
To find the maximum number of such rectangular blocks that can be packed into the above cube-shaped box;
Let ′n′ be the maximum number of rectangular blocks.
The volume of ′n′ rectangular blocks
$ \Rightarrow n ×l × b × h$.
$ \Rightarrow n ×3 × 2 × 1$
$ \Rightarrow 6 × n$
Volume of the above cube of side (s=6) = ${s^3}$ = ${6^3}$
Equating the volume of the above, we get
$ \Rightarrow 6 × n = ${6^3}$
$ \Rightarrow n = {6^2}$
$ \Rightarrow n = 36$

Therefore, the maximum number of such rectangular blocks that can be packed into the above cube-shaped box is ′36′.

Note: When you get to solve such problems you need to know that a cube is a three-dimensional solid object bounded by six square faces, facets, or sides, with three meetings at each vertex whereas cuboid is a 3D shape. Cuboids have six faces, which form a convex polyhedron. Broadly, the faces of the cuboid can be any quadrilateral. Whenever you need to calculate the number of small things fitted in a big thing then you have to assume the variables and get the total volume and get the value of the variable to get the number. Doing this will give you the right answers.