Question

A cube of each side 4 cm, has been painted black, red and green on pairs of opposite faces. It is then
cut into small cubes of each side 1 cm. How many small cubes will have atleast one face painted ?
Select one:
a. 56
O b. 24
O C. 16
d. 27​

Answer 1

Answer:

The cubes with 2 faces painted will be the cubes on the sides except the corner ones.

Therefore, face on the top and bottom will have 8+8=16 cubes and the front and the back face will have 4+4=8 cubes.

The cubes on the other two faces have been covered by the above four.

Therefore, total number of cubes =8+16=24.

Answer 2

Answer:

$\large{ \boxed{ \boxed{ \rm{56(A)}}}}$

Step-by-step Explanation:

Refer to the attachment for the shaded cube or the coloured cube.

First finding total no. of cubes:

$= \rm{ \dfrac{4 \times 4 \times 4}{1 \times 1 \times 1} }$

= 64

We have to count the cubes which have atleast 1 face painted i.e. 1 face coloured, 2 faces coloured and 3 faces coloured.

Cubes with 3 faces coloured:

These are the corner cubes because it is the junction or meeting point of three adjacent faces and we know that there are 8 vertices of cube.

• No. of 3 faces coloured = 8

Cubes with 2 faces coloured:

Between any two corner faces, are the ones having two faces coloured. Now we have to choose the lateral sides i.e. 4 lateral sides. (Don't count top and bottom because they will be already counted within the lateral faces).

• No. of 2 faces coloured = 8 + 6 + 6 + 4 = 24

Cubes with 1 face coloured:

The cubes excluding the edges and vertices, which lies on the central region are the one face coloured cubes. We can see that there are 4 such cubes in each face of the bigger cube. Hence,

• No. of 1 face coloured = 4 × 6 = 24

Hence,

The total no. of cubes with atleast one face coloured/painted is 8 + 24 + 24 = 56. And we are done! :D