Question

125 small but identical cubes are put together to form a large cube. this large cube is now painted on all six faces. (i) how many of the smaller cubes have no face painted at all. (a) 27 (b) 64 (c) 8 (d) 36 (ii) how many of the smaller cubes have exactly three faces painted? (a) 98 (b) 100 (c) 96 (d) 95 (iii) how many of the smaller cubes have atleast one side painted? (a) 4 (b) 8 (c) 9 (d) 27

Answer 1

The large cube formed is of :

5 cubes on the length by 5 cubes on the width by 5 cubes on the height.

Cubes with three faces painted are at the corners of the large cube.

The cube has 8 corners and hence they are 8.

The cubes with 2 faces painted are between each of the corners on each face.

Between each two corners we have three cubes.

When we literally count them we have :

12 on the top face, 12 on the bottom face and 12 for the side faces.

This gives a total of 36

For the ones painted on one face we have 9 on each face and since we have 6 faces we will have a total of 54.

For the ones with no face painted, we get the total cubes minus the sum of the 1 face painted, two face painted and three face painted.

This equals :

125 - (8 + 36 + 54) = 125 - 98

= 27

For better understanding, just draw the cube with its smaller cubes then try and shade it practically using different shades for two faces painted, three and one.

Answer 2

As there are 125 small identical cubes, the side of the larger cube will be cube root of 125

So, the side of the larger cube=5

We need to calculate interior part. Because only interior parts will have no face painted.

So, interior part will be a cube of side =5-2=3

Hence, number of cubes=3^3=27

Therefore, the answer of the first part will be (a) i.e. 27

The number of smaller cubes having exactly three faces painted= 8 because only corners of any cube has three faces painted and we know, there are 8 corners in each cubes.

For atleast one face painted, means cube with one face painted + cube with two faces painted+ cube with 3 faces painted

So,  

cubes with one face painted=6(n-2)^2=6X9=54

cubes with two faces painted=12(n-2)=12X3=36

cubes with 3 faces painted=8

Therefore, total number of cubes having atleast one face painted=54+36+8=98 cubes