Question

If all the surfaces of a cube of \( 15 \, \text{cm} \) side are coloured black and then cut into smaller cubes of sides \( 3 \, \text{cm} \) each, then find how many cubes will have only one surface coloured in black?

• A. 18
• B. 24
• C. 36
• D. 54

Hint:

Remember the formulas for a cut painted cube:
- 3 faces painted = \( 8 \) (always the corners)
- 2 faces painted = \( 12(n - 2) \) (on the edges)
- 1 face painted = \( 6(n - 2)^2 \) (on the face centers)
- 0 faces painted = \( (n - 2)^3 \) (the inner core)

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A large painted cube of side \( 15 \, \text{cm} \) is cut into smaller cubes of side \( 3 \, \text{cm} \). We need to determine the number of smaller cubes that have exactly one face painted.

Step 2: Key Formula or Approach:
When a larger cube of side \( L \) is cut into smaller cubes of side \( s \), the number of divisions along one edge is:
\[ n = \frac{L}{s} \] The number of smaller cubes having exactly one face painted is given by:
\[ N_1 = 6 \times (n - 2)^2 \]

Step 3: Detailed Explanation:
Given:
- Side of larger cube, \( L = 15 \, \text{cm} \)
- Side of smaller cubes, \( s = 3 \, \text{cm} \)
First, calculate \( n \):
\[ n = \frac{15}{3} = 5 \] Now, substitute \( n = 5 \) into the formula for one-face painted cubes:
\[ N_1 = 6 \times (5 - 2)^2 \] \[ N_1 = 6 \times (3)^2 \] \[ N_1 = 6 \times 9 = 54 \] Thus, there are \( 54 \) smaller cubes with only one surface painted black.

Step 4: Final Answer:
(D) 54