SSC CGL Reasoning Cube & Cuboid: Formulas and Important Questions

SSC CGL Reasoning Cube & Cuboid questions often become tricky when the same concept is framed in different ways, especially with changes in language and conditions. Many aspirants find it difficult to apply painted cube formulas, determine the value of n, and correctly interpret “at least” and “at most” cases. Strengthening these core concepts helps solve cube-based reasoning questions quickly and with better accuracy in the exam.

SSC CGL Reasoning Cube - Recap of Cube Formulas

Before solving cube-based reasoning questions, it is important to remember the key formulas used to calculate different types of coloured cubes after a larger cube is cut into smaller equal cubes.

• Three-Surface Coloured Cubes: Always 8

• Two-Surface Coloured Cubes: 12 × (n − 2)

• One-Surface Coloured Cubes: 6 × (n − 2)²

• Zero-Surface Coloured Cubes (Uncoloured Cubes): (n − 2)³

• Total Number of Cubes: n³

Here, n represents the ratio of the side length of the big cube to the side length of the small cube.

n = Length of Big Cube ÷ Length of Small Cube

Where:

• L₁ = Length of the Big Cube

• L₂ = Length of the Small Cube

Therefore, n = L₁ ÷ L₂

This value of n is the foundation of most cube and cuboid reasoning questions, as all major formulas are based on it.

Problem Solving: Basic Application

Let us apply the formula to a simple cube-cutting question.

Question 1: A coloured solid cube of length 12 cm is cut into smaller cubes of 3 cm each. How many cubes have two surfaces coloured?

Step 1: Calculate the value of n

n = Length of Big Cube ÷ Length of Small Cube

n = 12 cm ÷ 3 cm = 4

Step 2: Apply the formula for two-surface coloured cubes

Formula = 12 × (n − 2)

= 12 × (4 − 2)

= 12 × 2

= 24

Answer: 24 cubes have two surfaces coloured.

Problem Solving: "At Least" and "At Most" Concepts

Questions involving terms such as "at least" and "at most" often confuse candidates. Understanding these terms correctly helps in selecting the right cube categories for calculation.

Question 2: A coloured cube is cut into 125 smaller cubes. How many cubes have at least two surfaces painted?

Step 1: Calculate the value of n

When the total number of smaller cubes is given, n can be found by taking the cube root of the total number of cubes.

Total Number of Cubes = n³

125 = n³

n = Cube Root of 125

n = 5

Step 2: Understand the meaning of "At Least Two Surfaces Painted"

The term "at least two" means two or more.

In cube problems, this includes:

• Cubes with two surfaces painted

• Cubes with three surfaces painted

Since a cube cannot have more than three painted surfaces, these are the only categories that need to be counted.

Step 3: Apply the formulas

Two-Surface Coloured Cubes

Formula = 12 × (n − 2)

= 12 × (5 − 2)

= 12 × 3

= 36

Three-Surface Coloured Cubes

= 8

Total Cubes with At Least Two Surfaces Painted

= 36 + 8

= 44

Answer: 44 cubes have at least two surfaces painted.

Understanding "At Least" and "At Most" Conditions

Before solving cube questions, it is important to understand what terms such as "at least" and "at most" mean. These terms determine which categories of cubes should be included in the calculation.

Problem Solving: Fractional Size of Small Cubes

Question 3: How many small equal cubes of size 1/4 of the big cube can be cut?

Step 1: Determine n from the fraction

When the side of the small cube is given as a fraction of the big cube, the denominator directly gives the value of n.

Here, 1/4 means n = 4

Step 2: Find the total number of cubes

Total number of cubes = n³
= 4³
= 64

Question 4: How many small equal cubes of size 1/6 of the big cube can be cut?

Step 1: Determine n

1/6 means n = 6

Step 2: Find total cubes

Total number of cubes = n³
= 6³
= 216

Complex Language with Fractional n

Question 5: A cube is fully painted on all faces and then cut into smaller cubes where each side is 1/4 of the original cube’s side. Find the number of cubes with only one face painted.

Step 1: Determine n

Each side is 1/4 of original cube ⇒ n = 4

Step 2: Apply formula for one-surface coloured cubes

Formula = 6 × (n − 2)²
= 6 × (4 − 2)²
= 6 × (2)²
= 6 × 4
= 24

Multiple Questions from One Scenario

Scenario 1

A cube is painted on all faces and cut into 125 equal cubes.

Step 1: Calculate n

n³ = 125
n = 5

Question 6A: One face painted cubes

Formula = 6 × (n − 2)²
= 6 × (5 − 2)²
= 6 × 3²
= 6 × 9
= 54

Question 6B: No face painted cubes

Formula = (n − 2)³
= (5 − 2)³
= 3³
= 27

Scenario 2

A cube is painted on all faces and cut into 27 equal cubes.

Step 1: Calculate n

n³ = 27
n = 3

Question 7A: One face painted cubes

= 6 × (3 − 2)²
= 6 × 1²
= 6

Question 7B: No face painted cubes

= (3 − 2)³
= 1³
= 1

At Least / At Most Type

Question 8: A cube is cut into 216 smaller cubes. Find cubes with at least two surfaces painted.

Step 1: Find n

n³ = 216 ⇒ n = 6

Step 2: Apply concept

Two-surface cubes = 12 × (n − 2) = 12 × 4 = 48
Three-surface cubes = 8

Total = 48 + 8 = 56

Question 9: A cube is cut into 343 smaller cubes. Find cubes with at least one surface painted.

n³ = 343 ⇒ n = 7

One-surface = 6 × (n − 2)² = 6 × 25 = 150
Two-surface = 12 × (n − 2) = 60
Three-surface = 8

Total = 150 + 60 + 8 = 218

Finding Lengths

Question 10: A 15 cm cube is cut into 125 smaller cubes. Find side of small cube.

n³ = 125 ⇒ n = 5

n = Big cube ÷ Small cube
5 = 15 ÷ L₂
L₂ = 3 cm

Question 11: A cube is cut into 125 smaller cubes, each of volume 8 cm³. Find side of big cube.

n = 5

Side of small cube = cube root of 8 = 2 cm

n = L₁ ÷ L₂
5 = L₁ ÷ 2
L₁ = 10 cm