Cube MCQ Quiz - Objective Question with Answer for Cube - Download Free PDF

Calculation

Given:

Large cube side = 28 m

Small cube side = 7 m

Medium cube side = 14 m

First large cube gives: n small + 7 medium cubes

Second identical cube gives: 16 small + m medium cubes

We are to find n + m

Volume of large cube = 283 = 28 × 28 × 28 = 21952 m³

Volume of small cube = 73 = 343 m³

Volume of medium cube = 143 = 2744 m³

n × 343 + 7 × 2744 = 21952

n × 343 + 19208 = 21952

⇒ n × 343 = 2744

⇒n = 2744/343 = 8

16 × 343 + m × 2744 = 21952

⇒ 5488 + m × 2744 = 21952

⇒ m × 2744 = 16464

⇒ m = 16464/2744 = 6

So, n + m = 8 + 6 = 14

The correct answer is Option (2).

Given:

Side of cubical tank = 50 cm

Volume of water taken out = 61 litres = 61000 cm³

Formula used:

Volume of water removed = Area of base × Drop in water level

Area of base = Side × Side

Calculation:

Area of base = 50 × 50 = 2500 cm²

Volume of water removed = Area of base × Drop in water level

⇒ 61000 = 2500 × Drop in water level

⇒ Drop in water level = 61000 ÷ 2500

⇒ Drop in water level = 24.4 cm

∴ The correct answer is option (1).

Given:

The surface area of a cube = 1536 cm²

Formula used:

Surface Area of a Cube = 6a²

Volume of a Cube = a³

Where, a = side of the cube

Calculation:

6a² = 1536

⇒ a² = 1536 ÷ 6

⇒ a² = 256

⇒ a = √256

⇒ a = 16 cm

Now, Volume = a³

⇒ Volume = 16 × 16 × 16

⇒ Volume = 4096 cm³

∴ The correct answer is option (1).

Given:

Edge of first cube = 3 cm

Edge of second cube = 4 cm

Edge of third cube = 5 cm

All cubes are melted and recasted into a single cube.

Formula used:

Volume of a cube = (Edge)³

Total volume of the new cube = Sum of volumes of three cubes

Edge of the new cube = \(\sqrt[3]{\text{Total Volume}}\)

Calculation:

Volume of first cube = 3³ = 27 cm³

Volume of second cube = 4³ = 64 cm³

Volume of third cube = 5³ = 125 cm³

Total volume = 27 + 64 + 125 = 216 cm³

Edge of the new cube = \(\sqrt[3]{216}\)

⇒ Edge of the new cube = 6 cm

∴ The correct answer is option 2.

Given:

Surface Area (S) = 294 m²

Formula used:

Surface Area of a Cube = 6 × (Side)²

Where, Side = length of one edge of the cube

Calculation:

294 = 6 × (Side)²

⇒ (Side)² = 294 ÷ 6

⇒ (Side)² = 49

⇒ Side = √49

⇒ Side = 7 m

⇒ Side in cm = 7 × 100

⇒ Side = 700 cm

∴ The correct answer is option (2).

Given:

Each side of the cube = 8 cm

The rectangular container has a length of 16 cm, breadth of 8 cm, and height of 15 cm

Formula used:

The volume of cube = (Edge)3

The volume of a cuboid = Length × Breadth × Height

Calculation:

The volume of cube = The volume of the rectangular container with a length of 16 cm, breadth of 8 cm, and height of the water level rise

Let, the height of the water level will rise = x cm

So, 8³ = 16 × 8 × x

⇒ 512 = 128 × x

⇒ x = 512/128 = 4

∴ The rise of water level (in cm) is 4 cm

Given:

Three cubes with sides in the ratio of 3 ∶ 4 ∶  5 are melted to form a single cube whose diagonal is 18√3 cm.

Concept used:

The diagonal of a cube = √3a (where a is the sides)

Calculation: 

Let the s sides of the cubes will be 3x cm , 4x cm, and 5x cm

As per the question,

Volume of the new cube is 

(3x)³ +( 4x)³ +( 5x)³ = 216 x³.

⇒ side is = 6x

diagonal is 6x√3

⇒  6x√3 = 18√3

⇒ x = 3

The sides of the cubes will be 9 cm , 12 cm, and 15 cm

∴ The correct option is 2

Given:

The radius of the sphere, r = 15√3 cm

Concept:

The total surface area of a cube = 6 ×  (edge length)².

Length of the main diagonal of  cube = (edge length)√3

Solution:

The diameter of the sphere = Length of the main diagonal of the cube.

2 × \(15√ 3\) = a√3 

a = 30 cm

Total surface area of the cube = 6 ×  (edge length)²

Total surface area of the cube = 6 × (30)² = 5400 cm².

Hence, the total surface area of the largest possible cube cut out from the sphere will be 5400 cm².

Formula Used:

The volume of cube = (side)³

Calculation:

Let the sides of cubes be 2x cm, 3x cm, and 5x cm.

The volume of Cubes is (2x)³, (3x)³, (5x)³ respectively.

Total Volume = 54880 cm³

⇒(2x)3+ (3x)3 + (5x)3 = 54880

⇒8x3 + 27x3 + 125x3 = 54880

⇒ 160x³ = 54880

⇒ x³ = 54880 ÷ 160 = 343

⇒ x3 = 7³

⇒ x = 7

Hence, the side of the largest cube is 5 × 7 = 35 cm.

Given:

The ratio of the sides of three cubes = 3 ∶ 4 ∶ 5 

Diagonal of the resultant cube = \(12\sqrt 3 \)

Formula used:

The volume of a cube = side³

Diagonal of a cube = \(\sqrt 3 \) side

Concept used:

if three different cubes are melted down into a single cube then the sum of the volume of the three cubes is equal to the volume of the resultant cube

Calculation:

Diagonal of a cube = side\(\sqrt 3 \)

\(12\sqrt 3 \)  = side\(\sqrt 3 \)

Side of the resultant cube = 12 cm

The ratio of the sides of three cubes = 3 ∶ 4 ∶ 5 

let the sides of the three cubes be 3x, 4x, 5x

Sum of the volume of three cubes = Volume of the resultant cube

27x³ + 64x³ + 125x³ = 12³

⇒ 216x³ = 12³

⇒ 216x3 = 1728

⇒ x³ = 8

⇒ x = 2

then the sides of the cube are 6cm, 8cm, 10cm

The smallest side is 6cm

The answer is 6cm

Given:

Sum of all 12 edges of cube = 60 cm

Formula:

Number of edges in a cube = 12

Diagonal (body) of cube = √3a

Calculation:

Let the side of the cube be a cm, then

12a = 60

⇒ a = 60/12

⇒ a = 5 cm

∴ Largest rod that can be fit in cube  = √3a = 5√3 cm

Given:

Dimensions of metallic cuboid = 50 cm × 40 cm × 32 cm

The metallic cuboid is melted and recast into a cube. 

Formulas used:

Volume of cuboid = length × breadth × height 

Volume of cube with side 'a' = a³

Surface area of cube = 6a²

Calculation:

Volume of cuboid = Volume of cube 

50 cm × 40 cm × 32 cm = a³

⇒ 2 × 5² × 5 × 2³ × 2⁵ = a3

⇒ 64000 = a3

⇒ a = ∛64000 cm³

⇒ a = 40 cm 

∴ Surface area of cube = 6 × 40 × 40 

⇒ 9600 cm²

Given:

Area of the floor of a cubical room = 192 m² 

Formula used:

Diagonal of the cube = √3 × side of cube

Floor area = side² (floor of the cube is in the shape of a square)

Calculation:

According to the question,

The floor area of the cube = side² 

 ⇒ 192 = side²

⇒ side of cube = √192 = 8√ 3 m

Diagonal = √3 × side of the cube

⇒ √3 × 8 × √3 

⇒ 8 × 3 = 24 m

∴ The length of the longest rod that can be kept in that room is 24 m.

Let the length of the cube be x cm

⇒ The length after 2 cm increase = x + 2 m

⇒ Initial volume = x³

⇒ Volume after increasing side = (x + 2)³

⇒ (x + 2)³ – x³ = 488

⇒ x³ + 6x² + 12x + 8 – x³ = 488

⇒ 6x² + 12x – 480 = 0

⇒ x² + 2x – 80 = 0

⇒ (x – 8) (x + 10) = 0

⇒ x = 8, -10

Let the side of the cube is a.

The total surface area of the cube = 6a²

⇒ 6a² = 2904

⇒ a² = 2904/6 = 484

⇒ a = 22 cm