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Given:

Radius of cylinder = 14 cm

The curved surface area of the cylinder = 880 cm²

Formula used:

The curved surface area of cylinder = 2πrh

The volume of cylinder = πr²h

Where,

r = radius of the cylinder

h = height of the cylinder

Calculation:

Height of the cylinder

⇒ 2πrh = 880

⇒ 2 × (22/7) × 14 × h = 880

⇒ h = 10 cm

The volume of the cylinder

⇒ (22/7) × (14)² × 10

⇒ 6160 cm³

∴ The volume of the cylinder is 6160 cm³.

Given:

Radius = 14 cm

Volume = 6160 cm³

Formula used:

(i) Volume of cylinder = π × r² × h

(ii) Curved surface area = 2 × π × r × h

Calculations:

Volume = π × r² × h

⇒ 6160 = (22/7) × (14)² × h

⇒ h = (6160 × 7)/(22 × 196)

⇒ h = 10 cm

Curved surface area = 2 × π × r × h

⇒ 2 × (22/7) × 14 × 10

⇒ 2 × 22 × 2 × 10

⇒ 880 cm²

∴ The curved surface area is 880 cm²

Given:

The radius of a sphere is increased by 2.5 decimeter(dm),

then its surface area increases by 110 dm²

Concept used:

The surface area of sphere = 4πr²

The volume of sphere = (4/3) × πr³

Where r = radius and π = 22/7

a² - b² = (a + b)(a - b)

Calculation:

according to the question,

4π(r + 2.5)² - 4πr² = 110

⇒ 4π[(r + 2.5)² - r²] = 110

⇒ 4π[(r + 2.5 + r)(r + 2.5 - r)] = 110

⇒ 4 × 22/7 × (2r + 2.5) × 2.5 = 110

⇒ 10 × (2r + 2.5) = 35

⇒ 20r + 25 = 35

⇒ r = 1/2 dm

now,

Volume = 4/3 × π × (1/2)³

⇒ 4/3 × 22/7 × 1/8

⇒ 88/168 = 11/21 dm³

∴ The volume of the sphere is 11/21 dm³.

Given:

Side of cube = 9 cm

Cuboid Dimensions

Length = 5 cm, breadth = 13 cm, height = 31 cm

Formula used:

Volume of cube = (side)³

Volume of cuboid = length × breadth × height

Total surface area of cube = 6 × side²

Calculation:

Volume of Solid metallic cube

⇒ 9³

⇒ 729 cm³

Volume of Solid metallic cuboid

⇒ 5 × 13 × 31

⇒ 2015 cm³

Volume of Recast single cube

⇒ Volume of Solid metallic cube + Volume of Solid metallic cuboid

⇒ 729 + 2015

⇒ 2744 cm3

Side of Recast cube

⇒ (Side)³ = 2744

⇒ side = 14 cm

Total surface area of cube

⇒ 6 × 14 × 14

⇒ 1176 cm²

∴ The total surface area of recast cube is 1176 cm².

Given:

Curved surface area of cylinder = 308 cm²

Height = 14 cm

Formula used:

CSA (Curved surface area) = 2πrh

Volume = πr²h

Where r is radius and h is height

Calculation:

CSA = 2πrh

308 = 2 × (22/7) × r × 14

⇒ 308 = 88r

⇒ r = 7/2 = 3.5 cm

Volume = πr²h

⇒ Volume = (22/7) × (3.5)² × 14

⇒ Volume = 539 cm³ 

∴ Volume of the cylinder is 539 cm³

Given:

A solid metallic cuboid of dimensions 18 cm × 36 cm × 72 cm

where l = 18 cm, b = 36 cm and h = 72 cm

It is melted and recast into 8 cubes of the same volume.

Concept used:

The volume of cuboid = lbh

The total surface area of the cuboid = 2(lb + bh + hl)

Where l = length, b = breadth and h = height

The volume of cube = a³

The lateral surface area of cube = 4 × a²

Where a = side of cube

Explanation:

according to the question,

lbh = 8 × a³

⇒ 18 × 36 × 72 = 8 × a³

⇒ a³ = 18 × 36 × 9

⇒ a = √(9 × 2 × 9 × 2 × 2 × 9)

⇒ a = 18 cm

now,

according to the question,

2(lb + bh + hl) : 8 × 4 × a²

⇒ 2(18 × 36 + 36 × 72 + 72 × 18) : 8 × 4 × (18)²

⇒ 2 × 18 × 36(1 + 4 + 2) : 8 × 4 × 18 × 18

⇒ 36 × 36 × 7 : 32 × 18 × 18

⇒ 7 : 8

∴ The ratio is 7 : 8.

Given:

The ratio of the areas of two squares = 16 ∶ 1

Formula used:

Area of square = (Side of the square)²

Perimeter of square = 4 × (Side of the square)

Calculations:

Let the side of the two squares be a and b respectively

The ratio of the areas of two squares

⇒ a² ∶ b²  =16 ∶ 1

⇒ a ∶ b = 4 ∶ 1

The ratio of the perimeter of two squares

⇒ 4a ∶ 4b = (4 × 4) ∶ (4 × 1)

⇒ 16 ∶ 4

⇒ 4 ∶ 1

∴ The ratio of the perimeter of two squares is 4 ∶ 1

Given:

Increase in radius, R’ = 10%

Increase in height, H’ = 10%

Formula used:

Volume of cone = 1/3 × πr²h, where r is the radius of base and h is the height of the cone.

Calculation:

Let V be the volume of the original cone, R and H are the original radius and original height respectively.

Original volume, V = 1/3 × πR²H

New radius, R’ = (100 + 10)% of original radius

⇒ R’ = 110/100 × R

⇒ R’ = 11/10 × R

New height, H’ = (100 + 10)% of original height

⇒ H’ = 110/100 × H

⇒ H’ = 11/10 × H

New volume, V’ = 1/3 × π × (R’)2 × H’

⇒ V’ = 1/3 × π × (11/10 × R)² × (11/10 × H)

⇒ V’ = (1/3 × πR²H) × 1331/1000

⇒ V’ = V × 1331/1000

Increased volume = (V’ – V)/V × 100

⇒ Increased volume = (V × 1331/1000 – V)/V × 100

⇒ Increased volume = 33.1

∴ The volume is increased by 33.1%.

Given:

The perimeter of a rectangle is 24 cm. and the length of the rectangle is 7cm

Concept Used:

1.) Concept of the linear equation of single variable

2.) Perimeter of a rectangle is {2 × (Length + Breadth)} unit

3.) Area of a rectangle = (Length × Breadth) unit²

Calculation:

Let the breadth of the rectangle be x cm

Length of the rectangle is 24 cm and length is 7 cm

Accordingly,

2 × ( 7 + x) = 24

⇒ 7 + x = 12

⇒ x = 5

Breadth of the rectangle is 5 cm

Area of the rectangle = Length × Breadth

⇒ 7 × 5

⇒ 35 cm²

∴ Area of the rectangle is 35 cm²

Given :

Area of equilateral triangle is equal to the area of a regular hexagon 

Side of the equilateral triangle = 18 cm 

Formula used :

Area of equilateral triangle = (√3 × a²⁾/4  (where a is the side of the equilateral triangle)

Area of hexagon = (√3 × 3 × x²⁾/2 (where x is the side of hexagon)

Calculation :

Area of equilateral triangle = area of the hexagon 

(√3 × 18²)/4 = (√3 × 3 × x²)/2

⇒ x = 3√6 

Difference of side of triangle and hexagon = 18 - 3√6

∴ The difference of sides of triangle and hexagon is 18 - 3√6 cm

Given-

The perimeter of a regular hexagon and an equilateral triangle is same.

Concept Used-

Area of equilateral triangle = √3/4 × Side²

Area of a regular hexagon = 3√3/2 × Side²

Calculation- 

Let the perimeter of the triangle and hexagon be 6a

Each Side of Hexagon = a

Each Side of Triangle = 2a

Area of Hexagon = 3√3/2 × a²

Area of Triangle = √3/4 × (2a)2

⇒ √3a²

∴ The ratio of their areas = (3√3/2 × a²) : (√3a²)

⇒ 3 : 2

∴ The required ratio will be 3 : 2.

Given:

Area of hexagonal field is 3750√3 m² 

Cost of fencing is Rs. 29 per meter

Formula Used:

Area of hexagon = 6 × (√3)/4 × a²

Perimeter of a hexagon = 6 × a

Where a is a side of hexagon.

Cocnept Used:

Fencing is done around the boundaries of any field.

i.e. We need to calculate the perimeter of a field,

To find the length of fencing.

Calculation:

Area of hexagon = 6 × (√3)/4 × a2

⇒  3750√3 = 6 × (√3)/4 × a2

Given:

Height of cylinder = 11 cm

Curved surface area = 242 cm² 

Formula used:

Volume of cylinder = πr²h

CSA of cylinder = 2πrh

Calculation:

∵ CSA of cylinder = 2πrh

⇒ 242 = (2 × 22 × r × 11)/7

⇒ 242 × 7 = 2 × 22 × r × 11

⇒ (242 × 7)/(2 × 22 × 11) = r

⇒ r = 7/2 cm

∵ Volume of cylinder = πr²h

⇒ Volume of cylinder = (22/7) × (49/4) × 11

= 847/2 cm3

= 423.5 cm3

Given:

Ratio of Curved surface and total surface area of cylinder = 7 : 11

Curved surface area of cylinder = 704 cm² 

Formula Used:

Curved surface area = 2πrh ,  where r = radius of base and h = height of cylinder

Total Surface area = 2πrh + 2πr² = 2πr(h + r)

Volume of cylinder = πr²h

Calculation:

(2πrh)/[2πr(h + r)] = 7/11

⇒ h/(h + r) = 7/11

⇒ 11h = 7h + 7r

⇒ 4h = 7r

⇒ h/r = 7/4

Let the height be 7x and radius be 4x, then,

704 = 2 × (22/7) × 7x × 4x

⇒ x² = (704 × 7)/(28 × 22 × 2)

⇒ x² = 4

⇒ x = 2

Putting values back in 7x and 4x we get,

Radius = 4x = 4 × 2 = 8 cm

Height = 7x = 7 × 2 = 14 cm

Volume of cylinder = (22/7) × 8 × 8 × 14

⇒ V = 2816 cm³

∴ The volume of cylinder is 2816 cm³.

Given:

Volume of sphere = 606.375 m³.

Formula used:

Surface area of sphere = 4πr2

Volume of sphere = (4/3)π × r³

Calculation:

Volume of sphere = (4/3)πr³

⇒  (4/3)πr³ = 606.375 m³ 

⇒ (4/3)(22/7) × r³ = 606.375 m³

⇒ r3 = (606.375 × 7 × 3)/(22 × 4) 

⇒ r3 = 144.703125

⇒ r = 5.25 m

Surface area of sphere = 4πr²

⇒ Surface area of sphere = 4 × (22/7) × (5.25)²

⇒ (4 × 22 × 5.25 × 5.25)/7

⇒ (2425.5)/7

⇒ 346.5 m²

∴ The surface area of sphere is 346.5 m².

Given:

⇒ Side of a square = √144 = 12cm

Using pythgoras theorem,

⇒ Diagonal of a square = √288 = 12√2cm

⇒ Perimeter of rectangle = 2(length + breadth) = 4 × 12 = 48cm

⇒ Length = 12 + breadth

Solving,

⇒ Length = 18cm and Breadth = 6cm

Using pythagoras theorem,

⇒ Diagonal of rectangle = √360 = 6√10cm

∴ Required ratio = 12√2 : 6√10 = 2 : √5

Given:

⇒ (Radius of hemispherical pot)² = 882π × 1/2π

⇒ Radius of hemispherical pot = 21cm

⇒ Volume of hemispherical pot = 2/3 × π × 21³ = 6174π

⇒ Volume of cylinder = πr²h

⇒ πr²h = 6174π

⇒ h = 14cm

∴ The height of cylindrical can is 14cm.

Given:

Volume of the cube = 2744 cc

Formula used:

Volume of cube = \({a^3}\) and surface area = \(6{a^2}\)

Where, a = length of each side of the cube

Surface area of cuboid = 2(lb + bh + lh) sq.cm.

Where, l, b & h are the length, breadth and height of the cuboid respectively.

Calculations:

Volume of the cube = 2744 cc

∴ Length of each side = \(\sqrt[3]{{2744}}\) = 14 cm

And, surface area = 6 × 14² = 1176 sq.cm.

After the changes, dimensions are = (2 × 14), (2 × 14) and (1/4 × 14) ⇒ 28, 28, 7/2 cm respectively. [Since, one dimension is reduced by 300%, it will become 1/4 times of before]

∴ Surface area = 2(28 × 7/2 + 28 × 7/2 + 28 × 28) sq.cm.

= 980 sq.cm.

Required ratio = 1176 : 980 = 6 : 5

Given: 

The volume of the cube is 6859 cm³.

Concept: 

Side of cube = (Volume)^1/3

Surface area of cube = 6 × (Side)²

Calculation:

Side of the Cube = (6859)^1/3

⇒ (19)^{3 × 1/3}

⇒ 19 cm

Now, 

Surface area of the cube = 6 × (19)²

⇒ 6 × 361

⇒ 2166 cm²

∴ The required surface area of the cube is 2166 cm².

Given:

Diameter of wheel = 49 cm

∴ Radius of wheel = 49/2 = 24.5 cm

Total distance to cover = 770 m = 77000 cm

Concept:

The distance covered by the wheel is the number of revolutions the wheel makes. The distance covered by the wheel in one revolution is the measure of its circumference.

Formula used:

Circumference of circle = 2πr

Number of revolution taken = Total distance/Circumference

Calculation:

∵ Circumference of circle = 2πr

= 2 × (22/7) × (49/2)

= 154 cm

∴ Number of revolution taken by the wheel = 77000/154

= 500 revolutions 

Given:

One side of triangle = 7 cm

Perimeter of triangle = 18 cm

Area of triangle = √108 cm²

Formula Used:

If sides of triangle are a, b, and c

then,

Semi perimeter (S) = (a + b + c)/2

Area of triangle = √[s(s - a)(s - b)(s - c)

Calculation:

18 = (7 + b + c)

⇒ (b + c) = 11

⇒ c = 11 - b 

S = 18/2 = 9

√108 = √[(9)(9 - 7)(9 - b)(9 - c)]

Squaring both sides,

108 = 18 × (9 - b) × (9 - c)

Substituting Value of c as 11 - b

6 = (9 - b)[9 - (11 - b)]

⇒ 6 = (9 - b)(b - 2)

⇒ 6 = 9b + 2b - b² - 18

⇒ b² - 11b + 24 = 0

⇒  b2 - 8b - 3b + 24 = 0

⇒ b(b - 8) - 3(b - 8) = 0

⇒ (b - 8)(b - 3) = 0

⇒ b = 8, 3

If we Take side b as 8

Then,

c = 11 - 8 = 3

∴ The other two sides of triangle are 8 cm and 3 cm.

Given:

The difference between the exterior and interior angles at a vertex of a regular polygon is 150°

Formula Used:

Interior angle of a polygon having side n = (n – 2)180°/n

Calculations:

Let Ai and Ae be interior and exterior angles

Sum of Ai and Ae = 180°,

So, Ai + Ae = 180

∴ Ai - Ae = 150

So, solving for Ae & Ai

Ae = 15°

Ai = 165°

Interior angle of a polygon having side n = (n – 2)180°/n

∴ 165° = (n – 2)180°/n

⇒ 165n = 180n – 360

⇒ 360 = 15n

⇒ n = 24

Given-   

Ratio of radii of two cylinders = 3 : 4

Ratio of height of two cylinders = 8 : 5

Formula Used-

Volume of cylinder = πr²h       [where r = radius of base and h = height] 

Calculation- 

Let the radii of two cylinders be 3x and 4x and their heights be 8y and 5y 

Volume of 1st cylinder : Volume of 2nd cylinder = π(3x)² × 8y : π(4x)² × 5y

⇒ 72xy : 80xy

⇒ 9 : 10 

The ratio of their volumes is equal to 9 : 10

Given:

Radii of two cylinder are in the ratio = 3 : 4

Heights of two cylinder are in the ratio = 5 : 7

Formula:

Volume of cylinder = πr²h

Calculation:

r₁ = 3, r₂ = 4, h₁ = 5 and h₂ = 7

Volume of first cylinder = V₁

Volume of second cylinder = V₂

⇒ V₁ / V₂ = (π × 3² × 5) / (π × 4² × 7)

⇒ V₁ / V₂ = 45 / 112

Given:

Ratio of radius of two right circular cylinders = 3 : 2

Ratio of their volumes = 27 : 16

Formula used:

Volume of right circular cylinder = πr²h

Calculation:

let the height of both cylinder be h₁ and h₂

⇒ π3²h₁/π2²h₂ = 27 : 16

⇒ 9h₁/4h₂ = 27 : 16

⇒ 4h₁ = 3h₂

⇒ h₁ : h₂ = 3 : 4

∴ The ratio of their heights is 3 : 4

Given:

The capacity of cylinders A and B are equal.

Volume_(A) = Volume_(B) 

The ratio of diameters = A ∶ B = 1 ∶ 4

Formula Used:

Volume of cylinder = πr² h

Diameter = 2 × r

Where,

r → Radius of the cylinder 

h → Height of the cylinder 

Calculation:

r = D/2

The ratio of the radius of A and B

\(\Rightarrow \frac{{Diamater\;of\;A}}{{Diameter\;of\;B}} = 1/4\)

The capacity of cylinders A and B are equal.

Volume(A) = Volume(B) 

⇒ π× r_(A)²× h_(A) = π× r_(B)²× h_(B)

⇒ r(A)2× h(A) = r(B)2× h(B)

⇒ r(A)2/r(B)2 =  h(B)/ h(A)

⇒1²/4² = h(B)/ h(A)

⇒1/16 = h(B)/ h(A)

⇒  h_(A)/h_(B) = 16/1

⇒ h(A) : h(B) = 16 : 1

∴ The ratio of the heights of A and B is 16 ∶ 1.

Given:

The ratio of radii of two cylinders is 2 ∶ 3 and the ratio of their heights is 5 ∶ 3

Formula used:

The volume of the cylinder = π r² h

Here, r is the radius and h is the height of the cylinder

Calculation:

Let the radius be 2x and 3x respectively

And, let the height be 5y and 3y respectively

⇒ The ratio of the volume \(= {{r^2\times h} \over R^2\times H}\)

⇒ \( {{2x^2\times 5y} \over 3x^2\times 3y}\) = 20/27

∴ The ratio of their volumes is 20 ∶ 27

Given :

The area of a square = ABCD 

AB = 16 cm

Formula used :

The area of a square = side × side 

Calculation : 

Area of square = 16 cm × 16 cm = 256cm²

∴ The area of a square ABCD with AB = 16cm is 256cm².

Given:

The radius of in-circle of a triangle = 4 cm

The perimeter of the triangle = 20 cm

Formula used:

The area of the triangle = Semi perimeter of triangle × the radius of in-circle of a triangle

Calculation:

Semi perimeter of triangle = 20/2 = 10 cm

The area of the triangle = Semi perimeter of triangle × the radius of in-circle of a triangle

⇒ The area of the triangle = 10 × 4

⇒ 40 cm²

∴ The area of the triangle is 40 cm2

Given:

Total surface are of cube = 1944 cm²

Formula used:

TSA of cube = 6 × edge²

Volume of cube = edge³

Calculation:

Let the edge of cube be x m.

TSA of cube = 6 × edge²

⇒ 1944 = 6x²

⇒ x² = 324

⇒ x = √324

⇒ x = 18 m

Volume of cube = edge³

⇒ Volume = x³

⇒ Volume = 18³

⇒ Volume = 5832 m³ 

∴ The volume of cube is 5832 m³.

Given:

Radius = 4 cm

CSA = 20π cm²

Formula used:

CSA of cone = π × r × l where, l → slant height

Volume of cone = 1/3 × πr²h

Calculation:

CSA = 20π cm²

⇒ π × r × l = 20π 

⇒ 4 × l = 20

⇒ l = 5 cm

So, h = 3 cm                    [Using pythagorian triplet]

Volume = 1/3 × πr2h

⇒ Volume = 1/3 × π × 4 × 4 × 3

⇒ Volume = 16π cm³

∴ Volume of cone is 16π cm³.

Given:

Base area of cuboid = 34 sq cm.

height of cuboid = 3.5 cm

Formula used:

The volume of cuboid = base area of cuboid × height

Calculation:

Volume of cuboid = 34 × 3.5

⇒ Volume of cuboid = 119 cm³ 

Hence, the volume of the cuboid is 119 cm³.

Given:

Radius = 3.5 m

Number of revolution = 50

Formula used:

Circumference of circle = 2 π r

Number of revolution = Total distance covered/circumference of the wheel

Concept used:

The distance covered in one revolution is equal to the circumference of the circle

Calculation:

Circumference of circle = 2 π × 3.5 

⇒ 2 × (22/7) × 3.5

⇒ 22 m

Now, Total distance covered in 50 revolutions = 50 × 22 = 1100 m = 1.1 km

∴ The distance covered by the circular wheel is 1.1 km

Given:

The circumference of circular wire is 132 cm

Formula used:

Area of a square = Side × side

The perimeter of a square = 4 × side

Calculation:

The circumference of circular wire is 132 cm

Now, a square is formed with the same circular wire

∴ Circumference of circular wire = Perimeter of the square

⇒ 4 × side = 132

⇒ Side = 33 cm

Now, area of the square = side × side = 33 × 33 = 1089 square cm

Hence, option (1) is correct

Given:

Diameter = 56 cm

Number of revolution = 1000

Concept used:

Total distance = Number of revolution × Circumference

Circumference of circle = 2πr

Calculation:

Diameter = 56 cm

Radius = 56/2

⇒ Radius = 28 cm

Circumference of circle = 2πr

⇒ Circumference = 2 × 22/7 × 28

⇒ Circumference = 176 cm

Total distance = Number of revolution × Circumference

⇒ Total distance = 1000 × 176

⇒ Total distance = 176000 cm

⇒ Total distance = 176000 × 1/100000 km

⇒ Total distance = 1.76 km

∴ The distance covered by the wheel is 1.76 km.

Given:

Radius of circular wheel is 21 m and it need to cover the distance of 1.980 km

Formula used:

Circumference of circle = 2 π r

Number of revolution = Total distance covered/circumference of wheel

Calculation:

When radius of circle is 21 m

∴ Circumference of circle = 2 π × 21 = 132 m

Now, number of revolution required to cover 1.980 km

∴ Number of revolution = 1980/132 = 15

Hence, option (2) is correct

Given:

The diameter of wheel = 3 m

Number of revolutions in 1 min = 28

Distance to cover = 5.280 Km

Concept used:

Distance covered by wheel in 1 rotation = circumference of the wheel

Circumference = π × d

Calculation:

Circumference = π × d

⇒ Circumference = (22/7) × 3 = 66/7 m

Distance covered in 1 min = 28 rotations = 28 × 66/7 

⇒ Distance covered in 1 min = 264 m

Distance to be covered = 5.280 km = 5280m 

Time taken to cover 5280 m = 5280/264 = 20 mins

∴ Time taken to cover 5.280 km is 20 mins.

Given:

Ratio of radius = 2 : 5

Ratio of heights = 3 : 2

Formula used:

Volume of cylinders = πr²h

Where, r and h represents radius and height of cylinder.

Calculation:

Volume of 1^st cylinder/Volume of 2^nd cylinder = πr²₁h₁/πr₂²h₂

= π(2)²3/π(5)²2

Given:

Perimeter of rhombus = 68 cm

Diagonal, (d1) = 30 cm

Formula used:

Perimeter = 4 × side

Area of Rhombus = (d1 × d2)/2

Area of Rhombus = Side × Altitude

d12 + d22 = 4 × s2

Here, d1, d2, and s are diagonals and side of rhombus respectively.

Concept used:

All sides of a rhombus are equal and diagonals bisect each other at a right angle.

Calculation:

Perimeter = 4 × side

⇒ 68 = 4 × side

⇒ side = 17 cm

Now, d12 + d22 = 4 × s2

⇒ 302 + d22 = 4 × 172

⇒ d22 = 256

⇒ d2 = 16

Now, Area of Rhombus = (d1 × d2)/2

⇒ Area of Rhombus = (30 × 16)/2

∴ Area of Rhombus is 240 cm2

Given:

The volume of a cubical box = 512 cm³

Concept used:

Volume of a cube = (side)³

Detailed solution:

(side)³ = 512

⇒ side = \( \sqrt[3]{512}\)

⇒ side = 8

∴ the length of a side of the box is 8 cm 

Total cubes = 125

The only cubes present at the upper face and four corners of cube that will be visible to three sides of cube.

A cube is formed by 11 cube horizontally and 11 cube vertically and remaining 4 cube will be visible.

number of cubes required to make largest possible cube = 11 × 11 = 121

∴ Remaining cube = 125 - 121 = 4

Given:

Volume = 0.125 liters 

Formula used:

TSA = 6a²   (where a is the side of the cube)

Volume of cube = a³

1 liter = 1000 cm³

Calculations:

⇒ 0.125 liter = 0.125 × 1000

⇒ Volume = 125 cm²

⇒ a³ = 125

⇒ a = 5 cm

⇒ TSA = 6 × 5 × 5

⇒ TSA = 150 cm²

∴ The total surface area of the cube is 150 cm2

Given:
Length of a box = 84 cm
Breadth of a box = 116 cm
Height of a box = 172 cm

Concept used:
Volume of cuboid = l × b × h
Volume of cube = (Side)³
Number of cubes formed from large cuboid = Volume of cuboid/Volume of cube

Calculation:
Volume of cuboid = 84 × 116 × 172
H.C.F of 84, 116 and 172 = 4
The least side of the cube = 4 cm
Volume of cube = 4³
Number of cubes formed = (84 × 116 × 172)/(4 × 4 × 4)
⇒ 21 × 29 × 43
⇒ 26187
∴ The least number of cubes formed would be 26187.

Given

Area of equilateral triangle = 16√ 3

Perimeter of triangle = 3 × length of  side of  atriangle

Formula used

Area of equilateral triangle = (√3/4)(side)²

Calculation

⇒ (√ 3/4)(side)² = 16√ 3

⇒ (side)² = 64

⇒ side = 8 cm

⇒ Perimeter = 3 × 8

∴ Perimeter of triangle is 24cm

Given:

Depth of well = height = 40 m

Radius of well =  7 m

Length of platform = 22 m

Breadth of platform = 14 m

Formula Used:

Volume of cylinder = πr2h

Volume of cuboid = Length × Breadth × Height

Calculation:

Volume of cylinder = Volume of well

⇒ Volume of well = 22/7 × (7 m)² × (40 m)

⇒ Volume of well = (22 × 7 × 40) m²

Volume of platform = Volume of cuboid

Volume of platform = Volume of well

⇒ (22 × 7 × 40) m² = 22 m × 14 m × h

⇒ 20 m = h

∴ Height of the platform is 20 m

The correct option is 3 i.e. 20 m