Volume & Surface Area of Solids

Find the volume, the total surface area and the lateral surface of a rectangular solid having :

(i) length = 8.5 m, breadth = 6.4 m and height = 50 cm.

(ii) length = 5.6 dm, breadth = 2.5 dm and height = 1 m.

Answer

(i) Given,

Length = 8.5 m

Breadth = 6.4 m

Height = 50 cm = 0.5 m.

Volume of rectangular solid = l × b × h

= 8.5 × 6.4 × 0.5

= 27.2 m³.

Total surface area of the rectangular solid = 2(lb + bh + hl)

= 2(8.5 × 6.4 + 6.4 × 0.5 + 0.5 × 8.5)

= 2(54.4 + 3.2 + 4.25)

= 2(61.85)

= 123.70 m².

Lateral surface area of the solid = [2(l + b)h]

= [2(8.5 + 6.4) × 0.5]

= [2(14.9) × 0.5]

= 14.9 m².

Hence, volume = 27.2 m³, TSA = 123.7 m² and LSA = 14.9 m².

(ii) Given,

Length = 5.6 dm

Breadth = 2.5 dm

Height = 1 m = 10 dm

Volume of rectangular solid = l × b × h

= 5.6 × 2.5 × 10

= 140 dm³.

Total surface area of the solid = 2(lb + bh + hl)

= 2(5.6 × 2.5 + 2.5 × 10 + 10 × 5.6)

= 2(14 + 25 + 56)

= 2(95)

= 190 dm².

Lateral surface area of the solid = [2(l + b)h]

= [2(5.6 + 2.5) × 10]

= [2(8.1) × 10]

= 162 dm².

Hence, volume = 140 dm³, TSA = 190 dm² and LSA = 162 dm².

The volume of a rectangular wall is 33 m³. If its length is 16.5 m and height 8 m, find the width of the wall.

Answer

Given,

Volume = 33 m³

Length = 16.5 m

Height = 8 m

Volume of rectangular solid = l × b × h

⇒ 33 = 16.5 × b × 8

⇒ 33 = 132 × b

⇒ b = $\dfrac{33}{132}$

⇒ b = 0.25 m

⇒ b = 0.25 × 100 cm = 25 cm.

Hence, width = 25 cm.

Find the number of bricks, each measuring 25 cm × 12.5 cm × 7.5 cm, required to construct a wall 6 m long, 5 m high and 50 cm thick, while the cement and the sand mixture occupies $\dfrac{1}{20}\text{ th}$ of the volume of the wall.

Answer

Given,

Dimensions of wall :

Length = 6 m = 600 cm.

Height = 5 m = 500 cm

Thickness = 50 cm.

Dimension of brick :

Length = 25 cm

Breadth = 12.5 cm

Height = 7.5 cm

Calculating the volume of the wall,

Total volume of the wall = l × b × h

= 600 × 500 × 50

= 1,50,00,000 cm³.

Calculating the volume of bricks,

Volume of bricks = Volume of the wall - $\dfrac{1}{20}$ th volume of the wall (occupied by sand and cement)

= 1,50,00,000 - $\dfrac{1}{20}$ × 1,50,00,000

= 1,50,00,000 - 7,50,000

= 1,42,50,000 cm³.

Calculating volume of one brick,

Volume of one brick = l × b × h

= 25 × 12.5 × 7.5

= 2343.75 cm³.

Number of bricks = $\dfrac{\text{Total volume of the bricks}}{\text{Volume of one brick}}$

= $\dfrac{1,42,50,000}{2343.75}$

= 6080.

Hence, number of bricks = 6080.

A class room is 12.5 m long, 6.4 m broad and 5 m high. How many students can it accommodate, if each student needs 1.6 m² of floor area ? How many cubic metres of air would each student get?

Answer

Given,

Room dimensions :

Length = 12.5 m

Breadth = 6.4 m

Height = 5 m

Students per floor area = 1.6 m²

Calculating the floor area,

Floor area = l × b = 12.5 × 6.4 = 80 m².

Number of students = $\dfrac{\text{Total floor area}}{\text{Area per student}}$

= $\dfrac{80}{1.6}$

= 50.

Calculating the Volume of the room,

Volume = l × b × h

= 12.5 × 6.4 × 5

= 400 m³.

Air per student = $\dfrac{\text{Total volume}}{\text{Number of students}}$

= $\dfrac{400}{50}$

= 8 m³.

Hence, number of students = 50 and air per student = 8 m³.

Find the length of the longest rod that can be placed in a room measuring 12 m × 9 m × 8 m.

Answer

The dimensions of room :

Length = 12 m

Breadth = 9 m

Height = 8 m

Calculating the Diagonal of room,

$\text{Diagonal} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{12^2 + 9^2 + 8^2} \\[1em] = \sqrt{144 + 81 + 64} \\[1em] = \sqrt{289} \\[1em] = 17 \text{ m}.$

Length of the longest rod that can be placed in a room = Length of the diagonal of the room.

∴ Length of the rod = 17 m.

Hence, length of the rod = 17 m.

The volume of a cuboid is 14400 cm³ and its height is 15 cm. The cross-section of the cuboid is a rectangle having its sides in the ratio 5 : 3. Find the perimeter of the cross-section.

Answer

Given,

Volume of cuboid = 14400 cm³

Height = 15 cm

Given,

The cross-section of the cuboid is a rectangle having its sides in the ratio 5 : 3.

Let length = 5x and breadth = 3x.

We know that,

Volume of cuboid = l × b × h

⇒ 14400 = 5x × 3x × 15

⇒ 15x² = $\dfrac{14400}{15}$

⇒ 15x² = 960

⇒ x² = $\dfrac{960}{15}$

⇒ x² = 64

⇒ x = $\sqrt{64}$

⇒ x = 8.

∴ Sides of the rectangle :

⇒ Length = 5x = 5 × 8 = 40 cm.

⇒ Breadth = 3x = 3 × 8 = 24 cm

Perimeter of cross-section = 2(l + b)

= 2(40 + 24)

= 2(64)

= 128 cm.

Hence, perimeter of cross section = 128 cm.

The area of a path is 6500 m². Find the cost of covering it with gravel 14 cm deep at the rate of ₹ 5.60 per cubic metre.

Answer

Given,

Area of path = 6500 m².

Rate of gravelling = ₹ 5.60 per m³.

Depth = 14 cm = 0.14 m.

Calculating the volume of the gravel,

Volume = Area × Depth

= 6500 × 0.14

= 910 m³.

Calculating total cost of the gravel,

Total cost = Volume × Rate

= 910 × 5.60

= ₹ 5,096.

Hence, cost of covering the path with gravel = ₹ 5,096.

The cost of papering the four walls of a room 12 m long at ₹ 6.50 per square metre is ₹ 1,638 and the cost of matting the floor at ₹ 3.50 per square metre is ₹ 378. Find the height of the room.

Answer

Given,

Length = 12 m

Cost of papering (4 walls) = ₹ 1,638 at ₹ 6.50 per m².

Cost of matting (Floor) = ₹ 378 at ₹ 3.50 per m².

Calculating the floor area,

Floor area = $\dfrac{\text{Total cost of matting}}{\text{Rate}}$

= $\dfrac{378}{3.50}$

= 108 m².

Calculating width of the room,

Floor area = l × b

⇒ 108 = 12 × b

⇒ b = $\dfrac{108}{12}$ = 9 m

Calculating the area of four walls,

Area of four walls = $\dfrac{\text{Total cost of papering}}{\text{Rate}}$

= $\dfrac{1638}{6.5}$

= 252 m².

Calculating the height of the room,

Area of four walls = [2(l + b)h]

⇒ 252 = [2(12 + 9) × h]

⇒ 252 = 2(21) × h

⇒ 252 = 42 × h

⇒ h = $\dfrac{252}{42}$

⇒ h = 6 m.

Hence, height of the room = 6 m.

The dimensions of a field are 15 m × 12 m. A pit 7.5 m × 6 m × 1.5 m is dug in one corner of the field and the earth removed from it, is evenly spread over the remaining area of the field. Calculate, by how much the level of the field is raised.

Answer

Given,

Field dimension = 15 m × 12 m

Pit dimension = 7.5 m × 6 m × 1.5 m

Calculating the volume of the earth dug out,

Volume = 7.5 × 6 × 1.5

= 67.5 m³.

Calculating the remaining area,

Total area of the field = 15 × 12 = 180 m².

Area of pit opening = 7.5 × 6 = 45 m².

Remaining area = 180 - 45 = 135 m².

Calculating the rise in field level,

Rise in level = $\dfrac{\text{Volume of earth dug out}}{\text{Remaining area}}$

= $\dfrac{67.5}{135}$

= 0.50 m

= 0.50 × 100 = 50 cm.

Hence, rise in field level = 50 cm.

The sum of the length, breadth and depth of cuboid is 19 cm and the length of its diagonal is 11 cm. Find the surface area of the cuboid.

Answer

Given,

l + b + h = 19 cm ........(1)

Diagonal = 11 cm.

By formula,

Diagonal of a cuboid = $\sqrt{l^2 + b^2 + h^2}$

⇒ 11 = $\sqrt{l^2 + b^2 + h^2}$

⇒ 11² = l² + b² + h²

⇒ 121 = l² + b² + h² .........(2)

By formula,

(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

Substituting the values from equation (1) and (2) in above equation, we get :

⇒ (19)² = 121 + 2(lb + bh + hl)

⇒ 361 = 121 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 361 - 121

⇒ 2(lb + bh + hl) = 240

We know that,

Surface area of the cuboid = 2(lb + bh + hl)

Hence, surface area of the cuboid = 240 cm².

Three cubes, each of side 6 cm, are joined end to end. Find the surface area of resulting cuboid.

Answer

Given,

Side = 6 cm

When three identical cubes are joined end to end they form a cuboid.

∴ New length = 6 + 6 + 6 = 18 cm.

Breadth = 6 cm.

Height = 6 cm.

Total Surface area of cuboid = 2(lb + bh + hl)

= 2(18 × 6 + 6 × 6 + 6 × 18)

= 2(108 + 36 + 108)

= 2(252)

= 504 cm².

Hence, surface area of cuboid = 504 cm².

Find :

(i) the volume

(ii) the total surface area

(iii) the lateral surface area and

(iv) the length of the diagonal of a cube of side 10 cm.

Answer

Let 'a' be the side of a cube.

(i) Volume of cube = (a)³

= (10)³

= 1000 cm³.

Hence, volume = 1000 cm³.

(ii) By formula,

Total surface area of cube = 6a²

= 6 × 10²

= 600 cm².

Hence, total surface area of cube = 600 cm².

(iii) By formula,

Lateral surface area of cube = 4a²

= 4 × 10²

= 400 cm².

Hence, lateral surface area of cube = 400 cm².

(iv) By formula,

Diagonal of cube = $a\sqrt{3}$

= $10\sqrt{3}$

= 10 × 1.732

= 17.32 cm.

Hence, diagonal of cube = 17.32 cm.

The surface area of a cube is 1536 cm². Find :

(i) the length of its edge

(ii) its volume

(iii) the volume of its material whose thickness is 5 mm.

Answer

Let 'a' be the length of the edge (side of a cube).

(i) Total surface area of cube = 6a²

⇒ 1536 = 6a²

⇒ a² = $\dfrac{1536}{6}$

⇒ a² = 256

⇒ a = $\sqrt{256}$

⇒ a = 16 cm.

Hence, length of the edge = 16 cm.

(ii) By formula,

Volume of cube = a³

= 16³

= 4096 cm³.

Hence, volume of cube = 4096 cm³.

(iii) Material thickness = 5 mm = 0.5 cm

Outer edge = 16 cm

Inner edge = 16 - (0.5 + 0.5) = 16 - 1 = 15 cm.

Volume of the material = (outer edge)³ - (inner edge)³

= (16)³ - (15)³

= 4096 - 3375

= 721 cm³.

Hence, volume of the material = 721 cm³.

Two cubes, each of volume 512 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Answer

Let length of each side of cube be a cm.

Given,

Volume of each cube = 512 cm³.

Calculating the side of cube,

Volume of cube = a³

512 = a³

a = $\sqrt[3]{512}$

a = 8 cm.

Two cubes are joined end to end to form a cuboid.

∴ Length (l) = a + a = 8 + 8 = 16 cm

Breadth (b) = a = 8 cm

Height (h) = a = 8 cm

Total surface of cuboid = 2(lb + bh + hl)

= 2(16 × 8 + 8 × 8 + 8 × 16)

= 2(128 + 64 + 128)

= 2(320)

= 640 cm².

Hence, total surface area of cuboid = 640 cm².

(i) How many cubic cm of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout ?

(ii) If 1 cm³ of iron weighs 15 g, find the weight of the empty box in kg.

Answer

(i) Given,

External dimensions :

Length (L) = 36 cm

Breadth (B) = 25 cm

Height (H) = 16.5 cm

Internal dimensions:

Since the thickness = 1.5 cm

∴ Length (l) = 36 - (1.5 + 1.5) = 36 - 3 = 33 cm

Breadth (b) = 25 - (1.5 + 1.5) = 25 - 3 = 22 cm

Height (h) = 16.5 - 1.5 = 15 cm (Since, box is open)

Calculating the external volume,

Volume = L × B × H

= 36 × 25 × 16.5

= 14,850 cm³.

Calculating internal volume,

Volume = l × b × h

= 33 × 22 × 15

= 10,890 cm³

Calculating volume of iron,

Volume of iron = External volume - Internal volume

= 14,850 - 10,890

= 3,960 cm³.

Hence, volume of iron = 3,960 cm³.

(ii) Given,

Weight of iron = 15 g

Total weight of the box = Volume of iron × weight of iron.

= 3,960 × 15 g

= 59,400 g.

1000 g = 1 kg

∴ 59,400 g = $\dfrac{59,400}{1000}$ kg

= 59.4 kg

Hence, weight of the box = 59.4 kg.

A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, find the edge of third smaller cube.

Answer

Given,

Edge of largest cube = 12 cm

Edge of 1st smaller cube = 6 cm

Edge of 2nd smaller cube = 8 cm.

Let the edge of 3rd smaller cube = x cm

Calculating the volume of larger cube,

Volume = a³

= (12)³

= 1728 cm³.

Calculating the volume of smaller cubes,

Volume of 1st smaller cube = 6³ = 216 cm³.

Volume of 2nd smaller cube = 8³ = 512 cm³.

Volume of 3rd smaller cube = x³.

Since larger cube melts to form a smaller cubes,

∴ Volume of larger cube = Total volume of smaller cubes.

⇒ 1728 = 216 + 512 + x³

⇒ 1728 = 728 + x³

⇒ x³ = 1728 - 728

⇒ x³ = 1000

⇒ x = $\sqrt[3]{1000}$

⇒ x = 10 cm.

Hence, edge of third smaller cube = 10 cm.

The dimensions of a metallic cuboid are 100 cm × 80 cm × 64 cm. It is melted and recast into a cube. Find :

(i) the edge of the cube,

(ii) the surface area of the cube.

Answer

(i) Given,

Length = 100 cm

Breadth = 80 cm

Height = 64 cm

Calculating the volume of the cube,

Volume of cube = l × b × h

= 100 × 80 × 64

= 512000 cm³.

Let the edge of cube be a cm,

⇒ a³ = 512000

⇒ a = $\sqrt[3]{512000}$

⇒ a = 80 cm.

Hence, edge of the cube = 80 cm.

(ii) Surface area of the cube = 6a².

= 6 × (80)²

= 6 × 6400

= 38400 cm².

Hence, surface area of the cube = 38400 cm².

The square on the diagonal of a cube has an area of 1875 cm². Calculate :

(i) the side of the cube,

(ii) the surface area of the cube.

Answer

(i) Given,

Area = 1875 cm².

We know that,

Diagonal of cube (d) = $\sqrt{3}$ a

The square on the diagonal means a square whose side is equal to the diagonal of the cube.

∴ Area of square = (diagonal)²

⇒ 1875 = (a $\sqrt{3})$²

⇒ 1875 = 3a²

⇒ a² = $\dfrac{1875}{3}$

⇒ a² = 625

⇒ a = $\sqrt{625}$

⇒ a = 25 cm.

Hence, side of cube = 25 cm.

(ii) Given,

Total surface area of cube = 6a².

= 6 × (25)²

= 6 × 625

= 3750 cm².

Hence, surface area of the cube = 3750 cm².

The areas of three adjacent faces of a cuboid are x, y and z sq. units. If the volume is V cubic units, prove that V² = xyz.

Answer

Let the dimensions of cuboid be,

Length = l

Breadth = b

Height = h

We know that,

Volume of cuboid (V) = l × b × h

Areas of three adjacent faces :

x = l × b

y = b × h

z = h × l

Multiplying these three areas :

xyz = (l × b) × (b × h) × (h × l)

xyz = l² × b² × h²

xyz = (l × b × h)²

Substituting the value of V = l × b × h

∴ xyz = V².

Hence, proved that V² = xyz.

The diagonal of a cube is $16\sqrt{3}$ cm. Find its surface area and volume.

Answer

Let the side of a cube = a cm and diagonal = d cm.

Given,

Diagonal of cube = $16\sqrt{3}$ cm.

By formula,

Diagonal of cube (d) = $\sqrt{3}a$

∴ $\sqrt{3}a = 16\sqrt{3}$

a = 16 cm.

Calculating the total surface area of cube,

Total surface area of cube = 6a².

= 6 × 16²

= 6 × 256

= 1536 cm².

Calculating the volume of the cube,

Volume of a cube = a³

= 16³

= 4096 cm³.

Hence, surface area = 1536 cm² and volume = 4096 cm³.

Water flows into a tank 150 m long and 100 m broad through a rectangular pipe whose cross section is 5 dm × 3.5 dm, at the speed of 15 km/hr. In what time, will the water be 7 m deep?

Answer

Given,

Tank dimension :

Length = 150 m

Breadth = 100 m

Depth = 7 m

Cross section of pipe = 5 dm × 3.5 dm

= 0.5 m × 0.35 m.

Speed of water = 15 km/hr = 15000 m/hr.

Let time taken by pipe to fill the tank be x hours.

⇒ Volume of tank = Time × Volume of water discharged by pipe (per hour)

⇒ 150 × 100 × 7 = x × (Cross section of pipe × Speed of water)

⇒ 150 × 100 × 7 = x × 0.5 × 0.35 × 15000

⇒ 105000 = x × 2625

⇒ x = $\dfrac{105000}{2625}$

⇒ x = 40 hours.

Hence, time taken = 40 hours.

A rectangular tank is 25 m long and 7.5 m deep. If 540 m³ of water be drawn off the tank, the level of water in the tank goes down by 1.8 m. Calculate :

(i) the width of the tank

(ii) the capacity of the tank.

Answer

Given,

Tank dimension:

Length (l) = 25 m

Depth (h) = 7.5 m

Drop in level = 1.8 m

Volume removed = 540 m³.

(i) Volume removed = length × width × drop in level

⇒ 540 = 25 × width × 1.8

⇒ 540 = width × 45

⇒ Width = $\dfrac{540}{45}$ = 12 m.

Hence, width of the tank = 12 m.

(ii) Calculating the capacity of the tank,

Capacity of the tank = Length × Width × Depth

= 25 × 12 × 7.5

= 300 × 7.5

= 2250 m³.

Hence, capacity of tank = 2250 m³.

A swimming pool is 30 m long and 12 m broad. Its shallow and deep ends are 1.5 m and 3.5 m deep respectively. If the bottom of the pool slopes uniformly, how many litres of water will fill the pool?

Answer

From figure,

The swimming pool cross-section is in the shape of trapezium.

Area of cross-section = $\dfrac{1}{2} \times \text{ (Sum of // sides)} \times \text{ (Distance between them)}$

= $\dfrac{1}{2} \times (1.5 + 3.5) \times 30$

= 75 m².

Volume of pool = Area of cross-section × length

= 75 × 12

= 900 m³

= 900 × 1000 liters

= 900000 litres.

Hence, 900000 litres of water will fill the pool.

The cross-section of a piece of metal 2 m in length is shown in the adjoining figure. Calculate :

(i) the area of its cross-section;

(ii) the volume of piece of metal;

(iii) the weight of piece of metal to the nearest kg, if 1 cm³ of the metal weighs 6.5 g.

Answer

Given,

Length of the metal piece = 2 m = 200 cm.

(i) From figure,

Rectangle ABCD :

Length (AB = CD) = 12 cm

Breadth (BC = AD) = 13 cm.

Area of rectangle ABCD = length × breadth

= 12 × 13 = 156 cm².

Triangle DEF :

Height (DF) = DC - FC = 12 - 8 = 4 cm.

Base (DE) = AE - AD = 16 - 13 = 3 cm.

Area of triangle DEF = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × 3 × 4

= 6 cm².

Total cross-section area = 156 + 6 = 162 cm².

Hence, area of cross-section = 162 cm².

(ii) Calculating the volume of metal piece,

Volume of metal piece = Area of cross section × Length

= 162 × 200

= 32400 cm³.

Hence, volume of metal piece = 32400 cm³.

(iii) Given,

1 cm³ of metal = 6.5 g.

Calculating the weight of the metal piece,

Total weight of the metal piece = Volume × Weight of metal at 1 cm³

= 32400 × 6.5

= 210600 g.

1000 g = 1 kg

∴ 210600 g = $\dfrac{210600}{1000}$ kg

= 210.6 kg ≈ 211 kg.

Hence, weight of the piece of the metal = 211 kg.

The adjoining figure shows the cross-section of a concrete wall to be constructed. It is 1.9 m wide at the bottom, 90 cm wide at the top and 5 m high. If its length is 20 m, find :

(i) the cross-sectional area;

(ii) the volume of the concrete in the wall.

Answer

Let the bottom width be 'a' and top width be 'b'.

Given,

Bottom width (a) = 1.9 m

Top width (b) = 90 cm = 0.9 m

Height (h) = 5 m

Length = 20 m

(i) Cross-sectional area :

$\text{Area of trapezium} = \dfrac{1}{2} \times \text{ (sum of // sides)} \times \text{ (distance between them)} \\[1em] = \dfrac{1}{2} \times (1.9 + 0.9) \times 5 \\[1em] = \dfrac{1}{2} \times 2.8 \times 5 \\[1em] = 1.4 \times 5 \\[1em] = 7 \text{ m}^2.$

Hence, area of cross-section = 7 m².

(ii) Volume of concrete :

Volume of concrete = Area of cross-section × Length of the wall.

= 7 × 20

= 140 m³.

Hence, volume of concrete = 140 m³.

The cross-section of a 6 m long piece of metal is shown in the figure. Calculate :

(i) the area of the cross-section

(ii) the volume of the piece of metal in cubic centimeters.

Answer

From figure,

AD = BC = 8 cm

AB = CD = 6.5 cm

Triangle sides = AE = 5 cm & DE = 5 cm

Triangle base (AD) = 8 cm.

Length of metal piece = 6 m = 600 cm.

(i) Area of cross section :

Calculating the area of rectangle ABCD,

Area of rectangle ABCD = length × breadth

= 8 × 6.5

= 52 cm².

EAD is an isosceles triangle,

In an isosceles triangle, the perpendicular drawn from common vertex to the base, bisects the base.

AF = FD = $\dfrac{AD}{2} = \dfrac{8}{2}$ = 4 cm.

By using pythagoras theorem for the triangle EFD,

⇒ Hypotenuse² = Base² + Height²

⇒ ED² = FD² + EF²

⇒ 5² = 4² + EF²

⇒ 25 = 16 + EF²

⇒ EF² = 25 - 16

⇒ EF² = 9

⇒ EF = $\sqrt{9}$

⇒ EF = 3 cm.

∴ Height (EF) = 3 cm

Area of △ EAD = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × AD × EF

= $\dfrac{1}{2}$ × 8 × 3

= 4 × 3

= 12 cm².

Total area = Area of rectangle ABCD + Area of triangle EAD

= 52 + 12 = 64 cm².

Hence, area of cross-section = 64 cm².

(ii) Volume of metal:

Volume of metal = Area of cross-section × Length of the metal piece

= 64 × 600

= 38400 cm³.

Hence, volume of metal = 38400 cm³.

The square brass plate of side x cm is 1 mm thick and weighs 5.44 kg. If 1 cm³ of brass weighs 8.5 g, find the value of x.

Answer

Given,

Side of square = x cm.

Thickness of brass = 1 mm = 0.1 cm

Total weight = 5.44 kg = 5440 g

1 cm³ of brass weighs 8.5 g

Density = 8.5 g/cm³

Calculating the volume of the brass,

Volume of the brass = $\dfrac{\text{Total weight of brass}}{\text{Density of the brass}}$

= $\dfrac{5440}{8.5}$

= 640 cm³.

Calculating the area of the plate,

Area of the plate = length × breadth

= x × x = x²

We know that,

Volume = Area of the plate × Thickness of the brass.

⇒ 640 = x² × 0.1

⇒ x² = $\dfrac{640}{0.1}$

⇒ x² = 6400

⇒ x = $\sqrt{6400}$

⇒ x = 80 cm.

Hence, x = 80 cm.

The area of cross-section of a rectangular pipe is 5.4 cm² and water is pumped out of it at the rate of 27 kmph. Find, in litres, the volume of water which flows out of the pipe in 1 minute.

Answer

Given,

Area of cross-section of rectangular pipe = 5.4 cm².

Speed of water = 27 km/hr = 2700000 cm/hr

1 hr = 60 minutes

∴ Speed of water = $\dfrac{2700000}{60}$ = 45000 cm/min.

Calculating the distance traveled by water in 1 minute,

Distance = Speed × time

= 45000 × 1 = 45000 cm.

Calculating the volume of water,

Volume of water which flows out of pipe in 1 min = Area of cross-section × Distance traveled by water in 1 min

= 5.4 × 45000

= 243000 cm³.

1000 cm³ = 1 litre

∴ 243000 cm³ = $\dfrac{243000}{1000}$ liters = 243 litres.

Hence, 243 litres of water flows out of the pipe in 1 minute.

The adjoining figure shows a solid of uniform cross section. Find the volume of the solid. It is being given that all the measurements are in cm and each angle in the figure is a right angle.

Answer

Given solid figure consists of two cuboids.

Cuboid 1:

Length (l) = 4 cm

Breadth (b) = 6 cm

Height (h) = 3 cm

Cuboid 2:

Length (l) = 4 cm

Breadth (b) = 3 cm

Height (h) = 9 cm

Calculating the volume of cuboid 1,

Volume of cuboid 1 = length × breadth × height

= 4 × 6 × 3

= 72 cm³.

Calculating the volume of cuboid 2,

Volume of cuboid 2 = length × breadth × height

= 4 × 3 × 9

= 108 cm³.

∴ Total volume = 72 + 108

= 180 cm³.

Hence, volume of solid = 180 cm³.

The cross-section of a tunnel, perpendicular to its length is a trapezium ABCD in which AB = 8 m, DC = 6 m and AL = BM. The height of the tunnel is 2.4 m and its length is 40 m. Find :

(i) the cost of paving the floor of the tunnel at ₹ 16 per m².

(ii) the cost of painting the internal surface of the tunnel, excluding the floor at the rate of ₹ 5 per m².

Answer

Given,

AB = 8 m

DC = 6 m

Height = 2.4 m

Length of the tunnel = 40 m

AL = BM

From figure, LM = DC = 6 m (since perpendiculars drop from D and C).

From figure,

⇒ AB = AL + LM + MB

⇒ 8 = AL + 6 + BM

Since, AL = BM

∴ 8 = 2AL + 6

⇒ 2AL = 8 - 6

⇒ 2AL = 2

⇒ AL = $\dfrac{2}{2}$ = 1 m

∴ AL = BM = 1 m.

Calculating the lengths of the sloping sides AD and BC,

In right triangle ALD,

AL = 1 m

DL = 2.4 m

Using pythagoras theorem for the triangle ALD,

⇒ Hypotenuse² = Base² + Height²

⇒ AD² = AL² + DL²

⇒ AD² = 1² + (2.4)²

⇒ AD² = 1 + 5.76

⇒ AD² = 6.76

⇒ AD = $\sqrt{6.76}$

⇒ AD = 2.6 m.

∴ AD = BC = 2.6 m.

(i) Cost of paving the floor:

Area of tunnel = AB × length

= 8 × 40

= 320 m².

Total cost = Area × cost per m²

= 320 × 16

= ₹ 5,120.

Hence, cost of paving the floor = ₹ 5,120.

(ii) Cost of painting internal surface (excluding floor)

Area to be painted are :

Roof DC and Two sloping sides AD and BC.

Calculating the area of DC,

Area = length × DC

= 40 × 6 = 240 m².

Calculating the area of slope AD,

Area = length × AD

= 40 × 2.6 = 104 m².

Calculating the area of slope BC,

Area = Length × BC

= 40 × 2.6 = 104 m².

Total area = 240 + 104 + 104

= 448 m².

Cost rate = ₹ 5 per m²

Calculating the cost of painting the internal surface of the tunnel,

Total cost = Total area × Cost

= 448 × 5

= ₹ 2,240.

Hence, cost = ₹ 2,240.

If the length of each side of a cube is reduced by 25%, then the ratio of the volumes of the original and the new cube is :

1. 64 : 1

2. 4 : 3

3. 64 : 27

4. 32 : 9

Answer

Let the original side of cube be a units and side of cube after reduction be a'.

According to the question,

a' = a - 25% of a = a - 0.25a = 0.75a = $\dfrac{3}{4}a$.

We know that,

Volume of cube = (side)³.

Calculating original volume,

V₁ = a³.

Calculating the volume after reduction,

V₂ = (a')³

= $\Big(\dfrac{3}{4}a\Big)$³

= $\dfrac{27}{64}$ a³

Ratio of original volume to the new volume

V₁ : V₂

a³ : $\dfrac{27}{64}$ a³

$\dfrac{a^3}{\dfrac{27}{64}a^3}$

$\dfrac{64}{27}$

64 : 27.

Hence, option 3 is the correct option.

If the length of each side of a cube is reduced by 50%, then the ratio of the total surface area of the original and the new cube is :

1. 2 : 1

2. 4 : 1

3. 8 : 1

4. 8 : 3

Answer

Let the original side be a units and new side be a' units.

Reduction of 50% :

a' = a - 50% of a = a - $\dfrac{1}{2}a = \dfrac{a}{2}$ units.

We know that,

Total surface area of cube (TSA) = 6(side)².

Calculating the original total surface area of cube,

original TSA = 6a²

Calculating the new total surface area of a cube,

new TSA = 6(a')²

= 6 $\Big(\dfrac{a}{2}\Big)^2$

= 6 × $\dfrac{a^2}{4}$

= $\dfrac{3a^2}{2}$.

Ratio of TSA of original cube to the new cube:

original TSA : new TSA

6a² : $\dfrac{3a^2}{2}$

$\dfrac{6a^2}{\dfrac{3a^2}{2}}$

$\dfrac{4}{1}$

4 : 1.

Hence, option 2 is the correct option.

The length and the breadth of a cuboid are 60 cm and 50 cm respectively. If the total surface area of the cuboid is 14800 cm², then its height is :

1. 40 cm

2. 32 cm

3. 25 cm

4. 30 cm

Answer

Given,

Length = 60 cm

Breadth = 50 cm

Total surface area = 14800 cm².

By formula,

Total surface area of cuboid = 2(lb + bh + hl)

⇒ 14800 = 2(60 × 50 + 50 × h + h × 60)

⇒ 14800 = 2(3000 + 50h + 60h)

⇒ 3000 + 50h + 60h = $\dfrac{14800}{2}$

⇒ 3000 + 110h = 7400

⇒ 110h = 7400 - 3000

⇒ 110h = 4400

⇒ h = $\dfrac{4400}{110}$ = 40 cm.

Hence, option 1 is the correct option.

The base of a cuboid is a square and its height is 4 cm. If the volume of the cuboid is 200 cm³, then the length of the base is :

1. $4\sqrt{2}$ cm

2. 8 cm

3. 5 cm

4. $5\sqrt{2}$ cm

Answer

The base is a square having length and breadth both equal to x cm.

Given,

Height = 4 cm

Volume = 200 cm³.

We know that,

Volume = length × breadth × height

⇒ 200 = x × x × 4

⇒ 200 = x² × 4

⇒ x² = $\dfrac{200}{4}$

⇒ x² = 50

⇒ x = $\sqrt{50}$

⇒ x = $\sqrt{25 × 2}$

⇒ x = $5\sqrt{2}$ cm.

Hence, option 4 is the correct option.

The length of the diagonal of a cube is $16\sqrt{3} \text{ m}$. The volume of the cube is :

1. 4096 m³

2. 4196 m³

3. 3146 m³

4. 4036 m³

Answer

Given,

Diagonal of cube (d) = $16\sqrt{3} m$.

Let side of cube be a meters.

Diagonal (d) = $a\sqrt{3}$

⇒ $16\sqrt{3} = a\sqrt{3}$

⇒ a = 16 m.

Calculating the volume of cube,

Volume of cube = a³

= 16³

= 4096 m³.

Hence, option 1 is the correct option.

The total surface area of a cube is 96 cm². The length of a diagonal of the cube is :

1. 16 cm

2. $8\sqrt{3}$ cm

3. $6\sqrt{3}$ cm

4. $4\sqrt{3}$ cm

Answer

Given,

Total surface area of cube = 96 cm².

By formula,

Total surface of a cube = 6a²

⇒ 96 = 6a²

⇒ a² = $\dfrac{96}{6}$

⇒ a² = 16

⇒ a = $\sqrt{16}$

⇒ a = 4 cm.

Calculating the length of diagonal of a cube,

Diagonal of a cube = $a\sqrt{3} = 4\sqrt{3}$ cm.

Hence, option 4 is the correct option.

The length of the diagonal of a cube is $6\sqrt{6} \text{ cm}$. The total surface area of the cube is :

1. 216 cm²

2. 432 cm²

3. 1296 cm²

4. 1548 cm²

Answer

Given,

Length of the diagonal of a cube = $6\sqrt{6}$ cm

Calculating the side of a cube,

Diagonal of a cube = $a\sqrt{3}$

⇒ $6\sqrt{6} = a\sqrt{3}$

⇒ a = $\dfrac{6\sqrt{6}}{\sqrt{3}}$

⇒ a = $6\sqrt{2}$ cm.

Calculating the total surface area of a cube,

Total surface area of a cube = 6a²

= 6 × $(6\sqrt{2})^2$

= 6 × 36 × 2

= 432 cm².

Hence, option 2 is the correct option.

The volume of a cube is 125 m³. The total surface area of the cube is :

1. 150 m²

2. 250 m²

3. 375 m²

4. 432 m²

Answer

Given,

Volume of cube = 125 m³

Calculating the side of a cube,

Volume of cube = a³

⇒ 125 = a³

⇒ a = $\sqrt[3]{125}$

⇒ a = 5 m.

Calculating the total surface area of the cube,

Total surface area of the cube = 6a²

= 6 × 5²

= 6 × 25

= 150 m².

Hence, option 1 is the correct option.

460 cm² of metal sheet is needed to make a closed box of length 12 cm and height 5 cm. The breadth of the box is :

1. 10 cm

2. 12 cm

3. 14 cm

4. 16 cm

Answer

Given,

Length(l) = 12 cm

Height(h) = 5 cm

Total surface area of sheet required to make a closed box = 460 cm²

Let breadth of the box be b cm.

Calculating the breadth of the box,

Total surface area = 2(lb + bh + hl)

⇒ 460 = 2(12 × b + b × 5 + 5 × 12)

⇒ 460 = 2(12b + 5b + 60)

⇒ 460 = 2(17b + 60)

⇒ 17b + 60 = $\dfrac{460}{2}$

⇒ 17b + 60 = 230

⇒ 17b = 230 - 60

⇒ 17b = 170

⇒ b = $\dfrac{170}{17}$ = 10 cm.

Hence, option 1 is the correct option.

The length of the largest rod that can be kept in a room of length 5 m, breadth 4 m and height 3 m is :

1. $3\sqrt{2}\text{ m}$

2. $5\sqrt{2}\text{ m}$

3. $7\sqrt{2}\text{ m}$

4. $\sqrt{2}\text{ m}$

Answer

Given,

Length (l) = 5 m

Breadth (b) = 4 m

Height (h) = 3 m

Length of the largest rod that can fit in the room is the diagonal of the room.

Calculating the length of the diagonal of the room (cuboid),

$\text{Diagonal of room (d)} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{5^2 + 4^2 + 3^2} \\[1em] = \sqrt{25 + 16 + 9} \\[1em] = \sqrt{50} \\[1em] = \sqrt{25 × 2} \\[1em] = 5 \sqrt{2} \text{ m}.$

Hence, option 2 is the correct option.

The sum of the length, breadth and height of a cuboid is 24 cm, and the length of its diagonal is 15 cm. The area of its total surface is :

1. 348 cm²

2. 349 cm²

3. 350 cm²

4. 351 cm²

Answer

Given,

Sum of dimensions : l + b + h = 24

Length of diagonal (d) = 15 cm.

We know that,

Diagonal of cuboid (d) = $\sqrt{l^2 + b^2 + h^2}$

⇒ 15 = $\sqrt{l^2 + b^2 + h^2}$

Squaring on both sides,

⇒ 15² = $(\sqrt{l^2 + b^2 + h^2})^2$

⇒ 225 = l² + b² + h²

By formula,

(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

By substituting the values we get,

⇒ (24)² = 225 + 2(lb + bh + hl)

⇒ 576 = 225 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 576 - 225

⇒ 2(lb + bh + hl) = 351

Since, Total surface area of cuboid = 2(lb + bh + hl)

∴ Total surface area of cuboid = 351 cm².

Hence, option 4 is the correct option.

The volume of a cuboid whose length, breadth and height are 8 cm, 5 cm and 3 cm respectively is :

1. 120 cm³

2. 122 cm³

3. 124 cm³

4. 128 cm³

Answer

Given,

Length (l) = 8 cm

Breadth (b) = 5 cm

Height (h) = 3 cm

Calculating the volume of cuboid,

Volume of cuboid = l × b × h

= 8 × 5 × 3

= 120 cm³.

Hence, option 1 is the correct option.

The area of cross-section of a hosepipe is 3 cm². Water flows through it at a speed of 50 cm/sec. How many litres of water flows out of it in one minute?

1. 7 litres

2. 8 litres

3. 9 litres

4. 11 litres

Answer

Given,

Area of cross-section = 3 cm².

Speed of water = 50 cm/sec.

1 minute = 60 seconds

Calculating the distance traveled by water in 1 minute,

Distance = Speed × Time

Distance = 50 cm/sec × 60 sec

= 3000 cm.

Calculating the volume of water,

Volume = Area of cross-section × Distance by water in 1 minute

= 3 × 3000

= 9000 cm³.

1000 cm³ = 1 litre

∴ 9000 cm³ = $\dfrac{9000}{1000}$ litres

= 9 litres.

Hence, option 3 is the correct option.

The sum of the length, breadth and height of a cuboid is 41 cm. If the length of its diagonal is 25 cm, then its total surface area is :

1. 1050 cm²

2. 1052 cm²

3. 1054 cm²

4. 1056 cm²

Answer

Given,

Sum of sides : l + b + h = 41 cm.

Length of diagonal (d) = 25 cm.

We know that,

Diagonal of cuboid (d) = $\sqrt{l^2 + b^2 + h^2}$

⇒ 25 = $\sqrt{l^2 + b^2 + h^2}$

Squaring on both sides,

⇒ 25² = $(\sqrt{l^2 + b^2 + h^2})^2$

⇒ 625 = l² + b² + h²

By formula,

(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

By substituting the values we get,

⇒ (41)² = 625 + 2(lb + bh + hl)

⇒ 1681 = 625 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 1681 - 625

⇒ 2(lb + bh + hl) = 1056

Since, Total surface area of cuboid = 2(lb + bh + hl)

∴ Total surface area of cuboid = 1056 cm².

Hence, option 4 is the correct option.

The weight of a rectangular box with lid is 60 kg. The box filled with water weighs 600 kg. The weight of 1 litre of water is 1.2 kg. If the thickness of the box is 5 cm, and the external length and breadth of the box are 16 dm and 8.5 dm respectively, then the external height of the box is :

1. 5 dm

2. 6 dm

3. 7 dm

4. 8 dm

Answer

Given,

External length of rectangular box = 16 dm

External breadth of rectangular box = 8.5 dm

Thickness = 5 cm = 0.5 dm

Density of water = 1.2 kg/litre

Weight of box with lid = 60 kg

Weight of box filled with water = 600 kg

Calculating the weight of the water,

Weight of the water = Weight of filled box - Weight of empty box.

= 600 - 60 = 540 kg.

Calculating the volume of water,

Volume of water = $\dfrac{\text{Weight of the water}}{\text{Density of the water}}$

= $\dfrac{540}{1.2}$

= 450 litres.

1 litre = 1 dm³

∴ 450 litres = 450 dm³.

Calculating internal dimensions,

Internal length = External length - 2 × Thickness

= 16 - (2 × 0.5)

= 16 - 1 = 15 dm.

Internal breadth = External breadth - 2 × Thickness

= 8.5 - (2 × 0.5)

= 8.5 - 1 = 7.5 dm.

Calculating the internal volume of rectangular box,

Volume of cuboid = l × b × h

⇒ 450 = 15 × 7.5 × h

⇒ 450 = 112.5 × h

⇒ h = $\dfrac{450}{112.5}$

⇒ h = 4

∴ Internal height = 4 dm.

Calculating the external height of the box,

Since the box has thickness at both the top and bottom,

So,

External height = Internal height + 2(Thickness)

= 4 + 2(0.5)

= 4 + 1

= 5 dm.

Hence, option 1 is the correct option.

Case Study
The length, breadth and height of Kavita's bedroom are 6 m, 4 m and 3 m respectively. It has two equal windows, each of dimensions 1 m × 0.5 m. It also has a door of dimensions 2 m × 1 m.

Based on the above information, answer the following questions:

1. Area occupied by the door and the two windows is :
   (a) 1 m²
   (b) 2 m²
   (c) 2.5 m²
   (d) 3 m²

2. Kavita wants to whitewash the four walls of the room. Area to be whitewashed is :
   (a) 60 m²
   (b) 57 m²
   (c) 55 m²
   (d) 50 m²

3. Square tiles each of side 50 cm are laid on the floor of the room. The number of such tiles laid is :
   (a) 100
   (b) 98
   (c) 96
   (d) 72

4. Volume of air contained in the room is :
   (a) 72 m³
   (b) 70 m³
   (c) 60 m³
   (d) 52 m³

5. The length of the longest rod (to the nearest m) that can be placed in the room is :
   (a) 4 m
   (b) 5 m
   (c) 8 m
   (d) 7 m

Answer

Given,

Length (l) = 6 m

Breadth (b) = 4 m

Height (h) = 3 m

Dimension of each window = 1 m × 0.5 m

Dimension of door = 2 m × 1 m

1. Area of one window = 1 × 0.5 = 0.5 m²

∴ Area of two windows = 2 (1 × 0.5) = 1 m²

Area of one door = 2 × 1 = 2 m².

Total area = 2 + 1 = 3 m².

Hence, option (d) is the correct option.

2. Area to be whitewashed = Area of four walls - Area of two windows and a door.

Calculating the area of four walls,

We know that,

Area of four walls = 2h(l + b)

= 2 × 3 × (6 + 4)

= 6 × 10

= 60 m².

Area of two windows and a door = 3 m².

Area to be whitewashed = 60 - 3 = 57 m².

Hence, option (b) is the correct option.

3. Given,

Tile side = 50 cm = 0.5 m

Calculating the floor area,

Area of floor = Length × Breadth

= 6 × 4

= 24 m².

Area of square tile = (side)²

= (0.5)²

= 0.25 m²

Number of tiles = $\dfrac{\text{Area of floor}}{\text{Area of each tile}}$

= $\dfrac{24}{0.25}$

= 96.

Hence, option (c) is the correct option.

4. Volume of air contained in the room = l × b × h

= 6 × 4 × 3

= 72 m³.

Hence, option (a) is the correct option.

5. Length of the longest rod that can be placed in the room = Diagonal of room

$\text{Diagonal (d)} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{6^2 + 4^2 + 3^2} \\[1em] = \sqrt{36 + 16 + 9} \\[1em] = \sqrt{61} \\[1em] = 7.81 \approx 8 \text{ m}.$

Hence, option (c) is the correct option.

Case Study
Manish is a carpenter. One day, he made an open cubical box of internal edge 18 cm. The thickness of the plywood is 1 cm. He painted the inner surface of the box black and the outer lateral surfaces as green.

Based on the above information, answer the following questions:

1. Total area to be painted black is :
   (a) 1600 cm²
   (b) 1620 cm²
   (c) 1650 cm²
   (d) 1680 cm²

2. Length of the outer edge of the box is :
   (a) 20 cm
   (b) 19 cm
   (c) 18 cm
   (d) 18.5 cm

3. Total outer lateral surface area to be painted green is :
   (a) 1500 cm²
   (b) 1510 cm²
   (c) 1520 cm²
   (d) 1600 cm²

4. Total outer surface area of the box (excluding the top) is :
   (a) 1900 cm²
   (b) 1920 cm²
   (c) 2000 cm²
   (d) 2320 cm²

5. Length of the longest rod that can fit inside the box is :
   (a) 18 cm
   (b) 19 cm
   (c) 20 cm
   (d) 3 cm

Answer

Given,

Internal edge of cube = 18 cm

Thickness of plywood = 1 cm

1. Inside the open cubical box there are 5 faces (4 sides + 1 bottom)

Calculating area of one face,

Area = (side)²

= (18)²

= 324 cm².

Calculating total inner area,

Total inner area = 5 × 324 = 1620 cm².

Thus, the total area to be painted black is 1620 cm².

Hence, option (b) is the correct option.

2. Thickness 1 cm will be on both sides of the box.

∴ Outer edge = Internal edge of cube + 2(Thickness)

= 18 + 2(1)

= 18 + 2 = 20 cm.

Hence, option (a) is the correct option.

3. Outer edge = 20 cm

Lateral surface area refers to the 4 side walls only (not the bottom).

Outer lateral area = 4 × (outer edge)²

= 4 × (20)²

= 4 × 400

= 1600 cm².

Hence, option (d) is the correct option.

4. Total outer surface area of the box (excluding the top)

Total faces excluding the top = 5

Area of one face = (20)² = 400 cm².

Calculating the area of all the faces,

Area of all the faces = 5 × 400 = 2000 cm².

Hence, option (c) is the correct option.

5. The rod must lie completely inside the open box, so the maximum straight length inside the box equals the internal edge length = 18 cm.

Hence, option (a) is the correct option.

Assertion (A) : The length of the longest rod that can be put in a room of dimensions 10 m × 10 m × 5 m is 15 m.

Reason (R) : Length of the diagonal of a cuboid = $\sqrt{l^2 + b^2 + h^2}$

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

Room dimensions = 10 m × 10 m × 5 m

The longest rod that can be placed in a room = Diagonal of room.

We know that,

Diagonal of cuboid = $\sqrt{l^2 + b^2 + h^2}$

Diagonal of room = $\sqrt{10^2 + 10^2 + 5^2}$

= $\sqrt{100 + 100 + 25}$

= $\sqrt{225}$

= 15 m.

So length of the longest rod that can be placed in the room = 15 m.

∴ Assertion (A) is true.

By formula,

Length of the diagonal of cuboid = $\sqrt{l^2 + b^2 + h^2}$

∴ Reason (R) is true.

Both Assertion (A) and Reason (R) are true.

Hence, option 3 is the correct option.

Assertion (A) : The perimeter of one face of a cube is 20 cm. Its volume is 64 cm³.

Reason (R) : Volume of a cube of edge a is given by a³.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

Perimeter = 20 cm

Each face of a cube is a square.

Let the side of square be a cm.

Perimeter of a square = 4 × side.

⇒ 20 = 4a

⇒ a = $\dfrac{20}{4}$

⇒ a = 5 cm.

Calculating the volume of a cube,

Volume of cube = a³

= 5³

= 125 cm³.

∴ Assertion (A) is false.

Given,

Edge of a cube = a

By formula,

Volume of a cube = a³

∴ Reason (R) is true.

Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

The area of a base of a cuboidal water tank is 6500 cm² and the volume of water contained in it is 2.6 m³. The depth of the water is :

1. 5 m

2. 4 m

3. 3.5 m

4. 3 m

Answer

Given,

Volume of water (V) = 2.6 m³

Base area (A) = 6500 cm²

= $\dfrac{6500}{10000}$ = 0.65 m².

Let the depth of the water be 'd'.

Calculating the depth of the water,

Volume of water = Base area × Depth of water

⇒ 2.6 = 0.65 × d

⇒ d = $\dfrac{2.6}{0.65}$ = 4 m.

Hence, option 2 is the correct option.

The volume of a cuboid of dimensions x, y and z is V and its surface area is S. The value of $\dfrac{1}{V}$ is :

1. $\dfrac{2}{S}\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big)$

2. $\dfrac{S}{2} \Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big)$

3. $\dfrac{3}{S} \Big(\dfrac{1}{x} - \dfrac{1}{y} - \dfrac{1}{z}\Big)$

4. $\dfrac{S}{3} \Big(\dfrac{a + b + c}{abc}\Big)$

Answer

Given,

Volume of cuboid = V

Surface area of cuboid = S

Dimensions of cuboid = x, y, z

Volume of cuboid = length × breadth × height

V = xyz

Addition of the reciprocal values of the given dimensions,

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$

⇒ $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$ = $\dfrac{yz + xz + xy}{xyz}$

By substituting the value V = xyz, we get

⇒ $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$ = $\dfrac{yz + xz + xy}{V}$

⇒ V$\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big)$ = yz + xz + xy ..........(1)

We know that,

Surface area of cuboid = 2 (lb + bh + hl)

S = 2(xy + yz + zx)

⇒ $\dfrac{S}{2}$ = xy + yz + zx

By substituting the value of xy + yz + zx in equation (1), we get :

$\Rightarrow \dfrac{S}{2} = V\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big) \\[1em] \Rightarrow \dfrac{1}{V} = \dfrac{2}{S}\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big).$

Hence, option 1 is the correct option.

If a cube has surface area S and volume V, then the volume of the cube of surface area 2S is :

1. $\sqrt{2}V$

2. 2V

3. $2\sqrt{2}V$

4. $\dfrac{V}{\sqrt{2}}$

Answer

Let the side of the first cube be a units and volume be V cubic units.

We know that,

Surface area of the cube = 6a²

⇒ S = 6a²

⇒ a² = $\dfrac{S}{6}$

⇒ a = $\sqrt{\dfrac{S}{6}}$

Calculating the volume of first cube,

Volume of cube (V) = a³

Substituting the value of a, we get,

V = $\Big(\sqrt{\dfrac{S}{6}}\Big)^3$

V = $\Big(\dfrac{S}{6}\Big)^\dfrac{3}{2}$ .......(1)

Let the side of second cube be b units and volume be V' cubic units.

Surface area = 6b²

⇒ 2S = 6b²

⇒ S = 3b²

⇒ b² = $\dfrac{S}{3}$

⇒ b = $\sqrt{\dfrac{S}{3}}$

Volume of second cube (V') = b³

Substituting the value of b, we get

V' = $\Big(\sqrt{\dfrac{S}{3}}\Big)^3$

V' = $\Big(\dfrac{S}{3}\Big)^\dfrac{3}{2}$.......(2)

Taking the ratio of equation (1) and (2) we get,

$\Rightarrow \dfrac{V}{V'} = \dfrac{\Big(\dfrac{S}{6}\Big)^\dfrac{3}{2}}{\Big(\dfrac{S}{3}\Big)^\dfrac{3}{2}} \\[1em] \Rightarrow \dfrac{V}{V'} = \Big(\dfrac{\dfrac{S}{6}}{\dfrac{S}{3}}\Big)^{\dfrac{3}{2}} \\[1em] \Rightarrow \dfrac{V}{V'} = \Big(\dfrac{3}{6}\Big)^\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{V}{V'} = \Big(\dfrac{1}{2}\Big)^\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{V}{V'} = \dfrac{1}{2\sqrt{2}} \\[1em] \Rightarrow V' = 2 \sqrt{2}V$

Hence, option 3 is the correct option.

In the figure, four faces of the solid are shaded. The shaded area is :

1. 100 cm²

2. 150 cm²

3. 200 cm²

4. 300 cm²

Answer

Given,

Total height = 10 cm

Inner vertical step height = 6 cm

So lower height = 10 - 6 = 4 cm

Total front width = 10 cm

Top block width = 5 cm

Bottom block width = 5 cm.

Calculating the area of top face of the tall block,

Area of rectangle = length × breadth

= 5 × 10 = 50 cm².

Calculating the area of top face of the lower block,

Area of rectangle = 5 × 10 = 50 cm².

Calculating the vertical inner face (step)

Length = 6 cm

Breadth = 10 cm

Area = 6 × 10 = 60 cm².

Calculating the area of bottom side outer face,

Length = 10 cm

Breadth = 4 cm

Area = 4 × 10 = 40 cm².

Total shaded area = 50 + 50 + 60 + 40

= 200 cm².

Hence, option 3 is the correct option.

In the figure, there are fifty, 500 rupee notes in the bundle. Volume of the bundle is 98 cm³, length = 14 cm and breadth = 7 cm. The thickness of 1 note is :

1. 1 mm

2. 0.1 mm

3. 2 mm

4. 0.2 mm

Answer

Given,

Number of notes = 50

Volume of bundle = 98 cm³

Length (l) = 14 cm

Breadth (b) = 7 cm

Let the thickness of the bundle be h cm.

Calculating the thickness of the bundle,

Volume of cuboid = length × breadth × height

⇒ 98 = 14 × 7 × h

⇒ 98 = 98 × h

⇒ h = $\dfrac{98}{98}$ = 1 cm.

So total thickness of 50 notes = 1 cm.

∴ Thickness of 1 note = $\dfrac{1}{50}$ cm = 0.02 cm.

1 cm = 10 mm

∴ 0.02 cm = 0.02 × 10 = 0.2 mm.

Hence, option 4 is the correct option.

In the figure (i) below, a cuboid is shown. The surface area of three faces are marked as x, y and z. In figure (ii), two such cuboids are stacked. Find the surface area of the new cuboid in figure (ii) in terms of x, y and z.

Answer

Given,

Let the dimensions of the cuboid be :

Length = l

Breadth = b

Height = h

From the figure (i) the three faces have areas:

x = length × height = lh

y = breadth × height = bh

z = length × breadth = lb

When two identical cuboids, are stacked the height is doubled (2h), length (l) and breadth (b) will remain same.

In (ii) figure,

Surface area of top face = length × breadth = lb = z

Surface area of right vertical face = length × height = 2lh = 2x

Surface area of left vertical face = breadth × height = 2bh = 2y

Total surface area = 2x + 2y + z.

Hence, surface area of new cuboid = 2x + 2y + z.

A Rubik's cube is made up of several small cubes. Side lengths of each small cube is x. Find the outer surface area of the cube present at one of the corners of the Rubik's cube.

Answer

Given,

Side lengths of each small cube = x units.

From figure,

At each corner of a Rubik's cube is a small cube of length x units.

Outer surface area of cube = 6x²

Hence, outer surface area of the cube at one of the corners = 6x² sq. units.