Brilliant’s Composite Mathematics Class 6 Solutions Chapter 18

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Brilliant’s Composite Mathematics Class 6 Solutions Chapter 18 Volumes and Surface Area

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 18, Volumes and Surface Area. Here students can easily find step by step solutions of all the problems for Volumes and Surface Area, Exercise 18.1 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

Volumes and Surface Area Exercise 18.1 Solution

Question no – (1)

Solution :

In the question given,

Length = 3 cm

Breadth = 4 cm

Height = 5 cm

Volume of the cuboid

= 3 × 4 × 5

= 60 cm³

Surface Area of the cuboid,

= 2 [3 × 4 + 3 × 5 + 4 × 5] cm²

= 2 [12 +15+20]

= 2 × 47

= 94 cm²

Question no – (2)

Solution :

From the question we get,

Length = 36 m

Breadth = 12 m

Height = 1 m

∴ Volume of the Cuboid,

= (36 × 12 × 1)

= 432 m³

∴ Surface Area of the Cuboid,

= 2 [36 × 12 + 36× 1 + 12 × 1] cm²

= 2[432 + 36 + 12]

= 2 × 480

= 960 m²

Question no – (3)

Solution : 

From the question we get,

Length 12 cm,

Breadth = 16 cm,

Height = 8 cm

Now,

∴ Volume of the Cuboid,

= (12 ×16 × 8)

= 1536 cm³

∴ Surface Area of the Cuboid,

= 2 [12 × 16 + 12 × 8 + 16 × 8] cm²

= 2 × [192 + 96 + 128] cm²

= 2 × 416

= 832 cm² 

Question no – (4)

Solution :

In the given question,

Length = 4.6 cm

Breadth = 2.3 cm

Height = 5.7 cm

∴ Volume of the Cuboid,

= 4.6 × 2.3 × 5.7

= 60.306 cm³

∴ Surface Area of the Cuboid,

= 2 [4.6 × 2.3 + 4.6 × 5.7 + 2.3 × 5.7] cm²

= 2 × [10.58 + 26.22 + 13.11]

= 2 × 49.91

= 99.82 cm²

Question no – (5)

Solution :

(i) As we know that,

Volume of Cube = (edge)³

Surface Area of Cube = 6 × (edge)²

∴ Volume of the Cube,

= (7)³

= 343 cm³

∴ Surface Area of the Cube,

= 6 × 7²

= 6 × 49

= 294 cm²

(ii)  As we know that,

Volume of Cube = (edge)³

Surface Area of Cube = 6 × (edge)²

Now,

∴ Volume of the Cube,

= (4)³

= 64 m²

∴ Surface Area of the Cube,

= 6 × 4²

= 6 × 16

= 96 m²

(iii) As we know that,

Volume of Cube = (edge)³

Surface Area of Cube = 6 × (edge)²

Now,

∴ Volume of the Cube,

= (2.5)³

= 15.625 cm³

∴ Surface Area of the Cube,

= 6 × (2.5)²

= 6 × 6.25

= 37.5 cm²

(iv) As we know that,

Volume of Cube = (edge)³

Surface Area of Cube = 6 × (edge)²

Now, Volume of the Cube,

= (4.5)³

= 91.125 cm³

∴ Surface Area of the Cube,

= 6 × (4.5)²

= 6 × 20.25

= 121.5 cm²

Question no – (6)

Solution :

∴ Volume 4 cm cubes,

= (4)³

= 4 × 4 × 4 cm³

∴ Volume of 16 cm cubes,

= (16)³

= 16 × 16 × 16 cm³

∴ Cube can obtained,

= 16 ×  16 ×  16/ 4 ×   4 ×   4

= 64

Thus, 64 cubes can be obtained from the cube.

Question no – (7)

Solution :

As we know that,

Length = Volume/Breadth × Height

∴ Length

= 2912 × 10 × 10/12.8 × 6.5

= 35 cm

Therefore, the length will be 35 cm.

Question no – (8)

Solution :

As we know,

Volume = length × breadth × height

Volume of air,

= 9.5 × 8 × 4.2

= 319.2 m

Thus, the volume of air contained in room is 319.2 m³.

Question no – (9)

Solution :

Let x be the edge of the cube.

∴ 6 × x² = 384

x² = 384/6 = 64

x = √64

= 8

∴ Edge,

= 8 cm

∴ Volume,

= (8)³

= 512 cm³

Therefore, its volume is 512 cm³ 

Question no – (10)

Solution :

Surface Area of tank

= 2 × [12 ×9 + 12 ×4 + 9 ×4] m²

= 2 × [108 + 48 + 36] m²

= 2 × 192

= 384 m²

Now the Cost of sheet,

= 384/2 × 5

= 960 Rs

Therefore, the cost of iron sheet is Rs 960.

Question no – (11)

Solution :

First we need to find volume of the room,

= 12.25 × 100

= 1225 m³

Hence the height of the room,

= volume/length × breadth

= 1225 × 10 / 25 × 9.8

= 5 m

So, the height of the room is 5 m.

Question no – (12)

Solution :

∴ Volume of a match box,

= 4 × 2.5/10 × 1.5/10 cm³

= 15 cm³

∴ Volume of packet,

= 15 × 220

= 3300 cm³

∴ Number of boxes,

= 120 × 99 × 50 / 15 × 220

= 180

Hence, The volume of the packet will be 3300 cm3 and 180 such boxes can be placed.

Question no – (13)

Solution :

We know that,

Depth = volume/length × breadth

Depth of the water

= 65000/80 × 35 dm

= 20 dm

= 2 m ….(10d m = 1 m)

Therefore, the depth of the water will be 2 m.

Question no – (14)

Solution :

Area of field,

= 6 × 10000 m²

= 6 × 10000 × 10 × 10 dm²

Volume of rainfall,

= Area × depth

= 6 × 10000 × 10 × 10 × 6/10 dm³

= 36 × 100000 dm³

= 3600000 liters

Hence, 3600000 liters litres of water will fall.

Question no – (15)

Solution :

First,  dimension of each container

60 cm = 0.6 m,

45 m = 0.45 m

56 cm = 0.56 m

∴ Surface Area,

= 2 × [0.6 × 4.5 + 0.45 × 0.56 + 0.6 × 0.56]

= 2 × [0.27 + 0.252 + 0.336]

= 2 × 0.858

= 1.716 m²

∴ Surface Area of 240 Containers,

= 1.716 × 240

= 411.84 m²

∴ Cost of painting,

= 411.84 × 2.50

= 1029.6 Rs.

Therefore, the cost of the painting will be 1029.6 Rs.

Question no – (16)

Solution :

Surface Area of rectangular box,

= 2 × [12 × 8 + 12 × 6 + 18 × 6] cm²

= 2 [96 + 72 + 48] cm²

= 2 × 216

= 432 cm²

Hence, the area of the card board will be 432 cm²

Question no – (17)

Solution :

Area of granary,

= 8 × 6 × 3

= 144 m³

∴ Number of bags,

= 144 × 100/0.5

= 2880/13

= 221.53

Thus, 221 bags can be stored in the granary.

Question no – (18)

Solution :

As we know,

Area of iron sheet used in the tank = Area of four walls + area of base

= 2 (lh + bh)  + lb

= 2lh + 2bh + lb

= (2 × 12 × 5) + (2 × 8 × 5) + (12 × 8)

= 120 + 80 + 96

= 296 m²

∴ Length of iron sheet,

= area/width of sheet

= 296/4

= 74 m

∴ Cost of iron sheet,

= 74 × 3.5

= 259 Rs.

Therefore, the cost of iron sheet will be Rs 259

Question no – (19)

Solution :

Let,  the first edge = x

After 50% increased edge

= x × 50/100 + x

= 3x/2

First surface,

= 6x²

∴ After 50% increased surface

= 6 × (3x/2)²

= 6 × 9x² /4

= 27x²/2

∴ Increased,

= 27x²/2 – 6x²

= 15x²/2

∴ Increased percent (%) in surface area,

= 15x²/2/6x² × 100

= 15x²/2 ×6x² ×100

= 125%

Thus, 125% increase in the surface area of the circle.

Question no – (20)

Solution :

First Area of Cube block,

= (15)³

= 225

Now new Cubes Area,

= (3)³

= 27

∴ Number of Cubes,

= 15 × 15 × 15 /3 × 3 × 3

= 125

Therefore, 125 cubes can be formed.

Question no – (21)

Solution :

First we find Area of brick,

= 25 × 12 × 6

= 1500 cm²

Now Area of wall,

= 7.5 m × 3.6 m × 45 m

= 750 × 360 × 45 cm³

= 12150000 cm³

∴ Number of bricks

= 750 × 360 × 45/25 × 12 × 6

= 6750

Hence, 6750 bricks will be required to build the wall.

Next Chapter Solution : 

👉 Chapter 19 👈