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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 cm^{3}

**Surface Area of the cuboid,**

= 2 [3 × 4 + 3 × 5 + 4 × 5] cm^{2}

= 2 [12 +15+20]

= 2 × 47

= 94 cm^{2}

**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 m^{3}

**∴**** Surface Area of the Cuboid,**

= 2 [36 × 12 + 36× 1 + 12 × 1] cm^{2}

= 2[432 + 36 + 12]

= 2 × 480

= 960 m^{2}

**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 cm^{3}

**∴**** Surface Area of the Cuboid,**

= 2 [12 × 16 + 12 × 8 + 16 × 8] cm^{2}

= 2 × [192 + 96 + 128] cm^{2}

= 2 × 416

= 832 cm^{2 }

**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 cm^{3}

**∴**** Surface Area of the Cuboid,**

= 2 [4.6 × 2.3 + 4.6 × 5.7 + 2.3 × 5.7] cm^{2}

= 2 × [10.58 + 26.22 + 13.11]

= 2 × 49.91

= 99.82 cm^{2}

**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 cm^{3}

**∴** Surface Area of the Cube,

= 6 × 7^{2}

= 6 × 49

= 294 cm^{2}

**(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 m^{2}

**∴** Surface Area of the Cube,

= 6 × 4^{2}

= 6 × 16

= 96 m^{2}

**(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 cm^{3}

**∴** Surface Area of the Cube,

= 6 × (2.5)^{2}

= 6 × 6.25

= 37.5 cm^{2}

**(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 cm^{3}

**∴** Surface Area of the Cube,

= 6 × (4.5)^{2}

= 6 × 20.25

= 121.5 cm^{2}

**Question no – (6)**

**Solution :**

**∴** Volume 4 cm cubes,

= (4)^{3}

= 4 × 4 × 4 cm^{3}

**∴** Volume of 16 cm cubes,

= (16)^{3}

= 16 × 16 × 16 cm^{3}

**∴** 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 × x^{2} = 384

x^{2} = 384/6 = 64

x = √64

= 8

**∴** Edge,

= 8 cm

**∴** Volume,

= (8)^{3}

= 512 cm^{3}

Therefore, its volume is **512 cm**^{3 }

**Question no – (10)**

**Solution :**

Surface Area of tank

= 2 × [12 ×9 + 12 ×4 + 9 ×4] m^{2}

= 2 × [108 + 48 + 36] m^{2}

= 2 × 192

= 384 m^{2}

**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 m^{3}

**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 cm^{3}

= 15 cm^{3}

**∴** Volume of packet,

= 15 × 220

= 3300 cm^{3}

**∴** 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 m^{2}

= 6 × 10000 × 10 × 10 dm^{2}

**Volume of rainfall,**

= Area × depth

= 6 × 10000 × 10 × 10 × 6/10 dm^{3}

= 36 × 100000 dm^{3}

= 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 m^{2}

**∴**** Surface Area of 240 Containers,**

= 1.716 × 240

= 411.84 m^{2}

**∴**** 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] cm^{2}

= 2 [96 + 72 + 48] cm^{2}

= 2 × 216

= 432 cm^{2}

Hence, the area of the card board will be 432 cm^{2}

**Question no – (17)**

**Solution :**

**Area of granary,**

= 8 × 6 × 3

= 144 m^{3}

**∴**** 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 m^{2}

**∴** 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,

= 6x^{2}

**∴ **After 50% increased surface

= 6 × (3x/2)^{2}

= 6 × 9x^{2} /4

= 27x^{2}/2

**∴ **Increased,

= 27x^{2}/2 – 6x^{2}

= 15x^{2}/2

**∴**** Increased percent (%) in surface area,**

= 15x^{2}/2/6x^{2} × 100

= 15x^{2}/2 ×6x^{2} ×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 cm^{2}

**Now Area of wall,**

= 7.5 m × 3.6 m × 45 m

= 750 × 360 × 45 cm^{3}

= 12150000 cm^{3}

**∴**** 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 👈