**Rd Sharma Solutions Class 8 Chapter 21 Mensuration – II Volumes and Surface Areas of a Cuboid and a Cube**

Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 21, Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube). Here students can easily find Exercise wise solution for chapter 21, Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube). Students will find proper solutions for Exercise 21.1, 21.2, 21.3 and 21.4 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

**Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube) Exercise 21.1 Solution :**

**Question no – (1)**

**Solution : **

**(i)** Given in the question,

Length = 12 cm,

breadth = 8 cm,

height = 6 cm

∴ Volume is,

= (12 × 8 × 6) …(we know Volume = l × b × h)

= 576 cm^{3}

Therefore, the volume of a cuboid is 576 cm^{3}

**(ii)** In the given question,

Length = 1.2 m,

Breadth = 30 cm,

Height = 15 cm

∴ Volume is,

= (1200 × 80 × 15) (Volume = l × b × h)

= 540000 cm^{3}

Therefore, the volume of a cuboid is 540000 cm^{3}

**(iii)** In the question we get,

Length = 15 cm,

Breadth = 2.5 dm,

Height = 8 cm

∴ Volume is,

= (15 × 2500 × 8) (∵ Volume = l × b × h)

= 3000 cm^{3}

Therefore, the volume of a cuboid is 3000 cm^{3}

**Question no – (2) **

**Solution : **

**(i)** Cube whose side is = 4 cm

**∴** Volume is,

4 = 4 × 4 × 4

= 64 cm^{3}

Therefore, the volume of a cube is 64 cm^{3}

**(ii)** Cube whose side is = 8 cm

**∴** Volume is,

8 = 8 × 8 × 8

= 512 cm^{3}

So, the volume of a cube is 512 cm^{3}

**(iii)** Cube whose side is = 1.5 dm

**∴** Volume is,

= 1.5 dm = 1.5 × 1.5 × 1.5

= 3375 cm^{3}

Thus, the volume of a cube is 3375 cm^{3}

**(iv)** Cube whose side is = 1.2 m

**∴** Volume is,

= 1.2 m = 1.2 × 1.2 × 1.2

= 1.728 m^{3}

Hence, the volume of a cube is 1.728 m^{3}

**(v)** Cube whose side is = 25 mm

**∴** Volume is,

= 25 m = 25 × 25 × 25

= 15625 mm^{3}

Therefore, the volume of a cube is 15625 mm^{3}

**Question no – (3) **

**Solution :**

Given in the question,

Volume 100 cm^{3},

Length = 5 cm

Breadth = 4 cm

Let, height = h

**∴** According to questions,

h × l × b = 100

or, h = 100/l × b = 100/5 × 4 [∴ l = 5 cm b = 4 cm]

or, h = 5 cm

Therefore, the height of a cuboid will be 5 cm.

**Question no – (4) **

**Solution :**

In the question we get,

Length is = 10 cm

Breadth is = 5 cm

Let, height = h

**∴** According to questions,

= h × l × b = 300

or, h = 300/l × b = 300/10 × 5 [∴ l = 10, b = 5]

or, h = 6 cm

Therefore, the height of the vessel is 6 cm.

**Question no – (5) **

**Solution :**

As per the question,

Length is = 8 cm

Breadth is = 50 cm

Can hold = 4 litres of milk

**∴** 4 litres = 4 × 1000 = 4000 cm^{3}

Let, h × 8 × 50 = 400

or, h = 4000/8 × 50

= 10 cm

Hence, its height should be 10 cm.

**Question no – (6) **

**Solution :**

Given in the question,

Length is = 4 cm

Breadth is = 3 cm

Contain wood = 36 cm^{3}

Let, height = h

**∴** According to questions,

h × 4 × 3 = 36

or, h = 36/4 × 3

**∴** h = 3 cm

Therefore, the height cuboidal wooden block will be 3 cm.

**Question no – (7) **

**Solution :**

**(i)** The volume of a cube, if its edge is halved will be 1/8 times.

**(ii)** The volume of a cube, if its edge is trebled will be 27 times.

**Question no – (8) **

**Solution :**

**(i)** If Length is doubled, height is same and breadth is halved than the volume of a cuboid will be __Same.__

**(ii)** If length is doubled, height is doubled and breadth is same then the the volume will be__ 4 times.__

**Question no – (9) **

**Solution :**

Volume_{1} = 5 × 6 × 7 = 210 cm^{3}

Volume _{2} = 4 × 7 × 8 = 224 cm^{3}

Volumes_{3} = 2 × 3 × 13 = 78 cm^{3}

**∴ **Total volume = 512 cm^{3}

Method cubes volume = 512 cm^{3}

**∴ **method cubes h × l × b = 512 = × 8 × 8 × 8

**∴** Height = h = 8 cm

**Question no – (10) **

**Solution :**

Volume of iron piece,

= 50 × 40 × 10

= 20000 cm^{3}

1 cm^{3} of iron weights = 8 gm.

**∴** 20000 cm^{3} of iron weights,

= 8 × 20000

= 160000 gm.

= 160 kg

Therefore, the weight of the solid iron piece will be 160 kg.

**Question no – (11) **

**Solution :**

We know that,

3 m = 300 cm

**∴** Volume of wooden log,

= 300 × 75 × 50

= 1,125,000 cm^{3}

1 cubical volume,

= 25 × 25 × 25

= 15,625 cm^{3}

**∴** No of blocks volume,

= 1,125,000/15,625

= 72

Hence, 72 cubical block can be cut.

**Question no – (12) **

**Solution :**

Given in the question,

Length = 9 cm

Breadth = 4 cm

Height = 3.5 cm

Volume = 1.5 cm^{3}

**∴** Block’s volume,

= 9 × 4 × 3.5

= 126 cm^{3}

**∴** 1 beads volume = 1.5 cm^{3}

**∴** No of beads,

= 126/1.5

= 84

Thus, 84 beads can be made from the block.

**Question no – (13) **

**Solution :**

Box’s volume,

= 2 × 3 × 10

= 60 cm^{3}

Cartons volume,

= 40 × 36 × 24

= 34560 cm^{3}

∴ No of box,

= 24560/60

= 576

**Question no – (14) **

**Solution :**

Cuboidal block’s volume,

= 50 × 45 × 34

= 76500 cm³

Cuboids volume,

= 5 × 3 × 2

= 30 cm²

**∴** Number of cuboids,

= 76500/30

= 2550

Thus, 2550 cuboids can be obtained from this block.

**Question no – (15) **

**Solution :**

Length of cube = l

**∴** Length of cube A = L_{A} = 3l_{B}

= Cube_{A}/Cube_{B} = (l_{A})3/(l_{B})^{3} = (3l_{B} × l_{A} × lA)/(l_{B})^{3}

or, Cube_{A}/Cube_{B} = 27l_{B}/l_{B}^{3} = 27/1

**∴ **Cube A : Cube B = 27 : 1

Therefore, the ratio of the volume is 27 : 1

**Question no – (16) **

**Solution :**

Ice-cream brick’s volume,

= 20 × 10 × 7

= 1400 cm^{3}

Inner dimension of bricks stood volume,

= 100 × 50 × 42

= 210000 cm^{3}

No of bricks stood,

= 210000/1400

= 150

Hence, 150 such bricks can be stored in deep fridge.

**Question no – (17) **

**Solution :**

According to the question,

= V_{1} = (2)^{3} = 8 cm^{3}

= V_{2} = (4)^{3} = 64 cm^{3}

= 8 × V_{1}

**Question no – (18) **

**Solution :**

Tea packet volume,

= 10 × 6 × 4

= 240 cm^{3}

Cardboard volume,

= 50 × 30 × 20

= 30000 cm^{3}

Number of tea packets,

= 30000/240

= 125

Therefore, 125 such tea-packets can be placed in a cardboard box.

**Question no – (19) **

**Solution :**

Metal block’s volume,

= 5 × 4 × 3

= 60 cm^{3}

Another block’s volume,

= 15 × 8 × 3

= 360 cm^{3}

**∴ **60 cm^{3} blocks weight = 1 kg

**∴ **1 cm^{3} block’s weight = 1/60 kg

**∴ **360 cm^{3} block’s weight

= 360/60 kg

= 6 kg

Hence, the weight of a block of the same metal is 6 kg.

**Question no – (20) **

**Solution :**

Box’s volume,

= 56 × 40 × 25

= 56000 cm^{3}

Soap cake’s volume,

= 7 × 5 × 2.5

= 87.5 cm^{3}

**∴** No of cake’s volume,

= 56000/87.5

= 640

Thus, 640 soap cakes can be placed in the box.

**Question no – (21) **

**Solution :**

Let, breadth = b

According to questions,

= 3 × 4 × b = 48

or, b = 48/3 × 4

= 48/12

= 4 cm

Therefore, the breadth of cuboidal box will be 4 cm.

**Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube) Exercise 21.2 Solution :**

**Question no – (1) **

**Solution : **

**(i)** In the question we get,

Length = 12 m,

Breadth = 10 m,

Height = 4.5 m

**∴** Volume is,

= 12 × 10 × 4.5

= 540 m^{3}

Therefore, the volume in cube meter will be 540 m^{3}

**(ii)** Given in the question,

Length = 4 m,

Breadth = 2.5 m,

Height = 50 cm

**∴** Volume is,

= 4 × 2.5 × 0.05

= 0.5 m^{3}

Thus, the volume in cubic metre will be 0.5 m^{3}

**(iii)** In the question we get,

length = 10 m

breadth 25 dm

height = 25 cm

**∴** Volume is,

= 10 × 2.5 × 0.25

= 6.25 m^{3}

Therefore, the volume in cubic metre is 6.25 m^{3}

**Question no – (2) **

**Solution : **

**(i)** Cubes whose side is = 1.5 m.

Volume in cubic decimetre,

= 1.5 m = 15 dm

= (15)^{3}

= 3375 dm^{3}

Thus, volume of the cube is 3375 dm^{3}

**(ii)** Cubes whose side is = 75 cm

**∴** Volume in cubic decimetre,

= 75 cm = 7.5 dm

= (7.5)^{3}

= 421.875 dm^{3}

Hence, volume of the cube is 421.875 dm^{3}

**(iii) **Cubes whose side is = 2 dm 5 cm

**∴** Volume in cubic decimetre,

= 2 dm 5 cm = {(2 × 10) + 5}

= (20 + 5) = 25 cm

= 2.5 dm = (2.5)^{3}

= 15.625 dm^{3}

Therefore, volume of the cube is = 15.625 dm^{3}

**Question no – (3) **

**Solution : **

As per the given question,

Well measuring = 3 m by 2 m by 5 m

**∴** Clay dug out,

= 3 × 2 × 5

= 30 m^{3}

Therefore, 30 m^{3 }clay will dug out.

**Question no – (4) **

**Solution : **

Let, height = h

According to questions,

**∴** Area × h = 106

or, h = 168/Area

= h = 168/28

= h = 6 cm

Thus, the height of cuboid will be 6 cm.

**Question no – (5) **

**Solution : **

As per the question,

Length of Tank = 8 m

Breadth of Tank = 6 m

Height of Tank = 2 m

**∴** Water contain,

= 8 × 6 × 2

= 96 m^{3}

Therefore, tank can contain 96 m^{3} water.

**Question no – (6) **

**Solution : **

Let, breadth = B

According to questions,

= B × 10 × 2.5 = 50 …(∴ 50000 litres = 50 litres)

or, B = 50/10 × 2.5

= B = 2 m

Hence, the breadth of the tank will be 2 m

**Question no – (7) **

**Solution : **

According to the question,

L = 2 m

b = 2 m

h = 40 cm = 0.4 m

**∴** Diesel hold,

= 2 × 2 × 0.4

= 1.6 m^{3}

Thus, it can hold 1.6 m^{3 }diesel.

**Question no – (8) **

**Solution : **

Given in the question,

Length of room = 5 m,

Breadth of room = 4.5 m

Height of room = 3 m

**∴** Volume of Air contains,

= 5 × 4.5 × 3

= 67.5 m^{3}

Thus, the volume of the air in the room will be 67.5 m^{3}

**Question no – (9) **

**Solution : **

As per the given question,

length of tank = 3 m

Breadth of tank = 2 m

Depth of tank = 1 m .

**∴** Water hold,

= 3 × 2 × 1

= 6 m^{3}

= 6000 litres

Therefore, the tank can hold 6000 litres water.

**Question no – (10) **

**Solution : **

We know that,

3 m = 300 cm

**∴** Planks volume,

= 300 × 15 × 5

= 22500 cm^{3}

Block’s volume,

= 600 × 75 × 45

= 20,475000 cm^{3}

**∴** Number of planks,

= 20,475000/22500

= 910

Therefore, 910 planks cab be prepared.

**Question no – (11) **

**Solution : **

Brick’s volume,

= 25 × 10 × 8

= 2000 cm^{3}

Wall volume,

= 500 × 300 × 16

= 2400000 cm^{3}

**∴** Number of bricks,

= 2400000/2000

= 1200

Therefore, 1200 bricks will be required.

**Question no – (12) **

**Solution : **

Tank volume,

= 20 × 15 × 6

= 1800 m^{3}

= 1800000 litres

Water consumption,

= 4000 × 150

= 6000000 litres

**∴** Water of this tank will left,

= 1800000/600000

= 3 days

Therefore, water of this tank will last 3 days.

**Question no – (13) **

**Solution : **

Wells volume,

= 14 × 8 × 6

= 672 m^{3}

Let, the earth level rise = x m

According to questions,

= x × 70 × 60 = 672

or, x = 672/70 × 60 = 0.16

or, x = 0.16 m

= x = 16 cm

Therefore, earth level will rise 16 m.

**Question no – (14) **

**Solution : **

Let, rise in level of water = x m

According to questions,

= x × 250 × 130 = 3250

Or, x = 3250/250 × 130

= 32250/32500 = 0.1m

or, x = 0.1 m

Hence, the rise in the level of water will be 0.1 m.

**Question no – (15) **

**Solution : **

Let, thickness = h cm

Length is = 5 m

Breadth is = 0.4 m

According to questions,

= H × 5 × 0.4 = 0.6

or, H = 0.6/5 × 0.4

= 0.6/2

= 0.3 m

Therefore, the beam will be 0.3 m thick.

**Question no – (16) **

**Solution : **

Height = 6 cm 0.06 m

Area of field,

= 3 hectares = 3 × 10000

= 300000 m^{2}

Volume of water = area × height

= 30000 × 0.6

= 1800 m^{3}

= 1800000 litres.

Therefore, 1800000 litres will fall on there.

**Question no – (17) **

**Solution : **

Volume of 1 cube = 1 cm

= 0.1 m = (0.01)^{3 }m^{3}

**∴** Volume of 4000 cube,

= 400 × (0.01)^{3} m^{3}

Let, 3^{rd} edge of cuboidal beam = h

According to questions,

= h × 0.5 × 8 = 4000 × (0.01)^{3}

or, h = 4000 × 0.01 × 0.01 × 0.01/0.5 × 8 = 0.001

**∴** h = 0.001 m

Therefore, the third edge will be 0.001 m

**Question no – (18) **

**Solution : **

Volume of metal block,

= 2.25 × 1.5 × 0.27

= 0.911125 m^{3}

Volume of cube,

= 45 cm

= 0.45 m

= (0.45)^{3}

**∴** Number of cube,

= 2.25 × 1.5 × 0.27/0.45 × 0.45 × 0.45

= 0.91125/0.091125

= 10

Hence, 10 cubes will formed.

**Question no – (19) **

**Solution : **

Volume of iron,

= 600 cm × 6 cm × 2 cm

= 7200 cm^{3}

1 cm^{3} of iron weight = 8 gm

**∴** 7200 if iron height,

= 8 × 7200

= 57600

= 57.6 kg

Therefore, the weight of this piece is 57.6 kg.

**Question no – (20) **

**Solution : **

**(i)** 1 m^{3} = __10__^{6}^{ }cm^{3}

**(ii)** 1 litre = __1__ cubic decimetre

**(iii)** 1 kl = __1__ m^{3}

**(iv)** The volume of cube of side 8 cm is __512 cm__^{3}

**(v)** The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm^{3}. The height of the cuboid is __50__ cm.

= h 4000/8 × 0.10

= 50 cm

**(vi)** 1 cu.dm = __10__^{6}^{ }cu.mm

**(vii)** 1 cu. km = __10__^{9}^{ }cu. m

**(viii)** 1 litre = __10__^{3}^{ }cu.cm

**(ix)** 1 ml = __1__ cu.cm

**(x)** 1 kl = __1000__ dm^{3 }= __106__ cm^{3}

**Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube) Exercise 21.3 Solution :**

**Question no – (1) **

**Solution : **

**(i)** In the question we get,

length = 10 cm,

breadth = 12 cm,

height = 14 cm

Surface Area = 2 [(10 × 12)] + (10 × 14) + (12 × 14))]

= 2 [120 + 140 + 16]

= 2 × 428

= 856 cm^{2}

Therefore, the surface area of a cuboid is 856 cm^{2}

**(ii)** In the question we get,

Length = 6 dm,

Breadth = 8 dm,

Height = 10 dm

**∴** Surface Area,

= 2 [(6 × 8) + (6 × 10) + (8 × 10)]

= 2 [84 + 60 + 80]

= 2 × 188

= 376 dm^{2}

Hence, the surface area of a cuboid will be 376 dm^{2}

**(iii)** From the question we get,

Length = 2 m,

Breadth = 4 m,

Height = 5 m

**∴** Surface Area,

= 2 [(2 × 4) + (2 × 5) + (4 × 5)]

= 2 [8 + 10 + 20]

= 2 × 38

= 76 m^{2}

Thus, the surface area of a cuboid is 76 m^{2}

**(iv)** From the question we get,

Length = 3.2 m,

Breadth = 30 dm,

Height = 250 cm

Surface Area,

= 2 [(3.2 × 3) + (3.2 × 2.5)]

= 2 [9.6 + 8 + 7.5]

= 2 × 25.1

= 50.2 m^{2}

Therefore, the surface area of a cuboid is 50.2 m^{2}

**Question no – (2) **

**Solution : **

**(i) Cube whose volume is = 1.2 m**

Surface area of cube,

∴ 1.2 m = 6 (1.2)^{2}

= 6 × 1.44

= 8.64 m^{2}

Thus, the surface area of cube is 8.64 m^{2}

**(ii) Cube whose volume is = 27 cm**

Surface area of cube,

∴ 27 cm = 6(27)^{2}

= 6 × 729

= 4374 cm^{2}

Hence, the surface area of cube will be 4374 cm^{2}

**(iii) Cube whose volume is = 3 cm**

Surface area of cube,

∴ 3 cm = 6 (6)^{2}

= 6 × 9

= 54 cm^{2}

Thus, the surface area of cube will be 54 cm^{2}

**(iv) Cube whose volume is = 6 m**

Surface area of cube,

∴ 6 m = 6 (6)^{2}

= 6 × 36

= 216 m^{2}

Hence, the surface area of cube will be 216 m^{2}

**(v) Cube whose volume is = 2.1 m**

Surface area of cube,

∴ 2.1 m = 6 (2.1)^{2}

= 6 × 4.41

= 26.46 m^{2}

Therefore, the surface area of cube is 26.46 m^{2}

**Question no – (3) **

**Solution : **

Given, Cuboidal box is 5 cm by 5 cm by 4 cm.

**∴** Surface Area,

= 2[5 × 5) + (5 × 4) + (5 × 4)

= 2 [25 + 20 + 20]

= 2 × 65

= 130 cm^{2}

Therefore, the surface area cuboidal box 130 cm^{2 }

**Question no – (4) **

**Solution : **

**(i)** Let, one side of cube = l

According to questions,

= l^{3} = 343

or, l = 3√343

or, l = 7

**∴** Surface Area,

= 6 (7)^{2} = 6 × 72

= 294 cm^{2}

Thus, the surface area of cube will be 294 cm^{2}

**(ii)** Let, one side cube = l

According to questions,

= l^{3} = 216

or, = 3√216 = 6

**∴** Surface Area,

= 6 × (6)^{2}

= 216 dm^{2}

Hence, the surface area of cube will be 216 dm^{2}

**Question no – (5) **

**Solution : **

**(i)** Let, are edge is = l

According to questions,

= 6l^{2} = 96

or, l^{2} = 96/6 = 16

or, l = √16

= l = 4

**∴ **Volume,

= l^{3} = 4^{3}

= 64 cm^{2}

Thus, the volume of the cube will be 64 cm^{2}

**(ii) **Let, one edge = l

According to questions,

= 6l^{2} = 150

or, l^{2} = 150/6 = 25

or, l = √25 = 5

**∴** Volume,

= 6 (5)^{2} = 6 × 25

= 150 m^{3}

Therefore, the volume of a cube will be 150 m^{3}

**Question no – (6) **

**Solution : **

Let, common factor = x

Ratio of dimension = 5 : 3 : 1

Length = l = 5x

Breadth = b = 3x height = x

Now, according to the questions,

= 2 [(5x × 3x) + (5x × x) + (3x ×x) = 414

or, 15 x^{2} + 5x^{2} + 3x^{2} = 414/2 = 207

or, 23x^{2} = 207

or, x^{2} = 207/23 = 9

or, x = √9 = 3

**∴** Length = 5x = 5 × 3 = 15 m

**∴** Breadth = 3.x = 3 × 3 = 9 m

**∴** Height = x = 3 m

Therefore, the dimensions will be length = 15 m, breadth = 9 m and height = 3 m.

**Question no – (7) **

**Solution : **

Given, Length 25 cm, 0.5 m and Height 15 cm

**∴** Surface Area = 2 [(0.25 × 0.05) + (0.25 × 0.5) + (0.15 × 0.5)

= [0.0375 + 0.125 + 0.075]

= 2 × 2 0.2375

= 0.475 m^{2}

Therefore, the area of cardboard required will be 0.475 m^{2}

**Question no – (8) **

**Solution : **

Given, the edge of the box is = 12 cm

**∴** Surface Area,

= (6 (12)^{2 = 6 }× 144

= 864 cm^{2}

Hence, the surface area of a wooden box is 864 cm^{2}

**Question no – (9) **

**Solution : **

Area of tin required for 1 sheet,

= 2 [(26 × 26) + (26 × 45) + (26 × 45)

= 6032 cm^{2}

**∴** Area of tin required for 20 sheet,

= 20 × 6032 = 120660 cm^{2}

= 120640/100 × 100 m^{2}

= 12.064 m^{2}

**∴** Cost of tin,

= 12.064 × 10

= 120.64 Rs

Thus, the cost of tin sheet used is Rs. 120.64

**Question no – (10) **

**Solution : **

As per the given question,

Classroom is 11 m long, 8 m wide and 5 m high.

**∴** Area of the floor,

= (11 × 8)

= 88 m^{2}

**∴** Area of floor walls,

= [2 (11 + 8)] × 5

= 2 × 19 × 5

= 190 m^{2 }

Therefore, the sum of the areas will be 190 m^{2 }

**Question no – (11) **

**Solution : **

Area of floor,

= 200 × 15

= 300 m^{2}

Area of walls,

= 2 (20 + 15) × 3

= (2 × 35) × 3 = 70 × 3

= 210

**∴** Total area,

= 300 + 210

= 510 m^{2}

**∴** Total cost of repairing,

= 510 × 25

= 12750 Rs.

Therefore, the cost of repairing will be 12750 Rs.

**Question no – (12) **

**Solution : **

According to the question,

Perimeter of a floor of a room is = 30 m

Height is = 3 m

**∴** Area of wall = surface area × height

= 30 × 3

= 90m^{2}

Hence, the area of four walls will be 90m^{2}

**Question no – (13) **

**Solution : **

Let, length = l

Breadth = b

Height = h

**∴** Area of the floor = l × b

**∴** Area of one adjacent wall = l × h

**∴** Area of 2^{nd} adjacent wall = b × h

**∴** Total area = (l × b) × (l × h) × (b × h)

= 1b × 1h × bh

= l^{2} b^{2} h^{2}

**Question no – (14) **

**Solution : **

Given, Length = 4.5 m

Breadth = 3m

Height = 350 cm = 3.5 m.

**∴** Cost of plastering = 8/m^{2}

**∴** Area of walls,

= 2 × 3.5 (4.5 + 3)

= 7 × 7.5

= 52.2m^{2}

**∴** The cost of plastering,

= 65.7 × 8

= 528 Rs.

Thus, the cost plastering will be 528 Rs.

**Question no – (15) **

**Solution : **

Let, Length = l

Breadth = b

Height = h

**∴** S.A = 2(lb + bh + lh)

= 50 m^{2}

**∴** L.S.A = 2h (l + b) = 30 m^{2}

**∴** According to the question,

2 (lb + bh + lh) – 2h (l + b) = 50 – 30 = 20

or, 2 (lb + bh + lh) – 2h (l + b) = 20

or, lb + bh + lh + lh – bh = 20/2 = 10

or, lb = 10

Hence, the Area of base lb = 10m^{2}

**Question no – (16) **

**Solution : **

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

= 2 × 3.5 (7 + 6)

= 2 × 35 × 35

= 91 m^{2}

**∴** Area of four wall without Door and window,

= 90 – 17

= 74 m^{2}

**∴** Total cost of whit washing

= 74 × 1.50

= 111 Rs.

**Question no – (17) **

**Solution : **

Area of hall,

= 2385 – 60/1.20

= 1988

Let, breadth = b

Area of hall including door and window,

= 2 × 8 (80 + b)

= 16 (80 + b)

Area of hall 10 doors,

= 10 × (3 × 1.5)

= 45 m^{2}

Area of hall 10 windows,

= 10 (1.5 × 1m)

= 15 m^{2}

**According to questions,**

= 16 (80 + b) – (45 + 15) = 198

or, 16 (80 + b) = 1988 + 60 = 2048

or, 80 + b = 2048/16 = 128

or, b = 128 – 80

= b = 48 m

Therefore, the breadth of the hall will be 48 m.

**Next Chapter Solution : **

👉 Chapter 22 👈