Rs Aggarwal 2018 for Class 8 Math Chapter 1 - Rational Numbers

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Page No 222:

Question 1:

Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:
(i) length = 22 cm, breadth = 12 cm and height = 7.5 cm
(ii) length = 15 m, breadth = 6 m and height = 9 dm
(iii) length = 24 m, breadth = 25 cm and height = 6 m
(iv) length = 48 cm, breadth = 6 dm and height = 1 m

Answer:

Volume of a cuboid $= (L e n g t h \times B r e a d t h \times H e i g h t)$ cubic units
Total surface area $= 2 (l b + b h + l h)$ sq units
Lateral surface area $= [2 (l + b) \times h]$ sq units

(i) Length = 22 cm, breadth = 12 cm, height = 7.5 cm
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(22 \times 12 \times 7.5) = 1980 c m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(22 \times 12) + (22 \times 7.5) + (12 \times 7.5)] = 2 [264 + 165 + 90] = 1038 c m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (22 + 12) \times 7.5 = 510 c m^{2}$

(ii) Length = 15 m, breadth = 6 m, height = 9 dm = 0.9 m
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(15 \times 6 \times 0.9) = 81 m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(15 \times 6) + (15 \times 0.9) + (6 \times 0.9)] = 2 [90 + 13.5 + 5.4] = 217.8 m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (15 + 6) \times 0.9 = 37.8 m^{2}$

(iii) Length = 24 m, breadth = 25 cm = 0.25 m, height = 6 m
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(24 \times 0.25 \times 6) = 36 m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(24 \times 0.25) + (24 \times 6) + (0.25 \times 6)] = 2 [6 + 144 + 1.5] = 303 m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (24 + 0.25) \times 6 = 291 m^{2}$

(iv) Length = 48 cm = 0.48 m, breadth = 6 dm = 0.6 m, height = 1 m
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(0.48 \times 0.6 \times 1) = 0.288 m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(0.48 \times 0.6) + (0.48 \times 1) + (0.6 \times 1)] = 2 [0.288 + 0.48 + 0.6] = 2.736 m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (0.48 + 0.6) \times 1 = 2.16 m^{2}$

Page No 222:

Question 2:

The dimensions of a rectangular water tank are 2 m 75 cm by 1 m 80 cm by 1 m 40 cm. How many litres of water does it hold when filled to the brim?

Answer:

$1 m = 100 c m$
Therefore, dimensions of the tank are:
$2 m 75 c m \times 1 m 80 c m \times 1 m 40 c m = 275 c m \times 180 c m \times 140 c m$
∴ Volume = $Length \times Breadth \times Height = 275 \times 180 \times 140 = 6930000 {cm}^{3}$

Also, $1000 c m^{3} = 1 L$

∴ Volume $= \frac{6930000}{1000} = 6930 L$

Page No 222:

Question 3:

A solid rectangular piece of iron measures 1.05 m × 70 cm × 1.5 cm. Find the weight of this piece in kilograms if 1 cm³ of iron weighs 8 grams.

Answer:

$1 m = 100 c m$
∴ Dimensions of the iron piece = $105 cm \times 70 cm \times 1.5 cm$
Total volume of the piece of iron $= (105 \times 70 \times 1.5) = 11025 c m^{3}$
1 cm³ measures 8 gms.
∴Weight of the piece $= 11025 \times 8 = 88200 g = \frac{88200}{1000} = 88.2 k g (b e c a u s e 1 k g = 1000 g)$

Page No 222:

Question 4:

The area of a courtyard is 3750 m². Find the cost of covering it with gravel to a height of 1 cm if the gravel costs Rs 6.40 per cubic metre.

Answer:

$1 c m = 0.01 m$
Volume of the gravel used $= Area \times H eight = (3750 \times 0.01) = 37.5 m^{3}$
Cost of the gravel is Rs 6.40 per cubic meter.
∴ Total cost $= (37.5 \times 6.4) =$ Rs 240

Page No 222:

Question 5:

How many persons can be accommodated in a hall of length 16 m. breadth 12.5 m and height 4.5 m, assuming that 3.6 m³ of air is required for each person?

Answer:

Total volume of the hall $= (16 \times 12.5 \times 4.5) = 900 m^{3}$

It is given that $3.6 m^{3}$ of air is required for each person.
The total number of persons that can be accommodated in that hall $= \frac{Total volume}{Volume required by each person} = \frac{900}{3.6} = 250 people$

Page No 222:

Question 6:

A cardboard box is 1.2 m long, 72 cm wide and 54 cm high. How many bars of soap can be put into it if each bar measures 6 cm × 4.5 cm × 4 cm?

Answer:

Volume of the cardboard box $= (120 \times 72 \times 54) = 466560 c m^{3}$

Volume of each bar of soap $= (6 \times 4.5 \times 4) = 108 c m^{3}$

Total number of bars of soap that can be accommodated in that box $= \frac{Volume of the box}{Volume of each soap} = \frac{466560}{108} = 4320$ bars

Page No 222:

Question 7:

The size of a matchbox is 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 144 matchboxes? How many such packets can be placed in a carton of size 1.5 m × 84 cm × 60 cm?

Answer:

Volume occupied by a single matchbox $= (4 \times 2.5 \times 1.5) = 15 c m^{3}$

Volume of a packet containing 144 matchboxes $= (15 \times 144) = 2160 c m^{3}$

Volume of the carton $= (150 \times 84 \times 60) = 756000 c m^{3}$

Total number of packets is a carton $= \frac{Volume of the carton}{Volume of a packet} = \frac{75600}{2160} = 350$ packets

Page No 222:

Question 8:

How many planks of size 2 m × 25 cm × 8 cm can be prepared from a wooden block 5 m long, 70 cm broad and 32 cm thick, assuming that there is no wastage?

Answer:

Total volume of the block $= (500 \times 70 \times 32) = 1120000 c m^{3}$

Total volume of each plank $= 200 \times 25 \times 8 = 40000 c m^{3}$ $= 200 \times 25 \times 8 = 40000 c m^{3}$

∴ Total number of planks that can be made $= \frac{Total volume of the block}{Volume of each plank} = \frac{1120000}{40000} = 28$ planks

Page No 222:

Question 9:

How many bricks, each of size 25 cm × 13.5 cm × 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33 cm thick?

Answer:

Volume of the brick $= 25 \times 13.5 \times 6 = 2025 c m^{3}$
Volume of the wall $= 800 \times 540 \times 33 = 14256000 c m^{3}$

Total number of bricks $= \frac{Volume of the wall}{Volume of each brick} = \frac{14256000}{2025} = 7040$ bricks

Page No 222:

Question 10:

A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring 22 cm × 12.5 cm × 7.5 cm. If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?

Answer:

Volume of the wall $= 1500 \times 30 \times 400 = 18000000 c m^{3}$
Total quantity of mortar $= \frac{1}{12} \times 18000000 = 1500000 c m^{3}$
∴ Volume of the bricks $= 18000000 - 1500000 = 16500000 c m^{3}$

Volume of a single brick $= 22 \times 12.5 \times 7.5 = 2062.5 c m^{3}$

∴ Total number of bricks $= \frac{Total volume of the bricks}{Volume of a single brick} = \frac{16500000}{2062.5} = 8000$ bricks

Page No 222:

Question 11:

Find the capacity of a rectangular cistern in litres whose dimensions are 11.2 m × 6 m × 5.8 m. Find the area of the iron sheet required to make the cistern.

Answer:

Volume of the cistern $= 11.2 \times 6 \times 5.8 = 389.76 m^{3} = 389.76 \times 1000 = 389760$ litres

Area of the iron sheet required to make this cistern = Total surface area of the cistern
$= 2 (11.2 \times 6 + 11.2 \times 5.8 + 6 \times 5.8) = 2 (67.2 + 64.96 + 34.8) = 333.92 c m^{2}$

Page No 222:

Question 12:

The volume of a block of gold is 0.5 m³. If it is hammered into a sheet to cover an area of 1 hectare, find the thickness of the sheet.

Answer:

Volume of the block $= 0.5 m^{3}$
We know:
$1 h e c t a r e = 10000 m^{2}$
Thickness $= \frac{Volume}{Area} = \frac{0.5}{10000} = 0.00005 m = 0.005 cm = 0.05 mm$

Page No 222:

Question 13:

The rainfall recorded on a certain day was 5 cm. Find the volume of water that fell on a 2-hectare field.

Answer:

Rainfall recorded = 5 cm = 0.05 m
Area of the field = 2 hectare = $2 \times 10000 m^{2} = 20000 m^{2}$
Total rain over the field = $Area of the field \times Height of the field = 0.05 \times 20000 = 1000 m^{3}$

Page No 222:

Question 14:

A river 2 m deep and 45 m wide is flowing at the rate of 3 km/h. Find the quantity of water that runs into the sea per minute.

Answer:

Area of the cross-section of river $= 45 \times 2 = 90 m^{2}$

Rate of flow $= 3 \frac{k m}{h r} = \frac{3 \times 1000}{60} = 50 \frac{m}{m i n}$

Volume of water flowing through the cross-section in one minute $= 90 \times 50 = 4500 m^{3}$ per minute

Page No 222:

Question 15:

A pit 5 m long and 3.5 m wide is dug to a certain depth. If the volume of earth taken out of it is 14 m³, what is the depth of the pit?

Answer:

Let the depth of the pit be d m.
$Volume = Length \times w idth \times d epth = 5 m \times 3.5 m \times d m$
But,
Given volume = 14 m³
∴ Depth $= d = \frac{volume}{length \times width} = \frac{14}{5 \times 3.5} = 0.8 m$ = 80 cm

Page No 222:

Question 16:

A rectangular water tank is 90 cm wide and 40 cm deep. If it can contain 576 litres of water, what is its length?

Answer:

Capacity of the water tank $= 576 litres = 0.576 m^{3}$
Width = 90 cm = 0.9 m
Depth = 40 cm = 0.4 m

Length = $= \frac{capacity}{width×depth} = \frac{0.576}{0.9 \times 0.4} = 1.600 m$

Page No 223:

Question 17:

A beam of wood is 5 m long and 36 cm thick. It is made of 1.35 m³ of wood. What is the width of the beam?

Answer:

Volume of the beam $= 1.35 m^{3}$

Length = 5 m

Thickness = 36 cm = 0.36 m

Width = $= \frac{volume}{thickness×length} = \frac{1.35}{5 \times 0.36} = 0.75 m = 75 cm$

Page No 223:

Question 18:

The volume of a room is 378 m³ and the area of its floor is 84 m². Find the height of the room.

Answer:

Volume $= height × area$
Given:
Volume = 378 m³
Area = 84 m²

∴ Height $= \frac{volume}{area} = \frac{378}{84} = 4.5 m$

Page No 223:

Question 19:

A swimming pool is 260 m long and 140 m wide. If 54600 cubic metres of water is pumped into it, find the height of the water level in it.

Answer:

Length of the pool = 260 m
Width of the pool = 140 m

Volume of water in the pool = 54600 cubic metres

∴ Height of water $= \frac{volume}{length×width} = \frac{54600}{260 \times 140} = 1.5$ metres

Page No 223:

Question 20:

Find the volume of wood used to make a closed box of outer dimensions 60 cm × 45 cm × 32 cm, the thickness of wood being 2.5 cm all around.

Answer:

External length = 60 cm
External width = 45 cm
External height = 32 cm

External volume of the box $= 60 \times 45 \times 32 = 86400 {cm}^{3}$

Thickness of wood = 2.5 cm

∴ Internal length $= 60 - (2.5 \times 2) = 55$ cm
Internal width $= 45 - (2.5 \times 2) = 40$ cm
Internal height $= 32 - (2.5 \times 2) = 27$ cm

Internal volume of the box $= 55 \times 40 \times 27 = 59400 {cm}^{3}$

Volume of wood = External volume - Internal volume $= 86400 - 59400 = 27000 {cm}^{3}$

Page No 223:

Question 21:

Find the volume of iron required to make an open box whose external dimensions are 36 cm × 25 cm × 16.5 cm, the box being 1.5 cm thick throughout. If 1 cm³ of iron weighs 8.5 grams, find the weight of the empty box in kilograms.

Answer:

External length = 36 cm
External width = 25 cm
External height = 16.5 cm

External volume of the box $= 36 \times 25 \times 16.5 = 14850 {cm}^{3}$

Thickness of iron = 1.5 cm

∴ Internal length $= 36 - (1.5 \times 2) = 33$ cm
Internal width $= 25 - (1.5 \times 2) = 22$ cm
Internal height $= 16.5 - 1.5 = 15$ cm (as the box is open)

Internal volume of the box $= 33 \times 22 \times 15 = 10890 {cm}^{3}$

Volume of iron = External volume − Internal volume $= 14850 - 10890 = 3960 {cm}^{3}$

Given:
${1 cm}^{3} of iron = 8.5 grams$

Total weight of the box $= 3960 \times 8.5 = 33660 grams = 33.66 kilograms$

Page No 223:

Question 22:

A box with a lid is made of wood which is 3 cm thick. Its external length, breadth and height are 56 cm, 39 cm and 30 cm respectively. Find the capacity of the box. Also find the volume of wood used to make the box.

Answer:

External length = 56 cm
External width = 39 cm
External height = 30 cm

External volume of the box $= 56 \times 39 \times 30 = 65520 {cm}^{3}$

Thickness of wood = 3 cm

∴ Internal length $= 56 - (3 \times 2) = 50$ cm
Internal width $= 39 - (3 \times 2) = 33$ cm
Internal height $= 30 - (3 \times 2) = 24$ cm

Capacity of the box = Internal volume of the box $= 50 \times 33 \times 24 = 39600 {cm}^{3}$

Volume of wood = External volume − Internal volume $= 65520 - 39600 = 25920 {cm}^{3}$

Page No 223:

Question 23:

The external dimensions of a closed wooden box are 62 cm, 30 cm and 18 cm. If the box is made of 2-cm-thick wood, find the capacity of the box.

Answer:

External length = 62 cm
External width = 30 cm
External height = 18 cm

∴ External volume of the box $= 62 \times 30 \times 18 = 33480 c m^{3}$

Thickness of the wood = 2 cm

Now, internal length $= 62 - (2 \times 2) = 58$ cm
Internal width $= 30 - (2 \times 2) = 26$ cm
Internal height $= 18 - (2 \times 2) = 14$ cm

∴ Capacity of the box = internal volume of the box $= (58 \times 26 \times 14) c m^{3} = 21112 c m^{3}$

Page No 223:

Question 24:

A closed wooden box 80 cm long, 65 cm wide and 45 cm high, is made of 2.5-cm-thick wood. Find the capacity of the box and its weight if 100 cm³ of wood weighs 8 g.

Answer:

External length = 80 cm
External width = 65 cm
External height = 45 cm

∴ External volume of the box $= 80 \times 65 \times 45 = 234000 c m^{3}$

Thickness of the wood = 2.5 cm

Then internal length $= 80 - (2.5 \times 2) = 75$ cm
Internal width $= 65 - (2.5 \times 2) = 60$ cm
Internal height $= 45 - (2.5 \times 2) = 40$ cm

Capacity of the box = internal volume of the box $= (75 \times 60 \times 40) c m^{3} = 180000 c m^{3}$

Volume of the wood = external volume − internal volume $= (234000 - 180000) c m^{3} = 54000 c m^{3}$

It is given that $100 {cm}^{3} of wood weighs 8 g .$

∴ Weight of the wood $= \frac{54000}{100} \times 8 g = 4320 g = 4.32 kg$

Page No 223:

Question 25:

Find the volume, lateral surface area and the total surface area of a cube each of whose edges measures:
(i) 7 m
(ii) 5.6 cm
(iii) 8 dm 5 cm

Answer:

(i) Length of the edge of the cube = a = 7 m
Now, we have the following:
Volume $= a^{3} = 7^{3} = 343 m^{3}$
Lateral surface area $= 4 a^{2} = 4 \times 7 \times 7 = 196 m^{2}$
Total Surface area $= 6 a^{2} = 6 \times 7 \times 7 = 294 m^{2}$

(ii) Length of the edge of the cube = a = 5.6 cm
Now, we have the following:
Volume $= a^{3} = 5 . 6^{3} = 175.616 c m^{3}$
Lateral surface area $= 4 a^{2} = 4 \times 5.6 \times 5.6 = 125.44 c m^{2}$
Total Surface area $= 6 a^{2} = 6 \times 5.6 \times 5.6 = 188.16 c m^{2}$

(iii) Length of the edge of the cube = a = 8 dm 5 cm = 85 cm
Now, we have the following:
Volume $= a^{3} = 85^{3} = 614125 c m^{3}$
Lateral surface area $= 4 a^{2} = 4 \times 85 \times 85 = 28900 c m^{2}$
Total Surface area $= 6 a^{2} = 6 \times 85 \times 85 = 43350 c m^{2}$

Page No 223:

Question 26:

The surface area of a cube is 1176 cm². Find its volume.

Answer:

Let a be the length of the edge of the cube.
Total surface area $= 6 a^{2} = 1176 c m^{2}$
$\Rightarrow a = \sqrt{\frac{1176}{6}} = \sqrt{196} = 14 c m$
∴ Volume $= a^{3} = 14^{3} = 2744 c m^{3}$

Page No 223:

Question 27:

The volume of a cube is 729 cm³. Find its surface area.

Answer:

Let a be the length of the edge of the cube.

Then volume $= a^{3} = 729 c m^{3}$

Also, $a = \sqrt[3]{729} = 9 c m$

∴ Surface area $= 6 a^{2} = 6 \times 9 \times 9 = 486 c m^{2}$

Page No 223:

Question 28:

The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of side 45 cm. How many cubes are formed?

Answer:

1 m = 100 cm
Volume of the original block $= 225 \times 150 \times 27 = 911250 c m^{3}$

Length of the edge of one cube = 45 cm
Then volume of one cube $= 45^{3} = 91125 c m^{3}$

∴ Total number of blocks that can be cast $= \frac{volume of the block}{volume of one cube} = \frac{911250}{91125} = 10$

Page No 223:

Question 29:

If the length of each edge of a cube is doubled, how many times does its volume become? How many times does its surface area become?

Answer:

Let a be the length of the edge of a cube.
Volume of the cube $= a^{3}$
Total surface area $= 6 a^{2}$

If the length is doubled, then the new length becomes 2a.
Now, new volume $= (2 a)^{3} = 8 a^{3}$
Also, new surface area= $= 6 (2 a)^{2} = 6 \times 4 a^{2} = 24 a^{2}$
∴ The volume is increased by a factor of 8, while the surface area increases by a factor of 4.

Page No 223:

Question 30:

A solid cubical block of fine wood costs Rs 256 at Rs 500 per m². Find its volume and the length of each side.

Answer:

Cost of wood = Rs $500 / m^{3}$

Cost of the given block = Rs 256

∴ Volume of the given block $= a^{3} = \frac{256}{500} = 0.512 m^{3} = 512000 {cm}^{3}$

Also, length of its edge = a $= \sqrt[3]{0.512} = 0.8 m$ = 80 cm

Page No 227:

Question 1:

Find the volume, curved surface area and total surface area of each of the cylinders whose dimensions are:
(i) radius of the base = 7 cm and height = 50 cm
(ii) radius of the base = 5.6 m and height = 1.25 m
(iii) radius of the base = 14 dm and height = 15 m

Answer:

Volume of a cylinder = $π r^{2} h$
Lateral surface $= 2 π r h$
Total surface area $= 2 π r (h + r)$

(i) Base radius = 7 cm; height = 50 cm
Now, we have the following:
Volume $= \frac{22}{7} \times 7 \times 7 \times 50 = 7700 c m^{3}$
Lateral surface area $= 2 π r h$ $= 2 \times \frac{22}{7} \times 7 \times 50 = 2200 c m^{2}$
Total surface area $= 2 π r (h + r)$ $= 2 \times \frac{22}{7} \times 7 (50 + 7) = 2508 c m^{2}$

(ii) Base radius = 5.6 m; height = 1.25 m
Now, we have the following:
Volume $= \frac{22}{7} \times 5.6 \times 5.6 \times 1.25 = 123.2 m^{3}$
Lateral surface area $= 2 π r h$ $= 2 \times \frac{22}{7} \times 5.6 \times 1.25 = 44 m^{2}$
Total surface area $= 2 π r (h + r)$ $= 2 \times \frac{22}{7} \times 5.6 (1.25 + 5.6) = 241.12 m^{2}$

(iii) Base radius = 14 dm = 1.4 m, height = 15 m
Now, we have the following:
Volume $= \frac{22}{7} \times 1.4 \times 1.4 \times 15 = 92.4 m^{3}$
Lateral surface area $= 2 π r h$ $= 2 \times \frac{22}{7} \times 1.4 \times 15 = 132 m^{2}$
Total surface area $= 2 π r (h + r)$ $= 2 \times \frac{22}{7} \times 1.4 (15 + 1.4) = 144.32 c m^{2}$

Page No 227:

Question 2:

A milk tank is in the form of a cylinder whose radius is 1.5 m and height is 10.5 m. Find the quantity of milk in litres that can be stored in the tank.

Answer:

$r = 1.5 m h = 10.5 m Capacity of the tank = volume of the tank = π r^{2} h = \frac{22}{7} \times 1.5 \times 1.5 \times 10.5 = 74.25 m^{3} We know that 1 m^{3} = 1000 L ∴ 74.25 m^{3} = 74250 L$

Page No 227:

Question 3:

A wooden cylindrical pole is 7 m high and its base radius is 10 cm. Find its weight if the wood weighs 225 kg per cubic metre.

Answer:

Height = 7 m
Radius = 10 cm = 0.1 m
Volume $= {πr}^{2} h = \frac{22}{7} \times 0.1 \times 0.1 \times 7 = 0.22 m^{3}$
Weight of wood = 225 kg/m³
∴ Weight of the pole $= 0.22 \times 225 = 49.5 k g$

Page No 227:

Question 4:

Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm?

Answer:

Diameter = 2r = 140 cm
i.e., radius, r = 70 cm = 0.7 m

Now, volume $= {πr}^{2} h = 1.54 m^{3}$

$\Rightarrow \frac{22}{7} \times 0.7 \times 0.7 \times h = 1.54 ∴ h = \frac{1.54 \times 7}{0.7 \times 0.7 \times 22} = \frac{154 \times 7}{154 \times 7} = 1 m$

Page No 227:

Question 5:

The volume of a circular iron rod of length 1 m is 3850 cm³. Find its diameter.

Answer:

Volume $= {πr}^{2} h = 3850 {cm}^{3}$
Height = 1 m =100 cm

Now, radius, $r = \sqrt{\frac{3850}{π \times h}} = \sqrt{\frac{3850 \times 7}{22 \times 100}} = 1.75 \times 7 = 3.5 c m$
∴ Diameter =2(radius) $= 2 \times 3.5 = 7 c m$

Page No 227:

Question 6:

A closed cylindrical tank of diameter 14 m and height 5 m is made from a sheet of metal. How much sheet of metal will be required?

Answer:

Diameter = 14 m
Radius $= \frac{14}{2} = 7 m$
Height = 5 m

∴ Area of the metal sheet required = total surface area

$= 2 πr (h + r) = 2 \times \frac{22}{7} \times 7 (5 + 7) m^{2} = 44 \times 12 m^{2} = 528 m^{2}$

Page No 227:

Question 7:

The circumference of the base of a cylinder is 88 cm and its height is 60 cm. Find the volume of the cylinder and its curved surface area.

Answer:

Circumference of the base = 88 cm
Height = 60 cm

Area of the curved surface $= c i r c u m f e r e n c e \times h e i g h t = 88 \times 60 = 5280 c m^{2}$
Circumference $= 2 πr = 88 c m$
Then radius $= r = \frac{88}{2 π} = \frac{88 \times 7}{2 \times 22} = 14 c m$
∴ Volume $= {πr}^{2} h = \frac{22}{7} \times 14 \times 14 \times 60 = 36960 {cm}^{3}$

Page No 227:

Question 8:

The lateral surface area of a cylinder of length 14 m is 220 m². Find the volume of the cylinder.

Answer:

Length = height = 14 m
Lateral surface area $= 2 πrh = 220 m^{2}$
Radius $= r = \frac{220}{2 πh} = \frac{220 \times 7}{2 \times 22 \times 14} = \frac{10}{4} = 2.5 m$
∴ Volume $= {πr}^{2} h = \frac{22}{7} \times 2.5 \times 2.5 \times 14 = 275 m^{3}$

Page No 227:

Question 9:

The volume of a cylinder of height 8 cm is 1232 cm³. Find its curved surface area and the total surface area.

Answer:

Height = 8 cm
Volume $= {πr}^{2} h = 1232 {cm}^{3}$
Now, radius $= r = \sqrt{\frac{1232}{πh}} = \sqrt{\frac{1232 \times 7}{22 \times 8}} = \sqrt{49} = 7 c m$
Also, curved surface area $= 2 πrh = 2 \times \frac{22}{7} \times 7 \times 8 = 352 {cm}^{2}$
∴ Total surface area $= 2 πr (h + r) = (2 \times \frac{22}{7} \times 7 \times 8) + (2 \times \frac{22}{7} \times {(7)}^{2}) = 352 + 308 = 660 {cm}^{2}$

Page No 227:

Question 10:

The radius and height of a cylinder are in the ratio 7 : 2. If the volume of the cylinder is 8316 cm³, find the total surface area of the cylinder.

Answer:

We have: $\frac{r a d i u s}{h e i g h t} = \frac{7}{2}$
i.e., $r = \frac{7}{2} h$
Now, volume $= {πr}^{2} h = π {(\frac{7}{2} h)}^{2} h = 8316 {cm}^{3}$
$\Rightarrow \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h^{3} = 8316 \Rightarrow h^{3} = \frac{8316 \times 2}{11 \times 7} = 108 \times 2 = 216 \Rightarrow h = \sqrt[3]{216} = 6 c m$

Then $r = \frac{7}{2} h = \frac{7}{2} \times 6 = 21 c m$
∴ Total surface area $= 2 πr (h + r) = 2 \times \frac{22}{7} \times 21 \times (6 + 21) = 3564 {cm}^{2}$

Page No 227:

Question 11:

The curved surface area of a cylinder is 4400 cm² and the circumference of its base is 110 cm. Find the volume of the cylinder.

Answer:

Curved surface area $= 2 π r h = 4400 c m^{2}$
Circumference $= 2 πr = 110 cm$
Now, height $= h = \frac{c u r v e d s u r f a c e a r e a}{c i r c u m f e r e n c e} = \frac{4400}{110} = 40 c m$

Also, radius, $r = \frac{4400}{2 πh} = \frac{4400 \times 7}{2 \times 22 \times 40} = \frac{35}{2}$

∴ Volume $= {πr}^{2} h = \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 40 = 22 \times 5 \times 35 \times 10 = 38500 {cm}^{3}$

Page No 227:

Question 12:

A particular brand of talcum powder is available in two packs, a plastic can with a square base of side 5 cm and of height 14 cm, or one with a circular base of radius 3.5 cm and of height 12 cm. Which of them has greater capacity and by how much?

Answer:

For the cubic pack:
Length of the side, a = 5 cm
Height = 14 cm
Volume $= a^{2} h = 5 \times 5 \times 14 = 350 c m^{3}$

For the cylindrical pack:
Base radius $= r = 3.5 c m$
Height = 12 cm
Volume $= {πr}^{2} h = \frac{22}{7} \times 3.5 \times 3.5 \times 12 = 462 c m^{3}$

We can see that the pack with a circular base has a greater capacity than the pack with a square base.
Also, difference in volume $= 462 - 350 = 112 c m^{3}$

Page No 227:

Question 13:

Find the cost of painting 15 cylindrical pillars of a building at Rs 2.50 per square metre if the diameter and height of each pillar are 48 cm and 7 metres respectively.

Answer:

Diameter = 48 cm
Radius = 24 cm = 0.24 m
Height = 7 m

Now, we have:
Lateral surface area of one pillar $= πdh = \frac{22}{7} \times 0.48 \times 7 = 10.56 m^{2}$
Surface area to be painted = total surface area of 15 pillars $= 10.56 \times 15 = 158.4 m^{2}$
∴ Total cost $= Rs (158.4 \times 2.5) = Rs 396$

Page No 227:

Question 14:

A rectangular vessel 22 cm by 16 cm by 14 cm is full of water. If the total water is poured into an empty cylindrical vessel of radius 8 cm, find the height of water in the cylindrical vessel.

Answer:

Volume of the rectangular vessel $= 22 \times 16 \times 14 = 4928 c m^{3}$
Radius of the cylindrical vessel = 8 cm
Volume $= {πr}^{2} h$

As the water is poured from the rectangular vessel to the cylindrical vessel, we have:
Volume of the rectangular vessel = volume of the cylindrical vessel

∴ Height of the water in the cylindrical vessel $= \frac{v o l u m e}{{πr}^{2}} = \frac{4928 \times 7}{22 \times 8 \times 8} = \frac{28 \times 7}{8} = \frac{49}{2} = 24.5 c m$

Page No 227:

Question 15:

A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 11 cm. It is drawn out into a wire of diameter 1 mm. What will be the length of the wire so obtained?

Answer:

Diameter of the given wire = 1 cm
Radius = 0.5 cm
Length = 11 cm
Now, volume $= {πr}^{2} h = \frac{22}{7} \times 0.5 \times 0.5 \times 11 = 8.643 c m^{3}$
The volumes of the two cylinders would be the same.
Now, diameter of the new wire = 1 mm = 0.1 cm
Radius = 0.05 cm
∴ New length $= \frac{volume}{{πr}^{2}} = \frac{8.643 \times 7}{22 \times 0.05 \times 0.05} = 1100.02 c m$ ≅ 11 m

Page No 227:

Question 16:

A solid cube of metal each of whose sides measures 2.2 cm is melted to form a cylindrical wire of radius 1 mm. Find the length of the wire so obtained.

Answer:

Length of the edge, a = 2.2 cm
Volume of the cube $= a^{3} = {(2.2)}^{3} = 10.648 c m^{3}$
Volume of the wire $= {πr}^{2} h$
Radius = 1 mm = 0.1 cm
As volume of cube = volume of wire, we have:

$h = \frac{v o l u m e}{{πr}^{2}} = \frac{10.648 \times 7}{22 \times 0.1 \times 0.1} = 338.8 c m$

Page No 227:

Question 17:

How many cubic metres of earth must be dug out to sink a well which is 20 m deep and has a diameter of 7 metres? If the earth so dug out is spread over a rectangular plot 28 m by 11 m, what is the height of the platform so formed?

Answer:

Diameter = 7 m
Radius = 3.5 m
Depth = 20 m

Volume of the earth dug out $= {πr}^{2} h = \frac{22}{7} \times 3.5 \times 3.5 \times 20 = 770 m^{3}$
Volume of the earth piled upon the given plot $= 28 \times 11 \times h = 770 m^{3}$

$∴ h = \frac{770}{28 \times 11} = \frac{70}{28} = 2.5 m$

Page No 227:

Question 18:

A well of inner diameter 14 m is dug to a depth of 12 m. Earth taken out of it has been evenly spread all around it to a width of 7 m to form an embankment. Find the height of the embankment so formed.

Answer:

Inner diameter = 14 m
i.e., radius = 7 m
Depth = 12 m
Volume of the earth dug out $= {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 12 = 1848 m^{3}$

Width of embankment = 7 m
Now, total radius $= 7 + 7 = 14 m$

$Volume of the embankment = total volume - inner volume = {πr}_{o}^{2} h - {πr}_{i}^{2} h = πh ({r_{o}}^{2} - {r_{i}}^{2}) = \frac{22}{7} h (14^{2} - 7^{2}) = \frac{22}{7} h (196 - 49) = \frac{22}{7} h \times 147 = 21 \times 22 h = 462 \times h m^{3}$

Since volume of embankment = volume of earth dug out, we have:
$1848 = 462 h$
$\Rightarrow h = \frac{1848}{462} = 4 m$
∴ Height of the embankment = 4 m

Page No 227:

Question 19:

A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of the road roller is 84 cm and its length is 1 m.

Answer:

Diameter = 84 cm
i.e., radius = 42 cm
Length = 1 m = 100 cm
Now, lateral surface area $= 2 πrh = 2 \times \frac{22}{7} \times 42 \times 100 = 26400 {cm}^{2}$
∴ Area of the road $= lateral surface area \times no . of rotations = 26400 \times 750 = 19800000 {cm}^{2} = 1980 m^{2}$

Page No 228:

Question 20:

A cylinder is open at both ends and is made of 1.5-cm-thick metal. Its external diameter is 12 cm and height is 84 cm. What is the volume of metal used in making the cylinder? Also, find the weight of the cylinder if 1 cm³ of the metal weighs 7.5 g.

Answer:

Thickness of the cylinder = 1.5 cm
External diameter = 12 cm
i.e., radius = 6 cm
also, internal radius = 4.5 cm
Height = 84 cm

Now, we have the following:
Total volume $= {πr}^{2} h = \frac{22}{7} \times 6 \times 6 \times 84 = 9504 {cm}^{3}$
Inner volume $= {πr}^{2} h = \frac{22}{7} \times 4.5 \times 4.5 \times 84 = 5346 {cm}^{3}$
Now, volume of the metal = total volume − inner volume $= 9504 - 5346 = 4158 c m^{3}$
∴ Weight of iron $= 4158 \times 7.5 = 31185 g = 31.185 kg$ [Given: $1 c m^{3} = 7.5 g$ ]

Page No 228:

Question 21:

The length of a metallic tube is 1 metre, its thickness is 1 cm and its inner diameter is 12 cm. Find the weight of the tube if the density of the metal is 7.7 grams per cubic centimetre.

Answer:

Length = 1 m = 100 cm
Inner diameter = 12 cm
Radius = 6 cm
Now, inner volume $= {πr}^{2} h = \frac{22}{7} \times 6 \times 6 \times 100 = 11314.286 {cm}^{3}$
Thickness = 1 cm
Total radius = 7 cm

Now, we have the following:
Total volume $= {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 100 = 15400 {cm}^{3}$
Volume of the tube $= total volume - inner volume = 15400 - 11314.286 = 4085.714 c m^{3}$
Density of the tube = 7.7 g/cm³
∴ Weight of the tube $= volume \times density = 4085.714 \times 7.7 = 31459.9978 g = 31.459 kg$

Page No 228:

Question 1:

Tick (✓) the correct answer:
The maximum length of a pencil that can be kept in a rectangular box of dimensions 12 cm × 9 cm × 8 cm, is
(a) 13 cm
(b) 17 cm
(c) 18 cm
(d) 19 cm

Answer:

(b) 17

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

∴ $\sqrt{l^{2} + b^{2} + h^{2}} = \sqrt{12^{2} + 9^{2} + 8^{2}} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17 c m$

Page No 228:

Question 2:

Tick (✓) the correct answer:
The total surface area of a cube is 150 cm². Its volume is
(a) 216 cm³
(b) 125 cm³
(c) 64 cm³
(d) 1000 cm³

Answer:

(b) $125 c m^{3}$

Total surface area $= 6 a^{2} = 150 c m^{2}$ , where a is the length of the edge of the cube.
$\Rightarrow 6 a^{2} = 150$
$\Rightarrow a = \sqrt{\frac{150}{6}} = \sqrt{25} = 5 c m$
∴ Volume $= a^{3} = 5^{3} = 125 c m^{3}$

Page No 228:

Question 3:

Tick (✓) the correct answer:
The volume of a cube is 343 cm³. Its total surface area is
(a) 196 cm²
(b) 49 cm²
(c) 294 cm²
(d) 147 cm²

Answer:

(c) $294 c m^{2}$

Volume $= a^{3} = 343 c m^{3}$
$\Rightarrow a = \sqrt[3]{343} = 7 c m$
∴ Total surface area $= 6 a^{2} = 6 \times 7 \times 7 = 294 c m^{2}$

Page No 228:

Question 4:

Tick (✓) the correct answer:
The cost of painting the whole surface area of a cube at the rate of 10 paise per cm² is Rs 264.60. Then, the volume of the cube is
(a) 6859 cm³
(b) 9261 cm³
(c) 8000 cm³
(d) 10648 cm³

Answer:

(b) $9261 c m^{3}$

Rate of painting = 10 paise per sq cm = Rs 0.1/cm²
Total cost = Rs 264.60
Now, total surface area $= \frac{264.6}{0.1} = 2646 c m^{2}$
Also, length of edge, a $= \sqrt{\frac{2646}{6}} = \sqrt{441} = 21 c m$
$∴ Volume = 21^{3} = 9261 c m^{3}$

Page No 228:

Question 5:

Tick (✓) the correct answer:
How many bricks, each measuring 25 cm × 11.25 cm × 6 cm, will be needed to build a wall 8 m long, 6 m high and 22.5 cm thick?
(a) 5600
(b) 6000
(c) 6400
(d) 7200

Answer:

(c) 6400

Volume of each brick $= 25 \times 11.25 \times 6 = 1687.5 c m^{3}$
Volume of the wall $= 800 \times 600 \times 22.5 = 10800000 c m^{3}$
∴ No. of bricks = $\frac{10800000}{1687.5} = 6400$

Page No 228:

Question 6:

Tick (✓) the correct answer:
How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?
(a) 10
(b) 100
(c) 1000
(d) 10000

Answer:

(c) 1000

Volume of the smaller cube $= (10 c m)^{3} = 1000 c m^{3}$
Volume of box $= (100 c m)^{3} = 1000000 c m^{3}$ [1 m = 100 cm]
∴ Total no. of cubes $= \frac{100 \times 100 \times 100}{10 \times 10 \times 10} = 1000$

Page No 228:

Question 7:

Tick (✓) the correct answer:
The edges of a cuboid are in the ratio 1 : 2 : 3 and its surface area is 88 cm². The volume of the cuboid is
(a) 48 cm³
(b) 64 cm³
(c) 96 cm³
(d) 120 cm³

Answer:

(a) $48 c m^{3}$

Let a be the length of the smallest edge.
Then the edges are in the proportion a : 2a : 3a.
Now, surface area $= 2 (a \times 2 a + a \times 3 a + 2 a \times 3 a) = 2 (2 a^{2} + 3 a^{2} + 6 a^{2}) = 22 a^{2} = 88 c m^{2}$
$\Rightarrow a = \sqrt{\frac{88}{22}} = \sqrt{4} = 2$
Also, 2a = 4 and 3a = 6
∴ Volume $= a \times 2 a \times 3 a = 2 \times 4 \times 6 = 48 c m^{3}$

Page No 228:

Question 8:

Tick (✓) the correct answer:
Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is
(a) 1 : 3
(b) 1 : 9
(c) 1 : 27
(d) none of these

Answer:

(b) 1: 9

$\frac{Volume 1}{Volume 2} = \frac{1}{27} = \frac{a^{3}}{b^{3}} \Rightarrow a = \frac{b}{\sqrt[3]{27}} = \frac{b}{3} or b = 3 a or \frac{b}{a} = 3$
$Now, \frac{surface area 1}{surface area 2} = \frac{6 a^{2}}{6 b^{2}} = \frac{a^{2}}{b^{2}} = \frac{(b / 3)^{2}}{b^{2}} = \frac{1}{9} ∴ Ratio of the surface areas = 1 : 9$

Page No 228:

Question 9:

Tick (✓) the correct answer:
The surface area of a (10 cm × 4 cm × 3 cm) brick is
(a) 84 cm²
(b) 124 cm²
(c) 164 cm²
(d) 180 cm²

Answer:

Page No 228:

Question 10:

Tick (✓) the correct answer:
An iron beam is 9 m long, 40 cm wide and 20 cm high. If 1 cubic metre of iron weighs 50 kg, what is the weight of the beam?
(a) 56 kg
(b) 48 kg
(c) 36 kg
(d) 27 kg

Answer:

Page No 229:

Question 11:

Tick (✓) the correct answer:
A rectangular water reservoir contains 42000 litres of water. If the length of reservoir is 6 m and its breadth is 3.5 m, the depth of the reservoir is
(a) 2 m
(b) 5 m
(c) 6 m
(d) 8 m

Answer:

(a) 2 m

42000 L = 42 m³
$Volume = l b h$
$∴ Height (h) = \frac{volume}{l b} = \frac{42}{6 \times 3.5} = \frac{6}{6 \times 0.5} = 2 m$

Page No 229:

Question 12:

Tick (✓) the correct answer:
The dimensions of a room are (10 m × 8 m × 3.3 m). How many men can be accommodated in this room if each man requires 3 m³ of space?
(a) 99
(b) 88
(c) 77
(d) 75

Answer:

(b) 88

Volume of the room $= 10 \times 8 \times 3.3 = 264 m^{3}$
One person requires 3 m³.
∴ Total no. of people that can be accommodated $= \frac{264}{3} = 88$

Page No 229:

Question 13:

Tick (✓) the correct answer:
A rectangular water tank is 3 m long, 2 m wide and 5 m high. How many litres of water can it hold?
(a) 30000
(b) 15000
(c) 25000
(d) 35000

Answer:

(a) 30000

$Volume = 3 \times 2 \times 5 = 30 m^{3} = 30000 L$

Page No 229:

Question 14:

Tick (✓) the correct answer:
The area of the cardboard needed to make a box of size 25 cm × 15 cm × 8 cm will be
(a) 390 cm²
(b) 1390 cm²
(c) 2780 cm²
(d) 1000 cm²

Answer:

(b) $1390 c m^{2}$

Surface area $= 2 (25 \times 15 + 15 \times 8 + 25 \times 8) = 2 (375 + 120 + 200) = 1390 c m^{2}$

Page No 229:

Question 15:

Tick (✓) the correct answer:
The diagonal of a cube measures $4 \sqrt{3}$ cm. Its volume is
(a) 8 cm³
(b) 16 cm³
(c) 27 cm³
(d) 64 cm³

Answer:

(d) $64 c m^{2}$

Diagonal of the cube $= a \sqrt{3} = 4 \sqrt{3} c m$
i.e., a = 4 cm
∴ Volume $= a^{3} = 4^{3} = 64 c m^{3}$

Page No 229:

Question 16:

Tick (✓) the correct answer:
The diagonal of a cube is $9 \sqrt{3}$ cm long. Its total surface area is
(a) 243 cm²
(b) 486 cm²
(c) 324 cm²
(d) 648 cm²

Answer:

(b) 486 sq cm

Diagonal $= \sqrt{3} a c m = 9 \sqrt{3} c m$
i.e., a = 9
∴ Total surface area $= 6 a^{2} = 6 \times 81 = 486 c m^{2}$

Page No 229:

Question 17:

Tick (✓) the correct answer:
If each side of a cube is doubled then its volume
(a) is doubled
(b) becomes 4 times
(c) becomes 6 times
(d) becomes 8 times

Answer:

(d) If each side of the cube is doubled, its volume becomes 8 times the original volume.

Let the original side be a units.
Then original volume = a³ cubic units
Now, new side = 2a units
Then new volume = (2a)³sq units = 8 a³cubic units
Thus, the volume becomes 8 times the original volume.

Page No 229:

Question 18:

Tick (✓) the correct answer:
If each side of a cube is doubled, its surface area
(a) is doubled
(b) becomes 4 times
(c) becomes 6 times
(d) becomes 8 times

Answer:

(b) becomes 4 times.

Let the side of the cube be a units.
Surface area = 6a² sq units
Now, new side = 2a units
New surface area = 6(2a² ) sq units = 24a² sq units.
Thus, the surface area becomes 4 times the original area.

Page No 229:

Question 19:

Tick (✓) the correct answer:
Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. The edge of the new cube formed is
(a) 12 cm
(b) 14 cm
(c) 16 cm
(d) 18 cm

Answer:

(a) 12 cm

Total volume $= 6^{3} + 8^{3} + 10^{3} = 216 + 512 + 1000 = 1728 c m^{3}$
∴ Edge of the new cube $= \sqrt[3]{1728} = 12 c m$

Page No 229:

Question 20:

Tick (✓) the correct answer:
Five equal cubes, each of edge 5 cm, are placed adjacent to each other. The volume of the cuboid so formed, is
(a) 125 cm³
(b) 375 cm³
(c) 525 cm³
(d) 625 cm³

Answer:

(d) $625 c m^{3}$

Length of the cuboid so formed = 25 cm
Breadth of the cuboid = 5 cm
Height of the cuboid = 5 cm
∴ Volume of cuboid $= 25 \times 5 \times 5 = 625 c m^{3}$

Page No 229:

Question 21:

Tick (✓) the correct answer:
A circular well with a diameter of 2 metres, is dug to a depth of 14 metres. What is the volume of the earth dug out?
(a) 32 m³
(b) 36 m³
(c) 40 m³
(d) 44 m³

Answer:

(d) 44 m³

Diameter = 2 m
Radius = 1 m
Height = 14 m
$∴ Volume = {πr}^{2} h = \frac{22}{7} \times 1 \times 1 \times 14 = 44 m^{3}$

Page No 229:

Question 22:

Tick (✓) the correct answer:
If the capacity of a cylindrical tank is 1848 m³ and the diameter of its base is 14 m, the depth of the tank is
(a) 8 m
(b) 12 m
(c) 16 m
(d) 18 m

Answer:

(b) 12 m

Diameter = 14 m
Radius = 7 m
Volume = 1848 m³

$Now, volume = {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times h = 1848 m^{3} ∴ h = \frac{1848}{22 \times 7} = 12 m$

Page No 229:

Question 23:

Tick (✓) the correct answer:
The ratio of the total surface area to the lateral surface area of a cylinder whose radius is 20 cm and height 60 cm, is
(a) 2 : 1
(b) 3 : 2
(c) 4 : 3
(d) 5 : 3

Answer:

(c) 4 : 3

$Here, \frac{Total surface area}{Lateral surface area} = \frac{2 π r (h + r)}{2 π r h} = \frac{h + r}{h} = \frac{20 + 60}{60} = \frac{4}{3} = 4 : 3$

Page No 229:

Question 24:

Tick (✓) the correct answer:
The number of coins, each of radius 0.75 cm and thickness 0.2 cm, to be melted to make a right circular cylinder of height 8 cm and base radius 3 cm is
(a) 460
(b) 500
(c) 600
(d) 640

Answer:

(d) 640
$Total no . of coins = \frac{volume of cylinder}{volume of each coin} = \frac{π \times 3 \times 3 \times 8}{π \times 0.75 \times 0.75 \times 0.2} = 640$

Page No 230:

Question 25:

Tick (✓) the correct answer:
66 cm³ of silver is drawn into a wire 1 mm in diameter. The length of the wire will be
(a) 78 m
(b) 84 m
(c) 96 m
(d) 108 m

Answer:

(b) 84 m
$Length = \frac{volume}{π r^{2} 2} = \frac{66 \times 7}{22 \times 0.05 \times 0.05} = 8400 c m = 84 m$

Page No 230:

Question 26:

Tick (✓) the correct answer:
The height of a cylinder is 14 cm and its diameter is 10 cm. The volume of the cylinder is
(a) 1100 cm³
(b) 3300 cm³
(c) 3500 cm³
(d) 7700 cm³

Answer:

(a) 1100 cm³
Volume $= {πr}^{2} h = \frac{22}{7} \times 5 \times 5 \times 14 = 1100 {cm}^{3}$

Page No 230:

Question 27:

Tick (✓) the correct answer:
The height of a cylinder is 80 cm and the diameter of its base is 7 cm. The whole surface area of the cylinder is
(a) 1837 cm²
(b) 1760 cm²
(c) 1942 cm²
(d) 3080 cm²

Answer:

(a) 1837 cm²
Diameter = 7 cm
Radius =3.5 cm
Height = 80 cm
∴ Total surface area $= 2 πr (r + h) = 2 \times \frac{22}{7} \times 3.5 (3.5 + 80) = 22 (83.5) = 1837 {cm}^{2}$

Page No 230:

Question 28:

Tick (✓) the correct answer:
The height of a cylinder is 14 cm and its curved surface area is 264 cm². The volume of the cylinder is
(a) 308 cm³
(b) 396 cm³
(c) 1232 cm³
(d) 1848 cm³

Answer:

(b) 396 cm³
Here, curved surface area $= 2 πrh = 264 {cm}^{3}$
$\Rightarrow r = \frac{264 \times 7}{2 \times 22 \times 14} = 3 c m$
$∴ Volume = {πr}^{2} h = \frac{22}{7} \times 3 \times 3 \times 14 = 396 {cm}^{3}$

Page No 230:

Question 29:

Tick (✓) the correct answer:
The diameter of a cylinder is 14 cm and its curved surface area is 220 cm². The volume of the cylinder is
(a) 770 cm³
(b) 1000 cm³
(c) 1540 cm³
(d) 6622 cm³

Answer:

(a) 770 cm³
Diameter = 14 cm
Radius = 7 cm
Now, curved surface area $= 2 πrh = 220 {cm}^{2}$
$\Rightarrow h = \frac{220 \times 7}{2 \times 22 \times 7} = 5 c m$
$∴ Volume = {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 5 = 770 {cm}^{3}$

Page No 230:

Question 30:

Tick (✓) the correct answer:
The ratio of the radii of two cylinders is 2 : 3 and the ratio of their heights is 5 : 3. The ratio of their volumes will be
(a) 4 : 9
(b) 9 : 4
(c) 20 : 27
(d) 27 : 20

Answer:

(c) 20:27

$We have the following : \frac{r_{1}}{r_{2}} = \frac{2}{3} \frac{h_{1}}{h_{2}} = \frac{5}{3} ∴ \frac{V_{1}}{V_{2}} = \frac{{πr}_{1}^{2} h_{1}}{{πr}_{2}^{2} h_{2}} = \frac{20}{27}$

Page No 231:

Question 1:

Find the volume of a cube whose total surface area is 384 cm².

Answer:

Total surface area $= 6 a^{2}$
$\Rightarrow 6 a^{2} = 384$
$\Rightarrow a = \sqrt{\frac{384}{6}} = 8 c m$
$∴ Volume = a^{3} = 512 c m^{3}$

Page No 231:

Question 2:

How many soap cakes each measuring 7 cm × 5 cm × 2.5 cm can be placed in a box of size 56 cm × 40 cm × 25 cm?

Answer:

Volume of a soap cake $= 7 \times 5 \times 2.5 = 87.5 c m^{3}$
Volume of the box $= 56 \times 40 \times 25 = 56000 c m^{3}$

No. of soap cakes $= \frac{56000}{87.5} = 640 u n i t s$

∴ 640 cakes of soap can be placed in a box of the given size.

Page No 231:

Question 3:

The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm³. Find its radius and height.

Answer:

$\frac{Radius}{height} = \frac{r}{h} = \frac{5}{7} \Rightarrow r = \frac{5}{7} h$
Now, volume $= {πr}^{2} h = \frac{22}{7} \times \frac{5}{7} h \times \frac{5}{7} h \times h = 550 c m^{3}$
$∴ h = \sqrt[3]{\frac{550 \times 7 \times 7 \times 7}{22 \times 5 \times 5}} = 7 c m Also, r = \frac{5}{7} h = 5 c m$

Page No 231:

Question 4:

Find the number of coins, 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder with a height of 10 cm and a diameter of 4.5 cm.

Answer:

Volume of the coin $= {πr}^{2} h = \frac{22}{7} \times 0.75 \times 0.75 \times 0.2$
Volume of the cylinder $= {πr}^{2} h = \frac{22}{7} \times 2.25 \times 2.25 \times 10$
No. of coins $= \frac{volume of cylinder}{volume of coin} = \frac{2.25 \times 2.25 \times 10}{0.75 \times 0.75 \times 0.2} = 450 coins$

∴ 450 coins must be melted to form the required cylinder.

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Question 5:

Find the surface area of a chalk box, whose length, breadth and height are 18 cm, 10 cm and 8 cm respectively.

Answer:

Length = 18 cm
Breadth = 10 cm
Height = 8 cm
∴ Total surface area $= 2 (l b + l h + b h) = 2 (18 \times 10 + 18 \times 8 + 10 \times 8) = 2 (180 + 144 + 80) = 808 c m^{2}$

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Question 6:

The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. Find the diameter and height of the pillar.

Answer:

Curved surface area $= 2 πrh = 264 m^{2}$
$∴ r = \frac{264}{2 πh} = \frac{132}{πh} m$

Volume $= {πr}^{2} h = π \times \frac{132}{πh} \times \frac{132}{πh} \times h = 924 m^{3}$
$∴ h = \frac{132 \times 132 \times 7}{22 \times 924} = 6 m$

Now, $r = \frac{132}{πh} = \frac{132 \times 7}{22 \times 6} = 7 m$

i.e., diameter of the pillar, $d = 7 \times 2 = 14 m$

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Question 7:

Mark (✓) against the correct answer:
The circumference of the circular base of a cylinder is 44 cm and its height is 15 cm. The volume of the cylinder is
(a) 1155 cm³
(b) 2310 cm³
(c) 770 cm³
(d) 1540 cm³

Answer:

(b) 2310 cm³

Height = 15 cm
Circumference $= 2 πr = 44 cm$
$∴ r = \frac{44 \times 7}{2 \times 22} = 7 c m$
∴ Volume $= {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 15 = 2310 {cm}^{3}$

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Question 8:

Mark (✓) against the correct answer:
the area of the base of a circular cylinder is 35 cm² and its height is 8 cm. The volume of the cylinder is
(a) 140 cm³
(b) 280 cm³
(c) 420 cm³
(d) 210 cm³

Answer:

(b) 280 cm³
Area = 35 cm²
Height = 8 cm
$∴ Volume = base area \times height = 35 \times 8 = 280 {cm}^{3}$

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Question 9:

Mark (✓) against the correct answer:
A cuboid having dimensions 16 m × 11 m × 8 m is melted to form a cylinder of radius 4 m. What is the height of the cylinder?
(a) 28 m
(b) 14 m
(c) 21 m
(d) 32 m

Answer:

(a) 28 m
Volume of the cuboid $= 16 \times 11 \times 8 = 1408 m^{3}$
Volume of the cylinder $= {πr}^{2} h = 1408 m^{3}$
$∴ h = \frac{1408 \times 7}{22 \times 4 \times 4} = 28 m$

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Question 10:

Mark (✓) against the correct answer:
The dimensions of a cuboid are 8 m 6 m 4 m. Its lateral surface area is
(a) 210 m²
(b) 105 m²
(c) 160 m²
(d) 240 m²

Answer:

Lateral surface area $= 2 ((l + b) \times h) = 2 ((8 + 6) \times 4) = 2 (56) = 112 m^{2}$

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Question 11:

Mark (✓) against the correct answer:
The length, breadth and height of a cuboid are in the ratio 3 : 4 : 6 and its volume is 576 cm³. The whole surface area of the cuboid is
(a) 216 cm²
(b) 324 cm²
(c) 432 cm²
(d) 460 cm²

Answer:

(c) 432 sq cm
Volume $= l b h = 3 x \times 4 x \times 6 x = 72 x^{3} = 576 c m^{3}$
$\Rightarrow x = \sqrt[3]{\frac{576}{72}} = 2$
∴ Total surface area $= 2 (l b + b h + l h) = 2 (3 x 4 x + 4 x 6 x + 3 x 6 x) = 2 (48 + 96 + 72) = 432 {cm}^{2}$

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Question 12:

Mark (✓) against the correct answer:
The surface area of a cube is 384 cm². Its volume is
(a) 512 cm³
(b) 256 cm³
(c) 384 cm³
(d) 320 cm³

Answer:

(a) 512 cm³
Surface area $= 6 a^{2}$
$\Rightarrow 6 a^{2} = 384$
$\Rightarrow a = \sqrt{\frac{384}{6}} = \sqrt{64} = 8 cm$
∴ Volume $= a^{3} = 8^{3} = 512 {cm}^{3}$

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Question 13:

Fill in the blanks.
(i) If l, b, h be the length, breadth and height of a cuboid, then its whole surface area = (.......) sq units.
(ii) If l, b, h be the length, breadth and height of a cuboid, then its lateral surface area = (.......) sq units.
(iii) If each side of a cube is a, then its lateral surface area is ....... sq units.
(iv) If r is the radius of the base and h be the height of a cylinder, then its volume is (.......) cubic units.
(v) If r is the radius of the base and h be the height of a cylinder, then its lateral surface area is (......) sq units.

Answer:

(i) If l, b and h are the length, breadth and height of a cuboid, respectively, then its whole surface area is equal to $2 (l b + l h + b h)$ sq units.
(ii) If l, b and h are the length, breadth and height of a cuboid, respectively, then its lateral surface area is equal to $2 ((l + b) \times h)$ sq units.
(iii) If each side of a cube is a, then the lateral surface area is $4 a^{2}$ sq units.
(iv) If r and h are the radius of the base and height of a cylinder, respectively, then its volume is $π r^{2} h$ cubic units.
(v) If r and h are the radius of the base and height of a cylinder, then its lateral surface area is $2 π r h$ sq units.

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