# ICSE Solutions for Selina Concise Chapter 21 Solids Class 9 Maths

### Exercise 21(A)

1. The length breadth and height of a rectangular solid are in the ratio of 5 : 4 : 2 if the total surface area of 1216 cm². Find the length breadth and height of the solid

Let the length, breadth and height of rectangular solid are 5x, 4x, 2x.
Total surface area = 1216 cm2
2(5x. 4x + 4x.2x + 2x.5x) = 1216
⇒ 20x2 + 8x2 + 10x2  = 608
⇒ 38x2  = 608
⇒ x2 = (608/38) = 16
⇒ x = 4

Therefore, the length, breadth and height of rectangular solid are
5×4 = 20cm, 4×4 = 16cm, 2×4 = 8cm.

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

Let a be the one edge of a cube.
Volume  = a3
729 = a3
⇒ 93 = a3
⇒ 9 = a
⇒ a = 9 cm
Total surface area 6a2 = 6×92  = 486 cm2 .

3. The dimensions of a Cinema Hall are 100 m, 60 m and 15 m. How many persons can sit in the hall, if each requires 150 m3 of air?

Volume of cinema hall  = 100×60×15 = 90000 m3
⇒ 150m3 requires = 1 person
⇒ 90000m3 requires  = (1/150)×90000 = 600 persons
Therefore, 600 persons can sit in the hall.

4. 75 persons can sleep in a room 25 m by 9.6 m. If each persons requires 16 m3 of air; find the height of the room.

Let h be height of the room.
1 person requires 16m3
75 person requires 75×16 m3 = 1200m3
Volume of room is 1200 m3
1200 = 25×9.6×h
⇒ h = 1200/(25×9.6)
⇒ h = 5m

5. The edges of three cubes of metal are 3 cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube.

Volume of melted single cube = 33 + 43 + 53  cm3
= 27 + 64 + 125 cm3
= 216 cm3
Let a be the edge of the new cube.
Volume = 216 cm3
a3 = 216
⇒ a3 = 63
⇒ a = 6cm
Therefore, 6 cm is the edge of cube.

6. Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recasted into a single cube of edge 12 cm. Find ‘x’.

Volume of melted single cube x3 + 83 + 103 cm3
= x3 + 512 + 1000 cm3
= x3 + 1512 cm3
Given that 12cm is edge of the single cube.
123 = x3 + 1512 cm3
⇒ x3 = 123 - 1512
⇒ x3 = 1728 - 1512
⇒ x3 = 216
⇒ x3 = 63
⇒ x = 6 cm

7. Three equal cubes are placed adjacently in a row. Find the ratio of the total surfaced area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.

Let the side of a cube be 'a' units.
Total surface area of one cube = 6a2
Total surface area of 3 cubes = 3×6a2 = 18a2
after joining 3 cubes in a row, length of Cuboid = 3a
Breadth and height of cuboid = a
Total surface area of cuboid = 2 [3a2 + a2 + 3a2] = 14a2
Ratio of total surface area of cuboid to the total surface area of 3 cubes = 14a2/18a2 = 7/9.

8. The cost of papering the four walls of a room at 75 paisa per square meter Rs. 240. The height of the room is 5 metres. Find the length and the breadth of the room, if they are in the ratio 5 : 3.

Let the length and breadth of the room is 5x and 3x respectively.
Given that the four walls of a room at 75 paise per square met Rs. 240.
Thus,
240 = Area ×0.75
⇒ Area = 240/0.75
⇒ Area = 24000/75
⇒ Area = 320 m
⇒ Area = 2×Height (Length + Breadth)
⇒ 320 = 2×5(5x+ 3x)
⇒ 320 = 10×8x
⇒ 32 = 8x
⇒ x = 4
Length = 5x
= 5(4)m
= 20 m
= 3(4)m
= 12 m .

9. The area of a playground is 3650 m2. Find the cost of covering it with gravel 1.2 cm deep, if the gravel costs Rs. 6.40 per cubic metre.

The area of the playground is 3650 m2 and the gravels are 1.2 cm deep. Therefore the total volume to be covered will be:
3650×0.012 =43.8 m3.
Since the cost of per cubic meter is Rs. 6.40, therefore the total cost will be:
43.8×Rs.6.40 = Rs.280.32

10. A square plate of side ‘x’ cm is 8 mm thick. If its volume is 2880 cm3; find the value of x.

We know that
1mm = 1/10  cm
8mm = 8/10cm
Volume = Base area ×Height
⇒ 2880 cm3 = x × x × 8/10
⇒ 2880 ×(10/8) = x2
⇒ x2  = 3600
⇒ x = 60 cm

11. The external dimensions of a closed wooden box are 27 cm, 19 cm and 11 cm. If the thickness of the wood in the box is 1.5 cm; find:

(i) Volume of the wood in the box;

(ii) The cost of the box, if wood costs Rs. 1.20 per cm3;

(iii) Number of 4 cm cubes that could be placed into the box.

External volume of the box = 27×19×11 cm3 = 5643 cm
Since, external dimensions are 27 cm, 19cm, 11cm; thickness of the wood is 1.5 cm.
∴ Internal dimensions
= (27 - 2×1.5)cm, (19 - 2×1.5)cm, (11-2×1.5)cm
= 24cm, 16cm, 8cm
Hence interval volume of box = (24×16×8)cm3 = 3072 cm3
(i) Volume of wood in the box = 5643 cm3 - 3072 cm3 = 2571 cm3
(ii) Cost of wood = Rs. 1.20×2571 = Rs 3085.2
(iii) Vol. of 4 cm cube = 43 = 64cm3
Number of 4 cm cubes that could be placed into the box
= 3072/64 = 48

12. A tank 20 m long, 12 m wide and 8 m deep is to be made of iron sheet. If it is open at the top. Determine the cost of iron-sheet, at the rate of Rs. 12.50 per metre, if the sheet is 2.5 m wide.

Area of sheet = Surface area of the tank
⇒ Length of the sheet × its width = Area of 4 walls of the tank + Area of its base
⇒ length of the sheet × 2.5m = 2(20+12)×8 m2  + 20×12 m2
⇒ length of the sheet = 300.8 m
Cost of the sheet = 300.8 × Rs. 12.50 = Rs 3760

13. A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimetres. Calculate the exterior height of the box.

Let exterior height is h cm. Then interior dimensions are 78-3=75, 19-3=16 and h-3 (subtract two thicknesses of wood). Interior volume = 75 x 16 x (h-3) which must = 15 cu dm
= 15000 cm^3
(1 dm = 10cm, 1 cu dm = 103 cm3).
15000 cm3 = 75×16×(h – 3)
⇒ h -3 = 15000/(75×16) = 12.5cm
⇒ h =15.5cm.

14. The square on the diagonal of a cube has an area of 1875 sq. cm. Calculate:

(i) The side of the cube.

(ii) The total surface area of the cube.

(i) If the sides of the cube = a cm
The length of its diagonal = a√3 cm
And,
(a√3)2  = 1875
⇒ a = 25cm
(ii) Total surface area of the cube = 6a2
= 6(25)2  = 3750 cm2

15. A hollow square-shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm3 of iron in this tube. Find its thickness.

Given that the volume of the iron in the tube 192 cm2
Let the thickness of the tube = x cm
∴ Side of the external square = (5+2x)cm
∵ Ext. vol. of the tube - its internal vol. = volume of iron in the tube, we have,
(5+2x)(5+2x) × 8-5×5×8 = 192
⇒ (25+4x2 + 20x)×8 - 200 = 192
⇒ 200 + 32x2 + 160x - 200 = 192
⇒ 32x2 + 160x - 192 = 0
⇒ x2 + 5x - 6 = 0
⇒ x2 +6x- x- 6 = 0
⇒ x(x+6)-(x+6) =0
⇒ (x+6)(x-1) = 0
⇒ x- 1= 0
⇒ x = 1
therefore, thickness is 1 cm.

16. Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuboid as 648 m2; find the length of edge of each cube.

Also, find the ratio between the surface area of resulting cuboid and the surface area of a cube.

Let L be the length of the edge of each cube.
The length of the resulting cuboid = 4×L = 4L  cm
Let width(b) = L cm and its height(h) = L cm
∴ The total surface area of the resulting cuboid
= 2(l× b + b× h + h× l)
⇒ 648 = 2(4l × l + l× l + l ×4l)
⇒ 4l2 + l2 + 4l2  = 324
⇒ 9l2 = 324
⇒ l2 = 36
⇒ l = 6 cm
Therefore , the length of each cube is  6 cm.
(Surface area of the resulting cuboid)/(Surface area of cube) = 648/6l2
(Surface area of the resulting cuboid)/(Surface area of cube)= 648/6(6)2
(Surface area of the resulting cuboid)/(Surface area of cube) = 648/216 = 3/1 = 3: 1

### Exercise 21(B)

1. The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimetres.

Assume that all angles in the figures are right angles.

The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, 3 cm, and 9 cm respectively. Hence, volume of solid
= 9×4×3 + 6×4×3
= 108+72
= 180 cm3

2. A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.

Area of cross section of the solid = (1/2)(1.5 +3)×(40) cm2
= (1/2)(4.5)×40 cm2
= 90 cm2
Volume of solid = Area of cross section × length
= 90 × 15cm3
= 1350 cm3
= 1350000  liters  [since 1cm3 = 1000 lt.]

3. The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that:

AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate:

(i) The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m2 (sq. metre).

(ii) The cost of paving the floor at the rate of Rs. 18 per m2.

The cross section of a tunnel is of the trapezium shaped ABCD in which AB = 7m, CD = 5m and AM = BN. The height is 2.4 m and its length is 40m.
(i) AM = BN = (7-5)/2 =2/2 = 1 m
AD2  = AM2 + DM2 [Using pythagoras theorem]
= 12  + (2.4)2
= 1 + 5.76
= 6.76
= (2.6)2
Perimeter of the cross - section of the tunnel = (7+2.6 + 2.6 + 5)m = 17.2m
Length = 40 m
∴ Internal surface area of the tunnel (except floor)
= (17.2×40 - 40×7) m2
= (688 - 280)m2
= 408 m2
Rate of painting = Rs 5 per m2
Hence, total cost of painting  = Rs 5×408 = Rs 2040
(ii) Area of floor of tunnel l×b = 40×7 = 280 m2
Rate of cost of paving = Rs 18 per m2
Total cost = 280×18 = Rs 5040

4. Water is discharged from a pipe of cross-section area 3.2 cm2 at the speed of 5m/s. Calculate the volume of water discharged:

(i) In cm3 per sec.

(ii) In litres per minute.

(i) The rate of speed  =5 m/s = 500 cm/s
Volume of water flowing per sec = 3.2×500 cm3 = 1600 cm3
(ii) Vol. of water flowing per min = 1600×60 cm3 = 96000 cm3
Since 1000 cm3 = 1lt
Therefore, Vol. of water flowing per min = 96000/1000 = 96 litres

5. A hose-pipe of cross-section area 2 cm2 delivers 1500 litres of water in 5 minutes. What is the speed of water in m/s through the pipe?

Vol. of water flowing in 1sec = (1500×1000)/5×60 = 5000 cm3
Vol. of water flowing = area of cross section × speed of water
= 5000 cm3/s  = 2 cm2 × speed of water
⇒ speed of water = 5000/2 cm/s
⇒ speed of water = 2500 cm/s
⇒ speed of water = 25 m/s

6. The cross-section of a piece of metal 4 m in length is shown below. Calculate:

(i) The area of the cross-section;

(ii) The volume of the piece of metal in cubic centimetres.

If 1 cubic centimetre of the metal weighs 6.6 g, calculate the weight of the piece of metal to the nearest kg. (i)  Area of total cross section = Area of rectangle abce + area of △def
= (12×10) + (1/2)(16-10)(12-7.5)
= 120 + (1/2)×6×4.5 cm2
= 120 + 13.5 cm2
= 133.5 cm2
(ii) The volume of the piece of metal in cubic centimeters = Area of total cross section × length
= 133.5 cm2 × 400 cm  = 53400 cm3
1 cubic centimetre of the metal weighs 6.6g
53400 cm3 of the metal weighs 6.6× 53400 g = (6.6×53400)/1000  kg
= 352.440 kg
The weight of the piece of metal to the nearest Kg is 352 kg.

7. A rectangular water-tank measuring 80 cm 60 cm 60 cm is filled form a pipe of cross-sectional area 1.5 cm2, the water emerging at 3.2 m/s. How long does it take to fill the tank?

Vol. of rectangular tank = 80×60×60 cm3 = 288000 cm3
One liter = 1000 cm3
Vol. of water flowing in per sec =
1.5 cm2 ×3.2 m/s = 1.5 cm2 × (3.2 ×100)cm/s
= 480 cm3/s
Vol.  of water flowing in  1 min = 480×60 = 28800 cm3
Hence,
28800 cm3 can be filled = 1 min
⇒ 28800 cm3 can be filled = [(1/28800)×288000] min = 10 min

8. A rectangular card-board sheet has length 32 cm and breadth 26 cm. Squares each of side 3 cm, are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed. Length of sheet = 32 cm
Breadth of sheet = 26 cm
Side of each square = 3cm
∴ Inner length = 32 - 2×3 = 32 - 6 = 26 cm
Inner breadth = 26 - 2×3 = 26 -6 = 20 cm
By folding the sheet, the length of the container = 26 cm
Breadth of the container = 20 cm and height of the container = 3cm
∴ Vol. of the container = l ×b×h
= 26cm ×20cm×3cm = 1560 cm3

9. A swimming pool is 18 m long and 8 m wide. Its deep and shallow ends are 2 m and 1.2 m respectively. Find the capacity of the pool, assuming that the bottom of the pool slopes uniformly. Length of pool  = 18 m
Breadth of pool = 8 m
Height of one side = 2m
Height of second side  = 1.2 m
∴ Volume of pool = 18×8×(2+1.2)/2  m3
= (18×8×3.2)/2
= 230.4 m3

10. The following figure shows a closed victory-stand whose dimensions are given in cm.

Find the volume and the surface are of the victory stand.

Consider the box 1 Thus, the dimensions of box 1 are: 60cm, 40cm and 30 cm.
Therefore, the volume of box 1 = 60×40×30 = 72000 cm3
Surface area of box 1 = 2 (lb + bh + lh)
Since the box  is open at the bottom and from the given figure, we have,
Surface area of box 1 = 40 ×40 + 40×30 + 40×30+ 2(60×30)
= 1600 + 1200 +1200 +3600
= 7600 cm2
Consider the box 2 Thus, the dimensions of box 2 are : 40 cm , 30 cm and 30 cm.
Therefore, the volume of box 2 = 2(lb + bh + lh)
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 2 = 40×3 + 40×30 + 2(30×30)
= 1200 + 1200 + 1800
= 4200 cm2
Consider the box 3 Thus, the dimensions of box 2 are : 40 cm , 30 cm and 20 cm .
Therefore, the volume of box 3 = 40×30×20 = 24000 cm3
Surface area of box 3 = 2(lb + bh + lh)
Since the box is open at the bottom and from the given figure, we have
Surface area of  box 3 = 40×30 + 40×20 + 2(30×20)
= 1200 + 800 + 1200
= 3200 cm2
Total volume of the box = volume of box  1 + volume of box 2 + volume of box 3
= 72000 + 36000+ 24000
= 132000 cm3
Similarly, total surface area of the box
= surface area of box 1 + surface area of box 2 + surface area of box 3
= 7600 +4200 + 3200
= 15000 cm3

### Exercise 21 (C)

1. Each face of a cube has perimeter equal to 32 cm. Find its surface area and its volume.

The perimeter of a cube formula is, Perimeter = 4a where, a = length.
Given that perimeter of the face of the cube is 32 cm
⇒ 4a = 32cm
⇒ a = 32/4
⇒ a = 8cm
We know that surface area of a cube with side 'a'  = 6a2
Thus, Surface area = 6×82 = 6×64 = 384 cm2
We know that the volume of a cube with side 'a' = a3
Thus, volume = 83 = 512 cm3

2. A school auditorium is 40 m long, 30 m broad and 12 m high. If each student requires 1.2 m2 of the floor area; find the maximum number of students that can be accommodated in this auditorium. Also, find the volume of air available in the auditorium, for each student.

Given dimensions of the auditorium are: 40m × 30m ×12 m
The area of the floor = 40×30
Also given that each student requires 1.2 m2 of the floor area.
Thus, Maximum number of students = (40×30)/1.2 = 1000
Volume of the auditorium
= 40×30×12 m3
= Volume of air available for 1000 students
Therefore, Air available for each student = (40×30×12)/1000 m3 = 14.4 m3 .

3. The internal dimensions of a rectangular box are 12 cm x cm 9 cm. If the length of the longest rod that can be placed in this box is 17 cm; find x.

Length of longest rod = Length of the diagonal of the box ⇒ 172 = 122 + x2 + 92
⇒ x2 =  172 - 122 - 92
⇒ x2 = 289 - 144 - 81
⇒ x2  = 64
⇒ x = 8 cm

4. The internal length, breadth and height of a box are 30 cm, 24 cm, and 15 cm. Find the largest number of cubes which can be placed inside this box if the edge of each cube is

(i) 3 cm (ii) 4 cm (iii) 5 cm

(i) No. of cube which can be placed along length = 30/3 = 10
No. of cube along the breadth = 24/3 = 8
No. of cubes along the height = 15/3 = 5.
∴ The total no. of cubes placed = 10×8×5 = 400
(ii) Cubes along length = 30/4  = 7.5 = 7
Cubes along width = 24/4 = 6 and cubes along height = 15/4 = 3.75 = 3
∴ The total no. of cubes placed = 7×6×3 = 126
(iii) Cubes along length = 30/5 = 6
Cubes along width = 24/5 = 4.5 = 4 and cubes along height = 15/5 = 3
∴ The total no. of cubes placed = 6×4×3 =  72

5. A rectangular field is 112 m long and 62 m broad. A cubical tank of edge 6 m is dug at each of the four corners of the field and the earth so removed is evenly spread on the remaining field. Find the rise in level.

Vol. of the tank = vol. of earth spread
4×63 m3 = (112×62 -4×62) m2 × Rise in level
⇒ Rise in level = (4×63)/(112×62 - 4×62)
= 864/6800
= 0.127 m
= 12.7cm

6. When length of each side of a cube is increased by 3 cm, its volume is increased by 2457 cm3. Find its side. How much will its volume decrease, if length of each side of it is reduced by 20%?

Let a be the side of the cube.
Side of the new cube = a+ 3
Volume of the new cube = a3 + 2457
That is, (a+3)2 = a3 + 2457
⇒ a3 + 3×a×3(a+3) + 33 = a3 + 2457
⇒ 9a3 + 27a + 27 = 2457
⇒ 9a3 + 27a - 2430 = 0
⇒ a3 + 3a -270 = 0
⇒ a3 + 18a -  15a - 270 = 0
⇒ a(a+18) - 15(a+18) = 0
⇒ (a- 15)(a + 18) = 0
⇒ a -15 = 0 or a+18 = 0
⇒ a = 15 or a = -18
⇒ a = 15 cm [since side cannot be negative]
Volume of the cube whose side is 15 cm = 153 = 3375 cm3
Suppose the length of the given cube is reduced by 20%
Thus new side anew = a- ( 20/100)×a
= a[1-(1/5)]
= (4/5)×15
= 12 cm
Volume of the new cube whose side is 12 cm = 123 = 1728 cm3
Decrease in volume = 3375 - 1728 = 1647 cm3

7. A rectangular tank 30 cm × 20 cm × 12 cm contains water to a depth of 6 cm. A metal cube of side 10 cm is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in litres, that must be poured in the tank so that the metal cube is just submerged in the water.

The dimensions of rectangular tank = 30 cm× 20cm ×12cm
Side of the cube = 10cm
volume of the cube = 103 = 1000 cm3
The height of the water in the tank is  6 cm.
Volume of the cube till 6 cm = 10×10×6 = 600 cm3
Hence when the cube is placed in the tank,
then the volume of the water increase by  600 cm3 .

Out of this area, let us subtract the surface area of the cube.
Thus, the surface area of the shaded part in the above figure is  500 cm2
The displaced water is spread out in 500 cm2 to a height of 'h' cm.
And hence the volume of the water displaced is equal to the volume of the part of the cube in water.
Thus, we have ,
500 × h = 600 cm3
⇒ h = 600/500  cm
⇒ h = 1.2 cm
Thus, now the level of the water in the tank is = 6+1.2 = 7.2 cm
Remaining height of the water level,  so that the metal cube is just
submerged in the water = 10- 7.2 = 2.8 cm
Thus the volume of the water that must be poured in the tank so that the metal
cube is just submerged in the water = 2.8 ×500 = 1400 cm3
we know that 1000 cc = 1 litre
Thus, the required volume of water = 1400/1000 = 1.4 litres.

8. The dimensions of a solid metallic cuboid are 72 cm × 30 cm × 75 cm. It is melted and recast into identical solid metal cubes with each of edge 6 cm. Find the number of cubes formed.

Also, find the cost of polishing the surfaces of all the cubes formed at the rate Rs. 150 per sq. m.

The dimensions of a solid cuboid are: 72 cm, 30cm, 75cm
Volume of the cuboid = 72cm ×30 cm ×75 cm = 162000 cm3
Side of a cube = 6 cm
Volume of a cube = 63 = 216 cm3
The number of cubes = 162000/216 = 750
The surface area of a cube = 6a2 = 6×62 = 216 cm2
Total surface area of 750 cubes = 750 ×216 = 162000 cm2
Total surface area in square metres = 162000/10000
= 16.2 square metres
Rate of  polishing the surface per square metre = Rs. 150
Total cost of polishing the surface = 150×16.2 = Rs. 2430

9. The dimensions of a car petrol tank are 50 cm × 32 cm × 24 cm, which is full of petrol. If car’s average consumption is 15 km per litre, find the maximum distance that can be covered by the car.

The dimensions of a car petrol tank are = 50cm × 32 cm ×24 cm
Volume of the tank = 38400 cm3
We know that 1000 cm3 = 1 litre
Thus volume of the tank = 38400/1000 = 38.4 litres
The average consumption of the car = 15 km/litres
Thus, the total distance that can be covered by the car = 38.4×15 = 576 km.

10. The dimensions of a rectangular box are in the ratio 4 : 2 : 3. The difference between cost of covering it with paper at Rs. 12 per m2 and with paper at the rate of 13.50 per m2 is Rs. 1,248. Find the dimensions of the box.

Given dimensions of a rectangular box are in the ratio 4:2:3
Therefore, the total surface area of the box
= 2[4x ×2x + 2x ×3x + 4x ×3x]
= 2[8x2 + 6x2 + 12x2 ] m2
Difference between cost of covering the box with paper at Rs. 12 per m2  and with paper at Rs. 13.50 per m2 = Rs. 1,248
⇒ 52x2 [13.5 - 12] = 1248
⇒ 52× x2 ×1.5 = 1248
⇒ 78× x2 = 1248
⇒ x2 = 1248/78
⇒ x2 = 16
⇒ x = 4  [Length, width and height cannot be negative]
Thus, the dimensions of the rectangular box are = 4×4 m, 2×4m, 3×4 m
Thus, the dimensions are 16m, 8m and 12 m .