 # RD Sharma Solutions Class 8 Volumes Surface Area Cuboid Cube Exercise 21.1

## RD Sharma Solutions Class 8 Chapter 21 Exercise 21.1

#### Exercise 21.1

Q 1. Find the volume of a cuboid whose:

i) Length = 12 cm, breadth = 8 cm and height = 6 cm

ii) length = 1.2 m, breadth = 30 cm ,height = 15 cm

iii)  length = 15 cm, breadth = 2.5 dm, height = 8 cm

Soln:

i) For a cuboid, the given measurements are

Length = 12 cm

Height = 6 cm

We know that,

Volume of the cuboid = length x breadth x height Cubic units

Substitute the given values

V =12 x 8 x 6 = 576

Hence, Volume of the cuboid = 576 cm3

ii) For a cuboid, the given measurements are

Length = 1.2 m = 120 cm ( Since 1m= 100 cm)

Height = 15 cm

We know that,

Volume of the cuboid = length x breadth x height Cubic units

Substitute the given values

V =120 x 30 x 15 = 54000

Hence, Volume of the cuboid = 54000 cm3

iii) For a cuboid, the given measurements are

Length = 15 cm

Breadth = 2.5 dm = 25 cm (since 1dm=10cm)

Height = 8 cm

We know that,

Volume of the cuboid = length x breadth x height Cubic units

Substitute the given values

V =15 x 25 x 8 = 3000

Hence, Volume of the cuboid = 3000 cm3

Q2. Find the volume of cube whose side is:

i) 4 cm

ii) 8 cm

iii) 1.5 dm

iv) 1.2 m

v) 25 mm

Soln:

i) Given: Side of a cube,a = 4 cm

We know that,

Volume of the cube = ( side )3 =( a)3 Cubic units

V = ( 4)3 = 64 cm3

So, the Volume of the cube = 64 cm3

ii)Given: Side of a cube,a = 8 cm

We know that,

Volume of the cube = ( side )3 =( a)3 Cubic units

V = ( 8)3 = 512 cm3

So, the volume of the cube = 512 cm3

iii)Given: Side of a cube,a = 1.5 dm =15 cm ( since 1 dm = 10 cm)

We know that,

Volume of the cube = ( side )3 =( a)3 Cubic units

V = ( 15)3 = 3375 cm3

Hence, the volume of the cube = 3375 cm3

iv) Given: Side of a cube,a = 1.2 m =120 cm ( since 1 m = 100 cm)

We know that,

Volume of the cube = ( side )3 =( a)3 Cubic units

V = ( 120)3 = 1728000 cm3

Hence, the volume of the cube =1728000 cm3

v) Given: Side of a cube,a = 25 mm =2.5 cm ( since 1 mm = 0.1 cm)

We know that,

Volume of the cube = ( side )3 =( a)3 Cubic units

V = ( 2.5)3 = 15.625 cm3

Hence, the volume of the cube =15.625 cm3

Q3. Find the height of a cuboid of volume 100cm3, whose length and breadth are 5 cm and 4 cm respectively.

Soln:

Let us assume that, the height of the cuboid be “h” cm

The given data are:

Volume =  100 cm3

Length = 5 cm and

We know that, the  volume of the cuboid = length x breadth x height cubic units

100 = 5 x 4 x h

Therefore, h = 100/(4×5)

h = 100 / 20h= 5 cm

Hence, the height of the cuboid, h = 5 cm

Q 4. A cuboidal vessel is 10 cm long and 5 cm wide. How high must it be made to hold 300 cm3 of a liquid?

Soln:

Let us assume that, the height of the cuboid vessel be “h” cm

The given data are:

Volume =  300 cm3

Length = 10 cm and

We know that, the  volume of the cuboid = length x breadth x height cubic units

300 = 10 x 5 x h

Therefore, h = 300 /(10 x 5)

h = 300/50

h= 6 cm

Hence, the height of the cuboid vessel, h = 6 cm

Q5. A milk container is 8 cm long and 50 cm wide. What should be its height so that it can hold 4 litres of milk?

Soln:

The given data are:

Length = 8 cm

Let us assume that, the height of the cuboidal container be “h” cm

It is given that, the cuboidal container holds 4 litres of milk.

So, volume = 4 litres= 4 x 1000 cm3

=4000 cm3 ( Since, 1 L = 1000 cm3 )

We know that,

the volume of the cuboidal container = length x breadth x height cubic units

4000 = 8 x 50 x h

4000 = 400 x h

h = 4000/400= 10 cm

Hence, the height of the cuboidal shape milk container is 10 cm.

Q6. A cuboidal wooden block contains 36 cm3 wood. If it be 4 cm long and 3 cm wide, find its height.

Soln:

Given data:

The volume of the cuboidal wooden block, V =  36 cm3

Length= 4 cm

Assume that, the height of the wooden block be “h” cm

So, volume of a cuboid = length x breadth x height cubic units

36 = 4 x 3 x h

36 = 12 x h

h = 36/12 = 3 cm

So, the height of the wooden block = 3 cm.

Q 7. What will happen to the volume of the cube, if its edge is :

i) Halved

ii) Trebled?

Soln:

(i)  Assume that, the length of the edge of the cube be “a”.

We know that, the volume of the cube = ( side )3 = a3

If the length of the side is halved, then the length of the new edge will become a/2.

So, the volume of the new cube = ( side )3 = $(\frac{ a}{ 2 })^{3}$

= $\frac{a^{3}}{8}$

Therefore, it is observed that if the edge of a cube is halved, then the new volume of a cube will be

1/8  times the initial volume of a cube.

(ii) Assume that, the length of the edge of the cube be “a”.

We know that, the volume of the cube = ( side )3 = a3

If the length of the side is trebled, then the length of the new edge will become a x 3.

So, the volume of the new cube = ( side )3 = ( 3 x a )3 =  27 x a3

Therefore, it is observed that if the edge of a cube is trebled, then the new volume of a cube will be

27 times the initial volume of a cube.

Q8. What will happen to the volume of cuboid if its :

i) Length is doubled, height is same and breadth is halved?

ii) length is doubled, height is doubled and breadth is same?

Soln:

i) Assume that the length, breadth, and height of the cuboid be l, b and h, respectively.

We know that, the volume of cuboid = l x b x h cubic units

If the length of the cuboid is doubled, then the length will become 2 x l.

If the breadth of the cuboid is halved, then the breadth will become b/2

The height of the cuboid “h” remains the same.

so, the volume of the new cuboid = length x breadth x height

= 2 x l x h x (b/2)

= l x b x h .

So, it is observed that the new volume of the cuboid is the same as the initial volume of the cuboid.

Therefore, there is no change in the volume of the cuboid

ii) Assume that the length, breadth and height of the cuboid be l, b and h, respectively.

We know that, the volume of cuboid = l x b x h cubic units

If the length of the cuboid is doubled, then the length will become 2 x l.

If the height of the cuboid is doubled, then the height will become h x 2

The breadth of the cuboid “b” remains the same.

so, volume of the new cuboid = length x breadth x-height

= 2 x l x b x 2 x h = 4 x l x b x h

So, it is observed that the volume of the new cuboid is equal to four times the initial volume of the cuboid.

Q9. Three cuboids of dimensions 5 cm x 6 cm x 7 cm , 4 cm x 7 cm x 8 cm and 2 cm x 3 cm x 13 cm are melted and a cube is made. Find the side of the cube.

Soln:

Given:

Three cuboids dimensions are:

Cuboid 1 = 5 cm x 6 cm x 7 cm,

Cuboid 2 =4 cm x 7 cm x 8 cm

Cuboid 3 =2 cm x 3 cm x 13 cm

Given that, a new cube is formed by melting the given three cuboids.

so, volume of the new cube = sum of the volumes of three cuboids

V = ( 5 x 6 x 7 ) + ( 4 x 7  x 8  ) + ( 2  x 3  x 13 )

V = ( 210 ) + ( 224 ) + ( 78 ) = 512

So, the volume of cube after melting the three cuboids = 512 cm3

We know that, the volume of a cube = ( side )3 ,

So, it becomes

512 cm3 = ( side )3

Take cube root on both the sides,

( side ) = $\sqrt{ 512 } = 8 cm$

So, the side of the new cube = 8 cm.

Q 10. Find the weight of a solid rectangular iron piece of size 50 cm x 40 cm x 10 cm, if 1 cm3 of iron weighs 8 gm.

Soln :

Given, the rectangular iron piece dimension = 50 cm x 40 cm x 10 cm.

We know that, the volume of cuboid = l x b x h cubic units

So, volume = 50x 40x 10  = 20000 cm3

Given that, the weight of 1 cm3 of iron is 8 gm.

Therefore, the weight of the rectangular piece of iron for the given dimension = 20000 x 8 gm

Weight = 160000 gm

Weight= 160 x 1000 gm (1 kg = 1000 gm )

Therefore, the weight of the rectangular iron piece= 160 kg .

Q 11. How many wooden cubical blocks of side 25 cm can be cut from a log of wood of size 3 m by 75 cm by 50 cm, assuming that there is no wastage?

Soln :

Given that, the dimension of the wooden log =  3 m x 75 cm x 50 cm,

Convert all the metrics into cm, so it becomes

Dimension =300 cm x 75 cm x 50 cm ( Since 1 m = 100 cm )

So, the volume of the wooden cuboid is = 300 x 75 x 50

V = 1125000 cm3

Note that, the given information states that the side of each cubical block of wood,a = 25 cm.

Therefore,  the volume of one cubical block = ( a )3

= 253

So, the volume of cubical wooden block = 15625 cm3

so, the required quantity of cubical blocks is,

Volume of the wood log/volume of one cubical block  = 1125000 cm3 /156253 = 72

Q 12. A cuboidal block of silver is 9 cm long, 4 cm broad and 3.5 cm in height. From it, beads of volume 1.5 cm3 each are to be made. Find the number of beads that can be made from the block.

Soln:

Given:

Length of the cuboidal silver block = 9 cm

Height = 3.5 cm

we know that,  the volume of the cuboidal block = length x breadth x height cubic units

V= 9 x 4 x 3.5

V= 126 cm3

Therefore, the required number of beads with volume 1.5 cm3 that can be made from the block is,

= Volume of the silver block/ volume of one beads = 126 cm3 / 1.5 cm3

Q 13. Find the number of cuboidal boxes measuring 2 cm by 3 cm by 10 cm which can be stored in a carton whose dimensions are 40 cm, 36 cm, and 24 cm.

Soln :

Given: cuboidal box Dimension= 2 cm x 3 cm x 10 cm

V= ( 2 x 3 x 10 ) cm3

V= 60 cm3

Given that, the dimension of a carton box that can be stored = 40 cm x 36 cm x 24 cm

So, the Volume of the carton = ( 40 x 36 x 24 ) cm3

V= 34560 cm3

Hence, the number of cuboidal boxes stored in the carton is,

=  Volume of the carton /volume of one cuboidal box = 34560 cm3 /60 cm3

No. of Boxes= 576

Q 14. A cuboidal block of solid iron has dimensions 50 cm, 45 cm and 34 cm. How many cuboids of size 5 cm by 3 cm by 2 cm can be obtained from this block? Assume cutting causes no wastage.

Soln :

Given, cuboidal iron block dimension= 50 cm x 45 cm x 34 cm

We know that, the volume of the iron block = length x breadth x height cubic units

V= ( 50 x 45 x 34 ) cm3

V= 76500 cm3

The dimension of one small cuboids = 5 cm x 3 cm x 2 cm.

So, the volume of one small cuboid = l x b x h = 5 x 3 x 2 = 30 cm3

Hence, the required amount of small cuboids obtained from the iron block is,

= Volume of the iron block / volume of one small cuboid  =  76500 cm3/ 30 cm3= 2550

Therefore, the number of cuboids obtained is 2550.

Q 15. A cube A has side thrice as long as that of cube B. What is the ratio of the volume of cube A to that of cube B?

Soln:

Assume that,  the side length of the cube, B be “l” cm.

so, the length of the side of cube A is 3 x l cm.

Now, ratio =Volume of cube A /volume of cube B

= (3 * l)3/(l)3

= (27 x l3)/l3

Cancel l3.

Therefore, ration=27 / 1

Hence, the ratio of the volume of cube A to the volume of cube B is 27: 1

Q 16. An ice-cream brick measures 20 cm by 10 cm by 7 cm. How many such bricks can be stored in deep fridge whose inner dimensions are 100 cm by 50 cm by 42 cm?

Soln:

Given, An ice cream brick Dimension= 20 cm x 10 cm x 7 cm

So, V = l x b x h = ( 20 x 10 x 7 ) cm3 = 1400 cm3

Als given that, the deep fridge inner dimension = 100 cm x 50 cm x 42 cm.

So, V = lx b x h= ( 100 x 50 x 42 ) cm3 = 210000 cm3

So, the number of ice cream bricks stored in the fridge is,

ratio =   Volume of the fridge/volume of an ice-cream brick  = 210000cm3/(1400)cm3

Therefore, the number of ice-cream bricks= 150

Q 17. Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find volumes V1 and V2 of the cubes and compare them.

Soln:

Given, the edge of the cube 1 = 2 cm

the edge of the cube 2 = 4 cm.

So, the volume of the cube with side 2 cm, V1 = ( a )3

V1 = ( 2 )3

V1 = 8 cm3

The volume of the cube with side 4 cm,

V2 = ( side )3

V2= (4)3

V2= 64 cm3

It is observed that: V2 = 64 cm3 = 8 x 8 cm3 = 8 x V1

So, V2 = 8V1

Q 18. A tea – packet measures 10 cm x 6 cm x 4 cm. how many such tea – packets can be placed in a cardboard box of dimensions 50 cm x 30 cm x 0.2 m?

Soln:

Given:  A tea packet dimension is 10 cm x 6 cm x 4 cm.

Volume, V= length x breadth x height =  10 x 6 x 4

V = 240 cm3

Also, the cardboard box dimension = 50 cm x 30 cm x 0.2 m,

Change into common metric measurements in cm,

50 cm x 30 cm x 20 cm  ( Since, 1 m = 100 cm )

So, the volume of the cardboard box = lx b x h =  50 x 30 x 20

V = 30000 cm3

Hence, the required quantity of tea packets placed inside the cardboard box is

So,  ratio =Volume of the box / volume of a tea packet

=  30000 cm/ 240 = 125

Therefore, the number of tea packets= 125

Q 19. The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg. Find the weight of a block of the same metal of size 15 cm by 8 cm by 3 cm.

Soln:

Given: The metal block of dimension 5 cm x 4 cm x 3 cm has a weight of 1 kg.

We know that, volume = length x breadth x height cubic units

= 5 x 4 x 3

V = 60 cm3

Therefore, the weight of 60 cm3 of the metal is 1 kg

Again, the other block which is of same metal has the dimension of 15 cm x 8 cm x 3 cm.

So,  volume = 15 x 8 x 3

V = 360 cm3

Hence, the weight of the required block = 360 cm3

It can be written as 6 x 60 cm3

We know that the weight of 60 cm3 of the metal is 1 Kg

So, the weight of the metal block  with dimension 15 cm x 8 cm x 3 cm is 6 x 1 kg = 6 kg.

Q 20. How many soap cakes can be placed in a box of size 56 cm x 0.4 cm x 0.25 m, if the size of a soap cake is 7 cm x 5 cm x 2.5 cm?

Soln:

Given: A soap cake Dimension = 7cm x 5 cm x 2.5 cm

So, volume = 7 x 5 x 2.5

V = 87.5 cm3

Also, the dimension of the Soap cake box  is 56 cm x 0.4 m x 0.25 m,

Change into common metric measurements in cm,

i.e. , 56 cm x 40cm x 25 cm ( Since ,  1 m = 100 cm ).

So, the volume of the soap cake box = 56 x 40 x 25

V = 56000 cm3

Therefore, The no. of soap cakes that can be placed inside the box is

Volume of the box/ volume of a soap cake  = 56000 cm3 / 87.5  cm3 =640

No.of Soap cakes = 640

Q 21. The volume of a cuboidal box is 48 cm3. If its height and length are 3 cm and 4 cm respectively, find its breadth.

Soln:

Assume that, the breadth of the cuboidal box be “b” cm.

Given that, the volume of the cuboidal box is 48 cm3

Height = 3 cm

Length = 4 cm

We know that, volume of cuboidal box = length x breadth x height cubic units

48 = 4 x b x 3

48 = 12 x b

Therefore, b = 48/12

b= 4 cm

Hence, the breadth of the cuboidal box “b”is 4 cm.