## 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)Â length = 15 cm, breadth = 2.5 dm, height = 8 cm**

**Soln:**

**i)** For a cuboid, the given measurements are

Length = 12 cm

Breadth = 8 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 cm^{3}

**ii)** For a cuboid, the given measurements are

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

Breadth = 30 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 cm^{3}

**iii)Â **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 cm^{3}

**Â **

**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)**Â Given: Side of a cube,a = 4 cm

We know that,

Volume of the cube = ( side )^{3} =( a)^{3}Â Cubic units

V = ( 4)^{3} = 64 cm^{3}

So, the Volume of the cube = 64 cm^{3}

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

We know that,

Volume of the cube = ( side )^{3} =( a)^{3}Â Cubic units

V = ( 8)^{3} = 512 cm^{3}

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

**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}Â Cubic units

V = ( 15)^{3} = 3375 cm^{3}

Hence, the volume of the cube = 3375 cm^{3}

**iv)Â **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}Â Cubic units

V = ( 120)^{3} = 1728000 cm^{3}

Hence, the volume of the cube =1728000 cm^{3}

**v)Â **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}Â Cubic units

V = ( 2.5)^{3} = 15.625 cm^{3}

Hence, the volume of the cube =15.625 cm^{3}

**Q3. Find the height of a cuboid of volume 100cm ^{3}, 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 =Â 100 cm^{3}

Length = 5 cm and

Breadth = 4 cm

We know that, theÂ 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 cm ^{3} of a liquid?**

**Soln:**

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

The given data are:

Volume =Â 300 cm^{3}

Length = 10 cm and

Breadth = 5 cm

We know that, theÂ 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

Breadth = 50 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 cm^{3}

=4000 cm^{3} ( Since, 1 L = 1000 cm^{3} )

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 cm ^{3} 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 =Â 36 cm^{3}

Length= 4 cm

Breadth = 3 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 = 36/12Â = 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)Â **Â Assume that, the length of the edge of the cube be “a”.

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

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

1/8Â Â times the initial volume of a cube.

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

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

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} =Â 27 x a^{3}

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

27Â 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?**

**Â ****Soln:**

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

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

If the length of the cuboid is doubled, then the length will becomeÂ 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 =Â l x b x h cubic units

If the length of the cuboid is doubled, then the length will becomeÂ 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 7Â x 8Â ) + ( 2Â x 3Â x 13 )

V = ( 210Â ) + ( 224 ) + ( 78 ) = 512

So, the volume of cube after melting the three cuboids =Â 512 cm^{3 }

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

So, it becomes

512 cm^{3 }= ( side )^{3}

Take cube root on both the sides,

( side ) = \(\sqrt[3]{ 512 } = 8 cm\)

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 cm ^{3} of iron weighs 8 gm.**

**Soln :**

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

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

So, volume = 50x 40x 10Â = 20000 cm^{3}

Given that, the weight of 1 cm^{3} 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 =Â 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 cm^{3}

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

Therefore,Â the volume of one cubical block = ( a )^{3}

= 25^{3}

So, the volume of cubical wooden block = 15625 cm^{3}

so, the required quantity of cubical blocks is,

Volume of the wood log/volume of one cubical blockÂ = 1125000 cm^{3}Â /15625^{3} = 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 cm ^{3} 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

Breadth = 4 cm

Height = 3.5 cm

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

V= 9 x 4 x 3.5

V= 126 cm^{3}

Therefore, the required number of beads with volume 1.5 cm^{3} that can be made from the block is,

= Volume of the silver block/ volume of one beads = 126 cm^{3} / 1.5 cm^{3}Â

No of Beads= 84

**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 ) cm^{3}

V= 60 cm^{3}

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 ) cm^{3}

V= 34560 cm^{3}

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

= Â Volume of the carton /volume of one cuboidal box = 34560 cm^{3} /60 cm^{3}

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 ) cm^{3}

V= 76500 cm^{3}

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 2Â = 30 cm^{3}

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

= Volume of the iron block / volume of one small cuboidÂ =Â 76500 cm^{3}/ 30 cm^{3}= 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,Â 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 l^{3})/l^{3}

Cancel l^{3.}

Therefore, ration=27 / 1Â

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 ) cm^{3} = 1400 cm^{3}

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 ) cm^{3} = 210000 cm^{3}

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

ratio =Â Â Volume of the fridge/volume of an ice-cream brick Â =Â 210000cm^{3}/(1400)cm^{3}

Therefore, the number of ice-cream bricks= 150Â

**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 cm^{3}

The volume of the cube with side 4 cm,

V2 = ( side )^{3}

V2= (4)^{3}

V2= 64 cm^{3 }

It is observed that: V2 = 64 cm^{3} = 8 x 8 cm^{3} = 8 x V1

So, V2 = 8V1

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

**Soln:**

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

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

V = 240 cm^{3 }

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 Â ( Since, 1 m = 100 cm )

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

V = 30000 cm^{3}

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

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

=Â 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 cm^{3}

Therefore, the weight of 60 cm^{3} 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,Â volume = 15 x 8 x 3

V = 360 cm^{3}

Hence, the weight of the required block = 360 cm^{3}

It can be written as 6 x 60 cm^{3}

We know that the weight of 60 cm^{3} of the metal is 1 Kg

So, the weight of the metal blockÂ with dimensionÂ 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 cm^{3 }

Also, the dimension of the Soap cake boxÂ 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 , Â 1 m = 100 cm ).

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

V = 56000 cm^{3}

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

=Â Volume of the box/ volume of a soap cakeÂ = 56000 cm^{3} / 87.5Â cm^{3} =640

No.of Soap cakes = 640

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

**Soln:**

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

Given that, the volume of the cuboidal box is 48 cm^{3 }

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.