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# Class 8 RD Sharma Solutions – Chapter 21 Mensuration II (Volume and Surface Areas of a Cuboid and a Cube) – Exercise 21.1 | Set 2

### Question 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.

Solution:

The details given about cuboidal block of silver are –

Length of cuboidal block = 9 cm

Breadth of cuboidal block = 4 cm

Height of cuboidal block = 3.5 cm

Volume of a cuboid = l * b * h

= 9 * 4 * 3.5

= 126 cm3

= 126 / 1.5

### Question 13: Find the number of cuboidal boxes measuring 2 cm by 3 cm by 10 cm which can be stored in a carton whose dimensionsare 40 cm, 36 cm and 24 cm.

Solution:

The details given about cuboidal boxes are –

Length of cuboidal boxes = 2 cm

Breadth of cuboidal boxes = 3 cm

Height of cuboidal boxes = 10 cm

Volume of a cuboid = l * b * h

= 2 * 3 * 10

= 60 cm3

The details given about cartoon are –

Length of cartoon = 40 cm

Breadth of cartoon = 36 cm

Height of cartoon = 24 cm

Volume of a cartoon = l * b * h

= 40 * 36 * 24

= 34560 cm3

Number of boxes that can be stored in a cartoon = Volume of cartoon / Volume of cuboid

= 34560 / 60

= 576 boxes

### Question 14: A cuboidal block of solid iron has dimension 50 cm, 45 cm and 34 cm, how many cuboids of size 5 cm by 3 cm by 2 cm canbe obtained from this block. Assume cutting causes no wastage.

Solution:

The details given about cuboidal block are –

Length of cuboidal block = 50 cm

Breadth of cuboidal block = 45 cm

Height of cuboidal block = 34 cm

Volume of a cuboid = l * b * h

= 50 * 45 * 34

= 76500 cm3

The details given about cuboid are –

Length of cuboid = 5 cm

Breadth of cuboid = 3 cm

Height of cuboid = 2 cm

Volume of a cuboid = l * b * h

= 5 * 3 * 2

= 30 cm3

Number of boxes that can be stored in a cartoon = Volume of cartoon / Volume of cuboid

= 76500 / 30

= 2550 cuboids

### Question 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.

Solution:

Let side of cube B = x

Now side of cube A = 3x

Volume of cube A = (side)3

= (3x)3

= 27x3

Volume of cube B = (side)3

= (x)3

= x3

Volume of cube A / Volume of cube B = 27x3 / x3

= 27 : 1

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

Solution:

The details given about ice – cream brick are –

Length of ice – cream brick  = 20 cm

Breadth of ice – cream brick = 10 cm

Height of ice – cream brick = 7 cm

Volume of a ice – cream brick = l * b * h

= 20 * 10 * 7

= 1400 cm3

The details given about fridge are –

Length of fridge = 100 cm

Breadth of fridge = 50 cm

Height of fridge = 42 cm

Volume of a fridge = l * b * h

= 10 * 50 * 42

= 21000 cm3

Number of brick that can be stored in fridge = Volume of fridge / Volume of brick

= 21000 / 14000

= 150 bricks

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

Solution:

Volume of cube = (side)3

V1 = (2)3

= 8 cm3

Volume of cube = (side)3

V2 = (4)3

= 64

V1 / V2 = 8 / 64

V1 / V2 = 1 / 8

V2 = 8V1

### Question 18: A tea – packet measures 10 cm * 6 cm * 4 cm. How many such tea packets can be placed in a cardboard box of dimensions 50 cm * 30 cm * 0. 2 m?

Solution:

The details given about tea – packet are –

Length of tea – packet  = 10 cm

Breadth of tea – packet = 6 cm

Height of tea – packet = 4 cm

Volume of a tea – packet = l * b * h

= 10 * 6 * 4

= 240 cm3

The details given about cardboard box are –

Length of cardboard box = 50 cm

Breadth of cardboard = 30 cm

Height of cardboard = 0.2 m = 20 cm (1 m = 100 cm)

Volume of a cardboard box = l * b * h

= 50 * 30 * 20

= 30000 cm3

Number of tea – packets that can be placed in cardboard box = Volume of cardboard box / Volume of tea packet

= 30000 / 240

= 125 tea packet

### Question 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.

Solution:

The details given metal block are –

Length of block  = 5 cm

Breadth of block = 4 cm

Height of block = 3 cm

Volume of block = l * b * h

= 5 * 4 * 3

= 60 cm3

The details given about new box are –

Length of new block  = 15 cm

Breadth of new block = 8 cm

Height of new block = 3 cm

Volume of new block = l * b * h

= 15 * 8 * 3

= 360 cm3

Weight of 60 cm3 block = 1 kg

Weight of 360 cm3 block = 360 / 60

= 6 kg

### Question 20: How many soap cakes can be placed in a box of size 56 cm * 0.4 m * 0.25 m, if the size of a soap cake is 7 cm * 5 cm * 2.5 cm?

Solution:

The details given about box are –

Length of box  = 56 cm

Breadth of box = 0.4 m = 40 cm (1 m = 100 cm)

Height of box = 0.25 m = 25 cm (1m = 100 cm)

Volume of a box = l * b * h

= 56 * 40 * 25

= 56000 cm3

The details given about soap cake are –

Length of soap cake = 7 cm

Breadth of soap cake = 5 cm

Height of soap cake = 2.5 cm

Volume of a soap cake = l * b * h

= 7 * 5 * 2.5

= 87.5 cm3

Number of soap cakes that can be placed in a box = Volume of cardboard box / Volume of soap cake

= 56000 / 87.5

= 640 soap cakes

### Question 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.

Solution:

The details given about cuboidal box are

Volume of a cuboidal box = 48 cm3

Height of a cuboidal box = 3 cm

Length of a cuboidal box = 4 cm

Let height of cuboidal box = h

Volume of cuboid = l * b * h

48 = 3 * 4 * h

48 = 12 * h

h = 4 cm

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