RD Sharma solution class 8 chapter 21 Mensuration II(Volume and Surface Area of Cuboid and a Cube) Exercise 21.1

Exercise 21.1

Page-21.8

Question 1:

Find the volume of a cuboid whose
(i) length = 12 cm, breadth = 8 cm, 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.

Answer 1:

$$\begin{gathered} \text{(i)} \\ \text{In the given cuboid, we have:} \\ \text{length=12 cm, breadth=8 cm and height=6 cm} \\ \therefore \text{Volume of the cuboid = length×breadth×height} \\ \text{=12×8×6} \\ {\text{=576 cm }}^3 \\ \text{(ii)} \\ \text{In the given cuboid, we have:} \\ \text{length=1.2 m} \\ \text{=1.2×100 cm }(1 m = 100 cm) \\ \text{=120 cm} \\ \text{breadth=30 cm} \\ \mathrm{h}\text{eight=15 cm} \\ \therefore \text{Volume of the cuboid = length×breadth×height} \\ \text{=120×30×15} \\ {\text{=54000 cm }}^3 \\ \text{(iii)} \\ \text{In the given cuboid, we have:} \\ \text{length=1.5 dm} \\ \text{=1.5×10 } (1 dm = 10 cm) \\ \text{= 15 cm} \\ \text{breadth=2.5 dm=2.5×10 cm=25 cm} \\ \mathrm{h}\text{eight=8 cm} \\ \therefore \text{Volume of cuboid = length×breadth×height} \\ \text{=15×25×8} \\ {\text{=3000 cm}}^{ 3} \end{gathered}$$

Question 2:

Find the volume of a cube whose side is
(i) 4 cm
(ii) 8 cm
(iii) 1.5 dm
(iv) 1.2 m
(v) 25 mm

Answer 2:

$$\begin{gathered} \text{(i)} \\ \mathrm{The} \mathrm{s}\text{ide of the given cube is 4 cm.} \\ {\text{∴ Volume of the cube=(side)}}^3{\text{=(4)}}^3{\text{=64 cm}}^3 \\ \text{(ii)} \\ \mathrm{The} \mathrm{s}\mathrm{ide of the given cube is}\text{ 8 cm.} \\ {\text{∴ Volume of the cube=(side)}}^{3 }{\text{= (8)}}^{3 }{\text{= 512 cm}}^3 \\ \text{(iii)} \\ \mathrm{The} \mathrm{s}\mathrm{ide of the given cube }= \text{1.5 dm=1.5×10 cm=15 cm} \\ {\text{∴ Volume of the cube=(side)}}^3{\text{=(15)}}^3{\text{=3375 cm}}^3 \\ \text{(iv)} \\ \mathrm{The} \mathrm{s}\mathrm{ide of the given cube} \text{= 1.2 m=1.2×100 cm=120 cm} \\ {\text{∴ Volume of the cube=(side)}}^3{\text{=(120)}}^3{\text{=1728000 cm}}^3 \\ \text{(v)} \\ \mathrm{The} \mathrm{s}\mathrm{ide of the given cube }\text{= 25 mm = }\frac{25}{10}\text{ cm=2.5 cm} \\ {\text{∴ Volume of the cube=(side)}}^3{\text{=(2.5)}}^3{\text{=15.625 cm}}^3 \end{gathered}$$

Question 3:

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

Answer 3:

$$\begin{gathered} \mathrm{Let} \mathrm{us} \mathrm{suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboid} \mathrm{is} h \mathrm{cm}. \\ \mathrm{Given}: \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}= 100 {\mathrm{cm}}^3 \\ \mathrm{Length}=5 \mathrm{cm} \\ \mathrm{Breadth}=4 \mathrm{cm} \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 100=5\times 4\times h \\ \Rightarrow 100=20\times h \\ \therefore h=\frac{100}{20}=5 \mathrm{cm} \end{gathered}$$

Question 4:

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

Answer 4:

$$\begin{gathered} \mathrm{Let} h \mathrm{cm} \mathrm{be} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{vessel}. \\ \mathrm{Given}: \\ \mathrm{Length} =10 \mathrm{cm} \\ \mathrm{Breadth}=5 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{vessel}=300 {\mathrm{cm}}^3 \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 300=10\times 5\times h \\ \Rightarrow 300=50\times h \\ \therefore h=\frac{300}{50}=6 \mathrm{cm} \end{gathered}$$

Question 5:

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?

Answer 5:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{milk} \mathrm{container}=8 \mathrm{cm} \\ \mathrm{Breadth}=50 \mathrm{cm} \\ \mathrm{Let} h \mathrm{cm} \mathrm{be} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{container}. \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{container} \mathrm{can} \mathrm{hold} 4 \mathrm{L} \mathrm{of} \mathrm{milk}. \\ \mathrm{i}.\mathrm{e}., \mathrm{volume}=4 \mathrm{L}=4\times 1000 {\mathrm{cm}}^3=4000 {\mathrm{cm}}^3 (\because 1 \mathrm{L}=1000 {\mathrm{cm}}^3\mathrm{)} \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{container}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 4000=8\times 50\times h \\ \Rightarrow 4000=400\times h \\ \Rightarrow h=\frac{4000}{400}=10 \mathrm{cm} \\ \therefore \mathrm{The} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{milk} \mathrm{container} \mathrm{is} 10 \mathrm{cm}. \end{gathered}$$

Question 6:

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

Answer 6:

$$\begin{gathered} \mathrm{A} \mathrm{cuboidal} \mathrm{wooden} \mathrm{block} \mathrm{contains} 36 {\mathrm{cm}}^3 \mathrm{of} \mathrm{wood}. \\ \mathrm{i}.\mathrm{e}., \mathrm{volume}=36 {\mathrm{cm}}^3 \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{block}=4 \mathrm{cm} \\ \mathrm{Breadth} \mathrm{of} \mathrm{block}=3 \mathrm{cm} \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{block} \mathrm{is} h \mathrm{cm} \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cuboid}=\mathrm{lenght}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 36=4\times 3\times h \\ \Rightarrow 36=12\times h \\ \Rightarrow h=\frac{36}{12}=3 \mathrm{cm} \\ \therefore \mathrm{The} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{wooden} \mathrm{block} \mathrm{is} 3 \mathrm{cm}. \end{gathered}$$

Question 7:

What will happen to the volume of a cube, if its edge is
(i) halved
(ii) trebled?

Answer 7:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{cube} \mathrm{is} x. \\ \mathrm{Then}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cube}=(\mathrm{side}{)}^3=x^3 \\ \mathrm{When} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{side} \mathrm{is} \mathrm{halved}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{edge} \mathrm{becomes} \frac{x}{2}\mathrm{.} \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube}=(\mathrm{side}{)}^3={\left(\frac{x}{2}\right)}^3\mathrm{=}\frac{x^3}{2^3}\mathrm{=}\frac{x^3}{8}\mathrm{=}\frac{1}{8}\times x^3 \\ \mathrm{It} \mathrm{means} \mathrm{that} \mathrm{if} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{a} \mathrm{cube} \mathrm{is} \mathrm{halved}, \mathrm{its} \mathrm{new} \mathrm{volume} \mathrm{will} \mathrm{be} \frac{1}{8} \mathrm{times} \mathrm{the} \mathrm{initial} \mathrm{volume}. \\ \left(\mathrm{ii}\right) \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{cube} \mathrm{is} x. \\ \mathrm{Then}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cube}=(\mathrm{side}{)}^3=x^3 \\ \mathrm{When} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{side} \mathrm{is} \mathrm{trebled}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{edge} \mathrm{becomes} 3\times x\mathit{.} \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube}=(\mathrm{side}{)}^3=(3\times x{)}^3=3^3\times x^3=27\times x^3 \\ \mathrm{Thus}, \mathrm{if} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{a} \mathrm{cube} \mathrm{is} \mathrm{trebled}, \mathrm{its} \mathrm{new} \mathrm{volume} \mathrm{will} \mathrm{be} 27 \mathrm{times} \mathrm{the} \mathrm{initial} \mathrm{volume}. \end{gathered}$$

Question 8:

What will happen to the volume of a cuboid if its:
(i) Length is doubled, height is same and breadth is halved?
(ii) Length is doubled, height is doubled and breadth is sama?

Answer 8:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{length}, \mathrm{breadth} \mathrm{and} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboid} \mathrm{are} l, b \mathrm{and} h, \mathrm{respectively}. \\ \mathrm{Then}, \mathrm{volume} =l\times b\times h \\ \mathrm{When} \mathrm{its} \mathrm{length} \mathrm{is} \mathrm{doubled}, \mathrm{its} \mathrm{length} \mathrm{becomes} 2\times l\mathit{.} \\ \mathrm{When} \mathrm{its} \mathrm{breadth} \mathrm{is} \mathrm{halved}, its length becomes \frac{b}{2}. \\ \mathrm{The} \mathrm{height} h \mathrm{remains} \mathrm{the} \mathrm{same}. \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =2\times l\times \frac{b}{2}\times h \\ =l\times b\times h \\ \therefore \mathrm{It} \mathrm{can} \mathrm{be} \mathrm{observed} \mathrm{that} \mathrm{the} \mathrm{new} \mathrm{volume} \mathrm{is} \mathrm{the} \mathrm{same} \mathrm{as} \mathrm{th}e \mathrm{initial} \mathrm{volume}. \mathrm{So}, \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{change} \mathrm{in} \mathrm{volume}. \\ \left(\mathrm{ii}\right) \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{length}, \mathrm{breadth} \mathrm{and} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboid} \mathrm{are} l, b \mathrm{and} h, \mathrm{respectively}. \\ \mathrm{Then}, \mathrm{volume} =l\times b\times h \\ \mathrm{When} \mathrm{its} \mathrm{length} \mathrm{is} \mathrm{doubled}, \mathrm{its} \mathrm{length} \mathrm{becomes} 2\times l\mathit{.} \\ \mathrm{When} \mathrm{its} \mathrm{height} \mathrm{is} \mathrm{double}, \mathrm{it} \mathrm{becomes} 2\times h\mathit{.} \\ \mathrm{The} \mathrm{breadth} b\mathit{ }\mathrm{remains} \mathrm{the} \mathrm{same}. \\ \mathrm{Now}, v\mathrm{olume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =2\times l\times b\times 2\times h \\ =4\times l\times b\times h \\ \therefore \mathrm{It} \mathrm{can} \mathrm{be} \mathrm{observed} \mathrm{that} \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cuboid} \mathrm{is} \mathrm{four} \mathrm{times} \mathrm{the} \mathrm{initial} \mathrm{volume}. \end{gathered}$$

Question 9:

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

Answer 9:

$$\begin{gathered} \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{cuboids} \mathrm{are} 5 \mathrm{cm}\times 6 \mathrm{cm}\times 7 \mathrm{cm}, 4 \mathrm{cm}\times 7 \mathrm{cm}\times 8 \mathrm{cm} \mathrm{and} 2 \mathrm{cm}\times 3 \mathrm{cm}\times 13 \mathrm{cm}. \\ \mathrm{Now}, \mathrm{a} \mathrm{new} \mathrm{cube} \mathrm{is} \mathrm{formed} \mathrm{by} \mathrm{melting} \mathrm{the} \mathrm{given} \mathrm{cuboids}. \\ \therefore \mathrm{Voulume} \mathrm{of} \mathrm{the} \mathrm{cube}=\mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{volumes} \mathrm{of} \mathrm{the} \mathrm{cuboids} \\ =(5 \mathrm{cm}\times 6 \mathrm{cm}\times 7 \mathrm{cm})+(4 \mathrm{cm}\times 7 \mathrm{cm}\times 8 \mathrm{cm})+(2 \mathrm{cm}\times 3 \mathrm{cm}\times 13 \mathrm{cm}\mathrm{)} \\ =(210 {\mathrm{cm}}^3)+(224 {\mathrm{cm}}^3)+(78 {\mathrm{cm}}^3\mathrm{)} \\ =512 {\mathrm{cm}}^3 \\ \mathrm{Since} \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3, \mathrm{we} \mathrm{have}: \\ 512=(\mathrm{side}{)}^3 \\ \Rightarrow \left(\mathrm{side}\right)=\sqrt[3]{512}=8 \mathrm{cm} \\ \therefore \mathrm{The} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube} \mathrm{is} 8 \mathrm{cm}. \end{gathered}$$

Question 10:

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

Answer 10:

$$\begin{gathered} \mathrm{The} \mathrm{dimension} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{piece} \mathrm{of} \mathrm{iron} \mathrm{is} 50 \mathrm{cm}\times 40 \mathrm{cm}\times 10 \mathrm{cm}. \\ \mathrm{i}.\mathrm{e}., \mathrm{volume} = 50 \mathrm{cm}\times 40 \mathrm{cm}\times 10 \mathrm{cm}=20000 {\mathrm{cm}}^3 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{weight} \mathrm{of} 1 {\mathrm{cm}}^3 \mathrm{of} \mathrm{iron} \mathrm{is} 8 \mathrm{gm}. \\ \therefore \mathrm{The} \mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{piece} \mathrm{of} \mathrm{iron} =20000\times 8 \mathrm{gm} \\ =160000 \mathrm{gm} \\ =160\times 1000 \mathrm{gm} \\ =160 \mathrm{kg} (\because 1 \mathrm{kg}=1000 \mathrm{gm}) \end{gathered}$$

Question 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?

Answer 11:

$$\begin{gathered} \mathrm{The} \mathrm{dimension} \mathrm{of} \mathrm{the} \log \mathrm{of} \mathrm{wood} \mathrm{is} 3 \mathrm{m}\times 75 \mathrm{cm}\times 50 \mathrm{cm}, \mathrm{i}.\mathrm{e}., 300 \mathrm{cm}\times 75 \mathrm{cm}\times 50 \mathrm{cm} (\because 3 \mathrm{m}= 100 \mathrm{cm}). \\ \therefore \mathrm{Volume} =300 \mathrm{cm}\times 75 \mathrm{cm}\times 50 \mathrm{cm}=1125000 {\mathrm{cm}}^3 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{each} \mathrm{cubical} \mathrm{block} \mathrm{of} \mathrm{wood} \mathrm{is} \mathrm{of} 25 \mathrm{cm}. \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{cubical} \mathrm{block}=(\mathrm{side}{)}^3 \\ ={25}^3 \\ =15625 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{cubical} \mathrm{blocks}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{wood} \log }{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{cubical} \mathrm{block}} \\ =\frac{1125000 {\mathrm{cm}}^3}{15625 {\mathrm{cm}}^3} \\ =72 \end{gathered}$$

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.

Answer 12:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{block} \mathrm{of} \mathrm{silver}=9 \mathrm{cm} \\ \mathrm{Breadth}=4 \mathrm{cm} \\ \mathrm{Height}=3.5 \mathrm{cm} \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{block}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =9\times 4\times 3.5 \\ =126 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{beads} \mathrm{of} \mathrm{volume} 1.5 {\mathrm{cm}}^3 \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{made} \mathrm{from} \mathrm{the} \mathrm{block}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{silver} \mathrm{block}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{bead}} \\ =\frac{126 {\mathrm{cm}}^3}{1.5 {\mathrm{cm}}^3} \\ =84 \end{gathered}$$

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 dimensions are 40 cm, 36 cm and 24 cm.

Answer 13:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{one} \mathrm{cuboidal} \mathrm{box} = 2 \mathrm{cm}\times 3 \mathrm{cm}\times 10 \mathrm{cm} \\ \mathrm{Volume} =(2\times 3\times 10) {\mathrm{cm}}^3=60 {\mathrm{cm}}^3 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{dimension} \mathrm{of} \mathrm{a} \mathrm{carton} \mathrm{is} 40 \mathrm{cm}\times 36 \mathrm{cm}\times 24 \mathrm{cm}, \mathrm{where} \mathrm{the} \mathrm{boxes} \mathrm{can} \mathrm{be} \mathrm{sto}\mathrm{red}. \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{carton}=(40\times 36\times 24) {\mathrm{cm}}^3=34560 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{cuboidal} \mathrm{boxes} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{stored} \mathrm{in} \mathrm{the} \mathrm{carton}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{carton}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{cuboidal} \mathrm{box}}=\frac{34560 {\mathrm{cm}}^3}{60 {\mathrm{cm}}^3}=576 \end{gathered}$$

Question 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.

Answer 14:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{iron} \mathrm{block} = 50 \mathrm{cm}\times 45 \mathrm{cm}\times 34 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{iron} \mathrm{block}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(50\times 45\times 34) {\mathrm{cm}}^3=76500 {\mathrm{cm}}^3 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{dimension} \mathrm{of} \mathrm{one} \mathrm{small} \mathrm{cuboids} \mathrm{is} 5\mathrm{cm}\times 3 \mathrm{cm}\times 2 \mathrm{cm}. \\ \mathrm{Volume} \mathrm{of} \mathrm{one} \mathrm{small} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(5\times 3\times 2) {\mathrm{cm}}^3=30 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{small} \mathrm{cuboids} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{obtained} \mathrm{from} \mathrm{the} \mathrm{iron} \mathrm{block}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{iron} \mathrm{block}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{small} \mathrm{cuboid}}=\frac{76500 {\mathrm{cm}}^3}{30 {\mathrm{cm}}^3}=2550 \end{gathered}$$

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?

Answer 15:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{cube} \mathrm{B} \mathrm{is} l \mathrm{cm}. \\ \mathrm{Then}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{cube} \mathrm{A} \mathrm{is} 3\times l \mathrm{cm}. \\ \mathrm{Now}, \mathrm{ratio}=\frac{\mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{A}}{\mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{B}}\mathrm{=}\frac{(3\times l{)}^3 {\mathrm{cm}}^3}{(l{)}^3 {\mathrm{cm}}^3}\mathrm{=}\frac{{\mathrm{3}}^3\times l^3}{l^3}\mathrm{=}\frac{27}{1} \\ \therefore \mathrm{The} \mathrm{ratio} \mathrm{of} \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{A} \mathrm{to} \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{B} \mathrm{is} 27:1. \end{gathered}$$

Page-21.9

Question 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?

Answer 16:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{an} \mathrm{ice} \mathrm{cream} \mathrm{brick} = 20 \mathrm{cm}\times 10 \mathrm{cm}\times 7 \mathrm{cm} \\ \mathrm{Its} \mathrm{volume}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(20\times 10\times 7) {\mathrm{cm}}^3=1400 {\mathrm{cm}}^3 \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{inner} \mathrm{dimension} \mathrm{of} \mathrm{the} \mathrm{deep} \mathrm{fridge} \mathrm{is} 100 \mathrm{cm}\times 50 \mathrm{cm}\times 42 \mathrm{cm}. \\ \mathrm{Its} \mathrm{volume}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(100\times 50\times 42) {\mathrm{cm}}^3=210000 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{ice} \mathrm{cream} \mathrm{bricks} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{stored} \mathrm{in} \mathrm{the} \mathrm{fridge}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{fridge}}{\mathrm{volume} \mathrm{of} \mathrm{an} \mathrm{ice} \mathrm{cream} \mathrm{brick}}=\frac{210000 {\mathrm{cm}}^3}{1400 {\mathrm{cm}}^3}=150 \end{gathered}$$

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.

Answer 17:

$$\begin{gathered} \mathrm{The} \mathrm{edges} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{cubes} \mathrm{are} 2 \mathrm{cm} \mathrm{and} 4 \mathrm{cm}. \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cube} \mathrm{of} \mathrm{side} 2 \mathrm{cm}, {\mathrm{V}}_1=(\mathrm{side}{)}^3=(2{)}^3=8 {\mathrm{cm}}^3 \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cube} \mathrm{of} \mathrm{side} 4 \mathrm{cm}, {\mathrm{V}}_2=(\mathrm{side}{)}^3=(4{)}^3=64 {\mathrm{cm}}^3 \\ \mathrm{We} \mathrm{observe} \mathrm{the} \mathrm{following}: \\ {\mathrm{V}}_2=64 {\mathrm{cm}}^3=8\times 8 {\mathrm{cm}}^3=8\times {\mathrm{V}}_1 \\ \therefore {\mathrm{V}}_2=8{\mathrm{V}}_1 \end{gathered}$$

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?

Answer 18:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{a} \mathrm{tea} \mathrm{packet} \mathrm{is} 10 \mathrm{cm}\times 6 \mathrm{cm}\times 4 \mathrm{cm}. \\ \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{tea} \mathrm{packet}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(10\times 6\times 4) {\mathrm{cm}}^3=240 {\mathrm{cm}}^3 \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{dimension} \mathrm{of} \mathrm{the} \mathrm{cardboard} \mathrm{box} \mathrm{is} 50 \mathrm{cm}\times 30 \mathrm{cm}\times 0.2 \mathrm{m}, \mathrm{i}.\mathrm{e}., 50 \mathrm{cm}\times 30 \mathrm{cm}\times 20 \mathrm{cm} (\because 1 \mathrm{m}=100 \mathrm{cm}) \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cardboard} \mathrm{box}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(50\times 30\times 20) {\mathrm{cm}}^3=30000 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{tea} \mathrm{packets} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{placed} \mathrm{inside} \mathrm{the} \mathrm{cardboard} \mathrm{box}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{box}}{\mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{tea} \mathrm{packet}}=\frac{30000 {\mathrm{cm}}^3}{240 {\mathrm{cm}}^3}=125 \end{gathered}$$

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.

Answer 19:

$$\begin{gathered} \mathrm{The} \mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{block} \mathrm{of} \mathrm{dimension} 5 \mathrm{cm}\times 4 \mathrm{cm}\times 3 \mathrm{cm} \mathrm{is} 1 \mathrm{kg}. \\ \mathrm{Its} \mathrm{volume}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(5\times 4\times 3) {\mathrm{cm}}^3=60 {\mathrm{cm}}^3 \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{weight} \mathrm{of} 60 {\mathrm{cm}}^3 \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{is} 1 \mathrm{kg} \\ \mathrm{Again}, \mathrm{the} \mathrm{dimension} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{block} \mathrm{which} \mathrm{is} \mathrm{of} \mathrm{same} \mathrm{metal} \mathrm{is} 15 \mathrm{cm}\times 8 \mathrm{cm}\times 3 \mathrm{cm}. \\ \mathrm{Its} \mathrm{volume}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(15\times 8\times 3) {\mathrm{cm}}^3=360 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{required} \mathrm{block} =360 {\mathrm{cm}}^3 \\ =6\times 60 {\mathrm{cm}}^3 (\because \mathrm{Weight} \mathrm{of} 60 {\mathrm{cm}}^3 \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{is} 1\mathrm{Kg}) \\ =6\times 1 \mathrm{kg} \\ =6 \mathrm{kg} \end{gathered}$$

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?

Answer 20:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{a} \mathrm{soap} \mathrm{cake} = 7\mathrm{cm}\times 5 \mathrm{cm}\times 2.5 \mathrm{cm} \\ \mathrm{Its} \mathrm{volum}e=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(7\times 5\times 2.5) {\mathrm{cm}}^3=87.5 {\mathrm{cm}}^3 \\ \mathrm{Also}, \mathrm{the} \mathrm{dimension} \mathrm{of} \mathrm{the} \mathrm{box} \mathrm{that} \mathrm{contains} \mathrm{the} \mathrm{soap} \mathrm{cakes} \mathrm{is} 56 \mathrm{cm}\times 0.4 \mathrm{m}\times 0.25 \mathrm{m}, \mathrm{i}.\mathrm{e}., 56 \mathrm{cm}\times 40\mathrm{cm}\times 25 \mathrm{cm} (\because 1 \mathrm{m} = 100 \mathrm{cm}). \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{box}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(56\times 40\times 25) {\mathrm{cm}}^3=56000 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{soap} \mathrm{cakes} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{placed} \mathrm{inside} \mathrm{the} \mathrm{box}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{box}}{\mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{soap} \mathrm{cake}}=\frac{56000 {\mathrm{cm}}^3}{87.5 {\mathrm{cm}}^3}=640 \end{gathered}$$

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.

Answer 21:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{box} \mathrm{is} b \mathrm{cm}. \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{box}=48 {\mathrm{cm}}^3 \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{box}=3 \mathrm{cm} \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{box}=4 \mathrm{cm} \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{box}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 48=4\times b\times 3 \\ \Rightarrow 48=12\times b \\ \Rightarrow b=\frac{48}{12}=4 \mathrm{cm} \\ \therefore \mathrm{The} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{box} \mathrm{is} 4 \mathrm{cm}. \end{gathered}$$