RD Sharma solution class 8 chapter 21 Mensuration II(Volume and Surface Area of Cuboid and a Cube) Exercise 21.4

Exercise 21.4

Page-21.30

Question 1:

Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.

Answer 1:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{room}=12 \mathrm{m} \\ \mathrm{Breadth}=9 \mathrm{m} \\ \mathrm{Height}=8 \mathrm{m} \\ \mathrm{Since} \mathrm{the} \mathrm{room} \mathrm{is} \mathrm{cuboidal} \mathrm{in} \mathrm{shape}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{longest} \mathrm{rod} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{placed} \mathrm{in} \mathrm{the} \mathrm{room} \mathrm{will} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{diagonal} \mathrm{between} \mathrm{opposite} \mathrm{vertices}. \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{diagonal} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{using} \mathrm{the} \mathrm{Pythagorus} \mathrm{theorem} \\ =\sqrt{{\mathrm{l}}^2+{\mathrm{b}}^2} \\ =\sqrt{(12{)}^2+(9{)}^2} \\ \mathrm{=}\sqrt{144+81} \\ \mathrm{=}\sqrt{225} \\ =15 \mathrm{m} \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{longest} \mathrm{rod} \mathrm{would} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{diagonal} \mathrm{of} \mathrm{the} \mathrm{right} \mathrm{angle} \mathrm{triangle} \mathrm{of} \mathrm{base} 15 \mathrm{m} \mathrm{and} \mathrm{altitude} 8 \mathrm{m}. \\ \mathrm{Similarly}, \mathrm{using} \mathrm{the} \mathrm{Pythagorus} \mathrm{theorem}, \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{diagonal} \\ =\sqrt{{15}^2+8^2} \\ \mathrm{=}\sqrt{225+64} \\ =17 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{longest} \mathrm{rod} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{placed} \mathrm{in} \mathrm{the} \mathrm{room} \mathrm{is} 17 \mathrm{m}. \end{gathered}$$

Question 2:

If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that
$\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

Answer 2:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that}\mathit{ }V \mathrm{is} \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cuboid} \mathrm{of} \mathrm{length}=a, \mathrm{breadth}=b \mathrm{and} \mathrm{height}=c\mathit{.}\mathit{ }\mathrm{Also},\mathit{ }S \mathrm{is} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{cuboid}. \\ \mathrm{Then}, V=a\times b\times c \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ \Rightarrow S=2\times (a\times b+b\times c+a\times c) \\ \mathrm{Let} \mathrm{us} \mathrm{take} \mathrm{the} \mathrm{right}-\mathrm{hand} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{equation} \mathrm{to} \mathrm{be} \mathrm{proven}. \\ \frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{2}{2\times (a\times b+b\times c+a\times c)}\times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \\ \mathrm{=}\frac{1}{(a\times b+b\times c+a\times c)}\times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \\ \mathrm{Now}, \mathrm{multiplying} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{the} \mathrm{denominator} \mathrm{with} a\times b\times c, \mathrm{we} \mathrm{get}: \\ \frac{1}{(a\times b+b\times c+a\times c)}\times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\times \frac{a\times b\times c}{a\times b\times c} \\ \mathrm{=}\frac{1}{(a\times b+b\times c+a\times c)}\times (\frac{a\times b\times c}{a}+\frac{a\times b\times c}{b}+\frac{a\times b\times c}{c})\times \frac{1}{a\times b\times c} \\ \mathrm{=}\frac{1}{(a\times b+b\times c+a\times c)}\times (b\times c+a\times c+a\times b)\times \frac{1}{a\times b\times c} \\ \mathrm{=}\frac{1}{(a\times b+b\times c+a\times c)}\times (a\times b+b\times c+a\times c)\times \frac{1}{a\times b\times c} \\ \mathrm{=}\frac{1}{a\times b\times c} \\ \mathrm{=}\frac{1}{V} \\ \therefore \frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{V} \end{gathered}$$

Question 3:

The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V² = xyz.

Answer 3:

$$\begin{gathered} \mathrm{The} \mathrm{areas} \mathrm{of} \mathrm{three} \mathrm{adjacent} \mathrm{faces} \mathrm{of} \mathrm{a} \mathrm{cuboid} \mathrm{are} x,\mathit{ }y \mathrm{and} z\mathit{.} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{V} \\ \mathrm{Observe} \mathrm{that} x=\mathrm{length}\times \mathrm{breadth} \\ y=\mathrm{breadth}\times \mathrm{height}, \\ z=\mathrm{length}\times \mathrm{height} \\ \mathrm{Since} \mathrm{volume} \mathrm{of} \mathrm{cuboid} \mathrm{V}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}, \mathrm{we} \mathrm{have}: \\ {V}^2=V\times V \\ =(\mathrm{length}\times \mathrm{breadth}\times \mathrm{height})\times (\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}) \\ =(\mathrm{length}\times \mathrm{breadth})\times (\mathrm{breadth}\times \mathrm{height})\times (\mathrm{length}\times \mathrm{height}) \\ =x\times y\times z \\ =xyz \\ \therefore V^2=xyz \end{gathered}$$

Question 4:

A rectangular water reservoir contains 105 m³ of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.

Answer 4:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{water} \mathrm{reservoir}=12 \mathrm{m} \\ \mathrm{Breadth}=3.5 \mathrm{m} \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{reservoir}=h \mathrm{m} \\ \mathrm{Also}, \mathrm{it} \mathrm{contains} 105 {\mathrm{m}}^3 \mathrm{of} \mathrm{water}, \mathrm{i}.\mathrm{e}., \mathrm{its} \mathrm{volume}=105 {\mathrm{m}}^3 \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{water} \mathrm{reservoir}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 105=12\times 3.5\times h \\ \Rightarrow 105=42\times h \\ \Rightarrow h=\frac{105}{42}=2.5 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{depth} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{in} \mathrm{the} \mathrm{reservoir} \mathrm{is} 2.5 \mathrm{m}. \end{gathered}$$

Question 5:

Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D.

Answer 5:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{the} \mathrm{following}: \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{cube} \mathrm{A}=18 \mathrm{cm} \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{cube} \mathrm{B}=24 \mathrm{cm} \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{cube} \mathrm{C}=30 \mathrm{cm} \\ \mathrm{The} \mathrm{given} \mathrm{cubes} \mathrm{are} \mathrm{melted} \mathrm{and} \mathrm{moulded} \mathrm{into} \mathrm{a} \mathrm{new} \mathrm{cube} \mathrm{D}. \\ \mathrm{Hence}, \mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{D}=\mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{A}+\mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{B}+\mathrm{volume} \mathrm{of} \mathrm{cube} \mathrm{C} \\ =(\mathrm{side} \mathrm{of} \mathrm{cube} \mathrm{A}{)}^3+(\mathrm{side} \mathrm{of} \mathrm{cube} \mathrm{B}{)}^3+(\mathrm{side} \mathrm{of} \mathrm{cube} \mathrm{C}{)}^3 \\ ={18}^3+{24}^3+{30}^3 \\ =5832+13824+27000 \\ =46656 {\mathrm{cm}}^3 \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube} \mathrm{D}=\mathrm{x} \\ \Rightarrow {\mathrm{x}}^3=46656 \\ \Rightarrow \mathrm{x}=\sqrt[3]{46656}=36 \mathrm{cm} \\ \therefore \mathrm{The} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{bigger} \mathrm{cube} \mathrm{D} \mathrm{is} 36 \mathrm{cm}. \end{gathered}$$

Question 6:

The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.

Answer 6:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{room}=\mathrm{x} \mathrm{dm} \\ \mathrm{Since} \mathrm{breadth} \mathrm{is} \mathrm{twice} \mathrm{the} \mathrm{height}, \mathrm{breadth}=2\times \mathrm{height} \\ \mathrm{So}, \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{room}=\frac{\mathrm{breadth}}{2}\mathrm{=}\frac{\mathrm{x}}{2} \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{breadth} \mathrm{is} \mathrm{half} \mathrm{the} \mathrm{length}. \\ \mathrm{So}, \mathrm{breadth}=\frac{1}{2}\times \mathrm{length} \\ \mathrm{i}.\mathrm{e}., \mathrm{length}=2\times \mathrm{breadth}=2\times \mathrm{x} \\ \mathrm{Since} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{room}=512 \mathrm{cu} \mathrm{dm}, \mathrm{we} \mathrm{have}: \\ \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 512=2\times \mathrm{x}\times \mathrm{x}\times \frac{\mathrm{x}}{2} \\ \Rightarrow 512={\mathrm{x}}^3 \\ \Rightarrow \mathrm{x}=\sqrt[3]{512}=8 \mathrm{dm} \\ \mathrm{Hence}, \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{room}=2\times \mathrm{x}=2\times 8=16 \mathrm{dm} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{room}=\mathrm{x}=8 \mathrm{dm} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{the} \mathrm{room}=\frac{\mathrm{x}}{2}\mathrm{=}\frac{8}{2}=4 \mathrm{dm} \end{gathered}$$

Question 7:

A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per metre sheet, sheet being 2 m wide.

Answer 7:

$$\begin{gathered} \mathrm{A} \mathrm{closed} \mathrm{iron} \mathrm{tank} \mathrm{of} \mathrm{dimensions} 12 \mathrm{m} \mathrm{long}, 9 \mathrm{m} \mathrm{wide} \mathrm{and} 4 \mathrm{m} \mathrm{deep} \mathrm{is} \mathrm{to} \mathrm{be} \mathrm{made}. \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{tank}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (12\times 9+9\times 4+12\times 4) \\ =2\times (108+36+48) \\ =384 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{an} \mathrm{iron} \mathrm{sheet} \mathrm{is} \mathrm{Rs} 5 \mathrm{per} \mathrm{metre} \mathrm{and} \mathrm{the} \mathrm{sheet} \mathrm{is} 2 \mathrm{metres} \mathrm{wide}. \\ \mathrm{i}.\mathrm{e}., \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{sheet}=1 \mathrm{m}\times 2 \mathrm{m}=2 {\mathrm{m}}^2 \\ \mathrm{So}, \mathrm{the} \mathrm{cost} \mathrm{of} 2 {\mathrm{m}}^2 \mathrm{of} \mathrm{iron} \mathrm{sheet}=\mathrm{Rs} 5 \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{cost} \mathrm{of} 1 {\mathrm{m}}^2 \mathrm{of} \mathrm{iron} \mathrm{sheet}=\mathrm{Rs} \frac{5}{2} \\ \therefore \mathrm{Cost} \mathrm{of} 384 {\mathrm{m}}^2 \mathrm{of} \mathrm{iron} \mathrm{sheet}=384\times \frac{5}{2}=\mathrm{Rs} 960 \end{gathered}$$

Question 8:

A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12 m × 8 m × 6 m, find the cost of iron sheet at Rs 17.50 per metre.

Answer 8:

$$\begin{gathered} \mathrm{An} \mathrm{open} \mathrm{iron} \mathrm{tank} \mathrm{of} \mathrm{dimensions} 12 \mathrm{m}\times 8 \mathrm{m}\times 6 \mathrm{m} \mathrm{is} \mathrm{to} \mathrm{be} \mathrm{made}. \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{open} \mathrm{tank}=(\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{base})+(\mathrm{total} \mathrm{area} \mathrm{of} \mathrm{the} 4 \mathrm{walls}) \\ =(12\times 8)+2\times (8\times 6+12\times 6) \\ =\left(96\right)+2\times (48+72) \\ =336 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{the} \mathrm{iron} \mathrm{sheet} \mathrm{that} \mathrm{is} 4 \mathrm{m} \mathrm{wide} \mathrm{is} \mathrm{Rs} 17.50 \mathrm{per} \mathrm{metre}. \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{iron} \mathrm{sheet}=1 \mathrm{m}\times 4 \mathrm{m}=4 {\mathrm{m}}^2 \\ \mathrm{So}, \mathrm{the} \mathrm{cost} \mathrm{of} 4 {\mathrm{m}}^2 \mathrm{of} \mathrm{iron} \mathrm{sheet}=\mathrm{Rs} 17.50 \\ \therefore \mathrm{The} \mathrm{cost} \mathrm{of} \mathrm{iron} \mathrm{sheet} \mathrm{required} \mathrm{to} \mathrm{an} \mathrm{iron} \mathrm{tank} \mathrm{of} \mathrm{surface} \mathrm{area} 336 {\mathrm{m}}^2 =336\times \frac{17.50}{4}=\mathrm{Rs} 1470 \end{gathered}$$

Question 9:

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

Answer 9:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{cube}= \mathrm{x} \mathrm{cm} \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times {\mathrm{x}}^2=6{\mathrm{x}}^2 {\mathrm{cm}}^2 \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{surface} \mathrm{areas} \mathrm{of} \mathrm{three} \mathrm{such} \mathrm{cubes}=6{\mathrm{x}}^2+6{\mathrm{x}}^2+6{\mathrm{x}}^2=18 {\mathrm{x}}^2 {\mathrm{cm}}^2 \\ \mathrm{Now}, \mathrm{these} \mathrm{three} \mathrm{cubes} \mathrm{area} \mathrm{placed} \mathrm{together} \mathrm{to} \mathrm{form} \mathrm{a} \mathrm{cuboid}. \\ \mathrm{Then} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cuboid} \mathrm{will} \mathrm{be} 3 \mathrm{times} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{cube}=3\times \mathrm{x}=3\mathrm{x} \mathrm{cm} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{x} \mathrm{cm} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{x} \mathrm{cm} \\ \therefore \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (3\mathrm{x}\times \mathrm{x}+\mathrm{x}\times \mathrm{x}+3\mathrm{x}\times \mathrm{x}) \\ =2\times (3{\mathrm{x}}^2+{\mathrm{x}}^2+3{\mathrm{x}}^2\mathrm{)} \\ =2\times (7{\mathrm{x}}^2\mathrm{)} \\ =14{\mathrm{x}}^2 \mathrm{cm} \end{gathered}$$²

i.e., the ratio of the total surface area cuboid to the sum of the surface areas of the three cubes = $$\begin{gathered} 14 x^2 c{m}^{2 }: 18 x^2 c{m}^{2 } \\ = 7:9 \end{gathered}$$
Hence, the ratio is 7:9.

Question 10:

The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square metre.

Answer 10:

$$\begin{gathered} \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{room} \mathrm{are} 12.5 \mathrm{m}\times 9 \mathrm{m}\times 7 \mathrm{m}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{walls}=2\times (\mathrm{length}\times \mathrm{height}+\mathrm{breadth}\times \mathrm{height}) \\ =2\times (12.5\times 7+9\times 7) \\ =301 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{there} \mathrm{are} 2 \mathrm{doors} \mathrm{and} 4 \mathrm{windows} \mathrm{in} \mathrm{the} \mathrm{room}. \\ \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{door} \mathrm{are} 2.5 \mathrm{m}\times 1.2 \mathrm{m}. \\ \mathrm{i}.\mathrm{e}., \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{door}=2.5\times 1.2=3 {\mathrm{m}}^2 \\ \therefore \mathrm{Total} \mathrm{area} \mathrm{of} 2 \mathrm{doors}=2\times 3=6 {\mathrm{m}}^2 \\ \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{a} \mathrm{window} \mathrm{are} 1.5 \mathrm{m}\times 1 \mathrm{m}. \\ \mathrm{i}.\mathrm{e}., \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{window}=1.5\times 1=1.5 {\mathrm{m}}^2 \\ \therefore \mathrm{Total} \mathrm{area} \mathrm{of} 4 \mathrm{windows}=4\times 1.5=6 {\mathrm{m}}^2 \\ \mathrm{Hence}, \mathrm{the} \mathrm{total} \mathrm{area} \mathrm{to} \mathrm{be} \mathrm{painted}=301-(6+6)=289 {\mathrm{m}}^2 \\ \mathrm{The} \mathrm{rate} \mathrm{of} \mathrm{painting} 1 {\mathrm{m}}^2 \mathrm{of} \mathrm{wall}=\mathrm{Rs} 3.50 \\ \therefore \mathrm{The} \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{painting} 289 {\mathrm{m}}^2 \mathrm{of} \mathrm{wall}=\mathrm{Rs} 289\times 3.50=\mathrm{Rs} 1011.50 \end{gathered}$$

Question 11:

A field is 150 m long and 100 m wide. A plot (outside the field) 50 m long and 30 m wide is dug to a depth of 8 m and the earth taken out from the plot is spread evenly in the field. By how much is the level of field raised?

Answer 11:

$$\begin{gathered} \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{plot} \mathrm{dug} \mathrm{outside} \mathrm{the} \mathrm{field} \mathrm{are} 50 \mathrm{m}\times 30 \mathrm{m}\times 8 \mathrm{m}. \\ \mathrm{Hence}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{earth} \mathrm{dug}-\mathrm{out} \mathrm{from} \mathrm{the} \mathrm{plot}=50\times 30\times 8=12000 {\mathrm{m}}^3 \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{level} \mathrm{of} \mathrm{the} \mathrm{earth} \mathrm{rises} \mathrm{by} \mathrm{h} \mathrm{m}. \\ \mathrm{When} \mathrm{we} \mathrm{spread} \mathrm{this} \mathrm{dug}-\mathrm{out} \mathrm{earth} \mathrm{on} \mathrm{the} \mathrm{field} \mathrm{of} \mathrm{length} 150 \mathrm{m}, \mathrm{breadth} 100 \mathrm{m} \mathrm{and} \mathrm{height} \mathrm{h} \mathrm{m}, \mathrm{we} \mathrm{have}: \\ \mathrm{Volume} \mathrm{of} \mathrm{earth} \mathrm{dug}-\mathrm{out}=150\times 100\times \mathrm{h} \\ \Rightarrow 12000=15000\times \mathrm{h} \\ \Rightarrow \text{h=}\frac{12000}{15000}\text{=0.8 m } \\ \Rightarrow \mathrm{h}=80 \mathrm{cm} (\because 1 \mathrm{m}=100 \mathrm{cm}) \\ \text{∴ The level of the field will rise by 80 cm.} \end{gathered}$$

Question 12:

Two cubes, each of volume 512 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Answer 12:

$$\begin{gathered} \mathrm{Two} \mathrm{cubes} \mathrm{each} \mathrm{of} \mathrm{volume} 512 {\mathrm{cm}}^3 \mathrm{are} \mathrm{joined} \mathrm{end} \mathrm{to} \mathrm{end}. \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3 \\ \Rightarrow 512=(\mathrm{side}{)}^3 \\ \Rightarrow \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{cube}=\sqrt[3]{512}=8 \mathrm{cm} \\ \mathrm{If} \mathrm{the} \mathrm{cubes} \mathrm{area} \mathrm{joined} \mathrm{side} \mathrm{by} \mathrm{side}, \mathrm{then} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{resulting} \mathrm{cuboid} \mathrm{is} 2\times 8 \mathrm{cm}=16 \mathrm{cm}. \\ \mathrm{Breadth}=8 \mathrm{cm} \\ \mathrm{Height}=8 \mathrm{cm} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (16\times 8+8\times 8+16\times 8) \\ =2\times (128+64+128) \\ =640 {\mathrm{cm}}^2 \end{gathered}$$

Question 13:

Three cubes whose edges measure 3 cm, 4 cm, and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.

Answer 13:

$$\begin{gathered} \mathrm{Three} \mathrm{cubes} \mathrm{of} \mathrm{edges} 3 \mathrm{cm}, 4 \mathrm{cm} \mathrm{and} 5 \mathrm{cm} \mathrm{are} \mathrm{melted} \mathrm{and} \mathrm{molded} \mathrm{to} \mathrm{form} \mathrm{a} \mathrm{new} \mathrm{cube}. \\ \mathrm{i}.\mathrm{e}., \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube}=\mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{volumes} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{cubes} \\ =(3{)}^3+(4{)}^3+(5{)}^3 \\ =27+64+125 \\ =216 {\mathrm{cm}}^3 \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3 \\ \Rightarrow 216=(\mathrm{side}{)}^3 \\ \Rightarrow \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube}=\sqrt[3]{216}=6 \mathrm{cm} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (6{)}^2=216 {\mathrm{cm}}^2 \end{gathered}$$

Question 14:

The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room.

Answer 14:

$$\begin{gathered} \mathrm{The} \mathrm{cost} \mathrm{of} \mathrm{preparing} 4 \mathrm{walls} \mathrm{of} \mathrm{a} \mathrm{room} \mathrm{whose} \mathrm{length} \mathrm{is} 12 \mathrm{m} \mathrm{is} \mathrm{Rs} 340.20 \mathrm{at} \mathrm{a} \mathrm{rate} \mathrm{of} \mathrm{Rs} 1.35/{\mathrm{m}}^2. \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{four} \mathrm{walls} \mathrm{of} \mathrm{the} \mathrm{room}=\frac{\mathrm{total} \mathrm{cost}}{\mathrm{rate}}=\frac{\mathrm{Rs} 340.20}{\mathrm{Rs} 1.35}=252 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{matting} \mathrm{the} \mathrm{floor} \mathrm{at} 85 \mathrm{paise}/{\mathrm{m}}^2 \mathrm{is} \mathrm{Rs} 91.80. \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{floor}=\frac{\mathrm{total} \mathrm{cost}}{\mathrm{rate}}=\frac{\mathrm{Rs} 91.80}{\mathrm{Rs} 0.85}=108 {\mathrm{m}}^2 \\ \mathrm{Hence}, \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{room}=\frac{\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{floor}}{\mathrm{length}}\mathrm{=}\frac{108}{12}=9 \mathrm{m} \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{room} \mathrm{is} \mathrm{h} \mathrm{m}.\hspace{0.17em}\mathrm{Then}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{four} \mathrm{walls}=2\times (\mathrm{length}\times \mathrm{height}+\mathrm{breadth}\times \mathrm{height}) \\ \Rightarrow 252=2\times (12\times \mathrm{h}+9\times \mathrm{h}) \\ \Rightarrow 252=2\times \left(21\mathrm{h}\right) \\ \Rightarrow 21\mathrm{h}=\frac{252}{2}=126 \\ \Rightarrow \mathrm{h}=\frac{126}{21}=6 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{room} \mathrm{is} 6 \mathrm{m}. \end{gathered}$$

Question 15:

The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the wall.

Answer 15:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{hall}=18 \mathrm{m} \\ \mathrm{Its} \mathrm{width}=12 \mathrm{m} \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{wall} \mathrm{is} \mathrm{h} \mathrm{m}. \\ \mathrm{Also}, \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{and} \mathrm{the} \mathrm{flat} \mathrm{roof}=\mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{the} \mathrm{four} \mathrm{walls} \\ \Rightarrow 2\times (\mathrm{length}\times \mathrm{breadth})=2\times (\mathrm{length}+\mathrm{breadth})\times \mathrm{height} \\ \Rightarrow 2\times (18\times 12)=2\times (18+12)\times \mathrm{h} \\ \Rightarrow 432=60\times \mathrm{h} \\ \Rightarrow \mathrm{h}=\frac{432}{60}=7.2 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{height} \mathrm{of} \mathrm{wall} \mathrm{is} 7.2 \mathrm{m}. \end{gathered}$$

Question 16:

A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.

Answer 16:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{third} \mathrm{cube} \mathrm{be} \mathrm{x} \mathrm{cm}. \\ \mathrm{Three} \mathrm{small} \mathrm{cubes} \mathrm{are} \mathrm{formed} \mathrm{by} \mathrm{melting} \mathrm{the} \mathrm{cube} \mathrm{of} \mathrm{edge} 12 \mathrm{cm}. \\ \mathrm{Edges} \mathrm{of} \mathrm{two} \mathrm{small} \mathrm{cubes} \mathrm{are} 6 \mathrm{cm} \mathrm{and} 8 \mathrm{cm}. \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3 \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{big} \mathrm{cube}=\mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{volumes} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{small} \mathrm{cubes} \\ \Rightarrow (12{)}^3=(6{)}^3+(8{)}^3+(\mathrm{x}{)}^3 \\ \Rightarrow 1728=216+512+{\mathrm{x}}^3 \\ \Rightarrow {\mathrm{x}}^3=1728-728=1000 \\ \Rightarrow \mathrm{x}=\sqrt[3]{1000}=10 \mathrm{cm} \\ \therefore \mathrm{The} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{third} \mathrm{cube} \mathrm{is} 10 \mathrm{cm}. \end{gathered}$$

Question 17:

The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m³ of air?

Answer 17:

$$\begin{gathered} \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{a} \mathrm{cinema} \mathrm{hall} \mathrm{are} 100 \mathrm{m}\times 50 \mathrm{m}\times 18 \mathrm{m}. \\ \mathrm{i}.\mathrm{e}., \mathrm{volume} \mathrm{of} \mathrm{air} \mathrm{in} \mathrm{the} \mathrm{cinema} \mathrm{hall}=100\times 50\times 18=90000 {\mathrm{m}}^3 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{each} \mathrm{person} \mathrm{requires} 150 {\mathrm{m}}^3 \mathrm{of} \mathrm{air}. \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{persons} \mathrm{that} \mathrm{can} \mathrm{sit} \mathrm{in} \mathrm{the} \mathrm{cinema} \mathrm{hall}=\frac{\mathrm{volume} \mathrm{of} \mathrm{air} \mathrm{in} \mathrm{hall}}{\mathrm{volume} \mathrm{of} \mathrm{air} \mathrm{required} \mathrm{by} 1 \mathrm{person}}=\frac{90000}{150}=600 \end{gathered}$$

Question 18:

The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm × 3 cm × 0.75 cm can be put in this box?

Answer 18:

$$\begin{gathered} \mathrm{The} \mathrm{outer} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{closed} \mathrm{wooden} \mathrm{box} \mathrm{are} 48 \mathrm{cm}\times 36 \mathrm{cm}\times 30 \mathrm{cm}. \\ \mathrm{Also}, \mathrm{the} \mathrm{box} \mathrm{is} \mathrm{made} \mathrm{of} \mathrm{a} 1.5 \mathrm{cm} \mathrm{thick} \mathrm{wood}, \mathrm{so} \mathrm{the} \mathrm{inner} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{box} \mathrm{will} \mathrm{be} \text{(2×1.5=3)cm less.} \\ \mathrm{i}.\mathrm{e}., \mathrm{the} \mathrm{inner} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{box} \mathrm{are} 45 \mathrm{cm}\times 33 \mathrm{cm}\times 27 \mathrm{cm} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{box}=45\times 33\times 27=40095 {\mathrm{cm}}^3 \\ \mathrm{Also}, \mathrm{the} \mathrm{dimensions} \mathrm{of} \mathrm{a} \mathrm{brick} \mathrm{are} 6 \mathrm{cm}\times 3 \mathrm{cm}\times 0.75 \mathrm{cm}. \\ \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{brick}=6\times 3\times 0.75=13.5 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{bricks} \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{put} \mathrm{in} \mathrm{the} \mathrm{box}=\frac{40095}{13.5}=2970 \end{gathered}$$

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Question 19:

The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m² is Rs. 1248. Find the dimensions of the box.

Answer 19:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{dimensions} \mathrm{be} \mathrm{x} \mathrm{multiple} \mathrm{of} \mathrm{each} \mathrm{other}. \\ \mathrm{The} \mathrm{dimensions} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 2:3:4. \\ \mathrm{Hence}, \mathrm{length}=2\mathrm{x} \mathrm{m} \\ \mathrm{Breadth}=3\mathrm{x} \mathrm{m} \\ \mathrm{Height}=4\mathrm{x} \mathrm{m} \\ \mathrm{So}, \mathrm{total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{box}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (2\mathrm{x}\times 3\mathrm{x}+3\mathrm{x}\times 4\mathrm{x}+2\mathrm{x}\times 4\mathrm{x}) \\ =2\times (6{\mathrm{x}}^2+12{\mathrm{x}}^2+8{\mathrm{x}}^2\mathrm{)} \\ =2\times (26{\mathrm{x}}^2\mathrm{)} \\ =52{\mathrm{x}}^2 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{covering} \mathrm{the} \mathrm{box} \mathrm{with} \mathrm{paper} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{Rs} 8/{\mathrm{m}}^2 \mathrm{and} \mathrm{Rs} 9.50/{\mathrm{m}}^2 \mathrm{is} \mathrm{Rs} 1248. \\ \mathrm{Here}, \mathrm{the} \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{covering} \mathrm{the} \mathrm{box} \mathrm{at} \mathrm{a} \mathrm{rate} \mathrm{of} \mathrm{Rs} 8/{\mathrm{m}}^2=8\times 52{\mathrm{x}}^2=\mathrm{Rs} 416{\mathrm{x}}^2 \\ \mathrm{And} \mathrm{the} \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{covering} \mathrm{the} \mathrm{box} \mathrm{at} \mathrm{a} \mathrm{rate} \mathrm{of} \mathrm{Rs} 9.50/{\mathrm{m}}^2=9.50\times 52{\mathrm{x}}^2=\mathrm{Rs} 494{\mathrm{x}}^2 \\ \mathrm{Now}, \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{covering} \mathrm{the} \mathrm{box} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{Rs} 9.50/{\mathrm{m}}^2-\mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{covering} \mathrm{the} \mathrm{box} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{Rs} 8/{\mathrm{m}}^2=1248 \\ \Rightarrow 494{\mathrm{x}}^2-416{\mathrm{x}}^2=1248 \\ \Rightarrow 78{\mathrm{x}}^2=1248 \\ \Rightarrow {\mathrm{x}}^2\mathrm{=}\frac{1248}{78}=16 \\ \Rightarrow \mathrm{x}=\sqrt{16}=4 \\ \mathrm{Hence}, \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{box}=2\times \mathrm{x}=2\times 4=8 \mathrm{m} \\ \mathrm{Breadth}=3\times \mathrm{x}=3\times 4=12 \mathrm{m} \\ \mathrm{Height}=4\times \mathrm{x}=4\times 4=16 \mathrm{m} \end{gathered}$$