3.5 cm

Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the number of tiles required.
Edge of the cubical tank = 1.5 m = 150 cm (= a)
So, surface area of the tank = 5 x 150 x 150 cm²
Area of each square tile = side x side = 25 x 25 cm²
So, the number of tiles required \(=\frac{\text { surface area of the tank }}{\text { area of each tile }}\)
\(=\frac{5 \times 150 \times 150}{25 \times 25}=180\)
Cost of 1 dozen tiles, i.e., cost of 12 tiles = RS. 360
Therefore, cost of one tile \(=\text { RS. } \frac{360}{12}=\text { RS. } 30\)
So, the cost of 180 tiles = 180 x RS.30 = RS. 5400

Since Mary wants to paste the paper on the outer surface of the box; the quantity of paper required would be equal to the surface area of the box which is of the shape of a cuboid. The dimensions of the box are:
Length =80 cm, Breadth = 40 cm, Height = 20 cm.
The surface area of the box = 2(lb + bh + hl)
= 2[(80 x 40) + (40 x 20) + (20 x 80)] cm²
= 2[3200 + 800 + 1600] cm²
= 2 x 5600 cm² = 11200 cm²
The area of each sheet of the paper = 40 x 40 cm²
= 1600 cm²
Therefore, number of sheets required \(=\frac{\text { surface area of box }}{\text { area of one sheet of paper }}\)
\(=\frac{11200}{1600}=7\)
So, she would require 7 sheets

1300 cm²

Curved surface area of the tent = 552-2
= 550 m²
Radius(r) = 7 m
\(\therefore \ \pi \times 7\times l=550\)
\(\Rightarrow\) l = 25 m
\(\therefore \ h=\sqrt { { 25 }^{ 2 }-{ 7 }^{ 2 } } \)
= 24 m
Volume of the tent = \(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 7\times 7\times 24\)
= 1232 m³.

\(\frac { 231 }{ 16 } { m }^{ 3 }\)

(i) Diameter = 14 cm
Radius (r) = \(\frac { 14 }{ 2 } cm=7cm\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 7 \right) }^{ 2 }=616{ cm }^{ 2 }.\)
(ii) Diameter = 21 cm
Radius (r)   = \(\frac { 21 }{ 2 } cm\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
 \(=4\times \frac { 22 }{ 7 } \times { \left( \frac { 21 }{ 2 } \right) }^{ 2 }=1386{ cm }^{ 2 }.\)
(iii) Diameter = 3.5 m
Radius (r)   = \(\frac { 3.5 }{ 2 } m=1.75m\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 1.75 \right) }^{ 2 }\)
\(\\ =38.5{ m }^{ 2 }.\)

(i)   2r = 3.5 m
\(\Rightarrow\)  \(r=\frac { 3.5 }{ 2 } m\)
\(\Rightarrow\)   r = 1.75 m
h = 10 m
\(\therefore \) Inner curved surface area of the circular well = \(2\pi rh\)
\(=2\times \frac { 22 }{ 7 } \times 1.75\times 10=110{ m }^{ 2 }.\)
(ii) Cost of plastering the curved surface at the rate of Rs 40 per m² = Rs 110 \(\times\) 40 = Rs 4400.   

For a brick
l = 22.5 cm, b = 10 cm,
h = 7.5 cm
\(\therefore \)  Total surface area of a brick 
= 2 (lb + bh + hl)
= 2 (22.5 \(\times\) 10 + 10 \(\times\) 7.5 + 7.5 \(\times\) 22.5)
= 2 (225 + 75 + 168.75)
= 2(468.75) = 937.5 cm² = .09375 m²
\(\therefore \)  Number of bricks that can be painted out 
\(=\frac { 9.375 }{ .09375 } =100.\)

Vol. of hemispherical bowl = \(\frac { 2 }{ 3 } \pi r^{ 3 }\)
\(=\frac { 2 }{ 3 } \pi { \left( \frac { 36 }{ 2 } \right) }^{ 3 }{ cm }^{ 3 }\)
Vol. of cylindrical bottle = \(\pi\)r²h
= \(\pi\)(3)² \(\times\)6 cm³
Suppose required bottles are x.
\(\therefore \ x=\frac { \frac { 2 }{ 3 } \pi { (18) }^{ 3 } }{ \pi ({ 3) }^{ 3 }\times 6 } =24\)

0.31 km/hour
= 0.31 x 1000 m/hour
= 310 m/ hour
=\(\frac{310}{60}\) m/minute=\(\frac{31}{6}\) m/minute
ஃ  Volume of water that falls into the sea in a minute
\(=4 \times60\times\frac{31}{6}=1240 \ m^3\)

For cone
Base radius (r) = 8 m
Height (h)        = 6 m
\(\therefore \)  Slant height (l) = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { \left( 8 \right) }^{ 2 }+{ \left( 6 \right) }^{ 2 } } =10m\)
\(\therefore \)  Curved surface area = \(\pi rl\)
\(=\pi \left( 8 \right) \left( 10 \right) \)
\(\\ =80\pi { m }^{ 2 }\)
For cylinder
Base radius (R) = 8 m
Height (H) = 14 m
\(\therefore \)  Curved surface area = \(2\pi RH\)
 \(=2\pi \left( 8 \right) \left( 14 \right) \)
\(\\ =224\pi { m }^{ 2 }\)
\(\therefore \)  Total curved surface area = Curved surface area of the cone + Curved surface area of the cylinder
\(=80\pi +224\pi =304\pi { m }^{ 2 }\)
\(\\ =304\times 3.14{ m }^{ 2 }=954.56{ m }^{ 2 }\ \)
\(\therefore \)  Cost of canvas = 954.56 \(\times\) 50
= Rs 47728

Diameter = 50 cm
Radius (r) = \(\frac { 50 }{ 2 } cm=25cm\)
\( =\frac { 25 }{ 100 } m=\frac { 1 }{ 4 } m\)
Height (h) = 4 m
\(\therefore \)  Curved surface area of 1 pillar = \(2\pi rh\)
\(=2\times \frac { 22 }{ 7 } \times \frac { 1 }{ 4 } \times 4=\frac { 44 }{ 7 } { m }^{ 2 }\)
\(\therefore \) Curved surface area of 10 pillars
\(=\frac { 44 }{ 7 } \times 10{ m }^{ 2 }=\frac { 440 }{ 7 } { m }^{ 2 }\)
\(\therefore \)   Cost of painting at the rate of Rs 14 per square metre = \(\frac { 440 }{ 7 } \times 14=\)  Rs 880.

For classroom
l = 7 m, b = 6.5 m,
h = 4 m
\(\therefore\)  Area to be colour washed 
= 2(l + b) h - [3 \(\times\) 1.4 + 3 {2 \(\times\) 1}]  
= 2(7 + 6.5) 4 - (4.2 + 6)
= 108 - 10.2
= 97.8 m²
 \(\therefore\)  Cost of colour washing
= 97.8 \(\times\) 3.50
= Rs 342.30

Required number \(=\frac { 60\times 30\times 30 }{ 15\times 6\times 4 } =150\)

v = 5 \(\times\) (6 \(\times\) 2 \(\times\) 1.5)