6.615 cm²

7 : 3

126 cm²

8 m, 12 m, 6 m

28 cm

21.6 m

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

l = 5 m, b = 4 m,
h = 3 m
Area of the walls of the room = 2(l + b)h
= 2(5 + 4)3 = 54 m²
Area of the ceiling = lb
= (5) (4) = 20 m2
\(\therefore \) Total area of the walls of the room and the ceiling = 54 m² + 20 m² = 74 m²
\(\therefore \) Cost of white washing the walls of the room and the ceiling = 74 \(\times\) 7.50 = Rs 555.

Let the length, breadth and height of the rectangular hall be l m, b m and h m respectively.
Perimeter = 250 m
\(\Rightarrow\) 2(l + b) = 250
\(\Rightarrow\)  l + b = 125                ...(1)
Area of the four walls
\(=\frac { 15000 }{ 10 } =1500\)m²      
\(\Rightarrow\) 2(l + b)h = 1500
\(\Rightarrow\)  (l + b)h  = 750                  
\(\Rightarrow\)  125 h  = 750                Using (1)
\(\Rightarrow\)   \(h=\frac { 750 }{ 125 } \)
\(\Rightarrow\)   h = 6 m
Hence, the height of the hall is 6 m.            

(i) For herbarium 
l = 30 cm, b = 25 cm,
h = 25 cm
\(\therefore \)  Area of the glass = 2 (lb + bh + hl)
= 2[(30)(25) + (25)(25) + (25)(30)]
= 2[750 + 625 + 750] = 4250 cm².
(ii) The tape needed for all the 12 edges
= 4 (l + b + h)
= 4(30 + 25 + 25) = 320 cm.

For resulting cuboid
Length (l)    = 6 + 6 = 12 cm
Breadth (b) = 6 cm
Height (h)   = 6 cm
\(\therefore \) Surface area 
= 2(lb + bh + hl) 
= 2(12 \(\times\) 6 + 6 \(\times\) 6 + 6 \(\times\) 12)
= 360 cm²    

For tank
l = 12 m, b = 9 m,
h = 4 m
\(\therefore\) Total surface area of the tank 
= 2(l \(\times\) b + b \(\times\) h + h \(\times\) l)
= 2(12 \(\times\) 9 + 9 \(\times\) 4 + 4 \(\times\)12)
= 2(108 + 36 + 48)
= 2 (192) = 384 m²
Width of the iron sheet = 2 m
\(\therefore\)  Length of the iron sheet \(=\frac { 384 }{ 2 } =192m\)
\(\therefore\)  Cost of the iron sheet     = 192 \(\times\) 50
 = Rs 9600.

Let the height of the hall be h m.
Area of the floor   = l \(\times\) b
= 20  16 = 320 m²
Area of the flat roof   = l \(\times\) b
= 20 \(\times\) 16 = 320 m²
Sum of the areas of the four walls
= 2(l + b) h
= 2(20 +16) h = 72 h m²
According to the question,
72 h = 320 + 320
\(\Rightarrow \) 72h = 640  \(\Rightarrow\)  \(h=\frac { 640 }{ 72 } m\)
\(\Rightarrow \)   \(h=\frac { 80 }{ 9 } m\)
Hence, the height of the hall is \(\frac { 80 }{ 9 } m.\) 

For table cover
l = 4 - (0.25 + 0.25)
 \(\therefore\)  25 cm = 0.25 m
= 3.5 m
b = 2 - (0.25 + 0.25) = 1.5 m
\(\therefore\) Area of the cover = l \(\times\) b
= 3.5 \(\times\) 1.5 = 5.25 m²
\(\therefore\) Cost of polishing the top of the table at Rs 40 per square metre = 5.25 \(\times\) 40 = Rs 210.

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

Let the length and the breadth of the room be 4x m and x m respectively.
 \(\therefore\)  Area of the four walls of the room
= 2(l + b) h
= 2 (4x + x) 5 = 50 x m²
 \(\therefore\)  Cost of papering the four walls of the room @ 70 paise per square metre
= Rs 50x \(\times\) 0.70 = Rs 35 x
According to the question,
35x = 157.50
\(\Rightarrow\)  \(x=\frac { 157.50 }{ 35 } =4.5\)
\(\therefore\)  Breadth of the room = 4.5 m
\(\therefore\)  Length of the room = 4 \(\times\) 4.5 m = 18 m.

Let the dimensions of the box be 2k, 3k and 4k.
\(\therefore \)  Total surface area = 2(lb + bh + hl)
= 2(2k.3k + 3k.4k + 4k.2k)
= 52k²m²
Cost of covering at the rate of Rs 8 per m²
= 52k² \(\times\) 8 = Rs 416k²
Cost of covering at the rate of Rs 9.50 per m²
= 52k² \(\times\) 9.50 = Rs 494k²
Difference between the costs
= Rs 494k² - Rs 416k² = Rs 78k²
According to the question,
78k² = 1248
\(\Rightarrow\)  k² = 16 \(\Rightarrow\) k = 4
Hence, the dimensions of the box are 8 m, 12 m and 16 m.

L.H.S = \(\frac{1}{v}=\frac{1}{abc} \ \ \ \ \ \ \ \ \ .....(1)\)
R.H.S = \(\frac{2}{s}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)
=\(\frac{2}{2(ab+bc+ca)}(\frac{bc+ca+ab}{abc})\)
\(=\frac{1}{abc}\ \ \ \ \ \ \ ....(2)\)
(1) and (2) give
\(\frac{1}{v}=\frac{2}{s}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)

Let the length, breadth and height of the cuboid be l, b and h units respectively. then,
p = lb
q = bh
r = hl
ஃ pqr = (lb)(bh)(hl)
 =l²b²h²     .........(1)
Again   v = lbh
ஃ  v² = (lbh)²
= l²b²h²    .......(2) 
(1) and (2) give
v² = pqr

(i) Given edge of cubical box(a) = 20 cm
Thickness of wood = 2 cm
Volume of cubical box = a³
\(\therefore\) Volume of the wood=(22\(\times\)22\(\times\)22)-(20\(\times\)20\(\times\)20)
= 22³- 20³
= 2648 cm³
(ii) Volume of air trapped in it = 20\(\times\)20\(\times\)20
= 8000 cm³.

Thickness = \(\frac { Vol. }{ l\times b } =\frac { 1.104 }{ 2.3\times 0.75 } \)
= 0.64 m
No. of planks = \(\frac { 1.104 }{ 2.3\times 0.75\times 0.04 } =16\)

l = 140 cm
b = 110 cm
h = 80 cm
Surface area of open box = lb + 2(bh + hl)
Cost of painting box
\(=Rs\frac { 75 }{ 100\times 100 } [154+2(88+112)]\times 100\)
\(=\frac { 3 }{ 4 } [554]=3(138.5)\)
= Rs 415.5

Areas of four walls =\(\frac { Total\ Cost }{ Cost/{ m }^{ 2 } } \)
2(l + b)h  =\(\frac { 340.20 }{ 1.35 } =252\quad sq.m\)
2(l + b)h = 252      ........(i)
Area of floor = l\(\times\)b
\(=\frac { 91.80 }{ .85 } =108\ sq.m.\)
12\(\times\)b = 108  [l = 12 m, Given]
b = 9 m     ............(ii)
From eqn. (i) and (ii),
2(12 + 9)h = 252
\(h=\frac { 252 }{ 2\times 21 } =6m\)

l:b = 4:3, h = 550 cm,
l = 4y, b = 3y (taking y as constant)
h = 550 cm = 5.5 m
Total cost = L.S.A \(\times\)Rate per square metres
5082 = 2h(l + b)\(\times\)6.60
=2 \(\times\)5.5(4y + 3y) \(\times\)6.60
5082 = 2\(\times\)5.5\(\times\)7y\(\times\)6.60
10 = y
l = 4y = 4\(\times\)10 = 40 m
b = 3y = 3\(\times\)10 = 30 m

Let, Side of new cube = x
\(\therefore\) (12)³ = 8x³
or, \(\frac { { (12) }^{ 3 } }{ 8 } ={ x }^{ 3 }\)
\(\Rightarrow \ { \left( \frac { 12 }{ 2 } \right) }^{ 3 }={ x }^{ 3 }\)
or 6 = x
\(\therefore\) Side of new cube = 6 cm.
Ratio of surface areas of 8 new cubes to original cube
\(=\frac { 8\times { 6x }^{ 2 } }{ { 6(12 })^{ 2 } } \)
\(=\frac { 8\times 6(6)^{ 2 } }{ 6\times 12\times 12 } \)
\(=\frac { 8\times 6\times 6\times 6 }{ 6\times 12\times 12 } =\frac { 2 }{ 1 } \)
\(\therefore\) Ratio = 2:1.

Given, ratio = 2:3:4
Let length = 2x, breadth = 3x, height = 4x
Total surface are of the box = 2(lb + bh + hl)
= 2(2x\(\times\)3x + 3x\(\times\)4x + 4x\(\times\)2x)
= 2(6x²+12x²+8x²)
= 52 x²
\(\therefore\) Cost of wrapping at the rate of Rs 4 per m²
= Rs 4 \(\times\) 52 x²
= Rs 208 x²
Cost of wrapping at the rate of Rs 4.50 per m²
= Rs 4.50 \(\times\) 52 x²
= Rs 234 x²
According to question,
234x² - 208x² = 416
\(\Rightarrow\) 26x² = 416
\(\Rightarrow\) x² = 16
\(\Rightarrow\) x = 4
\(\therefore\) Length = 2x = 2\(\times\)4 = 8 m
Breadth = 3x = 3\(\times\)4=12 m
Height = 4x = 4\(\times\)4 = 16 m

Length of the rod \(=\sqrt { { \left( 10 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 } } \)

Volume = 15 \(\times\) 10 \(\times\) 8 = 1200 cm³

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

\(\frac { 2 }{ 3 } \times \left( 6\times 5\times 4 \right) 80{ m }^{ 3 }\)

Number of cubes = \(\frac { { \left( 20 \right) }^{ 3 } }{ { \left( 5 \right) }^{ 3 } } =64\)

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