(a) Given, radius of cone
\(r=\frac { 20 }{ 2 } =10\quad cm=0.1\quad m\)
Height of cone h=\(\frac { 1 }{ 2 } m\)
Slant height of cone l=\(\sqrt { { h }^{ 2 }+{ r }^{ 2 } } \)
\(l=\sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ (0.1) }^{ 2 } } \)
\(l=\sqrt { \frac { 26 }{ 100 } } =\frac { 5.1 }{ 10 } \)
= 0.51 cm
Curved surface Area of 100 cones = (100 \(\times\)\(\pi\)rl)m²
= (100\(\times\)3.14\(\times\)0.1\(\times\)0.51)m²
= 16.014 m²
Total cost of painting = 30 \(\times\)16.014
= Rs 480.42.
(b) Social values, environmental protection.

Total number blood donors = 85
(i) Number of donors whose blood is useful for Cheeku = 3
\(\therefore \) Number of favourable events = 85 - 3 = 82
Now, required probability\(=\frac{82}{85}\)
(ii) Responsible, rationality, love
(iii) Blood donation helps the needy. In fact it is a great act of charity and co-operation.

We have total number of colonies = 5
(i) Number of colonies in which more than 40 plants survived  = 3(B,C and E)
\(\therefore \) P(more than 40 plants survived in a colony)\(=\frac{3}{5}\)
(ii) Number of colonies in which less than 41 plants survived = 2(A and D)
P(less than 41 plants survived in colony)\(=\frac{2}{5}\)
(iii) In order to keep environment safe, we should grow more and more plants.

(i) P(sum is 7)\(=\frac{75}{500}=\frac{3}{20}\)
(ii) P(sum is more than 11)\(=\frac{70}{500}=\frac{7}{50}\)
(iii) P(sum is less than or equal to 6)\(=\frac{19+30+22+52+55}{500}\)
\(=\frac{178}{500}=\frac{89}{250}\)
(iv) P(sum is between 5 and 10)\(=\frac{52+75+70+53}{50}\)
\(=\frac{250}{500}=\frac{1}{2}\)

r_cone = 3 cm
h_cone = 4 cm
l_cone= 5 cm

Above given cone is formed with radius 3 cm, height 4 cm and slant height 5 cm when revolved about the fixed side of 4 cm.
\(V=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } .\frac { 22 }{ 7 } .(3)(3)(4)\)
= 37.71 cm³
Total surface area=\(\pi\)rl+\(\pi\)r²
\(=\frac { 22 }{ 7 } \times 3(5+3)\)
= 75.43 cm²

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

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

(a) The number of drivers in the age of 25-40 years and has more than 2 accidents in the year
= 15 + 10 = 25
(b) The number of drivers the age of whose is above 40 years and has accidents more than 1 but less than 3
= 13 + 17 = 30.

P(likes the detergent) = \(\frac { 375 }{ 500 } \)
= \(\frac { 3 }{ 4 } \)
(ii) P(does not like the detergent)
= 1-p(likes the detergent)
= \(1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)

\(\therefore\) Mean of 200 items = 50
\(\therefore\) Sum of items \(= 200 \times 50 = 10000\)
Corrected sum = 10000 - (92 + 8) + (192 + 88)
= 10180
\(\therefore\)  Correct mean \(= \frac {10180}{200}=50.9\)

Arranging the data in ascending order, we have,
17,17,17,18,19,23,23,23,23,25,26
Here, 23 occurs most frequency (4 times)
\(\therefore\)  Mode = 23
IF = f 4 is subtracted from each observation, then the new observations are
13,13,13,14,15,19,19,19,19,21,22
Here, 19 occurs most frequency (4 times)
\(\therefore\) Mode = 19

Number of observations (n) = 10, which is even,
\(\therefore\) Median
\(=\frac { { \left( \frac { n }{ 2 } \right) }^{ th }observation+{ \left( \frac { n }{ 2 } +1 \right) }^{ th }observation }{ 2 } \)
\(\Rightarrow 65=\frac { 5^{ th }observation+6^{ th }observation }{ 2 } \)
\(\Rightarrow\)   \(65 = \frac {x+\left(x+2\right)}{2}\)
\(\Rightarrow\)  2x + 2 = 130
\(\Rightarrow\)  2x = 130 - 2 = 128 
\(\Rightarrow\)  \(x = \frac {128}{2}=64\)

We shall first make the class intervals continuous. Then, the modified table is as follows:

For triangle   a = 15 cm, b = 14 cm, c = 13 cm
\( \therefore \ =\frac { a+b+c }{ 2 } s=\frac { 15+14+13 }{ 2 } =21\)c m
\(\therefore \)  Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-15)(21-14)(12-13) } \)
\(=\sqrt { 21(6)(7)(8) } =84\) cm²
Let the height of the parallelogram be h cm.
Then, area of the parallelogram = Base \(\times \) Height = 15 \(\times \) h = 15 cm²
According to the question, Area of the parallelogram = Area of the triangle
\(\Rightarrow \)  15h = 84
\(\Rightarrow \)  \(\frac { 84 }{ 15 } \) = 5.6 cm
Hence, the height of the parallelogram is 5.6 cm.

Let each of the equal sides of the rhombus be a cm.
Then,      Perimeter = a + a + a + a = 4a m
According to the question,  4a = 200
\(\Rightarrow \)   a= \(\frac { 200 }{ 4 } \) = 50 m

d₁= 80 m
a²=\({ \left( \frac { { d }_{ 1 } }{ 2 } \right) }^{ 2 }+{ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow \)   (50)² =( 40)²+\({ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow \) \({ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)= (50)^2-(40)² = 900= (30)²
\(\Rightarrow \)   \(\frac { { d }_{ 2 } }{ 2 } \)  = 30
\(\Rightarrow \)  d₂ = 60 m
\(\therefore \) Area of the rhombus = \(\frac { 1 }{ 2 } \) d₁d₂=\(\frac { 1 }{ 2 } \)\(\times \)80\(\times \)60 = 2400 m²

Let the depth be h m
Area of trapezium = 84 m²

\(\Rightarrow \)  Area of \(\Delta \)ABC +Area of \(\Delta \) ADC = 84 m²
\(\Rightarrow \)  \(\frac { 1 }{ 2 } \)(AB)(DE) +\(\frac { 1 }{ 2 } \) (DC)(DE) = 84 
\(\Rightarrow \) \(\frac { 1 }{ 2 } \) (12)(h) + (8)(h) = 84
\(\Rightarrow \)  6h + 4h = 84
\(\Rightarrow \)  10h = 84
\(\Rightarrow \) h =\(\frac { 84 }{ 10 } \) = 8.4
Hence, the depth of the canal is 8.4 m.

For one black colour sheet  a= 4 cm, b = 6 cm
\(\therefore \) Area \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)     | Sheet is an isosceles triangle
 \(=\frac { 4 }{ 4 } \sqrt { 4{ (6) }^{ 2 }-4^{ 2 } } \)  \(=\sqrt { 128 } =8\sqrt { 2 } \) cm²
\(\therefore \) Area of 2 black colour sheets = 2\(\times 8\sqrt { 2 } \) = \( 16\sqrt { 2 } \) cm²
Similarly, area of 2 white colour sheets =  \( 16\sqrt { 2 } \) cm²
\(\therefore \) Total area of black and white colour sheets =  \( 16\sqrt { 2 } \) + \( 16\sqrt { 2 } \) =  \( 32\sqrt { 2 } \) cm²

In right triangle PSQ,  PQ\(\frac { 1 }{ 2 } \)² = PS² + QS²     |By Pythagoras Theorem
= (12)² + (16)²
= 144 + 256 =400
\(\Rightarrow \)  PQ = \(\sqrt { 400 } \) = 20 cm
Now, for  \(\Delta \)PQR
a = 20cm, b = 48cm, c = 52cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 20+48+52 }{ 2 } =60\) cm
\(\therefore \)  Area of  \(\Delta \)PQR \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 60(60-20)(60-48)(60-52) }\)
\( \\ =\sqrt { (60)(40)(12)(8) } \)
 \(=\sqrt { \left( 6\times 10 \right) \left( 4\times 10 \right) \left( 6\times 2 \right) \left( 8 \right) } \)
\(=6\times 10\times 8=480\) cm²
Area of \(\Delta \) PSQ = \(\frac { 1 }{ 2 } \)\(\times \)Base\(\times \)Altitude
=\(\frac { 1 }{ 2 } \)\(\times \)16\(\times \)12=96 cm²
\(\therefore \) Area of the shaded portion =Area of \(\Delta \)PQR - Area of \(\Delta \)PSQ
= 480 - 96 = 384 cm²

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.

Height of the cone(h) = 24 cm
Let r cm be the radius of the base and l cm an can be the slant height of the cone, then
\(l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+{ 24 }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+576 } \)
Now, curved surface area=\(\pi\)rl
\(\Rightarrow \ \frac { 22 }{ 7 } \times r\times \sqrt { { r }^{ 2 }+576 } =550\)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =550\times \frac { 7 }{ 22 } \)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =175\)
Squaring both the sides we get
r² (r² + 576) = 30625
(r²)² + 576r²-30625 = 0
Let r² = x
x² + 576x - 30625 = 0
\(\Rightarrow\) x² + 625x - 49x - 30625 = 0
\(\Rightarrow\) x(x + 625)-  49(x + 625) = 0
\(\Rightarrow\) (x + 625)(x - 49) = 0
\(\Rightarrow\) x + 625=0 or x - 49 = 0
\(\Rightarrow\) x = 625 or x = 49
not possible x = 49
\(\therefore\) r²= 49
\(\Rightarrow\) r = 7 cm
Volume & the cone = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+24\)
= 1232 cm³.