Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm
Capacity of the bucket = volume of the frustum
\(\Rightarrow \frac { 1 }{ 3 } \times \pi h[{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 }]\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 45[{ 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7)]\)
= 22 x 15 x (28 x 4 + 7 + 28)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

Let assumed mean A = 30
C = 5

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10 
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

D= b²-4ac
⇒ 4(ac + bd)² - 4(a² + b²)(c² + d²)
⇒ 4[(ac + bd)² - (a² + b²)(c² + d²)]
⇒ 4(a²c² + b²d² + 2acbd - a²c²b²c² - a²d² - b²d²]
⇒ 4[2acbd - a²d² - b²c²]
⇒ 4[a²d² + b²c² - 2adbc]
⇒-4[ ad - bc]²
We have ad≠ bc
∴ ad- be of 0
⇒ (ad - bc)² > 0
⇒ 4(ad - bc)² < 0 ⇒ D < 0
Hence the given equation has no real roots.

Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with 'a' and 'b' = 4.
Since 0  ≤ r < 4, the possible remainders are 0,1,2,3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3

Let P(x, y) be equidistant from the points A (7, 1) and B (3, 5).
We are given that AP = BP. So, AP² = BP²
(x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
x²- 14x + 49 + y - 2y + 1 = x²- 6x + 9 + y -10y + 25
x - y = 2
Which is the required relation.

\(\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=2 \sec \theta \)
\(\text { LHS }=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \)
\( =\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}} \)
\(=\sqrt{\frac{(1+\sin \theta)^{2}}{1^{2}-\sin ^{2} \theta}}+\sqrt{\frac{(1-\sin \theta)^{2}}{1^{2}-\sin ^{2} \theta}} \)
\(=\sqrt{\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}} \)
\(=\frac{1+\sin \theta}{\cos \theta}+\frac{1-\sin \theta}{\cos \theta} \)
\(=\frac{1+\sin \theta+1-\sin \theta}{\cos \theta} \)
\(=2 \times \frac{1}{\cos \theta} \)
\(=2 \sec \theta \)
= RHS

Here a = 256, \(r=\frac{64}{256}=\frac{1}{4} \neq 1\)
Sum upto n terms
\( \mathrm{S}_{\mathrm{n}}=\frac{a\left(r^{n}-1\right)}{r-1} \)
\(= \frac{256\left[\left(\frac{1}{4}\right)^{n}-1\right]}{\frac{1}{4}-1}=\frac{256\left[\left(\frac{1}{4}\right)^{n}-1\right]}{-\frac{3}{4}} \)
\(= \frac{-1024}{3}\left[\left(\frac{1}{4}\right)^{n}-1\right]=\frac{1024}{3}\left[1-\left(\frac{1}{4}\right)^{n}\right] \)

\(\frac { x+2 }{ x+3 } +\frac { x-1 }{ x-2 } =\frac { (x-2)(x+2)+(x+3)(x-1) }{ (x+3)(x-2) } \)
\(=\frac { { x }^{ 2 }-4+{ x }^{ 2 }+2x-3 }{ (x+3)(x-2) } \)
\(=\frac { 2{ x }^{ 2 }+2x-7 }{ \left( x+3 \right) \left( x-2 \right) } \)

Given m = \(\sqrt 3\) , let θ be the inclination of the line
tan θ = \(\sqrt 3\)
We get, θ = 60⁰

The standard deviation of a given data is 4.5. If we subtract some fixed constant from all the data, the standard deviation will not change.
Each value of the data decreased by 5, the new standard deviation will not change.
New standard deviation = 4.5

Let r be the radius of the sphere. Given that, surface area of sphere = 154 m²
4\(\pi\)r² = 154
\(4\times \frac { 22 }{ 7 } \times { r }^{ 2 }=154\)
gives \({ r }^{ 2 }=154\times \frac { 1 }{ 4 } \times \frac { 7 }{ 22 } \)
hence, \({ r }^{ 2 }=\frac { 49 }{ 4 } \)We get r = \(\frac{7}{2}\)
Therefore, diameter is 7 m

In Is \(\triangle\)ABC ~ \(\triangle\)PQR
\(\frac { PQ }{ AB } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } ;\frac { QR }{ BC } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Since \(\frac { 1 }{ 2 } \neq \frac { 2 }{ 5 } ,\frac { PQ }{ AB } \neq \frac { QR }{ BC } \)
The corresponding sides are not proportional.
Therefore \(\triangle\)ABC  is not similar to \(\triangle\)PQR.

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47b₁ + 54.10 = 2.47 b₂ + 54.10
2.47b₁ = 2.47b₂ ⇒ b₁ = b₂
Thus, h(b₁) = h(b₂) ⇒ b₁ = b₂, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) =147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.

From \(\Delta ADC\) we have
AC² = AD² + CD²       ....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB² = AD² + BD²     ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² - AC² = BD² - CD²
AB² + CD² = BD² + AC²

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

Volume of a cylinder \(=\pi { r }^{ 2 }h\)
Volume of a cone \(=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
Volume of a sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
Their ratio V₁ : V₂ : V₃

\(h:\frac { h }{ 3 } :\frac { 4r }{ 3 } \)
3h: h: 4r
3h : h : 2(2r)    (where 2r = h)
∴  V₁ : V₂ : V₃ = 3: 1 : 2

Σ(x -10)² = 79 = Σx²- 20x + 100 = 79
= Σx²- 20Σx + 100 x 9 = 79
= Σx²- 20 x 99 + 900 = 79
Σx² = 79 + 1980 - 900 = 1159
Σ(x - \(\bar { x } \))² = Σ(x - 11)² = Σ(x² - 22x + 121)
= Σx² - 22Σx + 121 x 9
= 1159 - 22 x 99 + 1089 = 70
∴ Σx² = 1159, Σ(x - \(\bar { x } \))² = 70

Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
X²+x-20 = 0
(x + 5)(x - 4) = 0
x=-5,4
∴ x=4
(∵ x≠-5, x is a natural number
The next number = 4 + 1= 5
Two consecutive numbers are 4,5.

We have
a₃ = a + (3 - 1)d = a + 2d = 5  (1)
a₇ = a + (7 - 1)d = a + 6d = 9 (2)
(1) - (2) ⇒ -4d -4 ⇒ d = 1.
Sub, d = 1 in (1), we get
a + 2(1) = 5
a = 3
Hence the required A.P. is 3, 4, 5, 6, 7.

A = {5, 6, 8, to}, B = {19, 15,9, 11}
f(x) = 2x - 1
f(5) = 2(5) - 1 = 9
f(8) = 2(5)-1 = 15
\(\therefore\) a = 9; b = 15

A = {1,2,3,4,5}
B = {1,2,3,4, ... }
f: A \(\rightarrow\)B, f(x) = x²
\(\therefore\)  f(1) = 1² = 1
f(2)  = 2²= 4
f(3) = 3² = 9
f(4)  = 4²= 16
f(5) = 5²= 25
\(\therefore\) Range of f = {1, 4, 9, 16, 25}
Elements in A have been connected with different elements in B. Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula above, is given by
= \(\frac { 1 }{ 2 } \)[1(6 + 5) +(-4) (-5 + 1) + (-3)(-1 - 6)]
= \(\frac { 1 }{ 2 } \)[11 + 16 + 21] = 24 square units.

f(x) = 2x - k, g(x) = 4x + 5
f o g = f(g(x)) = f{4x + 5)
= 2(4x + 5) - k
= 8x + 10 - k
g o f = g(f(x)) = g(2x - k) = 4(2x - k) + 5
= 8x - 4k + 5
Given f o g = g o f
8x + 10 - k = 8x - 4k + 5
4k - k = 5 - 10
3k = - 5
k = \(\frac{-5}{3}\)

Let AB be the lamp post and CD be the apartment given CD = 66 m = EB.
\( \angle A C E=60^{\circ} \)
\(\angle E C B=\angle C B D=30^{\circ} \)
In the right triangle \(\triangle\)BDC
\(\tan 30^{\circ}=\frac{C D}{B D}\)
\( \frac{1}{\sqrt{3}}=\frac{66}{B D} \)
\(B D=66 \sqrt{3}\)
= 66 x 1.732 = 114.312 m
The distance between the lamp post and the apartment = 114.31 m
Now BD = EC = 114.31 m
In the right triangle \(\triangle\)ACE
\( \tan 60^{\circ} =\frac{A E}{C E} \)
\(\sqrt{3} =\frac{A E}{66 \sqrt{3}} \)
\(A E =66 \sqrt{3} \times \sqrt{3}[\text { From (1)] }\)
= 66 x 3 =198m
Height of the lamp post = AB
= AE + EB = 198 + 66 = 264m

1³ + 2³ + 3³ + ... + n³ = \(\left[\frac{n(n+1)}{2}\right]^{2}\)
1³ + 2³ + 3³ + ... + n³ = (120)²
\(\frac{n(n+1)}{2}=120\)
n(n + 1) = 120 x 2
n(n + 1) = 240
n² + n - 240 = 0
(n - 15)(n + 16) = 0
n = 15 (or) (- 16)
Number of terms cannot be negative
Number of terms to be taken = 15

Arranging the values in ascending order we get, 4, 7, 8, 10, 11 and n = 5

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 350 }{ 5 } -\left( \frac { 40 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45
When we add 3 to all the values, we get the new values as 7, 10, 11, 13, 14.

 Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 635 }{ 5 } -\left( \frac { 55 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45

Given that intercepts are in the ratio 2 : 5
\(\frac{a}{b} =\frac{2}{5} \)
\(a =\frac{2 b}{5} \)
Equation of the line in Intercepts form is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\frac{x}{\left(\frac{2 b}{5}\right)}+\frac{y}{b}=1 \)
\(\frac{5 x}{2 b}+\frac{y}{b}=1 \)
5x + 2y = 2b
This passes through ( 1, - 4)
5(1) + 2(-4) = 2b
\(5-8=2 b \Rightarrow b=-\frac{3}{2} \)
\(a =\frac{2 b}{5}=\frac{2\left(-\frac{3}{2}\right)}{5}=-\frac{3}{5} \)
Equation of a straight line is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{\left(-\frac{3}{5}\right)}+\frac{y}{\left(-\frac{3}{2}\right)}=1\)
\(\frac{5 x}{-3}+\frac{2 y}{-3}=1\)
5x + 2y + 3 = 0.

Let r be the radius of the hemisphere.
Given that, diameter of the hemisphere = side of the cube = l
Radius of the hemisphere = \(\frac{l}{2}\)
TSA of the remaining solid = Surface area of the cubical part + C.S.A. of the hemispherical part − Area of the base of the hemispherical part
= 6 x (Edge)² + 2\(\pi\)r²−\(\pi\)r²
= 6 x (Edge)² +  \(\pi\)r²
\(=6{ \times (l) }^{ 2 }+\pi { \left( \frac { l }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)
Total surface area of the remaining solid \(=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)sq. units

\(\frac { 4{ x }^{ 2 }y }{ 2{ x }^{ 2 } } \times \frac { 6x{ z }^{ 3 } }{ 20{ y }^{ 4 } } =\frac { { 3x }^{ 3 }z }{ 5{ y }^{ 3 } } \)

Given, vertices of a quadrilateral arc
A(- 4, - 2), B(5, - 1), C(6, 5) and D(- 7 ,6).
Let B Q, R and S be the mid points of the sides
AB, BC, CD and AD respectively
Mid point of 
\(\mathrm{AB}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) =\mathrm{P}\left(\frac{-4+5}{2}, \frac{-2-1}{2}\right) \)
\(=\mathrm{P}\left(\frac{1}{2},-\frac{3}{2}\right) \)
Mid point of BC
\(=Q\left(\frac{5+6}{2}, \frac{-1+5}{2}\right)=Q\left(\frac{11}{2}, 2\right)\)
Mid point of CD
\(=R\left(\frac{6-7}{2}, \frac{5+6}{2}\right)=R\left(-\frac{1}{2}, \frac{11}{2}\right)\)
Mid point of AD
\(=S\left(\frac{-4-7}{2}, \frac{-2+6}{2}\right)=S\left(-\frac{11}{2}, 2\right)\)
Slope of PQ \(=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-\frac{3}{2}-2}{\frac{1}{2}-\frac{11}{2}}=\frac{-\frac{7}{2}}{-\frac{10}{2}}=\frac{7}{10}\)
Slope of QR  \(= \frac{2-\frac{11}{2}}{\frac{11}{2}+\frac{1}{2}}=\frac{-\frac{7}{2}}{\frac{12}{2}}=-\frac{7}{12} \)
Slope of RS  \(= \frac{\frac{11}{2}-2}{-\frac{1}{2}+\frac{11}{2}}=\frac{\frac{7}{2}}{\frac{10}{2}}=\frac{7}{10} \)
Slope of PS  \(= \frac{-\frac{3}{2}-2}{\frac{1}{2}+\frac{11}{2}}=\frac{-\frac{7}{2}}{\frac{12}{2}}=-\frac{7}{12} \)
Slope of PQ = Slope of RS = PQ || RS
Slope of QR = Slope of PS = QR || PS
Hence, the mid points form a parallelogram.

(i) From the given graph
(a) f(0) = 9
(b) f(7) = 6
(c) = f(2)
(d) = f(10) = 0
(ii) From the graph, it is known that
when x = 9.5, f(x) = 1
(iii) (a) Domain = {x|0 ≤ x ≤ 10, x \(\in \) R}
(b) Range = {x|0 ≤ x ≤ 9, x \(\in \) R}
(iv) The image of '6' under f is '5'.