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)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have 
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 75 }{ 5 } \)=15
d'=\(\frac { x-\bar { x } }{ c } =\frac { x-A }{ c } \)
A is assumed mean c is common factor.
Here A= 15, 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 c
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
If 3 is added to each value, we get 8, 13, 18,23, 28 as new values.

∴ 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\(\sqrt { 2 } \)
S.D. doesn't change when a number is added or subtracted to the values.

A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

\(\frac { (k-12) }{ a } { x }^{ 2 }+\frac { 2(k-12) }{ b } x+\frac { 2 }{ c } =0\)
D² = b²- 4ac = (2(k - 12))² - 4(k - 12)(2)
= 4(k - 12)[(k - 12) - 2]
= 4(k-12)(k- 14)
The given equation will have equal roots, if D = 0
⇒ 4(k-12)(k-14) 0
k - 12 = 0 or k - 14 0
k 12, 14

Let us assume the opposite, (1) \(\sqrt { 3 } \) is irrational.
Hence \(\sqrt { 3 } =\frac { p }{ q } \)
Where p and q (q ≠ 0) are co-prime (no common factor other than 1)
Hence, \(\sqrt { 3 } =\frac { p }{ q } \)
\(\sqrt { 3 } \)q = p
Squaring both side
\({ (\sqrt { 3 }q ) }^{ 2 }={ p }^{ 2 }\)
3q² = p²
\({ q }^{ 2 }=\frac { p }{ 3 } \)
Hence, 3 divides p² So 3 divides p also   .....(1)
Hence we can say
\(\frac{p}{3}\) = c where c is some integer
s, p =p²
Putting p = 3c
3q² = (3c)²
3q² = 9c²
q2 = \(\frac13\) x 9c²
q² = 3c²
\(\frac{9^2}{3}\) = c²
Hence 3 divides q²
So, 3 divides q also  ...(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 } \) is irrational.

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.

Through O, draw PQIIBC so that P lies on AB and Q lies on DC
Now, PQ II BC
\(PQ\bot AB\quad PQ\bot OC\)
\(\left( \because \angle B={ 90 }^{ 0 }and\angle C={ 90 }^{ 0 } \right) \)
So, \(\angle BPQ={ 90 }^{ 0 }\quad \angle CQP={ 90 }^{ 0 }\)
Therefore BPQC and APQD are both rectangles. Now from \(\Delta OPB\)
OB² = BP² + OP²
Similarly from \(\Delta OQD\)
OD² = OQ² + DQ²
From \(\Delta OQC\)
OC²  = OQ² + CQ²
\(\Delta OAP\) we have
OA² = AP² +OP²
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ²
(As BP = CQ and DQ = AP)
= CQ² + OP² + OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC²+ OA
[From (3) and (4)]

A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.

Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30^o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45^o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m

BL and CM are medians at the \(\triangle\)ABC in which
\(A=\angle { 90 }^{ 0 }\)
From \(\triangle\)ABC
BC² = AB² + AC²
(Pythagoras theorem)

From \(\Delta ABL\)
BL² = AL² + AB²
\({ BL }^{ 2 }=\left( \cfrac { { AC }^{ 2 } }{ 2 } \right) +{ AB }^{ 2 }\)
(L is the mid-point at AC)
\({ BL }^{ 2 }=\cfrac { { AC }^{ 2 } }{ 4 } +{ AB }^{ 2 }\)
4BL² = AC² + 4AB²
From \(\Delta CMA\)
CM² = AC² + AM²
\({ CM }^{ 2 }={ Ac }^{ 2 }+\left( \cfrac { AB }{ 2 } \right) ^{ 2 }\)
(M is the mid-point at AB)
\({ CM }^{ 2 }={ AC }^{ 2 }+\cfrac { { AB }^{ 2 } }{ 4 } \)
4CM² = 4AC²+ AB²
Adding (2) and (3), we have
4(BL² + CM²) = 5(AC² + AB²)
4(BL² + CM²) = 5BC²

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr² = 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

\(\bar { { x }_{ 1 } } \) =\(\frac { 140 }{ 5 } \) =28
σ₁ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 880 }{ 5 } } \)
= \(\sqrt { 176 } \)
= 13.27
CV₁ = \(\frac { \sigma }{ \bar { x_{ 1 } } } \) x 100
CV₁ = \(\frac { 13.27 }{ 28 } \) x 100
= 47.39%
CV₁ < CV₂

\(\bar { { x }_{ 2 } } =\frac { 150 }{ 5 } \) =30
σ₂ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 1600 }{ 5 } } \)
= \(\sqrt { 320 } \)
 =17.89
CV₁ = \(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100
CV₂  = \(\frac { 17.89 }{ 30 } \) x 100
= 59.63%
∴ Team A is more consistent.

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.

Seven years ago, let Swathi's age be x years .
Seven years ago, let Varun's age was 5x² years.
Swathi's present age = x + 7 years
Varun's present age = (5x² + 7) years
3 years hence, we have
Swathi's age = x + 7 + 3 years
=x + 10 years
Varun's age = 5x² + 7 + 3 years
= 5x² + 10 years
It is given that 3 years hence Swathi's age will
be \(\frac{2}{5}\) of Varun's age.
∴ x+10=\(\frac{2}{5}\)(5x²+10)
⇒ x+10=2x²+4
⇒ 2x²-x-6=0
⇒ 2x(x-2)+3(x-2)=0
⇒(2x+3)(x-2)=0
⇒ x-2=0
⇒ x=2(∵2x+3≠0 as x>0)
Hence Swathi's present age = (2 + 7) years
= 9 years
Varun's present age = (5 x 2² + 7) years
= 27 years

a_n = 3 +2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 x 2 = 7
a₃ = 3 + 2 x 3 = 9
List of numbers becomes 5, 7, 9, 11,.........
Here, 7 - 5 = 9 - 7 = 11 - 9 = 2 and so on. So, it forms an A.P. with common difference d = 2.
To find S₂₄ we have n = 24, a = 5, d = 2.
S₂₄ = \(\frac{24}{2}\) [2 x 5 + (24 - 1) x ]
= 12 [10 + 46] = 672.
So, sum of first 24 terms of the list of numbers is 672.

1,-1,-3, -5, ...
t₂ - t₁ = -1 - 1 = -2
t₃ - t₂ = -3 - (-1)= -2
t₄ - t₃ = -5 - (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

f(x) = 1 + x
g(x) = (2x - 1)
fog(x) = f(g(x) =  f(2x - 1)
= 1 + 2x - 1 = 2x ...(1)
gof(x) = g(f(x) = g(1 + x) = 2(1 + x) - 1
= 2 + 2x - 1
= 2x + 1      ...(2)
(1) \(\neq \)(2)
\(\therefore\) fog(x) \(\neq \) gof(x)
It is verified

We know that diagonals at a parallelogram bisect each other.
So, the coordinates of the mid-point at AC = coordinates of the mid-point at BD.
i.e., \(\left[ \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 2+3 }{ 2 } \right] \)
\(\left[ \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 5 }{ 2 } \right] \)
\(\frac { 15 }{ 2 } =\frac { 8+P }{ 2 } \)
P = 7