Mean \(\bar { x } \)  = \(\frac { 16+13+17+21+18 }{ 5 } =\frac { 85 }{ 5 } \)  = 17

σ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 34 }{ 5 } } =\sqrt { 638 } \)
σ = 2.61
Co-efficient of variation
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 2.61 }{ 17 } \) x 100
= 15.35%

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 = {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.

\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \)
For f(1) & f(3)
Since 1 & 3 lies between -5 ≤ x ≤2, We take, f(x) = x +5
f(1) = 1 + 5 = 6
f(-3) = -3 +5 = 2
For f(4): Since 4 lies between 2 < x < 6 We take, f(x) = x -1
f(4) = 4 -1 = 3
For f(-6): Since -6 lies between -7≤ x < -5 ,We take, f(x) = x² + 2x +1
f(-6) = (-6)² + 2(-6) + 1 = 36 -12 + 1 = 24 +1 = 25
Now ,\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) =\(\frac{ 4(2) +2(3) }{ 25 - 3(6)}\)
= \(\frac{ 8 + 6 }{ 25 -18}\)
= \(\frac{14 }{ 7}\) = 2
Hence, the value of \(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) = 2

f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 - 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) - f(-3) = 36 + 2 = 38

(i) If D is (0,k) then D is a point on the Cross Road.
Therefore, substituting x = 0, y = k in the equation of Cross Road,
we get, 0 - 3k + 18 = 0
Value of k = 6
Therefore, D is (0, 6)
(ii) To find E, let E be (q, 2)
Put y = 2 in the equation of the Cross Road,
we get, 4q - 6 + 18 = 0
4q = - 12 gives q = - 3
Therefore, The point E is (-3, 2)
Thus the Cross Road meets the North Street at D(0, 6) and
East Avenue at E (-3, 2)

2y + z = 3(-x + 1)
3x + 2y + z = 3        .....(1)
-x + 3y - z = -4       .....(2)
3x + 2y + z = -\(\frac{1}{2}\)        ....(3)

0 = 7
which is a contradiction.
The system of equations has no solution

F^o = (c^o x 1.8) + 32
σ_c = 5°C
σ_F = (1.8 x 5°C) . 9°F
Adding value to data doesn't affect standard deviation.
New variance = σ²_F = 81°F.

Total students = 35
n(S) = 35
Boys : girls = 4 : 3
Let the number of boys = 4x
and number of girls = 3x
4x + 3x = 35
7x = 35
\(x=\frac{35}{7}=5\)
Number of boys = 4 x 5 = 20
Number of girls = 3 x 5 = 15
Boys are numbered = {1, 2,3,4,5,6,7,8, 9,10, 11, 12,13, 14, 15,16, 17, 18,19, 20}
Girls are numbered = {21, 22,23, 24, 25, 26,27 , 29, 29, 30, 31, 32, 33, 34, 35}
Let A be the event of getting a boy with prime roll number.
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
\(\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{8}{35}\)
Let B be the event of getting a girl with composite roll number
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34,35}
n(B) = 12
\(\mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{12}{35}\)
Let C be the event of getting an even roll number
C = {2, 4, 6,8, 10, 12, 14, 16, 19, 20, 22, 24, 26, 29, 34, 32, 34}
n(C) = 17
\(\mathrm{P}(\mathrm{C})=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{17}{35}\)
Since A and B are mutually exclusive events
P(A n B) = 0
B n C = {22, 24, 26, 29, 30, 32, 34}
n ( B n C ) = 7
\(P(B \cap C) =\frac{n(B \cap C)}{n(\mathrm{~S})}=\frac{7}{35} \)
\(A \cap C =\{2\} \)
\(\mathrm{n}(A \cap C) =1 \)
\(P(A \cap C) =\frac{n(A \cap C)}{n(S)}=\frac{1}{35} \)
\(P(A \cap B \cap C) =0[\because n(A \cap B)=0] \)
\(P(A \cup B \cup C)=P(A)+P(B)+P(C)- P(A \cap B)-P(B \cap C)-P(A \cap C) + P(A \cap B \cap C) \)
\(=\frac{8}{35}+\frac{12}{35}+\frac{17}{35}-0-\frac{7}{35}-\frac{1}{35}+0 \)
\(=\frac{8+12+17-7-1}{35}=\frac{29}{35} \)
Required probabiliry = \(\frac{29}{35}\)

Let the assumed mean A = 34.5

Standard deviation \(\sigma=C \times \sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{N}-\left(\frac{\Sigma f_{i} d_{i}}{N}\right)^{2}} \)
\(\sigma=2 \times \sqrt{\frac{1052}{84}-\left(\frac{-158}{84}\right)^{2}} \)
\(\sigma=2 \times \sqrt{\frac{22092}{1764}-\frac{6241}{1764}} \)
\(\sigma=2 \times \sqrt{\frac{15851}{1764}}=2 \times \sqrt{8.98} \)
\(\sigma=2 \times 2.99 \simeq 5.99 \)
Standard deviation \(\sigma \simeq 5.99 \simeq 6\)

Let CD be the statue of tall 1.6 m.
BC be the pedestal.
From the right triangle \(\triangle\)ABC
\( \tan 40^{\circ} =\frac{B C}{A B} \)
\(0.8391 =\frac{B C}{A B} \)
\(A B =\frac{B C}{0.8391}\)          ....(1)
From the right triangle \(\triangle\)ABD
\( \tan 60^{\circ} =\frac{B D}{A B} \)
\(\sqrt{3} =\frac{B C+C D}{A B} \)
\(1.732 =\frac{B C+1.6}{A B} \)
\(A B =\frac{B C+1.6}{1.732} \)        ....(2)
From (1) and (2)
\(\frac{B C}{0.8391}=\frac{B C+1.6}{1.732}\)
1.732 BC = 0.8391 (BC + 1.6)
1.732 BC = 0.8391 BC + (0.8391) (1.6)
1.732 BC - 0.8391 BC = 1.34256
0.8929 BC = 1.34256
\(\mathrm{BC}=\frac{1.34256}{0.8929}=\frac{13425.6}{8929}=1.5 \mathrm{~m}\)
Height of the pedestal = 1.5 m

Given AB || CD || EF
AB = 6 cm, BD = 5 cm, EF = 4 cm, CD = x cm, DE = y cm
\(\text { In } \triangle E C D \text { and } \triangle E A B\)
\( \angle C E D=\angle A E B \)            [common]
\(\angle E C D=\angle E A B \)             [corresponding angles]
\(\triangle E C D \sim \triangle E A B\)
[By AA similarity criteria]      ...(1)
\(\therefore \ \frac{E C}{E A}=\frac{C D}{A B}\)
[ Corresponding parts of similar triangles are proportional]
\(\frac{E C}{E A}=\frac{x}{6}\)            ....(2)
In \(\triangle A C D\ and\ \triangle A E F \)
\(\angle C A D=\angle E A F\)          [Common]
\( \angle A C D =\angle A E F \)        [Corresponding angles]
\(\triangle A C D \sim \triangle A E F \)      [By AA similarity]
\(\frac{A C}{A E} =\frac{C D}{E F}\)
[Corresponding parts of similar triangle are proportional]
\(\therefore \quad \frac{A C}{A E}=\frac{x}{4}\)          ....(3)
Adding (2) and (3)
\( \frac{E C}{E A}+\frac{A C}{A E} =\frac{x}{6}+\frac{x}{4} \)
\(\frac{E C+A C}{A E} =\frac{4 x+6 x}{24} \)
\(\frac{A E}{A E} =\frac{10 x}{24} \)
\(1 =\frac{10 x}{24} \)
\(\mathrm{x} =\frac{24}{10}=\frac{12}{5} \mathrm{~cm}\)
From (1) \(\triangle E C D \sim \triangle E A B\)
\( \frac{D C}{A B} =\frac{E D}{E B} \)
\(\frac{x}{6} =\frac{y}{5+y} \)
\(\because \mathrm{x}=\frac{12}{5} \Rightarrow \frac{12 / 5}{6} =\frac{y}{5+y} \)
\(\frac{12}{5 \times 6} =\frac{y}{5+y} \)
12(5 + y) = 30 y
60 + 12y = 30 y
60 = 30y - 12y
18y = 60
\( y=\frac{60}{18} \)
\(y=\frac{10}{3} \mathrm{~cm}\)

Diameter of the bowl = 14 cm
Radius r = 7 cm
Volume of hemisphere  \(=\frac{2}{3} \pi r^{3} \text { cu. units } \)
\(=\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \)
\(=\frac{2156}{3}=1718.67 \mathrm{~cm}^{3} \)
Radius of cylinder 'r' = 7 cm
Height 'h' = 6 cm
Volume of cylinder \(=\pi r^{2} h \text { cu. units } \)
\(=\frac{22}{7} \times 7 \times 7 \times 6 \)
= 924 cm³
capacity of the vessel = Volume of hemisphere + Volume of cylinder
= 718.67 + 924
= 1642.67 cm³

Let R,r be the outer and inner radii respectively
Given outer surface area = 576 cm²
\(4 \pi R^{2}=576 \pi\)
R² = 144
R = 12 cm 
Inner surface area = \(324 \pi \mathrm{cm}^{2}\)
\(4 \pi r^{2}=324 \pi\)
r² = 81
r = 9 cm
Volume of the material required
\(=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \text { cu. units } \)
\(=\frac{4}{3} \times \frac{22}{7} \times\left(12^{3}-9^{3}\right) \)
\(=\frac{4}{3} \times \frac{22}{7} \times 999 \)
= 4186.285 = 4186.29 cm³

Let the internal and external radii of the hemispherical shell be r and R
respectively.
Given that, R = 5 m, r = 3 m
C.S.A. of the shell = 2\(\pi\)(R² + r²) sq. units
\(=2\times \frac { 22 }{ 7 } \times \left( 25+9 \right) =213.71\)
T.S.A. of the shell = \(\pi\)(3R² + r²) sq. units
\(=\frac { 22 }{ 7 } (75+9)=264\)
Therefore, C.S.A. = 213.71 m² and T.S.A. = 264 m².

The vertices are A(1, - 4), B(2, - 3) and C(4, - 7).
The slope of AB = \(\frac { -3+4 }{ 2-1 } =\frac { 1 }{ 1 } =1\)
The slope of BC = \(\frac { -7+3 }{ 4-2 } =\frac { -4 }{ 2 } =-2\)
The slope of AC = \(\frac { -7+4 }{ 4-2 } =\frac { -4 }{ 3 } =-1\)
Slope of AB x slope of AC = (1) (−1) = −1
AB is perpendicular to AC ㄥA = 90⁰
Therefore, ΔABC is a right angled triangle.

Given ABC - DEF
then we have
\(\frac{\operatorname{Area}(\Delta \mathrm{ABC})}{\text { Area }(\Delta \mathrm{DEF})}=\frac{A B^{2}}{D E^{2}}=\frac{B C^{2}}{E F^{2}}=\frac{A C^{2}}{D F^{2}} \)
\(\frac{9}{16}=\frac{B C^{2}}{E F^{2}} \)
\(\frac{9}{16} =\frac{2.1 \times 2.1}{E F^{2}} \)
\(\mathrm{EF}^{2} =\left(\frac{2.1 \times 4}{3}\right)^{2} \)
\(\mathrm{EF} =\frac{2.1 \times 4}{3}=2.8 \)
EF = 2.8 cm

From the diagram,
The solid is a cuboid volume of cuboid = length x breadth x height
where l = 24 - 2x, b - 24 - 2x,. h = x
Volume V (x) = (24 - 2x) (24 - 2x) x
V(x) = x(24 - 2x)², x > 0
= 4x³ - 96x² + 576x, x > 0
So, the domain is 0 < x < 12

(i) Given Set - Builder form
{(x, y)|x = 2y, x \(\in \) {2,3,4,5}, y \(\in \) {1,2,3,4}
x = 2y
y = 1 ⇒ x = 2
y = 2 ⇒ x = 4
y = 3 ⇒ x = 6
y = 4 ⇒ x = 8
Relation R - {(2, 1), (4,2)}
(a) Arrow diagram

b) Graph

(c) Roster form R =  {(2, 1), (4, 2)}
ii. Given set
{(x, y)|y = x + 3, x, y are natural numbers < 10}.
When x = 1, y = 1 + 3 = 4
When x = 2, y = 2 + 3 = 5
When x = 3, y = 3 + 3 = 6
When x = 4, y = 4 + 3 = 7
When x = 5, y = 5 + 3 = 8
When x = 6, y = 6 + 3 = 9
When x = 7, y = 7 + 3 = 10 is not possible
Since x and y are less than 10
Relation R = {(1,4), (2,5), (3,6), (4,7), (5,8), (6,9)}
(a) Arrow diagram

(b) Graph

(c) Roster form
{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}