Let W be the point on the window where the angles of elevation and depression are measured. Let PQ be the house on the opposite side.
Then WA is the width of the street.
Height of the window = h metres
= AQ (WR = AQ)
Let PA = x metres
In right triangle \(\triangle\)PAW, \(tan{ \theta }_{ 1 }=\frac { AP }{ Aw } \)
gives \(tan{ \theta }_{ 1 }=\frac { x }{ AW } \)
so, \(AW=\frac { x }{ tan{ \theta }_{ 1 } } \)
we get, AW = x cot θ          ...(1)
In right triangle \(\triangle\)QAW, tan θ₂ = \(\frac { AQ }{ AW } \)
gives  tan θ₂ = \(\frac { h }{ AW } \)
we get, AW = h cot θ₂ ...(2)
From (1) and (2) we get, x cot θ₁ = h cot θ₂
gives, x = h\(\frac { cot{ \theta }_{ 2 } }{ { cot\theta }_{ 1 } } \)
Therefore, height of the opposite house = PA + AQ = x + h = h\(\frac { cot{ \theta }_{ 2 } }{ { cot\theta }_{ 1 } } \)\(\left( 1+\frac { cot{ \theta }_{ 2 } }{ { cot\theta }_{ 1 } } \right) \)
Hence Proved.

(A - B)^T= A^T-B^T
L.H.S = \((A-B)=\left[ \begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix} \right] -\left[ \begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix} \right] =\left[ \begin{matrix} -3 & 2 \\ 0 & -2 \end{matrix} \right] ...(2)\)
\({ (A-B) }^{ T }=\left[ \begin{matrix} -3 & 0 \\ 2 & -2 \end{matrix} \right] ...(1)\)
\({ A }^{ T }=\left[ \begin{matrix} 1 & 1 \\ 2 & 3 \end{matrix} \right] ,{ B }^{ T }=\left[ \begin{matrix} 4 & 1 \\ 0 & 5 \end{matrix} \right] \)
\(=\left[\begin{array}{rr} -3 & 0 \\ 2 & -2 \end{array}\right]\)
From (1) and (2)
(A - B)^T= A^T-B^T
Hence verified.

Let A be the initial position of the bird.
B be the position after travelling 30 km at an angle of 35^o from A.
Let C be the position after travelling 32 km at an angle of 48^o from B.
[complementary angle]
\(\sin 42^{\circ}=\frac{P C}{B C} \)
\(0.6691=\frac{P C}{32} \)
PC = 32 x 0.6691 = 21.41 km
C is 21.41 km to the North of B.

Let A be the initial position of the bird.
B be the position after travelling 30 km at an angle of 35^o from A.
Let C be the position after travelling 32 km at an angle of 48^o from B.
[complementary angle]
From the right triangle AOB
\(\cos 55^{\circ}=\frac{O A}{A B} \)
\(0.5736=\frac{O A}{30} \)
OA = 30 x 0.5736 = 17.21 km
B is 17 .21 km to the West of A

Given that intercepts are equal.
a = b
Equation of the line in intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{a}+\frac{y}{a}=1 \)
x + y = a
This passes through (- 8, 4)
-8 + 4 = a
a = -4
b = -4
Equation of the straight line is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{-4}+\frac{y}{-4}=1 \)
x + y + 4 = 0.

y = -0.1x + 1
Initially, time x = 0  y = 1
Total MB of the song is 1.

On subtracting equation (2) from equation (1), equation (3) from equation (2)and equation (1) from equation (3), we get
x - y = (l - m)d
y - z = (m - n)d
z - x = (n -l)d
(x - y)n + (y - z)l + (z - x)m = [(l -m)n + (m -n)l + (n -l)m]d
= [ln - mn + lm - nl + nm - lm]d = 0

(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)

Use the slope and y intercept to write an equation 
The equation is y =  \(\frac { 9 }{ 5 } x\) + 32

Given 
A = {1,2,3,4,5,6,7}
B = {2,3,5,7}
C = {2}
A x ( B - C) = (A x B) - (A x C)
B - C = {2,3,5,7} - {2}
=  {3,5,7}
A x (B-C) = {1,2,3,4,5,6,7} x {3.5.7}
={(1,3),(1,5),(1,7),(2,3),(2,5),(2,7) (3,3),(3,5),(3,7),(4,3),(4,5),(4,7) (5,7),(5,3),(5,5) (6,3),(6,5),(6,7),(7,3),(7,5),(7,7)}    ....(1)
A x B = {1,2,3,4,5,6,7} x {2,3,5,7}
= {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7) (3,2),(3,3),(3,5),(3,7),(4,2),(4,3),(4,5),(4,7) (5,2),(5,3),(5,5),(5,7),(6,2),(6,3),(6,5),(6,7) (7,2),(7,3),(7,5),(7,7)}
 A x C = {1,2,3,4,5,6,7} x {2}
= {(1,2),(2,2),(3,2),(4,2),(5,2)(6,2),(7,2)}
(A x B) - (A x C) = {(1,3),(1,5),(1,7),(2,3) (2,5),(2,7),(3,3),(3,5) (3,7),(4,3),(4,5),(4,7), (5,3),(5,5),(5,7),(6,3) (6,5),(6,7),(7,3),(7,5),(7,7)}                     ..(2)
From (1) and (2), it is clear that
A x ( B - C) = (A x B) - (A x C)
Hence verified

\(\frac { y+z }{ 4 } =\frac { z+x }{ 3 } =\frac { x+y }{ 2 } \)
x + y + z = 27           
\(\frac { y+z }{ 4 } =\frac { z+x }{ 3 }\)
We get, 4x - 3y+ z = 0        .(1)
Equating \(\frac{y+z}{4}=\frac{x+y}{2}\)
On simplifying 2x + y - z = 0      ........(2)


Substituting the value x = 2 in (4)
6(3) - 2y = 0
-2y = -18
\(y=\frac{18}{2}=9\)
Substituting x = 3,y = 9 in (3)
3 + 9 + z  = 27
z = 27 - 12
z = 15
Solution: x = 3, y = 9, z = 15

x	f	fx	d = x - \(\overline { x } \)	d2
k	2	2k	\(\frac { -15 }{ 7 } k\)	\({ \left( \frac { -15 }{ 7 } k \right) }^{ 2 }\)
2k	1	2k	\(\frac { -8 }{ 7 } k\)	\({ \left( \frac { -8 }{ 7 } k \right) }^{ 2 }\)
3k	1	3k	\(\frac { -1 }{ 7 } k\)	\({ \left( \frac { -1 }{ 7 } k \right) }^{ 2 }\)
4k	1	4k	\(\frac { 6k }{ 7 } \)	\({ \left( \frac { 6k }{ 7 } \right) }^{ 2 }\)
5k	1	5k	\(\frac { 13 }{ 7 } k\)	\({ \left( \frac { 13 }{ 7 } k \right) }^{ 2 }\)
6k	1	6k	\(\frac { 20 }{ 7 } k\)	\({ \left( \frac { 20 }{ 7 } k \right) }^{ 2 }\)

\(\overline { x } =\frac { \sum { fx } }{ \sum { f } } =\frac { 22k }{ 7 } \)
\({ \sigma }^{ 2 }=\frac { \sum { { f }_{ i }{ { { d }^{ 2 } }_{ 1 } } } }{ \sum { f } } \)
\(=\frac { { k }^{ 2 } }{ { 7 }^{ 2 } } \frac { [{ 1 }^{ 2 }+{ 6 }^{ 2 }+{ 8 }^{ 2 }+{ 13 }^{ 2 }+{ 15 }^{ 2 }+{ 20 }^{ 2 }] }{ 7 } \)
\(160=\frac { { k }^{ 2 } }{ { 7 }^{ 2 } } \times 1120\)
\({ k }^{ 2 }=\frac { 160\times { 7 }^{ 3 } }{ 1120 } \)
= 49 ⇒ k 土 7
k = 7 since k is a +ve number .

When two dice are rolled once, the Sample space
S = \(\left\{ \begin{matrix} (1,1) \\ (2,1) \\ \begin{matrix} (3,1) \\ (4,1) \\ \begin{matrix} (5,1) \\ (6,1) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,2) \\ (2,2) \\ \begin{matrix} (3,2) \\ (4,2) \\ \begin{matrix} (5,2) \\ (6,2) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,3) \\ (2,3) \\ \begin{matrix} (3,3) \\ (4,3) \\ \begin{matrix} (5,3) \\ 6,3) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,4) \\ (2,4) \\ \begin{matrix} (3,4) \\ (4,4) \\ \begin{matrix} (5,4) \\ (6,4) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,5) \\ (2,5) \\ \begin{matrix} (3,5) \\ (4,5) \\ \begin{matrix} (5,5) \\ (6,5) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,6) \\ (2,6) \\ \begin{matrix} (3,6) \\ (4,6) \\ \begin{matrix} (5,6) \\ (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)
n(S) = 36
Let 'A' be the event of getting an even number on the first die.
\(A=\left\{ \begin{matrix} (2,1), & (2,2), & \begin{matrix} (2,3), & (2,4), & \begin{matrix} (2,5), & (2,6) \end{matrix} \end{matrix} \\ (4,1), & (4,2), & \begin{matrix} (4,3), & (4,4), & \begin{matrix} (4,5), & (4,6) \end{matrix} \end{matrix} \\ (6,1), & (6,2), & \begin{matrix} (6,3), & (6,4), & \begin{matrix} (6,5), & (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)
n(A) = 18
\(P(A)=\frac { n(A) }{ n(S) } =\frac { 18 }{ 36 } \)
Let B be the event of getting total face sum 8.
B = {(2, 6), (3,5), (4, 4), (5, 3), (6, 2)}
n(B) = 5
\(P(A)=\frac { n(B) }{ n(S) } =\frac { 5 }{ 36 } \)
(A ∩ B) = {(2, 6), (4,4), (6, 2)}
n(A ∩ B) = 3
p(A ∩ B) = \(\frac{(A∩B)}{n(S)}=\frac{3}{36}\)
∴ P(A U B) = P(A) + P(B)+ P(A ∩ B)
\( =\frac{18}{36}+\frac{5}{36}-\frac{3}{36} \)
\(\mathrm{P}(A \cup B)=\frac{18+5-3}{36}=\frac{23-3}{36}=\frac{20}{36}=\frac{5}{9} \)
Probability of getting even number in the first die or a total face sum 8 is \(\frac{5}{9} .\)

If Priya and Vidhya are visiting the shop in the same week, the sample space
S = {(Mon, Mon) (Mon, Tue) (Mon, Wed) (Mon, Thur) (Mon, Fri) (Mon, Sat) (Tue, Mon) (Tue, Tue) (Tue, Wed) (Tue, Thur) (Tue, Fri) (Tue, Sat) (Wed, Mon) (Wed, Tue) (Wed, Wed) (Wed, Thur) (Wed, Fri) (Wed, Sat) (Thur, Mon) (Thur, Tue) (Thur, Wed) (Thur, Thur) (Thur, Fri) (Thur, Sat) (Fri, Mon) (Fri, Tue) (Fri, Wed) (Fri, Thur) (Fri, Fri) (Fri, Sat) (Sat, Mon) (Sat, Tue) (Sat, Wed) (Sat, Thur) (Sat, Fri) (Sat, Sat)}
n(S) = 36
(i) Let 'A' be the event that both will visit the shop on the same day
4 = {(Mon, Mon) (Tue, Tue) (Wed, Wed) (Thur, Thur) (Fri, Fri) (Sat, Sat)}
n(A) = 6
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Both will visit the shop on different days
\(\mathrm{P}(\bar{A})=1-\mathrm{P}(\mathrm{A})=1-\frac{1}{6} \)
\(\mathrm{P}(\bar{A})=\frac{6-1}{6}=\frac{5}{6} \)
i.e., probability that both will visit the shop on different days  \(=\frac{5}{6}\)
(iii) Let 'B' be the event that both will visit the shop on consecutive days
B = {(Mon, Tue) (Tue, Wed) (Wed, Thur) (Thur, Fri) (Fri, Sat)}
n(B) = 5
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{5}{36}\)

When three fair coins are tossed together, the sample space
S = {(HHH), (THH), (HTH),(HHT), (TTH), (THT), (HTT), (TTT)}
N(s) = 8
(i) Let A be the event of getting all heads
A = {HHH}
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{7}{8}\)
(iii) Let C be the event of getting at most one head
C = {HTT, THT, TTH, TTT}
n(C) = 4
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)
(iv) Let D be the event of getting at most two tails
P = {HHH, HHT, HTH, HTT, THH, THT, TTH}
n(D) = 7
\(\mathrm{P}(\mathrm{D})=\frac{n(D)}{n(S)}=\frac{7}{8}\)

When we roll two dice, the sample space is given by
S = \(\{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \)
n(S) = 36
(i) Let A be the event of getting the sum of outcome values equal to 4.
Then A = {(1, 3),(2, 2),(3, 1)}; n(A) = 3.
Probability of getting the sum of outcomes equal to 4 is  P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(ii) Let B be the event of getting the sum of outcome values greater than 10.
Then B = {(5,6),(6,5),(6,6)}; n(B) = 3
Probability of getting the sum of outcomes greater than 10 is P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(iii) Let C be the event of getting the sum of outcomes less than 13. Here all the outcomes have the sum value less than 13. Hence C = S
Therefore, n(C) = n(S) = 36
Probability of getting the sum value less than 13 is P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 36 } \) = 1

First we find the coefficient of variation for guavas and oranges separately.
Number of guavas, n=7

xi	xi2
3	9
5	25
6	36
4	16
3	9
5	25
4	16
Σxi=30	Σxi2=136

Mean \(\bar { { x }_{ 1 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₁=\(\sqrt { \frac { 136 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 19.43-18.40 } \) ≃ 1.01
Coefficient of variation for guavas
C.V₁=\(\frac { { \sigma }_{ 1 } }{ \bar { { x }_{ 1 } } } \times 100\)% = \(\frac { 1.01 }{ 4.29 } \) x 100% =23.54%

xi	xi2
1	1
3	9
7	49
9	81
2	4
6	36
2	4
Σxi=30	Σxi2=184

 Number of oranges n=7
Mean \(\bar { { x }_{ 2 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₂=\(\\ \sqrt { \frac { 184 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 26.29-18.40 } \)=2.81
Coefficient of variation for oranges
C.V₁=\(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100% = \(\frac { 2.81 }{ 4.29 } \) x 100% = 65.50%
CV₁=23.54%, CV₂=65.50%. Since, C.V₁ < C.V₂,  we can conclude that the consumption of guavas is more consistent than oranges.

Let C and D are two ships.
Let AB be the height of the light house.
\(\mathrm{CD}=200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right) m\)
In the right ABD
\(\tan 60^{\circ}=\frac{A B}{B D} \)
\(\sqrt{3}=\frac{A B}{B D} \)
\(\mathrm{BD}=\frac{A B}{\sqrt{3}} \)
In the right triangle ABC
\(\tan 45^{\circ} =\frac{A B}{B C} \)
\(1 =\frac{A B}{B C} \)
\(B C =A B \)
\((1)+(2) \Rightarrow B D +B C=\frac{A B}{\sqrt{3}}+A B \)
\(\mathrm{CD}=A B\left(\frac{1}{\sqrt{3}}+1\right) \quad[\because \mathrm{CB}+\mathrm{BD}=\mathrm{CD}] \)
\(\frac{C D}{\left(\frac{1}{\sqrt{3}}+1\right)}=\mathrm{AB} \)
\(A B=\frac{200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}\)
AB = 200 m
Height of the light house is 200 m.

Let AB denote the height of the peak and be 'h'.
In ΔABC, 
tan 8^o = \(\frac { AB }{ BC } =\frac { h }{ m } \)
m = \(\frac { AB }{ BC } =\frac { h }{ m } \) ...(1)

In Δ ABC,
tan 40^o =\(\frac { AB }{ BC } =\frac { h }{ m } \)
m+1 = \(\frac { h }{ tan4 } \) ..(2)
From (1) and (2)
\(\frac { h }{ tan8 } +1=\frac { h }{ tan4 } \)
\(1=\frac { h }{ tan4 } -\frac { h }{ tan8 } \)
\(h\left[ \frac { tan8-tan4 }{ tan4tan8 } \right] =1\)
h = \(\frac { tan4\times tan8 }{ tan8-tan4 } \)
= 0.14 mile (approx)

Given points (-2, -1), (4, 0), (3, 3) and (-3, 2) let the points be A (-2, -1), B (4, 0), C (3, 3) and D (-3,2)
Slope = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \)
Slope of AB = \(\frac{-1-0}{-2-4}=\frac{1}{6} \)
Slope of BC = \(\frac{0-3}{4-3}=\frac{-3}{1}=-3 \)
Slope of CD = \(\frac{3-2}{3+3}=\frac{1}{6} \)
Slope of DA = \(\frac{2+1}{-3+2}=\frac{3}{-1}=-3 \)
Slope of AB = Slope of CD = AB is parallel to CD
Slope of BC = Slope of DA = BC is parallel to DA
Hence, the given points form a parallelogram

n^th term of a G.P = ar^n-1
Given \(2^{\text {nd }} \text { term } =\sqrt{6} \)
\(a r^{2-1} =\sqrt{6} \)
\(a r =\sqrt{6} \)
\(6^{\text {th }} \text { term } =9 \sqrt{6} \)
\(a r^{6-1} =9 \sqrt{6} \)
\(a r^{5} =9 \sqrt{6} \)
\(\therefore \frac{a r^{5}}{a r} =\frac{9 \sqrt{6}}{\sqrt{6}} \)
r⁴ = 9
r² = 3
\( r=\sqrt{3} \)
\(Also\ ar =\sqrt{6} \)
\(a=\frac{\sqrt{6}}{r}=\frac{\sqrt{2} \sqrt{3}}{\sqrt{3}} \)
\(a=\sqrt{2} \)
The G.P. is a, ar, ar²,...
\(=\sqrt{2}, \sqrt{6}, \sqrt{2}(\sqrt{3})^{2} \ldots \ldots \)
\(=\sqrt{2}, \sqrt{6}, 3 \sqrt{2} \ldots . \)

Let the assumed mean, A = 35, c = 10

Marks	Mid value (xi)	fi	di = xi-A	di = \(\frac { x_{ i }-A }{ c } \)	fidi	fidi2
0-10	5	8	-30	-3	-24	72
10-20	15	12	-20	-2	-24	48
20-30	25	17	-10	-1	-17	17
30-40	35	14	0	0	0	0
40-50	45	9	10	1	9	9
50-60	55	7	20	2	14	28
60-70	65	4	30	3	12	36
		N = 71			Σfidi = -30	Σfidi2 = 210

Standard deviation σ = \(c\times \sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \)
σ = \(10\times \sqrt { \frac { 210 }{ 71 } -\left( \frac { 30 }{ 71 } \right) ^{ 2 } } =10\times \sqrt { \frac { 210 }{ 71 } -\frac { 900 }{ 5041 } } \)
= 10 x \(\sqrt { 2.779 } \); σ ≃ 16.67

1² + 2² + 3² + ... + n² = 2025
\({\left[\frac{n(n+1)}{2}\right]^{2}=2025=(45)^{2}} \)
\(\Rightarrow \frac{n(n+1)}{2}=45 \)
n² + n = 90
n² + n - 90 = 0
(n - 9) (n + 10) = 0
n = 9, - 10
n = -10 is not possible
n = 9

Given points A(- 4, 2) and B( 6, - 4)
Mid point of AB \(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \)
\(=\left(\frac{-4+6}{2}, \frac{2-4}{2}\right)=(1,-1) \)
slope of AB = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{2+4}{-4-6}=\frac{6}{-10}=\frac{-3}{5}\)
Slope of perpendicular line = \(\frac{5}{3}\)
Perpendicular bisector is passing through (1, -1) and having slope \(\frac{5}{3}\)
The equation of perpendicular bisector is
y - y₁ = m(x - x₁)
y + 1 = \(\frac{5}{3}(x-1)\)
3y + 3 = 5x - 5
5x - 3y - 8 = 0

Volume of a cone = 1005 \(\frac{5}{7}\) cu.cm
\(\text { i.e., } \frac{1}{3} \pi r^{2} h=1005 \frac{5}{7}\)
area of base = area of circle
\( =201 \frac{1}{7} \text { sq. units } \)
\(i.e \ \pi r^{2}=201 \frac{1}{7} \Rightarrow r^{2}=64 \)
Substituting in (1), r = 8 cm
\(\frac{1}{3}\left(201 \frac{1}{7}\right) \mathrm{h}=1005 \frac{5}{7} \)
\(\frac{1}{3}\left(\frac{1408}{7}\right) \mathrm{h}=\frac{7040}{7} \)
\(h=\frac{7040}{7} \times \frac{7}{1408} \times 3=15 \mathrm{~cm}\)
Slant height of cone \(l =\sqrt{h^{2}+r^{2}} \)
\(=\sqrt{15^{2}+8^{2}}=\sqrt{225+64} \)
\(=\sqrt{289}=17 \mathrm{~cm} \)

Slant height l = 4 m
Perimeter of larger circle = 2nR = 18 cm
\(R=\frac{9}{\pi}=\frac{63}{22} \mathrm{~m}\)
Perimeter of smaller circle = \(2 \pi r\) = 16 m
\(r=\frac{8}{\pi}=\frac{56}{22} \mathrm{~m}\)
C.S.A of Frustum of cone = \(\pi l(\mathrm{R}+r) \text { sq. units }\)
\(=\frac{22}{7} \times 4 \times\left(\frac{63}{22}+\frac{56}{22}\right) \)
\(=\frac{4}{7}(119)=68 \mathrm{~m}^{2} \)
Cost of painting per sq. m = Rs. 100
Cost of painting for 68 sq. m = 68 x 100
= Rs.6800

Hollow metallic cylinder
External radius R = 4.3 cm
Internal radius r = 1.1 cm
Length = height = h = 4 cm
Volume \(=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right) \mathrm{h} \text { cu. units } \)
\(=\pi\left((4.3)^{2}-(1.1)^{2}\right)(4) \)
\(=\pi(18.49-1.21) 4 \)
\(=69.12 \pi \mathrm{cm} 3 \)
Solid cylinder
height h = 12 cm
radius r = ?
volume \(=\pi r^{2} h\ sq. units \)
\(=\pi r^{2}(12) \)
Given, Hollow cylinder is melted to form solid cylinder
Volume of cylinder = Volume of hollow cylinder
\(\pi r^{2}(12) =69.12 \pi \)
\(r^{2} =\frac{69.12}{12}=5.76 \)
r = 2.4 cm
Diameter of cylinder  = 2r = 4.8 cm

Let AB be the height of the kite above the ground. Then, AB = 75.
Let AC be the length of the string.
In right triangle ABC,\(\angle \)ACB = \(60°\)
\(sin\theta =\frac { AB }{ AC } \)
\(sin60°=\frac { 75 }{ AC } \)
gives \(\frac { \sqrt { 3 } }{ 2 } =\frac { 75 }{ AC } \) so, AC = \(\frac { 150 }{ \sqrt { 3 } } =50\sqrt { 3 } \) 
Hence, the length of the string is 50\(\sqrt { 3 } m\)

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

A = {1, 2),B = {1, 2, 3, 4}, C = {5, 6},D = {5, 6, 7, 8}
A x C = {1, 2} x {5, 6}
\(\mathrm{A} \times \mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\)
B x D = { 1, 2, 3, 4} x { 5, 6, 7, 8}
\(\begin{array}{r} \mathrm{B} \times \mathrm{D}=\{({1,5}),(1,6),(1,7),(1,8) ({2,5}),(2,6),(2,7),(2,8) (3,5),(3,6),(3,7),(3,8) (4,5),(4,6),(4,7),(4,8)\} \end{array}\)
(A x C) is a subset of (B x D)

f(x) = x², g(x) = 3x, h(x) = x - 2
f o g = f [g(x)] = f (3x)
= (3x)² - 9x²
(f o g) o h = (f o g) [h (x)] = (f  o g) [x - 2]
= 9 ( x - 2)²
g o h = g [h (x)] = g[x - 2] = 3 (x - 2)
f o (g o h) = f [g (h(x))]
= f [3(x - 2)] = [3(x - 2))]² = 9 (x - 2)²
(f o g) o h = f o( g o h)
Hence proved.

Let TR = y. Since, OT is perpendicular bisector of PQ
PR = QR = 4 cm
In\(\triangle\)ORP, OP² = OR² + PR²
OR² = OP² - PR²
OR² = 5² - 4² = 25 - 16 = 9 \(\Rightarrow\) OR = 3cm
OT = OR + RT = 3 + y          ..(1)
In \(\triangle\)PRT, TP² + TR² + PR²      ..(2)
and \(\triangle\)OPT we have, OT² = TP² + OP²
OT² = (TR² + PR²) + OP² (substitute for TP² from (2))
(3 + y)² = y² + 4² + 5² (substitute for OT from (1))
9 + 6y² + 16 + 25 therefore \(y=TR=\frac { 16 }{ 3 } \)
6y = 41 - 9 we get  \(y=\frac { 16 }{ 3 } \)
From (2), TP² = TR² + PR²
\(TP2=\left( \frac { 16 }{ 3 } \right) +4^{ 2 }=\frac { 256 }{ 9 } +16=\frac { 400 }{ 9 } \)so, \(TP=\frac { 20 }{ 3 } \)

f(x) = \(\sqrt { 1+\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } } } \)
\( f(x)=\sqrt{1-t} \)
\(where\ t=\sqrt{1-\sqrt{1-x^{2}}}\)
\(1-t \geq 0 \)
\(t \leq 1 \)
\(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
Squaring \(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
\(-\sqrt{1-x^{2}} \leq 0 \)
\(\sqrt{1-x^{2}} \geq 0 \)
\(1-x^{2} \geq 0 \)
\(x^{2} \leq 1 \)
= x [-1, 1] i.e., {- 1, 0, 1}

Total number of steps n = 30.
Bottom step requires a = 100 bricks
Each successive steps requires 2 less than the previous step.
Number of bricks in each step form an A.P. 100, 98, 96, ....upto 30 terms
a = 100,d = -2,n = 30
(i) Number of bricks required for the topmost step is l
l = -a + (n - 1)d
= 100 + (30 - 1)(-2)
= 100 + 29 (-2)
= 100 - 58 = 42.
Topmost step requires 42 bricks
(ii) Number of bricks required to build the staircase,
\(\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(l+a)=\frac{30}{2}(100+42)\) 
= 15 x 142 = 2130
Total number of bricks required = 2130

\(\left\{\begin{array}{l} \text { Sum of natural numbers between } \\ 602 \text { and } 902 \text { not divisible by } 4 \end{array}\right\}=\) \(\left\{\begin{array}{l} \text { Sum of all natural } \\ \text { numbers between } \\ 602 \text { and } 902 \end{array}\right\}-\left\{\begin{array}{l} \text { Sum of all natural } \\ \text { numbers between } \\ 602 \text { and } 902 \text { divisible } \\ \text { by } 4 \end{array}\right\}\)
\(\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 a+(n-1) d]\)
Sum of Natural numbers upto \(\mathrm{n}=\frac{n(n+1)}{2}\)
603 + 604 +..... + 901 = (1 + 2+ ... + 901) - (1 + 2 + 3 +...+ 602)
\(=\frac{901 \times(901+1)}{2}-\frac{602 \times(602+1)}{2} \)
\(=\frac{901 \times 902}{2}-\frac{602 \times 603}{2} \)
= 4,06,351 - 1,81,503
= 2,24,848
Again numbers divisible by 4 between 602 and 902 are 604 + 608 + .... + 900
\(\mathrm{n} =\left(\frac{l-a}{d}\right)+1=\left(\frac{900-604}{4}\right)+1 \)
\(=\left(\frac{296}{4}\right)+1=74+1=75 \)
\(\mathrm{~S}_{\mathrm{n}} =\frac{75}{2}[2(604)+(75-1)(4)] \)
\(=\frac{75}{2} \times 2[604+(74 \times 2)] \)
= 75 x 752 = 56400
Required sum =. 224848 - 56400
= 1,68,448

Let Sarah's house is at A and James's house is at 'B' from the picture.
Distance between Sarah's house to James house through Street B and C
= 1.5 miles + 2 miles = 3.5 miles
Distance through street C is AC² = AB² + BC²

AC² = = (1.5)² + (2),
= 2.25 + 4
= 6.25
\(A C=\sqrt{6.25}=2.5\)
AC = 2.5 miles
Difference between two paths = 3.5 - 2.5 = 1 mile
Direct path along C street is 1 mile shorter

Let r, R and h be the internal radius, external radius and height of the hollow cylinder respectively.
Given that, r = 21cm, R = 28 cm, h = 9 cm
Now, volume of hollow cylinder = \(\pi\)(R² − r²)h cu. units
\(=\frac { 22 }{ 7 } \left( { 28 }^{ 2 }-21^{ 2 } \right) \times 9\)
\(=\frac { 22 }{ 7 } (784-441)\times 9=9702\)
Therefore, volume of iron used = 9702 cm³

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.

Given points A (2, 2), B (- 2,- 3), C (1, - 3) and D (x, y)

Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \)
Slope of AB = \(\frac{2+3}{2+2}=\frac{5}{4} \)
Slope of BC = \(\frac{-3+3}{-2-1}=0 \)
Slope of CD = \(-\frac{-3-y}{1-x} \)
Slope of AD = \(\frac{2-y}{2-x} \)
Since, the points form a parallelogram
AB is parallel to CD and BC is parallel to AD
Slope of AB = Slope of CD
\(\frac{5}{4}=\frac{-3-y}{1-x}\)
5(1 - x) = 4(-3 -y)
5 - 5x = -12 -4y
5x - 4y = 17
Slope of BC = Slope of AD
\(0=\frac{2-y}{2-x}\)
2 - y = 0
y = 2
Substituting in (1)
5x - 4(2) = 17
5x = 17 + 8 = 25
\(x=\frac{25}{5}=5\)
x = 5, y = 2

From the figure
r = 6 cm
R = 12 cm
h = 8 cm
\(l =\sqrt{h^{2}+(\mathrm{R}-\mathrm{r})^{2}} \)
\(=\sqrt{8^{2}+(12-6)^{2}} \)
\(=\sqrt{64+36} \)
\(=\sqrt{100}=10 \mathrm{~cm} \)
Area to be painted = C.S.A + area of top circular region
\(=\pi(R+r) l+\pi r^{2} \)
\(=\frac{22}{7}(12+6)(10)+\frac{22}{7}(6)^{2} \)
\(=\frac{22}{7}(180)+\frac{22}{7}(36) \)
\(=\frac{22}{7}(180+36) \)
\(=\frac{22}{7}(216)=\frac{4752}{7}=678.86 \)
Cost of painting per sq. cm = Rs. 2
Total cost = 678.86 x 2 = Rs. 1357.72

Let h, l, R and r be the height, slant height, outer radius and inner radius of the frustum.
Given that, diameter of the top = 10 m; radius of the top R = 5 m.
diameter of the bottom = 4 m; radius of the bottom r = 2 m, height h = 4 m
Now, \(l=\sqrt { { h }^{ 2 }+\left( R-{ r } \right) ^{ 2 } } \)
\(=\sqrt { { 4 }^{ 2 }+(5-2)^{ 2 } } \)
\(l=\sqrt { 16+9 } =\sqrt { 25 } =5m\)
Here, C.S.A. = \(\pi\)(R + r)l sq. units
\(\frac { 22 }{ 7 } (5+2)\times 5={ 110m }^{ 2 }\)
T.S.A. = \(\pi\)(R + r)l + \(\pi\)R² + \(\pi\)r² sq. units
\(\frac { 22 }{ 7 } \left[ (5+2)5+25+4 \right] =\frac { 1408 }{ 7 } =201.14\)
Therefore, C.S.A. = 110 m² and T.S.A. = 201.14 m²

In ABC , we have DE || BC
\(\frac{A B}{A D}=\frac{A C}{A E}\) [By Thales Theorem]        ..(1)
In ADC, we have
\(\frac{A D}{A F}=\frac{A C}{A E}\) [By Thales Theorem]        .....(2)
From (1) and (2) we get
\(\frac{A B}{A D}=\frac{A D}{A F}\)
AD² = AB x AF

Given: ABCD is a trapezium in which DC || AB and EF || AB
To prove that \(\frac { AE }{ ED } =\frac { BF }{ FC } \)
Construction : join AC meeting EF at G
Proof:
In ADC, we have
EG || DC
\(\Rightarrow \frac{A E}{E D}=\frac{A G}{G C}\) [By Thales theorem]               ...(1)
In ABC , we have
\(\frac{A G}{G C}=\frac{B F}{F C}\) [By Thales theorem]        ....(2)
From (1) and (2), we get
\(\frac{A E}{E D}=\frac{B F}{F C}\)

Construction
Step 1 : Draw a line segment QR = 5 cm.
Step 2 : At Q draw QE such that \(\angle\)RQE = 30^o.
Step 3 : At Q draw QF such that \(\angle EQF\) = 90^o
Step 4 : Draw the perpendicular bisector XY to QR which intersects QF at O and QR at G.
Step 5 : With O as centre and OQ as radius draw a circle.
Step 6: From G mark an arc in the line XY at M, such that GM = 42. cm.
Step 7 : Draw AB through M which is parallel to QR.
Step 8 : AB meets the circle at P and S.
Step 9 : Join QP and RP. Then\(\triangle\)PQR is the required triangle

The function f is defined by three values in intervals I, II, III as shown by the side.
For a given value of x = a, find out the interval at which the point a is located, there after find
f(a) using the particular value defined in that interval.
(i) First, we see that, x = 4 lie in the third interval.
Therefore, f(x) = 3x - 2; f(4) = 3(4) = 10
(ii) x = -2 lies in the second interval
Therefore, f(x) = x² - 2; f(-2) = (-2)² - 2 = 2
(iii) From (i), f(4) =10.
To find f(1) first we see that x = 1 lies in the second interval.
Therefore, f(x) = x²-2 ⇒ f(1) = 1² - 2 = -1
So, f(4) + 2f(1) = 10 + 2(-1) = 8
(iv) We know that f(1) = -1 and f(4) = 10
For finding f(-3), we see that x = −3, lies in the first interval.
Therefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1
Hence, \(\frac { f(1)-3f(4) }{ f(-3) } =\frac { -2-3(10) }{ 1 } \) = - 31

Given cosec\(\theta \) + cot\(\theta \) = p   ...(1)
cosec²\(\theta \) - cot²\(\theta \) = 1 (identity)
\(\operatorname{cosec} \theta-\cot \theta=\frac{1}{\operatorname{cosec} \theta+\cot \theta}\)
cosec\(\theta \) - cot\(\theta \) =\(\frac { 1 }{ { p } } \)  .... (2)
Adding(1) and (2) we get, 2cosec\(\theta \) = \(p+\frac { 1 }{ p } \) 
2cosec\(\theta \)\(\frac { { p }^{ 2 }+1 }{ p } \) ....(3) 
Subtracting (2) from (1), we get, 2cot\(\theta \) = \(p-\frac { 1 }{ p } \) 
2cot\(\theta \) = \(\frac { { p }^{ 2 }-1 }{ p } \)   ...(4)
Dividing (4) by (3) we get,\(\frac { 2cot\theta }{ 2cosec\theta } =\frac { { p }^{ 2 }-1 }{ p } \times \frac { p }{ { p }^{ 2 }+1 } gives,cos\theta =\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

Let the 100t digit be 'x', 10's be 'y' and Unit digit be 'z'.
The three digit number is 100x + 10y + z.
Now given, x + y + z = 11 .       ..(1)
100z +10y + x = 5 (100x + 10y + z) + 46
Simplifying
499x + 40y - 95z = - 46          .........(2)
x + 2y = z
x + 2y - z = 0    ....(3)

Substituting the value x = 1 in(4)
2(1) + 3y = 11
3y = 11 - 2 = 9
\(y=\frac{9}{3}=3\)
Substituting x = 1, y = 3 in (1)
1 + 3 + z = 11
z = 11 - 4 = 7
x = 1, y = 3, z = 7
The original three digit number is 137
i.e., 100(1) + 10(3) + 1(7) = 100+ 30 + 7 =137

x + 2y - z = 6         ...(1)
-3x - 2y + 5z = -12     ....(2)
x-  2z = 3          ....(3)

The system of equations has infinitely many solutions

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47b₁ + 54.10 = 2.47b₂ + 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.
2. 47 = 147. 96 - 54. 10 = 93. 86
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cm.

Let the three numbers be x, y, z
From the given data we get the following equations,
3x + y + 2z = 5 .....(1)   
x + 3z - 3y = 2  .....(2)   
2x + 3y - z = 1  .....(3)

Substituting y = 2 in (5), -14 + 7x = 7 gives, z = 3
Substituting y = 2 and z = 3 in (1),  3x + 2 + 6 = 5 we get x = -1
Therefore, x = –1, y = 2, z = 3.

Now,cos\(\theta \) + sin\(\theta \) =\(\sqrt { 2 } \) cos\(\theta \) 
Squaring both sides,
(cos\(\theta \) + sin\(\theta \) )² =(\(\sqrt { 2 } \) cos\(\theta \) )²
cos² \(\theta \) + sin² \(\theta \) + 2sin\(\theta \) cos\(\theta \)  = 2cos²\(\theta \) 
2cos²\(\theta \) - cos²\(\theta \) - sin²\(\theta \)  = 2sin\(\theta \) cos\(\theta \) 
cos²\(\theta \)  - sin²\(\theta \)  = 2sin\(\theta \) cos\(\theta \) 
(cos\(\theta \) + sin\(\theta \) ) (cos\(\theta \) + sin\(\theta \) ) = 2sin\(\theta \) cos\(\theta \) 
cos\(\theta \) - sin\(\theta \)  = \(\frac { 2sin\theta cos\theta }{ cos\theta +sin\theta } \) =\(\frac { 2sin\theta cos\theta }{ \sqrt { 2 } cos\theta } \) [since cos\(\theta \)  + sin\(\theta \) =\(\sqrt { 2 } \) cos\(\theta \) ]
=\(\sqrt { 2 } \) cos\(\theta \) 
Therefore cos\(\theta \) - sin\(\theta \)  =\(\sqrt { 2 } \) cos\(\theta \)

First take 2⁵ ≡ 15 (mod 17)
(25)² ≡ 15² (mod 17)
≡ 4 (mod 17)
2¹⁰ ≡ 4 (mod 17)
(2¹⁰)⁴ ≡ 4⁴ (mod 17)  
2⁴⁰ ≡ 1 mod 17
(2⁴⁰)² ≡ 1² mod 17
2.2⁸⁰ ≡ 2 x 1 (mod 17)
2⁸¹ ≡ 2 mod 17
Remainder when 2⁸¹ is divided by 17 is 2.

Let Arul's speed of working be x
Let Mohan's speed of working be y
Let Ram's speed of working be z
given that they are working together.
Let 'w' be the quantum of work.
Also given that Mohan takes twice the time as Arul for finishing the work.
\(\therefore \frac { w }{ y } =2\times \frac { w }{ x } \therefore x=2y\)
\(\therefore y=\frac { x }{ 2 } \)   (2)
Also Ram takes 3 times the time as Arul for finishing the work.
\(\therefore \frac { w }{ z } =3\times \frac { w }{ x } \)
\(\therefore x=3z\quad \therefore z=\frac { x }{ 3 } \)
Substitute (2) and (3) in (1),
\(x+\frac { x }{ 2 } +\frac { x }{ 3 } =\frac { w }{ 6 } \)
∴ 6x + 3x + 2x = w
11x = w
\(x=\frac { w }{ 11 } ,y=\frac { w }{ 22 } ,z=\frac { w }{ 33 } \)
Working alone time taken as
\(Arul:\frac { w }{ x } =\frac { w }{ w/11 } =11hrs.\)
\(Mohan:\frac { w }{ y } =\frac { w }{ w/22 } =22hrs.\)
\(Ram:\frac { w }{ z } =\frac { w }{ w/33 } =33hrs\)

Let the 
100's digit be 'x'
10's digit be 'y'
Unit's digit be 'z'
Given 100y+ 10x + z - 54 = 3(100x + 10y + z)
Substituting 290x - 70y + 22 = -54 ( /2)
145x - 35y + z = -27          .........(1)
l00x + 10y + z + 198 = 100z + 10y + x
Substituting 99x - 992 = - 198 (99)
x - z = - 2              ...........(2)
y = x + 2 (y - z)
x + y - 2z = 0          ..........(3)
Consider (1) and (3)
145x - 35y + z = -27        .......(1)
35x + 35y - 70z = 0          .......(4)
180x - 69z = -27            .......(5)
Consider (5) and (2)

\(x=\frac{111}{111}=1\)
Substituting x = 1 in         .........(2)
1 - z = -z
z = 1 + 2 = 3
Substituting x - 1, z = 3 in (3)
1 + y - 6 = 0
y = 5
solution: x = 1, y = 5, z = 3
The number is 153.

Let the number of students in sections A, B and C be 'x', 'y' and 'z' respectively
Given x + y + z = 150           ........(1)
x - 6 = z + 6
x - z = 12            .....(2)
4z = x + y
x + y - 4z = 0         ..........(3)
consider (1) and (3)

\(z=\frac{150}{5}=30\)
substituting in (2)
x - 30 = 12
x = 30 + 12
x = 42
substituting x = 42, z = 30 in (1)
42 + y + 30 = 150
y = 150 - 72
y = 78
The number of students in sections A, B and C are 42, 78, 30 respectively.

Given: A = {1,2,3} , B = {2,3,5} , C = {3,4} ,D = {1,3,5}
\(A\cap C\) = {3}
\(B\cap D\) = {3,5}
\((A\cap C)\times(B\cap D)=\{ 3\} \times \{ 3,5\} \)
= {(3,3),(3,5)}          ..(1)
A x B = {(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)}
C x D = {(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)}
\((A\times B)\cap (C\times D)\)  = {(3,3),(3,5)}                ....(2)
From (1) and (2),it is clear that
(A ∩ C) x (B ∩ D) = (A x B) ∩ (C x D)
Hence it is true.

A =  {x \(\in \) N| 1 < x < 4} = {2,3), B = {x \(\in \) W| 0 ≤ x < 2) = (0,1), C = {x \(\in \) N| x < 3} = (1,2)
(i)  A x (B U C) = (A x B) U (A x C)
B U C = (0,1) U (1,2) = {0,1,2}
A x (B U C) = {2,3) x {0,1,2} = {(2,0),(2,1)(2,2)(3,0)(3,1),(3,2)         ..(1)
A x B = {2,3} x {0,1} = {(2,0),(2,1),(3,0),(3,1)}
A x C = {2,3} x {1,2} = {(2,1),(2,2),(3,1)(3,2)}
(A x B) U (A x C) = {(2,0),(2,1),(3,0),(3,1)} U {(2,1),(2,2),(3,1),(3,2)}
= {(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)}                ...(2)
From (1) and (2), A x (B U C) = (A x B) U  (A x C) is verified.
(ii)  A x (B ∩ C) = (A x B) ∩ (A x C)
(B ∩ C) = {0,1} ∩ {1,2} =  {1}
A x (B ∩ C) = {2,3} x {1} = {(2,1),(3,1)}     .... (3)
A x B = {2,3} x {0,1} = {(2,0),(2,1),(3,0),(3,1)}
A x C = {2,3} x {1,2} = {(2,1),(2,2),(3,1),(3,2)
(A x B) ∩ (A x C) = {(2,0),(2,1),(3,0),(3,1)} ∩ {(2,1),(2,2),(3,1),(3,2)}
= {(2,1),(3,1)}                .... (4)
From (3) and (4), A x (B ∩ C) = (A x B) ∩ (A x C) is verified.

 Given a triangle \(\triangle\)PQR. We have to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the given \(\triangle\)PQR.

Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex P.
3. Joined Q₃ to R and drawn a line through Q₇ parallel to Q₃R, intersecting the extended line segment QR at R'
4. Drawn a line through R' parallel to RP intersecting the extended line segment QP at P.
5. Then PQR' is the required triangle each of whose sides is seven-thirds of the corresponding sides of PQR.

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 7 }{ 4 } \) of the corresponding sides of the triangle PQR.
Steps of construction
1. Construct a DPQR with any measurement.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 4 in \(\frac { 7 }{ 4 } \))
Q_1,Q_2,Q_3,Q_4,Q_5,Q₆ and Q₇ on QX so that 
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇
4. Join Q₄ (the 4th point, 4 being smaller of 4 and 7 in \(\frac { 7 }{ 4 } \)) to R and draw a line through Q₇ parallel to Q₄R, intersecting the extended line segment QR at R'.
5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'.
Then \(\triangle\)P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of \(\triangle\)PQR.

P(A) = 0.5
P(B) =  0.3
Since A and B are mutually exclusive events
\(P(A \cap B)=0\)
\( \mathrm{P}(\text { either } \mathrm{A} \text { or } \mathrm{B}) =\mathrm{P}(A \cup B) \)
\(\mathrm{P}(A \cup B) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cup B) =0.5+0.3-0=0.8\)
P(neither A nor B) = \(\mathrm{P}(\overline{A \cup B})\)
\(\mathrm{P} \overline{(A \cup B)}=1-\mathrm{P}(A \cup B)=1-0.8=0.2\) 
Probability of neither A nor B happen = 0.2

1³ + 2³ + 3³ +...K³ = \(\left[\frac{k(k+1)}{2}\right]^{2}=44100=(210)^{2}\)
1 + 2 + 3 +...+ k = \(\frac{k(k+1)}{2}=210\)
1 + 2 + 3 +...+ k = 210

Sum of first k natural numbers  = \(\frac{k(k+1)}{2}=325\)
Sum of cube of first k natural numbers
\(=\left[\frac{k(k+1)}{2}\right]^{2} \)
\(=(325)^{2}=1,05,625 \)
\(1^{3}+2^{3}+3^{3}+\ldots+k^{3}=1,05,625 \)

Let r₁, r₂ be the radii of two spheres
\(\text { Given } \frac{r_{1}}{r_{2}}=\frac{4}{7} \Rightarrow \mathrm{r}_{1}=\frac{4 r_{2}}{7}\)
Ratio of the volumes = \(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}\)
\(=\frac{\left(\frac{4 r_{2}}{7}\right)^{3}}{r_{2}^{3}}=\frac{4^{3}}{7^{3}}\)
Ratio of volumes \(V_{1}: V_{2}=64: 343=\frac{64}{343}\)

Let x be the length of the ladder. BC = 4 ft, AC = 7 fit.
By Pythagoras theorem we have, AB² = AC² + BC²
x² = 7² + 4² gives x² = 49 + 16
x² = 65, Hence \(x=\sqrt { 65 } \)
The number \(\sqrt { 65 } \) is between 8 and 8.1.
82 = 64 < 65.61 = 8.1²
Therefore, the length of the ladder is approximately 8.1ft