Sample space S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number greater than 4
E = {5, 6}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

(i) We know that cosec A =  sec(90⁰ A)
sec(90⁰ - A) sec(34⁰)
90⁰ - A =34⁰
We get A = 90° − 34°
A = 56⁰
(ii) We know that tan B = cot(90⁰ - B)
cot(90⁰ - B) = cot 47⁰
90⁰ - B = 47⁰
We get B = 90° − 47°
B = 43⁰

For clarity, let us draw a rough figure indicating the measurements:

(i) The semi-perimeter of Fig.1, s \(\frac{30+40+50}{2}\)= 60 m
Fig. 2, s = \(\frac{35+40+45}{2}\) = 60 m
Fig. 3, s = \(\frac{40+40+40}{2}\) = 60 m
Note that all the semi-perimeters are equal.
(ii) Area of triangle using Heron’s formula:
In fig. 1, Area of triangle = \(\sqrt { 60(60-30)(60-40)(60-50) } \)
= \(\sqrt { 60\times 30​​\times 20\times 10 } \)
= \(\sqrt { 30\times 2​​\times 30\times 2\times 10\times 10 } \)
= 600 m²
In fig. 2, Area of triangle = \(\sqrt { 60(60-35)(60-40)(60-45) } \)
= \(\sqrt { 60\times 25\times 20\times 15 } \)
= \(\sqrt { 20\times 3​​\times 5\times 5\times 20\times 3 \times 5 } \)
= 300\(\sqrt{5}\) (since \(\sqrt{5}\) = 2.236)
= 670.8 m²
In fig. 3, Area of triangle = \(\sqrt { 60(60-40)(60-40)(60-40) } \)
= \(\sqrt { 60\times 22\times 20\times 20 } \)
= \(\sqrt { 3​​\times 20\times 20\times 20\times 20 } \)
= 400\(\sqrt{3}\) (since \(\sqrt{3}\) = 1.732)
= 692 .8 m²
We find that though the perimeters are same, the areas of the three triangular plots are different. The area of triangle in fig. 3 is the greatest among these; the buyer can be suggested to choose this since it is more spacious.

Let P, Q, R be the points on the line segment joining A(−11, 4) and B(9, 8) such that AP = PQ = QR = RB .
Here Q is the mid-point of AB, P is the mid-point of AQ and R is the mid-point of QB.
Q is the mid-point of AB =\(\left( \frac { -11+9 }{ 2 } ,\frac { 4+8 }{ 2 } \right) =\left( \frac { -2 }{ 2 } ,\frac { 12 }{ 2 } \right) \)= (-1, 6)
P is the mid-point of AQ =\(\left( \frac { -11-1 }{ 2 } ,\frac { 4+6 }{ 2 } \right) =\left( \frac { -12 }{ 2 } ,\frac { 10 }{ 2 } \right) \) = (-6, 5)
R is the mid-point of QB =\(\left( \frac { -1+9 }{ 2 } ,\frac { 6+8 }{ 2 } \right) =\left( \frac { 8 }{ 2 } ,\frac { 14 }{ 2 } \right) \) = (4, 7)
Hence the points which divides AB into four equal parts are P(–6, 5), Q(–1, 6) and R(4, 7).

We know that,
(a + b + c)²= a² +b² +c² + 2ab + 2bc + 2ca
Substituting a = x, b = 2y and c = 3z
(x +2y +3z)² = x² + (2y)² + (3z)² + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x² + 4y² + 9z² + 4xy + 12y² + 6zx

\(\sqrt { 44 } +\sqrt { 99 } -\sqrt { 275 } \) = \(\sqrt { 4\times 11 } +\sqrt { 9\times 11 } -\sqrt { 11\times 25 } \)
= \(\left( 2\sqrt { 11 } +3\sqrt { 11 } \right) -5\sqrt { 11 } =5\sqrt { 11 } -5\sqrt { 11 } =0\)

A = {x : x ∈ Z, -2 < x ≤ 4} = {-1,0,1,2,3,4}
B = {x : x ∈ W, x ≤ 5} = {0,1,2,3,4,5}
C = {-4,-1,0,2,3,4}
AU(B⋂C)
B⋂C = {0,1,2,3,4,5} ∩ {-4,-1,0,2,3, 4} = {0,2, 3, 4}
AU(B∩C)
B⋂C = {0,1,2,3,4,5} ∩ {-4,-1,0,2,3,4} = {0,2,3,4}
AU(B⋂C) = {-1,0,1,2,3,4} U (0,2,3,4}
= {-1,0,1,2,3,4}........(1)
(A∩B)U(A∩C)
A⋂B = {0,1,2,3,4}
A⋂C = {-1,0,2,3,4}
(A⋂B)⋃(A⋂C) = {0,1,2,3,4} U {-1,0,2,3,4}
= {-1,0,1,2,3,4}..........(2)
From (1) and (2), it is verified that
AU(B∩C) = (AUB)∩(A⋃C)

\(\bar { x } =20.2\) 
\( \bar{x}=\cfrac{\Sigma fx}{\Sigma f}\)
\(=\cfrac { 10\times 6+15\times 8+20p+25\times 10+30\times 6 }{ 6+8+p+10+6 } \)
\(20.2=\cfrac { 60+120+20p+250+180 }{ 30+p } \)
(30+p)20.2 = 610+20 p
606+20.2p = 610+20 p
20.2 p-20 p = 610-606 = 4
0.2 p = 4
\(\Rightarrow \) \(p=\cfrac { 4\times 10 }{ 0.2\times 10 } =\cfrac { 40 }{ 2 } =20\)

(i) \( (300000)^{2} \times(20000)^{4}\)
\(=\left(3.0 \times 10^{5}\right)^{2} \times\left(2.0 \times 10^{4}\right)^{4} \)
\(=3^{2} \times 10^{10} \times 2^{4} \times 10^{16} \)
\(=9 \times 16 \times 10^{10+16} \)
\(=144 \times 10^{26} \)
\(=1.44 \times 10^{28} \)
(ii) (0.000001)¹¹ ÷ (0.005)³
\( =\left(1.0 \times 10^{-6}\right)^{11} \div\left(5.0 \times 10^{-3}\right)^{3} \)
\( =\frac{1.0 \times 10^{-66}}{125.0 \times 10^{9}} \)
\( =\frac{1000.0 \times 10^{-69}}{125.0 \times 10^{-9}} \)
\( =8.0 \times 10^{-69} \times 10^{9} \)
\( =8.0 \times 10^{-60} \)
(iii) \(\left\{(0.00003)^{6} \times(0.00005)^{4}\right\} \div\left\{(0.009)^{3} \times(0.05)^{2}\right\}\)
=\(\frac{(3.0\times10^{-5})^6\times(5.0\times10^{-5})^4}{(9.0\times10^{-3})^3\times(5.0\times10^{-2})^2}\)
=\(\frac{3^6\times10^{-30}\times5^4\times10^{20}}{9^3\times10^{-9}\times5^{2}\times10^{-4}}=\frac{3^6\times5^4\times10^{-30-20}}{(3^{2})^3\times10^{-9-4}\times5^{2}}=\frac{ ̶3̶^6̶\times5^4\times10^{-50}}{ ̶3̶^6̶\times5^2\times10^{-13}}\)
\( =5^{4-2} \times 10^{-50+13} \)
\(=5^{2} \times 10^{-37} \)
\(=25 \times 10^{-37} \)
\(=2.5 \times 10^{1} \times 10^{-37}\)
\(=2.5 \times 10^{-36} \)

Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M U S) = 40, n (M) = 26, n (S) = 24
n (M U S) = n (M) + n (S) - n (M n S)
40 = 26 + 24 - n (M n S)
n (M n S) = 26 + 24 - 40
= 50 - 40 = 10
∴ Number of students opted for Mathematics and Science = 10

Let x = 17.2151515 ......  (1)
Here period of decimal is 3, multiply equation (1) by 1000
1000 x = 17215.151515
(2) - (1)
999x = 17197.936
\( x=\frac{17197.936}{999} \\ x=\frac{17197936}{999000} \\ x=\frac{5681}{330} \)

h(z) - f(z) = z⁵ - 6z⁴ + z - (6z² + 10z -7)
= z⁵ -6z⁴ + z - 6z² - 10z + 7
= z⁵ -6z⁴ -6z² + z - 10z + 7
= z⁵– 6z⁴– 6z²– 9z+ 7
The degree of the polynomial is 5

Since B is on the x-axis, the y-coordinate of B is 0.
So, the coordinates of the point B is (11, 0)
By the distance formula the distance between the points A (7, 3), B (11, 0) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(AB=\sqrt { \left( 11-7 \right) ^{ 2 }+\left( 0-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( 4 \right) ^{ 2 }+\left( -3 \right) ^{ 2 } } =\ \sqrt { 16+9 } =\sqrt { 25 } =5\)​​​​

Let p (x) 2x³ - x² - 12x - 9
Sum of the co-efficients = 2 - 1- 12- 9 = -20 \(\neq \) 0
Hence x-1 is not a factor
Sum of co-efficients of even powers with constant = -1 - 9 = -10
Sum of co-efficients of odd powers = 2 - 12= -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.

Then p (x) = (x + 1)(2x² - 3x - 9)
Now 2x² - 3x - 9 = 2x² - 6x + 3x - 9 = 2x (x - 3) + 3 (x - 3)
= (x - 3)(2x + 3)
Hence 2x³ - x² - 12x - 9 (x + 1) (x - 3) (2x + 3)

(i) x²+10x + 24 = x²+ 6x + 4x + 24
= x(x + 6) + 4(x + 6)
= (x + 6)(x + 4)

(ii) z²+ 4z -12
= z²+ 6z - 2z -12
= z(z + 6) -2(z + 6)
= (z + 6)(z - 2)

(iii) p²- 6p -16 = p²- 8p + 2p -16
= p(p - 8)+2(p - 8)
= (p - 8)(p + 2)

(iv) t²+ 72 -17t = t²-17t + 72
= t²- 9t - 8t + 72
= t(t - 9) - 8 (t - 9)
= (t - 9)(t - 8)

(v) y² -16y- 80 = y² - 20y + 4y - 80
= y(y - 20) + 4 (y - 20)
= (y - 20) (y + 4)

(vi) a² + 10a - 600
= a² + 30a - 20a - 600
= a(a + 30)- 20(a + 30)
= (a + 30)(a - 20)
​​​​​

We have a parallelogram ABCD in which diagonals AC bisect ㄥA.
ㄥDAC = ㄥBAC
(i) To prove that AC bisects LC
∵ ABCD is a parallelogram
∴ AB II DC and AC is a transversal
∴ ㄥ1 = ㄥ3 (Alternate interior angle)     .........(1)
.Also BC II AD and AC is a transversal.
∴  ㄥ2 = ㄥ4 (Alternate interior angle)  ...........(2)
But AC bisects ㄥA
∴ ㄥ1 = ㄥ2
From (1), (2) and (3) we get
ㄥ3 = ㄥ4
∴ AC bisects ㄥC.
(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ㄥ1 = ㄥ4 [ ∵ ㄥ1 = ㄥ2 = ㄥ4]
∴ BC = AB (side opposite to equal angles are equal) (4)
Similarly AD = DC            ....... (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram)        .......(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

In the given \(\triangle ABC\)
\(\angle A=64^0\ and\ \angle B=58^0\)
\(\angle C=180^0-(64^0+58^0)\)
= 180° - 122°
= 58°
Since OC is the bisector of \(\angle C\)
\(y=\frac{58^0}{2}=29^0\)
Given \(\triangle OBC\)
\(\angle OBC=\frac{58^0}{2}=29^0\)
\(\angle OCB=29^0\)
\(\therefore\angle BOC=180^0-(29^0+29^0)\)
x = 180° - 58°
x = 122°
\(\angle x=122^0\ and\ \angle y=29^0\).

\(\cfrac { cos{ 63 }^{ 0 }20' }{ sin{ 26 }^{ 0 }40' } =\cfrac { cos63^{ 0 }20' }{ cos{ 63 }^{ 0 }20' } =1\)

\( \frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}} \)
\(= \frac{\tan \left(90^{\circ}-65^{\circ}\right)}{\cot 65^{\circ}}+\frac{\sin \left(90^{\circ}-50^{\circ}\right)}{\cos 50^{\circ}} \)
\(= \frac{\cot 65^{\circ}}{\cot 65^{\circ}}+\frac{\cos 50^{\circ}}{\cos 50^{\circ}}=1+1=2 \)

TSA = 2 (lb + bh + lh)
= 2 (10 \(\times\)12 + 12 \(\times\) 14 +10 \(\times\) 14)
= 2 (120 + 168 + 140)
= 856 cm²
LSA = 2 (bh + lh)
= 2 (12 \(\times\) 14 + 10 \(\times\) 14)
= 2 (168 + 140)
= 2 (308) = 616 cm².

\(G\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) =\left( \frac { 2+5+9 }{ 3 } ,\frac { -5+11+9 }{ 9 } \right) =\left( \frac { 16 }{ 3 } ,5 \right) \)

\(120=\frac { 8(50)+4(y) }{ 12 } \)
1440 = 400 + 4y
4y = 1040
\(y=\frac { 1040 }{ 4 } =260\ km\)

Number of cards n(S) = 52
No. of King cards n(A) = 4
No. of Queen cards n(B) = 4
No. of Jack cards n(C) = 4
Probability of drawing a King card
\(\frac { n(A) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Queen card
=\(\frac { n(B) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Jack card
=\(\frac { n(C) }{ n(S) } =\frac { 4 }{ 52 } \)
∴ The Probability of drawing a King or a Queen or a Jack from a deck of cards
= p(A) + P(B)+ P(C) =\(\frac { 4 }{ 52 } +\frac { 4 }{ 52 } +\frac { 4 }{ 52 } =\frac { 4+4+4 }{ 52 } =\frac { 12 }{ 52 } =\frac { 3 }{ 13 } \).

D (x, y) is the Mid point BC
\(\therefore \ D(x,y)=\left( \frac { 1+5 }{ 2 } ,\frac { -1+1 }{ 2 } \right) \)
\(=\left( \frac { 6 }{ 2 } ,\frac { 0 }{ 2 } \right) \)
= (3, 0)
AD is the median through A
Here A x₂ y₂ D x₃ y₃
(-1,3)     (3,0)
\( \therefore\) Length of AD \(=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-(-1)) }^{ 2 }+{ (0-3) }^{ 2 } } \)
\(=\sqrt { { (3+1) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =5\) units 

Two rational numbers between the given two irrational numbers are 0.5152 and 0.5352

(i) \(3\over 4\) = 0.75

(ii) \(5\over 8\) = 0.625

(iii) \(9\over 25\) = 0.36

6 has the maximum frequency 9. Therefore 6 is the mode.

OD = DP =\(\cfrac { 4cm }{ 2 } \) = 2 cm
AD = BD =\(\sqrt { { 5 }^{ 2 }-{ 4 }^{ 2 } } \)
= \(\sqrt { 25-19 } \)
= \(\sqrt { 9 } \)
= 3cm
\(\therefore\) The length of the common chrod AB = AD + BD = (3 + 3) cm = 6 cm

\((a+b)^{2}=a^{2}+b^{2} +2 a b \)
\(\left(x+\frac{1}{x}\right)^{2} =x^{2}+\frac{1}{x^{2}}+2 x \frac{1}{x} \)
 = 23 + 2 = 25 
\( \therefore x+\frac{1}{x} =\pm 5\) 
\( \left(x+\frac{1}{x}\right)^{3} =x^{3}+\frac{1}{x^{3}}+3(x)\left(\frac{1}{x}\right)(x+\frac{1}{x}) \)
\(5^{3} =x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) \)
\(125 =x^{3}+\frac{1}{x^{3}}+3(5) \)
\(125-15 =x^{3}+\frac{1}{x^{3}} \)
\(x^{3}+\frac{1}{x^{3}} =\pm 110\)

\((a)\frac { 2 }{ 3 } \)
\((b)\frac { 43 }{ 90 } \)
\((c)\frac { 1 }{ 999 } \)

Equivalent sets

In \(\triangle \)LMN
LM = 7.5 cm,
MN = 5 cm,
LN = 8 cm

Construction :
Step 1: Draw \(\triangle \)LMN with LNM = 8 cm, MN = 5 cm, LM = 7.5 cm
Step 2: Construct perpendicular bisectors for any two sides (LN and MN) to find the mid points of LM and MN.
Step 3: Draw the medians LD, ME. Let them meet at G.
Step 4: G is the centroid of the triangle LMN.