Total no. of attempts n(S) = 40
Total no. of attempts by A team n(A) = 32
Total no. of attempts by the opponent team B = n(B) = 40 - 32 = 8
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 8 }{ 40 } =\frac { 1 }{ 5 } \).

When two dice are rolled
Sample space
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) Event of the sum is equal to 1 = 0
∴ Probability = \(\frac { 0 }{ n(S) } \) = 0
(ii) Event of the sum is equal to 4
B = {(1, 3), (2, 2), (3, 1)}
n(B) = 3
P(B)= \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 12 } \)
(iii) Event of the sum is equal to less than 13
C= {(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(C) = 36
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 6 } \).

Hypotenuse = 10 cm
One of the acute angle = 24° 24'
sin 24° 24' = 0.4131'
\(\cfrac { x }{ 10 } =0.4131\)
x = 0.4131 \(\times\) 10
x = 4.131
cos 24° 24' = 0.9107
\(\cfrac { y }{ 10 } =0.9107\)
y = 9.107
\(\therefore\) Area of the triangle = \(\cfrac { 1 }{ 2 } bh\)
= \(\cfrac { 1 }{ 2 } \times y\times x\)
\(=\cfrac { 1 }{ 2 } \times 9.107\times 4.131=18.81sq.cm\)

Area of the parallelogram = 2 \(\times\) Area of the Δ ABC
s = \(\frac { 34+20+42 }{ 2 } =\frac { 96 }{ 2 } \)  = 48 m
Area of Δ ABC =\(\sqrt { 48(48-34)(48-20)(48-42) } \)
=\(\sqrt { 48\times 14\times 28\times 6 } =\sqrt { 112896 } \)=  336 m²
∴ Area of the parallelogram = 2 \(\times\) 336 m² = 672 m².

Perimeter of the rhombus land = 160 m
4a = 160 m
a = 40 m
One of the diagonal = 48 m
∴ Area of the land = 2 \(\times\) Area of the Δ ABC
s = \(\frac { 40+40+48 }{ 2 } =\frac { 128 }{ 2 } \)= 64 m
Area of Δ ABC =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 64(64-40)(64-40)(64-48) } \)
=\(\sqrt { 64\times 24\times 24\times 16 } =\sqrt { 589824 } \)
= 768 m²
∴ Area of the land = 2 \(\times\) 768 m² = 1536 m²

\(\cfrac { x }{ y } =\cfrac { 5 }{ 6 } \)
\(\Rightarrow\) 6x = 5y
6x - 5y = 0 .......(1)
\(\cfrac { x-8 }{ y-8 } =\cfrac { 4 }{ 5 } \)
\(\Rightarrow\) 5(x - 8) = 4(y - 8)
5x-40 = 4y-32
5x -4y = 40 - 32
5x - 4y = 8 ------ (2)
(1) x 5 \(\Rightarrow\) 30x-25y = 0
(2) x 6 \(\Rightarrow\) \(\cfrac { 30x-24y=48 }{ -y=-48 } \)    ...(2)
Substitute y = 48 in (1)
6x- 5(48) = 0
6x-240 = 0
6x = 240
\(x=\cfrac { 240 }{ 6 } =40\)
y = 48
\(\cfrac { x }{ y } =\cfrac { 40 }{ 48 } =\cfrac { 5 }{ 6 } \)
\(\therefore\) The number are in the Ratio 5 : 6

In the figure,
\(\triangle\)OPM, \(\triangle\)OQN are similar triangles. In similar triangles the sides are in the same proportional.
\(\cfrac { QN }{ PM } =\cfrac { QO }{ PO } \)
\(\cfrac { h }{ 5 } =\cfrac { 35 }{ 25 } \)
\(h=\cfrac { 5\times 35 }{ 25 } \)

h = 7m

x₁ y₁
Here (-5,11)
x₂ y₂
(4,-7)
m  n
7 : 2
\(=P(x,y)\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 7\times 4+2\times -5 }{ 7+2 } ,\frac { 7\times -7+2\times 11) }{ 7+2 } \right) \)
\(=\left( \frac { 28-10 }{ 9 } ,\frac { -49+22 }{ 9 } \right) \)
\(=\left( \frac { 18 }{ 9 } ,\frac { -27 }{ 9 } \right) =(2,-3)\)

In a parallelogram diagonals bisect each other and diagonals are not equal.
Mid point of DE \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
(-3, 6)\(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { x_{ 1 }+{ x }_{ 2 } }{ 2 } =-3\ \ \ \ \ \frac { y_{ 1 }+y_{ 2 } }{ 2 } =6\)
x₁ + x₂ = -6........(1)   
y₁ + y₂ = 12.......(2)
Mid point of EF\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
(0, 7)=\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
\(\frac { x_{ 2 }+{ x }_{ 3 } }{ 2 } =0\ \ \ \frac { y_{ 2 }+{ y }_{ 3 } }{ 2 } =7\)
x₂+x₃= 0..............(3)
Mid point of FD=\(\left( \frac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } ,\frac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } \right) =(-3,-2)\)
x₃ + x₁ = -6.......(5)     
y₃ + y₁ = -4..........(6)

∴ D (x₁, y₁) = (-6, -3)
Mid points of the diagonals are equal in parallelogram
∴ We have to prove this
Mid point of AC \(=\left( \frac { (-3)+(-3) }{ 2 } ,\frac { 6+(-2) }{ 2 } \right) =\left( \frac { -6 }{ 2 } ,\frac { 4 }{ 2 } \right) =(-3,2)\)
Mid Point of BD \(=\left( \frac { -6+0 }{ 2 } ,\frac { -3+7 }{ 2 } \right) =\left( \frac { -6 }{ 2 } ,\frac { 4 }{ 2 } \right) =(-3,2)\)
∴ Mid point of AC = Mid point of BD
∴ ABCD is a parallelogram

(i) \(\cfrac { 2x+2 }{ 3 } =\cfrac { 3x-6 }{ 5 } \)
5(2x+2) = 3(3x - 6) [By cross multiplication]
10x + 10 = 9x - 18
10x = 9x - 18 - 10
10x = 9x-28
10x- 9x = -28
x = -28
(ii)

2 = 4 (x + 1)-x
2 = 4x+4-x
3x + 4 = 2'
3x+4-4 = 2-4
3x =-2

Total number of centum scorers = 1184
Therefore n = 1184
(i) Let E₁ be the event of getting a centum scorer in Mathematics.
Therefore n(E₁) = 233, That is, r₁ = 233
\(P({ E })_{ 1 }=\frac { { r }_{ 2 } }{ n } =\frac { 233 }{ 1184 } \)
(ii) Let E₂ be the event of getting a centum scorer in Science.
Therefore n(E₂ ) = 106, That is, r₂ = 106
\(P({ E })_{ 2 }=\frac { { r }_{ 2 } }{ n } =\frac {106 }{ 1184 } \)
P(E^'₂) = 1− P(E₂)
\(=1-\frac { 106 }{ 1184 } \)
\(=\frac { 1078 }{ 1184 } \)

In this experiment, each trial is a match where Team I faces Team II.
We are concerned about the winning status of Team I.
There are 10 trials in total; out of which Team I wins in the 1^st, 6^th and 9^th matches.
The relative frequency of Team I winning the matches = \(\frac { 3 }{ 10 } \)or 0.3

Side of a cube = 7 cm
Now length of the resulting cuboid (l) = 7+7 =14 cm
Breadth (b) = 7 cm, Height (h) = 7 cm
So, Total Surface Area = 2(lb + bh + lh)
= 2[(14 \(\times\) 7)+(7 \(\times\) 7)+(14 \(\times\) 7)]

= 2(98 + 49 + 98)
= 2 × 245
= 490 cm²
Lateral Surface Area = 2(l + b) × h
= 2(14 + 7) × 7 = 2 × 21× 7
= 294 cm²

Here, length (l) = 25 m, breadth (b) =15 m, height (h) = 5 m.
Area of four walls = LSA of cuboid
= 2(l + b) × h
= 2(25 +15) × 5
= 80 × 5 = 400 m²
Area of the floor = l × b

= 25 ×15
= 375 m²
Total renovating area of the hall = (Area of four walls + Area of the floor) = (400 + 375) m² = 775 m²
Therefore, cost of renovating at the rate of Rs.80 per m² = 80 × 775
= Rs. 62,000

write 64⁰34' = 64⁰30' + 4'
From the table we have, sin64⁰30' = 0.9026 

(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

(i) sin74° = sin(90⁰ -16⁰)    (since, 90⁰ -16⁰ = 74⁰ )
RHS is of the form sin(90⁰ - \(\theta\)) = cos\(\theta\)
Therefore sin74° =  cos16⁰
(ii)  tan12° =  tan(90⁰ - 78⁰) (since, 12⁰= 90⁰ = 78⁰ )
RHS is of the form tan(90⁰-  \(\theta\)) = cot \(\theta\)
Therefore tan12° = cot 78⁰
(iii) cosec 39° = cosec(90⁰ -  51⁰) (since, 39⁰ = 90⁰ - 51⁰ )
RHS is of the form cosec(90⁰ - \(\theta\)) = sec\(\theta\)
Therefore cosec39° = sec51⁰

Let a = 28 m, b = 15 m and c = 41 m
Then, s = \(\frac{a+b+c}{2}=\frac{28+15+41}{2}=\frac{84}{2}\) = 42m
Area of triangular field =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 42(42-28)(42-15)(42-41) } \)
=\(\sqrt { 42\times 14​​\times 17\times 1 } \)
=\(\sqrt { 2\times 3\times 7\times 7\times 2\times 3\times 3\times 3\times 1 } \)
\(=2 \times 3 \times 7 \times 3\)
= 126 m²
Given the cost of levelling is Rs. 20 per m².
The total cost of levelling the field = 20 \(\times\)126 = Rs. 2520.

Given, 2x + 3y = 14 ...(1)
            3x − 4y = 4 ...(2)
To eliminate y:

Therefore, x = 4
Substitute x = 4 in (1) to get 2x + 3y = 14
2(4) + 3y = 14
8 + 3y = 14
y = 2
Thus the solution is x = 4, y = 2.
Verification :
2x+3y = 14 ...(1)
2(4)+3(2) = 14
8 + 6 = 14
14 = 14 True
3x – 4y = 4 ...(2)
3(4) – 4(2) = 4
12 – 8 = 4
4 = 4 True

Let the vertices of a triangle be A(1, −6), B(−5, 2) and C(x₃, y₃)
Given the centroid of a triangle as (−2, 1) we get,
\(\frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } \) = -2
\(\frac { 1-5+{ x }_{ 3 } }{ 3 } \) = -2
−4 + x₃ = −6
x₃ = −2
\(\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \) = 1
\(\frac { 6+2+{ y }_{ 3 } }{ 3 } \) = 1
−4 + y₃ = 3
y₃ = 7
Therefore, third vertex is (−2, 7).

Given x – 2y = 7 …(1)
2x + 3y = 7 …(2)
When x = 5, y = −1 we get
From (1) x – 2y = 5 – 2(−1) = 5 + 2 = 7 which is RHS of (1)
From (2) 2x + 3y = 2(5) + 3(−1) = 10−3 = 7 which is RHS of (2)
Thus the values x = 5, y = −1 satisfy both (1) and (2) simultaneously. Therefore (5,−1) is a solution of the given equations.

The given system of equations are 8x − 3y = 5xy ...(1)
6x − 5y = −2xy ...(2)
Observe that the given system is not linear because of the occurrence of xy term. Also note that if x =0, then y =0 and vice versa. So, (0,0) is a solution for the system and any other solution would have both x \(\neq \) 0 and y \(\neq \) 0
Let us take up the case where x \(\neq \) 0 and y \(\neq \) 0
Dividing both sides of each equation by xy,
\(\frac { 8x }{ xy } -\frac { 3y }{ xy } =\frac { 5xy }{ xy } \)
\(\frac { 6x }{ xy } -\frac { 5y }{ xy } =\frac { -2xy }{ xy } \)
Let \(a=\frac { 1 }{ x } ,b=\frac { 1 }{ y } \)
We get, \(\frac { 8 }{ y } -\frac { 3 }{ x } =5\)  .....(3)
\(\frac { 6 }{ y } -\frac { 5 }{ x } =-2\)  .......(4)
(3)&(4) respectively become, 8b − 3a = 5 ...(5)
b − 5a = −2 ...(6)
which are linear equations in a and b.
To eliminate a, we have,  (5) \(\times\) 5\(\Rightarrow\) 40b −15a = 25 .....(7)
(6) × 3\(\Rightarrow\) 18b −15a = −6 .....(8)
Now proceed as in the previous example to get the solution\(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \).
Thus, the system have two solutions \(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \) and (0,0).

Given points are A(–3, 6) and B(1, –2), P(–2, 4) divide AB internally in the ratio m : n.
By section formula,
\(P(x, y)=P\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)
= P(−2, 4) .......(1)
Here x₁ = −3, y₁ = 6, x₂ = 1, y₂ = −2
\((1)\Rightarrow \left( \frac { m(1)+n(-3) }{ m+n } ,\frac { m(-2)+n(6) }{ m+n } \right) \)
Equating x-coordinates, we get
\(\frac{m-3 n}{m+n}=-2\) or m− 3n = −2m− 2n
3m = n
\(\frac{m}{n}=\frac{1}{3}\)
m : n = 1: 3
Hence P divides AB internally in the ratio 1 : 3.