Modal class = 5-7
l = 5 , f₁=80,f₀=45,h=2,f₂=55
Mode = \(l+\frac { (f_{ 1 }f_{ 0 }) }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(=5+\frac { 80-45 }{ 160-45-55 } \times 2=5+\frac { 35\times 2 }{ 60 } \)
= 6.17

\(a=-\frac { 3 }{ 2 } \)

Here, (x₁, y₁)=(3, -7), (x₂, y₂) = (-8, 6) and (x₃, y₃) = (5, 10)
\(\therefore \) Coordinates of the centroid of a triangle are \(\left( \frac { 3-8+5 }{ 3 } ,\frac { -7+6+10 }{ 3 } \right) \) i.e. (0, 3).
\(\therefore\) centroid of triangle\(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)

Required probability=1 - p

4 cm

Yes

\(20(\sqrt{3}+2)m\)

3 cm

Let the five numbers in A.P. be
a-2d, a-d, a, a+d, a+2d
Here, sum of five numbers = 40
\(\therefore \) a-2d + a-d + a + a+d + a+2d = 40 
\(\Rightarrow \) 5a = 40;   \(\Rightarrow \) a = 8
Hence, the middle term is 8.

Internal radius=\(\frac{24}{2}\)cm
External radius=\(\frac{25}{2}\)cm
Area to be painted=outer C.S.A. + inner C.S.A. + area of the ring.
=2πR²+2πr²+π[R²-r²]
=2x\(\frac{22}{7}\)x\(\frac{25}{2}\)x\(\frac{25}{2}\)+2x\(\frac{22}{7}\)x\(\frac{24}{2}\)x\(\frac{24}{2}\)+\(\frac{22}{7}\)\(\left[\left(\frac{25}{2}\right)^{2}-\left(\frac{24}{2}\right)^{2}\right]\)
=2x\(\frac{22}{7}\)x\(\frac{1}{4}\)[625+576]+\(\frac{22}{7}\)\(\left[\frac{25}{2}-\frac{24}{2}\right]\)\(\left[\frac{25}{2}-\frac{24}{2}\right]\)
=\(\frac{22}{7}\)\(\left[\frac{1201}{2}+\frac{1}{2}\times \frac{49}{2}\right]\)
=\(\frac{22}{7}\)x\(\frac{1}{4}\)[2402+49]
=\(\frac{22}{7}\)x\(\frac{1}{4}\)x2451 cm²
Cost of painting=Rs.0.05x\(\frac{22}{7}\)x\(\frac{1}{4}\)x2451=Rs.96.28

\(l=6\pi ,r=6cm,\theta =?\)
\(l=\frac { \theta \pi r }{ { 180 }^{ ° } } \Rightarrow 6\pi =\frac { \theta \times \pi \times 6 }{ { 180 }^{ ° } } \Rightarrow \theta ={ 180 }^{ ° }\)

Distance between the points P(x,2) and Q(3,-6) is 10.Using distance formula PQ=10
\(\Rightarrow \quad \sqrt { { ({ x }-3) }^{ 2 }-{ \{ 2-(-6) }^{ 2 }\} ​​ } =10\)
\(\Rightarrow \quad \sqrt { { x }^{ 2 }+{ 3 }^{ 2 }-2\times 3\times x+\left( 2+6 \right) ^{ 2 } } =10\)
\(\Rightarrow \quad \sqrt { { x }^{ 2 }+9-6x+{ 8 }^{ 2 } } =10\)
\(\Rightarrow \quad \sqrt { { x }^{ 2 }+9-6x+64 } =10\)
\(\Rightarrow \quad \sqrt { { x }^{ 2 }-6x+73 } =\quad 10\)
Squaring both sides, we get
\(\quad { x }^{ 2 }-6x+73=100\)
\(\Rightarrow \quad { x }^{ 2 }-6x+73-100=0\)
\(\\ \Rightarrow \quad { x }^{ 2 }-6x-27=0\)
\(\\ \Rightarrow \quad { x }^{ 2 }-9x+3x-27=0\)
\(\Rightarrow x(x-9)+3(x-9)=0\)
\(\Rightarrow \quad (x-9)(x+3)=0\)
\(\Rightarrow \quad either\quad x-9=0\quad or\quad x+3=0\)
\(\Rightarrow \quad x=9\quad or\quad x=-3\)
gnoring x = —3 as it is given that x is a positive Integer.
Thus, only solution is x = 9. 

Let number of camels be x
ATQ \(\frac { 1x }{ 4 } +2\sqrt { x } +15=x\)

\(\Rightarrow \frac { x+8\sqrt { x } +60 }{ 4 } =x\)
\(\Rightarrow x+8\sqrt { x } +60=4x\)
\(\Rightarrow 3x-8\sqrt { x } -60=0\)
Let \(\sqrt { x } =1\)
\(\therefore 3t^{ 2 }-8t-60=0\)
\(\Rightarrow 3t^{ 2 }-18t+10t-60=0\)
\(\Rightarrow 3t(t-6)+10(t-6)=0\)
(3t+10)=0 or (t-6)=0
\(\Rightarrow t=\frac { -10 }{ 3 } \) or t=6
Rejecting t = \(\frac { -10 }{ 3 } \Rightarrow t=6\)
\(\Rightarrow \sqrt { x } =6\Rightarrow x=36\)

Let the two APs be a₁,a₂,a₃,.....,a_n ... and b₁, b₂, b₃, ...., b_n.....
Also, let d be the same common difterence of two AP's. Then, the nth term of first AP is
a_n = a₁ + (n - 1)d
and the nth term of second AP is
b_n = b₁ + (n - 1) d
Now, a_n - b_n = [a₁ + (n - 1)d] - [b₁ + (n - 1) d]
\(\Rightarrow\) a_n - b_n = a₁ - b₁, \(\forall\) n \(\in\)N
\(\therefore\) a₁₀₀ - b₁₀₀ = a₁ - b₁ = 100    [given]
So, a₁₀₀₀ - b₁₀₀₀ = a₁ - b₁
\(\Rightarrow\) a₁₀₀₀ - b₁₀₀₀ = 100    [\(\because\) a₁ - b₁ = 100]
Hence, the difference between their 1000 th terms is also 100 for all n \(\in\) N.

In MDC, By Pythagoras theorem,
AC₂ = AD² + CD²
= AD² + (BC + BD)²
= AD² + BC² + 2BC x BD + BD²
= AB² + BC² + 2BC x BD

Since the points are collinear
The area of triangle = 0
Area of triangle \(=\frac { 1 }{ 2 } \left[ { { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) } \right] \)
\(\Rightarrow \ \frac { 1 }{ 2 }[5(4-y)+(-3)(y-2)+x(2-4)]=0\)
\(\Rightarrow \ \frac { 1 }{ 2 }[20-5y-3y+6+(-2x)]=0\)
\(\Rightarrow \ \frac { 1 }{ 2 }[-2x-8y+26]=0\)
\(\Rightarrow \ x+4y-13=0\)

Let O be the centre and P be the mid-point of BC.Then, OP丄BC.

Since, ∆ABC is an isosceles triangle and P is the mid-point of BC. Therefore, AP .L BC as median from the vertex in an isosceles triangle is perpendicular to the base.
Let AP = x
and PB = CP = y
In ΔAPB and ΔOPB,
AB² = BP² + AP²
36 = y² + x²
and OB²=OP²+BP²
81= (9- x)² + y²
On subtracting Eqq.(i) from Eq.(ii) we get
81- 36 = [(9 - x)² + y²]- (y² + x2)
45 =81-18x+x²+y² -y² -x²
45=81-18x
18x=81-45
18x = 36
\(x={36\over 18}\)
x=2cm
On putting x = 2 in Eq.(i), we get
36=y²+4
y²=32
\(y=4\sqrt{2}cm\)
BC=2BP
=2y=\(8\sqrt{2}cm\)
Now, area of ∆ABC=\({1\over2}\times BC\times AP\)
\(={1\over2}\times8\sqrt{2}\times2\)
\(=8\sqrt{2}cm^2\)
Hence, the area of ∆ABC is \(=8\sqrt{2}cm^2\)

Area of shaded region = 2 x [Area of semi-circle with diameter AQ -Area of semi-circle with diameter AP].
= 37.68\(\pi\) cm²

60°

(i) \(\frac { 3 }{ 8 } \)
(ii) \(\frac { 1 }{ 2 } \)
(iii) \(\frac { 12 }{ 365 } \)
(iv) \(\frac { 59 }{ 365 } \)

Here, S_q = 63q - 3q²             ....(i)
Also,  S_q-1 = 63 ( q - 1 ) - 3 ( q - 1 )²
= 63q - 63 - 3 ( q2 + 1 - 2q )
\(\Rightarrow\)   500 + 100d + 50 + 50 - d = 500 - 100d + 50 + 5 + d - 594
\(\Rightarrow\)   198d = - 594
\(\Rightarrow\)    d = - 3
Thus, the three digits are 5 - (  - 3 ), 5 and 5 - 3, (i.e., 8, 5 and 2 )
Hence, the required number is 852.

As, \(\sqrt { 3 } \sin { \theta } =\cos { \theta } \Rightarrow \tan { \theta } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore \frac { \sin { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \sin { \theta } +\cos { \theta } } =\frac { \tan { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \tan { \theta } +1 } \)
\(\frac { 1+\sqrt { 3 } }{ 4 } \)

Quotient=2x²+5, remainder=x+2

Consider 2 and \(\sqrt{2}\)
Consider \(2+\sqrt{3}\) and \(2-\sqrt{3}\) 
Also, consider \(\sqrt{2} \text { and } \sqrt{3}\)
No, either rational or irrational.

Area of triangular shaded region
\(=\frac { \angle P }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+\frac { \angle Q }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+\frac { \angle C }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }\)
\(\left[ \because Arc\ of\ sector=\frac { \theta }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 } \right] \)

\(=\frac { (\angle P+\angle Q+\angle C) }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }\)
\(\left[ \because \angle P+\angle Q+\angle R={ 180 }^{ 0 } \right] \)
=11 x 7=77 cm²
Area of the rectangular shaded region
\(=\frac { 90 }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+\frac { 90 }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+\frac { 90 }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }\)
\(=2\times \frac { 90 }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times { 7 }\times 7=77\quad { cm }^{ 2 }\)
The farmer has allowed Rani and Ramu to cut equal amount of grass (77 cm² ) without any gender bias.

1.5 cm, 2.5 cm

When two dice are thrown simultaneously, then sample space contain ( 6 X6=36) outcomes
Sum 2 i.e., [(1,1),(1,1)] = two outcomes
P( Sum 2) = 2 / 36 = 1 / 18
Sum 3 i.e., (1,2), (1,2), (2,1), (2,1) =4 outcomes.
P( Sum 3) = 4 / 36 = 1 / 9
Sum 4 i.e., [(1,3), (1,3), (2,2), (2,2),(3,1), (3,1)] =6 outcomes
P( Sum 4) = 6 / 36 = 1 / 6
Sum 5 i.e., [(2,3), (2,3), (3,2), (3,2),(4,1), (4,1)] =6 outcomes
P( Sum 5) = 6 / 36 = 1 / 6
Sum 6 i.e., [(3,3), (3,3), (4,2), (4,2),(5,1), (5,1)] =6 outcomes
P( Sum 6) = 6 / 36 = 1 / 6
Sum 7 i.e., [(4,3), (4,3), (5,2), (5,2),(6,1), (6,1)] =6 outcomes
P( Sum 7) = 6 / 36 = 1 / 6
Sum 8 i.e., [(5,3), (5,3), (6,2), (6,2)] =4 outcomes
P( Sum 8) = 4 / 36 = 1 / 9
Sum 9 i.e., [(6,3), (6,3)] =2 outcomes.
P( Sum 9) = 2 / 36 = 1 / 18​​​

273.2