\(\Sigma fx=630\) , \(\Sigma f=40\)
Mean  = \(\frac { 630 }{ 40 } =15.75\)

Consider a triangle ABC with each side equal to 2a.
\(\angle A=\angle B =\angle C\) = 60°
Draw AD is perpendicular to BC
\(\triangle BDA\cong \triangle CDA\) by RHS
BD = CD
\(\angle BAD=\angle CAD\) = 30° by CPCT
By Pythagoras theorem,
\(AD=\sqrt{3}a\)
In \(\triangle BDA\), cosec 30° = \(\frac{AB}{BD}=\frac{2a}{a}=2\)
and cos 60° = \(\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}\)

x²-6x+7

Here, 327.7081 is terminating. So, it represents a rational number.
Thus, \(327.7081=\frac { 3277081 }{ 10000 } =\frac { p }{ q } \)
Here, q = 10⁴ = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2⁴ x 5⁴ = (2 x 5)⁴
So, the prime factors of q are 2 and 5.

\(P\ge -\frac { 4 }{ 3 } \)

Given endpoints of diameter of a circle are (x₁,y₁)=(24, 1) and (x₂, y₂)=(2, 23).
Now, diameter of a circle, d = \(\sqrt { { \left( 2-24 \right) }^{ 2 }+{ (23-1) }^{ 2 } } \) \([ \therefore distance=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } }] \)
= \(\sqrt { { \left( -22 \right) }^{ 2 }+{ (22) }^{ 2 } } =\sqrt { { (22) }^{ 2 }+(1+1) } \)
 =\(22\sqrt { 1+1 } =22\sqrt { 2 } \) units
\(\therefore \)   Radius of a circle, r = \(\frac { d }{ 2 } \) = \(\frac { 22\sqrt { 2 } }{ 2 } \) = \(11\sqrt { 2 } \) units.

0.95

1 : 9

3344 cm²

Perfect square numbers from 1 to 10 are 1, 4 and 9
Therefore, Required probability = \(\frac { 3 }{ 10 } \)

55^∘

7 cm

As , n-2, 4n-1 and 5n+2 are in AP.
\(\therefore \) 4n-1-(n-2)=5n+2-(4n-1)
\(\Rightarrow \) 4n-1 -n+2=5n+2-4n+1
\(\Rightarrow \) 3n+1=n+3
\(\Rightarrow \) 2n=2
\(\Rightarrow \) n=1

\({ \pi r }^{ 2 }=2\pi r\times 2\)
\(\Rightarrow \quad r=4\Rightarrow \quad d=2r=8\)

\(5\sqrt { 2 }\ \ units\)

0

Let PQ is tower and QR is its shadow. 

In right \(\Delta\) PQR, 
\(\frac { PQ }{ QR } =\tan { { 60 }^{ o } } \)
\(\Rightarrow\)  \(\frac { 30 }{ QR } =\sqrt { 3 } \)
\(\Rightarrow\)  \(QR=\frac { 30 }{ \sqrt { 3 } } \)
\(=10\sqrt { 3 } \) m

Given equation is,  x²-3x+5=(x+5)²
x²-3x+5=x²+10x+25
13x+20=0
Which is not of the form ax²+bx+c=0
Hence it is not a quadratic equation

On dividing by 2 ,we  get
\(x^{ 2 }+\frac { x }{ 2 } -2=0\)
Adding and subtracting \(\left( \frac { 1 }{ 4 } \right) ^{ 2 }\) we get
\(x^{ 2 }+2x\left( \frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 4 } \right) ^{ 2 }-\left( \frac { 1 }{ 4 } \right) ^{ 2 }-2=0\)
\(\Rightarrow \left( x+\frac { 1 }{ 4 } \right) ^{ 2 }-\frac { 33 }{ 16 } =0\)
\(\Rightarrow \left( x+\frac { 1 }{ 4 } \right) ^{ 2 }=\frac { 33 }{ 16 } =0\)
Taking the square root we get
\(\left( x+\frac { 1 }{ 4 } \right) =\pm \frac { \sqrt { 33 } }{ 4 } \)
\(x=\frac { -1+\sqrt { 33 } }{ 4 } ,\frac { -1-\sqrt { 33 } }{ 4 } \)

S_n = 3n² - 4n
S₁ = 3(1)² - 4(1) = - 1
S₂ = 3(2)²- 4(2) = 4
a₁ = S₁ =- 1
a₂ = S₂- S₁ = 4 - (- 1) = 5
d = a₂ - a₁ = 5 - (-1) = 6
A.P. is - 1, 5, 11,............
a₁₂ = a + 11d
=-1 +11 \(\times\) 6
= 65.

We know that, tangents drawn from an exterior point to a circle are equal in length.
AD = AF = x cm
BD = BE = y cm
CE = CF = z cm
Given, AB = 8 cm
⇒ AD + BD = 8cm
⇒ x + y = 8
BC = 10 cm ⇒ BE + CE = 10 cm
⇒ y + z = 10
and CA = 12 cm ⇒ CF + AF = 12 cm
⇒ z + x = 12
On adding Eqs. (i), (ii) and (iii), we get
2(x + y + z) =30
⇒ x + y + z = 15
On subtracting Eq. (ii) from Eq. (iv), we get
x =15 - 10 = 5
On subtracting Eq. (iii) from Eq. (iv), we get
y = 15-12 = 3
On subtracting Eq. (i) from Eq. (iv), we get
z=15-8=7
AD = xcm = 5 cm
BE= ycm = 3 cm
and CF = z cm = 7 cm
Hence, the length of AD, BE and CE are 5 cm, 3 cm and 7 cm, respectively.

Let OP intersect AB at a point C. Here, PA and PB are the two tangents from a point P lying outside the circle, to the circle with centre O. 

\(\angle \)APO = \(\angle \)BPO [∵ O lies on the bisector of ZAPB] 
Now, in \(\triangle\)ACP and \(\triangle\)BCP, we have 
AP = BP [tangents from an external pointl 
PC = PC [commonl]
\(\angle \)APO = \(\angle \)BPO [proved above] 
⇒ \(\triangle\)ACP ≅ \(\triangle\)BCP   [by SAS congruence axiom]
⇒ AC = BC   [c.p.c.t.] 
and  \(\angle \)ACP = \(\angle \)BCP 
= \(1\over2\) x 180° = 90° [c.p.c.t.]
Hence, OP is the perpendicular bisector of AB. 

Given : A ΔABC, in which AB = 5 cm, BC = 6 cm and CA = 7 cm. 

Required: ΔA'BC' ∼ ΔABC with \(\frac { 4 }{ 5 } \) (reduced) scale-factor.
Steps of Construction : 
1. Construct ΔABC, such that AB=5cm, BC = 6 cm and CA = 7 cm. 
2. Through B, construct an acute ㄥCBX, 90^o).
3. Mark five points on BX such that BB₁ = B₁B₂ = B₂B₃ - B₃B₄ = B₄B₅.
4. Join B₅C. 
5. Through B₄, draw B₄C'||B₅C, intersecting BC in C'. 
6. Through C', draw C'A'||CA, intersecting BA in A'.
Hence, ΔA'BC' is the required triangle.

The three-digit numbers which are divisible by 11 are 110, 121, 132, ..., 990
Let there are n, 3-digit numbers divisible by 11
Clearly, it forms an A.P. with a = 110, d = 11
So,         a_n = 990
\(\Rightarrow\)  a + ( n - 1 )d = 990
110 + ( n - 1 )d  = 990
110 + 11n - 11 = 990
11n + 99 = 990  \(\Rightarrow\)  11n = 891
\(\Rightarrow\)        n = 81
Thus, there are 81 numbers of three-digit which are divisible by 11.

Let speed of the stream be x km/h
Speed of the boat in still water = I I km/h 
\(\therefore \) Upstream speed be (11-x) km/h and downstream speed be (11+x) km/h
Distance = 12 km
Time taken for downstream direction = \(\frac { 12 }{ 11+x } \) hours
Time taken for upstream direction = \(\frac { 12 }{ 11-x } \) hours
ATQ \(\frac { 12 }{ 11+x } +\frac { 12 }{ 11-x } =2\frac { 3 }{ 4 } \Rightarrow \frac { 12(11-x)+12(11+x) }{ (11+x)(11-x) } =\frac { 11 }{ 4 } \)
\(\Rightarrow \frac { 132+132 }{ 121-x^{ 2 } } =\frac { 11 }{ 4 } \Rightarrow 4\times 264=11(121-x^{ 2 })\)
\(\Rightarrow \frac { 4\times 264 }{ 11 } =121-x^{ 2 }\Rightarrow 4\times 24=121-x^{ 2 }\)
\(\Rightarrow x^{ 2 }=25\Rightarrow x=\pm 5\)
Hence speed of the stream is x=5 km/h

It is given that n = 100
So, 76 + x + y = 100, i.e., x + y = 24
The median is 525, which lies in the class 500 – 600
So, l = 500, f = 20, cf = 36 + x, h = 100
Using the formula : Median \(=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) h, \text { we get }\)
\(525=500+\left(\frac{50-36-x}{20}\right) \times 100\)
i.e. 525 - 500 = (14 - x) \(\times\) 5
i.e. 25 = 70 - 5 x
i.e. 5 x = 70 - 25 = 45
So, x = 9
Therefore, from (1), we get 9 + y = 24
i.e. y = 15

BC and OX bisect each other
So BXCO is a parallelogram, BE II XC and BX II CF
In \(\triangle\)ABX, by B.P.T.,
\(\frac { AF }{ FB } =\frac { AO }{ OX } ...(i)\)
In \(\triangle\)AXC,
\(\frac { AE }{ EC } =\frac { AO }{ OX } ...(ii)\)
(i) and (ii) give,
\(\frac { AE }{ FB } =\frac { AE }{ EC } \quad \)
So by converse of B.P.T.,
EF II BC
(i) give \(\frac { OX }{ OA } =\frac { AB }{ AF } \)
Adding 1 on both sides
\(\frac { AX }{ OA } =\frac { AB }{ AF } \)
or OA : AX = AF : AB

LHS = \({ (\sin { A } +cosecA) }^{ 2 }+{ (\cos { A } +\sec { A } ) }^{ 2 }\)
\(=\sin ^{ 2 }{ A } +{ cosec }^{ 2 }A+2\sin { A } cosecA+\cos ^{ 2 }{ A } +\sec ^{ 2 }{ A } +2\cos { A } \sec { A } \)
\(\left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=(\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } )+(1+\cot ^{ 2 }{ A } )+2\sin { A } \frac { 1 }{ \sin { A } } +(1+\tan ^{ 2 }{ A } )+2\cos { A } \frac { 1 }{ \cos { A } } \)\(\left[ \because { cosec }^{ 2 }A=1+\cot ^{ 2 }{ A } ,\sec ^{ 2 }{ A } =1+\tan ^{ 2 }{ A } ,cosecA=\frac { 1 }{ \sin { A } } and\sec { A } =\frac { 1 }{ \cos { A } } \right] \)\(=1+1+\cot ^{ 2 }{ A } +2+1+\tan ^{ 2 }{ A } +2\quad \left[ \because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1 \right] \)
\(=7+\tan ^{ 2 }{ A } +\cot ^{ 2 }{ A } \)
= RHS
Hence proved.

Given In figure, ABC is a triangle in which \(\angle ABC>{ 90 }^{ ° }\) and \(AD\bot CB\) produced.
To prove AC² = AB² + BC² + 2 BC . BD
Proof In right angled \(\triangle ADC\), \(\angle D={ 90 }^{ ° }\)
\(\therefore\)  AC² = AD² + DC²         [by Pythagoras theorem]
= AD² + (BD + BC)²         [\(\because\)  DC = DB + BC]
= (AD² + BD²) + BC² + 2 BD . BC
[\(\because\)  (a + b)² = a² + b² + 2ab]
\(\Rightarrow\)  AC² = AB² + BC² + 2 BC . BD
[\(\because\)  in right angled \(\triangle ADB\), AB² =  AD² + DB²]

Let the number of days in which work was finished be n.
Number of workers on Ist day = 150
Number of workers on IInd day = 146
Number of workers on IIIrd day = 142 and so on
One day equivalent of all the worker
= 150 + 146 + 142 + ... upto to n workers
\(={n\over2}[2\times150+(n-1)\times(-4)]\)
\(={n\over2}(304-4n)=152n-{2n}^{2}\)        ....(i)
If 150 workers would have worked every day then number of days required to finish the work = ( n - 8 )
\(\therefore\)  One day equivalent of workers = 150 ( n - 8) = 150n - 1200                      ...(ii)
From (i) and (ii), we have
152n - 2n² = 150n - 1200
\(\Rightarrow\)   2n² - 2n - 1200 = 0
\(\Rightarrow\)  n² - n - 600 = .0
\(\Rightarrow\)  ( n - 25 ) ( n + 24 ) = 0
\(\Rightarrow\)  n = - 24, n = 25
\(\therefore\) Work has completed in 25 days.