Mean  \(= \frac { \Sigma fx }{ \Sigma f } \)
\(50=\frac { \Sigma fx }{ 100 } \)
\(\Sigma fx=5000\)
Correct \(\Sigma fx^{ ' }=5000-100+110\)
=5010
Correct Mean  \(= \frac { 5010 }{ 100 } \)
= 50.1
Median Will remain same median = 52

In \(\triangle\)ABC and \(\triangle\)DEF,
AB = 3DE and AB = 3DF
\(\Rightarrow \frac { AB }{ DE } =3; \frac { AC }{ DF } =3;\)
P₁ = 3P₂
BC = 3EF
\(\Rightarrow \frac { AB }{ DE } =\frac { AC }{ DF } =\frac { BC }{ EF } =3\)
\(\triangle\)ABC~\(\triangle\)DEF
\(\Rightarrow\frac { ar(\triangle ABC) }{ ar(\triangle DEF) } ={ \left( \frac { AB }{ DE } \right) }^{ 2 }={ (3) }^{ 2 }=9.\)

Here, \(a=\sqrt { 2 } ,a+d=\sqrt { 8 } =2\sqrt { 2 } \)
 \(d=2\sqrt { 2 } -\sqrt { 2 } =\sqrt { 2 } \\ \therefore Next\quad term=\sqrt { 32 } +\sqrt { 2 } \\ =4\sqrt { 2 } +\sqrt { 2 } \\ =5\sqrt { 2 } \\ =\sqrt { 50 } \)

123.4 kg

17.5 cm

We know that area of sector of a circle =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)
Given, radius of circle, r = 6 cm
and angle of sector, \(\theta\)= 60°
\(\therefore\) Area of sector of a circle \(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^2=\frac{132}{7} \mathrm{~cm}^2\)

Let a be the first term and d be the common difference of given AP.
Now, as the nth term of an AP is a_n = a + (n-1) d
\(\therefore\) a₃= a + 2d = 12   [\(\because\) a₃ = 12, given] ...(i)
and a₅₀ = a + 49 d =106    ...(ii)
[\(\because\) a₅₀ = 106, given]
On subtracting Eq. (i) from Eq. (ii), we get
47 d = 94 \( \Rightarrow d=\frac{94}{47}=2\)
On putting the value of d in Eq. (i), we get
a + 2 \(\times\) 2 = 12 \(\Rightarrow\) a = 12 - 4 = 8
Now, 29th term, a₂₉ = a + (29 - 1) d
= 8 + 28 \(\times\) 2 = 8 + 56 = 64

∠LOM= 90^o

Volume of cone = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
Volume of metal in 504 cones
\(=504\times \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 3.5 }{ 2 } \times \frac { 3.5 }{ 2 } \times 3\)
Volume of Sphere = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
\(= \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 3 }\)
Volume of sphere= Volume of 504 cones
\(\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 3 }=504\times \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 3.5 }{ 2 } \times \frac { 3.5 }{ 2 } \times 3\)
\(\Rightarrow { r }^{ 3 }={ \left( \frac { 21 }{ 2 } \right) }^{ 3 }\)
\(\Rightarrow\) r = 10.5 cm
\(\therefore\) Diameter = 21 cm
Surface area = 4\(\pi { r }^{ 2 }\)
= 4 x \(\frac {22}{7}\) x 10.5 x 10.5
= 1386 cm².

Mean = \(\frac { \Sigma f_{ i }x_{ i } }{ \Sigma f_{ i } } =\frac { 1640 }{ 50 } \times 32.8\)
Modal class = 30-40
l=30, f₁=15 , f₂=7, f₀=10 , h=10
Mode = \(l+\frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }f_{ 0 }-f_{ 2 } } \times h\)
\(30+\frac { 15+10 }{ 30-10-7 } \times 10\)
30+3.85
=33.85

\(\because AB\parallel EW\)  [given]
\(\therefore \frac { DA }{ AE } =\frac { DB }{ BW } \) [by basic proportionality theorem]
\(\Rightarrow \frac { DA }{ DE-DA } =\frac { DB }{ DW-DB } \)
\(\Rightarrow \frac { 4 }{ 12-4 } =\frac { DB }{ 24-DB } \Rightarrow \frac { 4 }{ 8 } =\frac { DB }{ 24-DB } \)
\(\Rightarrow 24-DB=2DB\Rightarrow 24=3DB\Rightarrow DB=\frac { 24 }{ 3 } =\) 8 cm

Given, radii of two concentric circles are 13 cm and 8 cm.
Produced BD to meet the bigger circle at E. Join AE.
Then, ㄥAEB = 90° 
OD丄BE
and BD = DE

OD || AE
In Δ AEB, O and D are the mid-points of AB and BE, respectively
Therefore, by mid-point theorem, we have
\(OD={1\over2}AE\Rightarrow\ AE=2\times8=16\)
In right angled ∆ODB,
OB² = OD² + BD² [by Pythagoras theorem)
13² = 8² + BD²
BD² = 169 - 64 = 105
BD = \(\sqrt{105}cm\)
DE= \(\sqrt{105}cm\)
Now, in right angled ∆AED,
AD² = AE² + ED²
AD² = (16)² + (\(\sqrt{105}\))² = 256 + 105
AD² = 361 ⇒ AD =\(\sqrt{361}\)
AD = 19cm
Hence, the length of AD is 19 cm

56 cm

Volume of the cube = (5 x 5 x 1 + 2) cm³ = 27 cm³
Length of edge of the cube = \(\sqrt [ 3 ]{volume}=\sqrt [ 3 ]{ 27} \)= 3 cm

37

We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact  Fig. We need to prove that
 
\(\angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}\) 
Let \(\angle \mathrm{PTQ}=\theta\) 
Now, TP = TQ. So, TPQ is an isosceles triangle.
Therefore, \(\angle \mathrm{TPQ}=\angle \mathrm{TQP}=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{1}{2} \theta\) 
Also, \(\angle \mathrm{OPT}=90^{\circ}\) 
So, \(\angle \mathrm{OPQ}=\angle \mathrm{OPT}-\angle \mathrm{TPQ}=90^{\circ}-\left(90^{\circ}-\frac{1}{2} \theta\right)\) 
\(=\frac{1}{2} \theta=\frac{1}{2} \angle \mathrm{PTQ}\) 
This gives \(\angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}\)

Given \(\Delta P M S \sim \Delta Q M R\) andPQ II SR.
To show PS = QR
\( \because \Delta P M S \sim \Delta Q M R \)
\(\therefore \frac{P S}{Q R}=\frac{P M}{Q M}=\frac{M S}{M R} \) ........(i)
[∴ corresponding sides of similar triangles are proportional]
Now, consider ΔPMQ and ΔRMS. In these triangles, we have 
\(\angle P M Q=\angle R M S \text { [vertically opposite angles] }\)and \(\angle M P Q=\angle M R S \text { [alternate angles] }\)
\(\therefore \Delta P M Q \sim \Delta R M S\) [by AA similarity criterion]
\(\Rightarrow \frac{P M}{R M}=\frac{M Q}{M S}\) [∴ corresponding sides of similar triangles are proportional]
\(\Rightarrow \frac{P M}{Q M}=\frac{M R}{M S}\)  .........(ii)
From Eqs. (i) and (ii), we get
\(\frac{M S}{M R}=\frac{M R}{M S} \Rightarrow M S^{2}=M R^{2} \Rightarrow M S=M R\)
From Eq. (i), we get
\(\frac{P S}{Q R}=1 \Rightarrow P S=Q R\) Hence proved.

The table for given distribution is

Here, \(\sum { f_{ i }x_{ i } } =1.60\) and \(\sum { f_{ i } } =24\)
Mean concentration of SO₂\(=\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\frac { 1.60 }{ 24 } \)
=0.066 (approx.)
Here, n=24 ⇒\(\frac { n }{ 2 } =12\)
The cumulative frequency just greater than 12 is 14 and corresponding class interval is 0.06-0.08.Here, l=0.06, cf=11, f=3 and h=0.02
Median\(=l+\left\{ \frac { \frac { n }{ 2 } -cf }{ f } \right\} \times h\)
\(=0.06+\left\{ \frac { \frac { 24 }{ 2 } -11 }{ 3 } \right\} \times 0.02=0.06+\left\{ \frac { 12-11 }{ 3 } \right\} \times 0.02\\ =0.06+\frac { 0.02 }{ 3 } =0.06+0.007=0.067(approx).\)
Promoting awarness regarding protection of environment and taking suggestive measures to make it pollution free.

Given, Diameter of hemisphere=Diameter of a cylinder=Height of building (H) let
Radius of hemisphere=Radius of cylinder=\(\frac { H }{ 2 } \) and height of cylinder\(=H-\frac { H }{ 2 } =\frac { H }{ 2 } \)
Now, Volume of building =Volume of hemisphere+Volume of cylinder
4 m

Given sector is OAPB 
Area of the sector \(=\frac{\theta}{360} \times \pi r^{2}\)
\(=\frac{30}{360} \times 3.14 \times 4 \times 4 \mathrm{~cm}^{2}\)
\(=\frac{12.56}{3} \mathrm{~cm}^{2}=4.19 \mathrm{~cm}^{2}(\text { approx. })\)
Area of the corresponding major sector
\(\begin{aligned} &=\pi r^{2}-\text { area of sector } \mathrm{OAPB}\\ &=(3.14 \times 16-4.19) \mathrm{cm}^{2}\\ &=46.05 \mathrm{~cm}^{2}=46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{aligned}\)
Alternatively, area of the major sector  \(=\frac{(360-\theta)}{360} \times \pi r^{2}\)
\(\begin{array}{l} =\left(\frac{360-30}{360}\right) \times 3.14 \times 16 \mathrm{~cm}^{2} \\ =\frac{330}{360} \times 3.14 \times 16 \mathrm{~cm}^{2}=46.05 \mathrm{~cm}^{2} \\ =46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{array}\)

Since, Jaspal Singh repays his loan of Rs.118000, with first instalment of Rs 1000 and increases each instalment by Rs 100.
\(\therefore \) His instalments are Rs 1000, Rs 1100, Rs 1200, Rs 1300, ... which forms an A.P.
Here, first term is Rs 1000 and common difference is Rs 100.
\(\therefore \) 30th instalment = a₃₀ = a + 29d
 = Rs (1000 + 29 \(\times \) 100)
= Rs (1000 + 2900) = Rs 3900
Amount paid in 30 instalments = S₃₀
\(\Rightarrow \) S₃₀ =  Rs \(\quad \frac { 30 }{ 2 } (2\times 1000+29\times 100)\)
 \(=\ Rs\ 15 (2000+2900)\\ =\ Rs\ 15(4900)\ =\ Rs\ 73500\)
Amount of loan still have to pay
 =  Rs (118000 - 73500)
 =  Rs 44500 

(c) If Assertion is correct but Reason is incorrect.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(i) Let the production of TV sets in first year be a units.Then, production in the next consecutive years are a + d, a + 2d, ......
Thus, we get the sequence,
a, a + d, a + 2d, ...
This is an AP sequence, whose first term = a and common difference = d.
Given, T₆ = 16000 and T₉ = 22600
\(\therefore\) a + (6 - 1)d = 16000
and a + (9 - 1)d = 22600
[\(\because\) T_n = a + (n - 1)d]
\(\Rightarrow\) a + 5d = 16000  ....(i)
and  a + 8 d = 22600   ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
3d = 22600 - 16000
\(\Rightarrow\) 3d = 6600
\(\Rightarrow\) d = 2200
Putting d = 2200 in Eq. (i), we get
a + 5 \(\times\) 2200 = 16000
\(\Rightarrow\)  a = 16000 - 11000 = 5000
Hence, the production during first year is 5000 sets.
(ii) The production during 8th yr is
T₈ = a + (8 - 1) d
= 5000 + 7 \(\times\) 2200
= 5000 + 15400 = 20400
Hence, the production during 8th yr is 20400 sets.
(iii) The production during first 3 yrs,
\(\begin{aligned} S_3 & =\frac{3}{2}[2 a+(3-1) d] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{3}{2}[2 \times 5000+2 \times 2200] \end{aligned}\)
= 3(5000 + 2200]
= 3 \(\times\) 7200 = 21600
(iv) Let in n th year, the production is 29200.
\(\because\)   T_n = a + (n - 1)d
\(\therefore\)  29200 = 5000 + (n - 1) 2200
\(\Rightarrow\)(n - 1)2200 = 24200
\(\Rightarrow \quad(n-1)=\frac{24200}{2200}\)
\(\Rightarrow\) n - 1 = 1
\(\Rightarrow\) n = 12
(v) The difference of the production during 7th yr and 4th yr
= T₇ - T₄
= a + (7 - 1) d - [a + (4 - 1)d]
= 6d - 3d = 3d
=3 \(\times\) 2200 = 6600

(a) (i) Length of the engine = 11 x 200 = 2200 cm
= 22 m
(b) (iv) They are not the mirror image of one another.
(c) (iv) Actual width of the door = 0.35 x 200 = 70 cm
= 0.7m
(d) (ii) Their altitudes have a ratio 25 : 15.
(e) (iii) \(\triangle A B C \sim \triangle A D E\) ( AA Similarity)

\(\frac{A B}{A D}=\frac{B C}{D E}\)
\(\rightarrow \frac{x}{x+4}-\frac{3}{6}-\frac{1}{2}\) 
\(\Rightarrow 2 x=x+4 \Rightarrow x=4\) 

(i) (b): 

Here, OS the is radius of circle.
Since radius at the point of contact is perpendicularto tangent.
So, \(\angle\)OSA = 90°
(ii) (d): Since, length of tangents drawn from an external point to a circle are equal.
\(\therefore\) AS=AP,BP=BQ,
CQ = CR and DR = DS
(iii) (a): AP = AS = AD _ DS = AD _ DR (Using (1)
= 11 - 7 = 4 cm
(iv) (b): In quadrilateral OQCR,

\(\angle\)QCR = 60° (Given)
And \(\angle\)OQC = \(\angle\)ORC = 90° [Since, radius at the point of contact is perpendicular to tangent.]
\(\therefore\) \(\angle\)QOR = 360° - 90° - 90° - 60° = 120°
(v) (c): From (1), we have AS = AP, DS = DR,
BQ = BP and CQ = CR
Adding all above equations, we get
AS + DS + BQ + CQ = AP + DR + BP + CR
\(\Rightarrow\) AD + BC = AB + CD