Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 50² + 80² ≠ 100²
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.

\(-\frac { 145 }{ 12 } \)

(i)  \(\frac{3}{5}\)   (ii) \(\frac{4}{5}\)

Here, \(a=\sqrt { 3 } ,\ d-\sqrt { 12 } -\sqrt { 3 } =2\sqrt { 3 } -\sqrt { 3 } =\sqrt { 3 } \)
Next term to \(\sqrt { 27 } =\sqrt { 27 } +\sqrt { 3 } =3\sqrt { 3 } +\sqrt { 3 } =4\sqrt { 3 } \)
\(=\sqrt { 3\times 16 } =\sqrt { 48 } \)

Here \(\sqrt { 2x+9 } +x=13\)
\(\Rightarrow \sqrt { 2x+9 } =13-x\)
On squaring both side, we get
\(\left( \sqrt { 2x+9 } \right) ^{ 2 }=(13-x)^{ 2 }\)
\(\Rightarrow 2x+9=169+x^{ 2 }-26x\)
\(\Rightarrow x^{ 2 }-28x+160=0\)
\(\Rightarrow x^{ 2 }-20x-8x+160=0\)
\(\Rightarrow (x-20)(x-8)=0\)
\(\Rightarrow \) x = 20 or x = 8

In ΔDCA = ΔHGF,
∠DAC = ∠HFG ......(i)
\(\left[\begin{array}{l} \because \quad \Delta A B C \sim \Delta F E G \\ \therefore \angle C A B=\angle G F E \\ \Rightarrow \angle C A D=\angle G F H \text { or } \angle D A C=\angle H F G \end{array}\right]\)
and ∠DCA = ∠HGF   ........(ii)
\(\left[\begin{array}{l} \because \Delta A B C \sim \Delta F E G \\ \therefore \angle A C B=\angle F G E \\ \Rightarrow \frac{1}{2} \angle A C B=\frac{1}{2} \angle F G E \Rightarrow \angle D C A=\angle H G F \end{array}\right]\)
From Eqs. (i) and (ii),
ΔDCA - ΔHGF [by AA similarity criterion)

\(N=600\Rightarrow \frac { N }{ 2 } =300\)
Median class= 1000-2000
\(Median\quad =\quad l+\left( \frac { \frac { N }{ 2 } -c.f }{ f } \right) \times h\)
\(Median\quad =\quad 1000+\left( \frac { 300-250 }{ 190 } \right) \times 1000\)
\(=1000+\frac { 50 }{ 190 } \times 1000\)
=1000+263.16=1263.16
Median = Rs 1263.16

In \(\triangle\)ABC DE || AC,    (Given)
\(\frac { BD }{ DA } =\frac { BE }{ EC } \quad \quad \quad (BPT)\quad ...(i)\)
In \(\triangle\)ABE, DF || AE,    (Given)
\(\frac { BD }{ DA } =\frac { BF }{ FE } \quad \quad \quad (BPT)\quad ...(ii)\)
From (i) and (ii), we have
\(\frac { BF }{ FE } =\frac { BE }{ EC } \)
Hence proved.

a=p²,b=p²-q²,c=-q²
D=b²-4ac
= (p² - q²)²-4(p)²(-q)²
= p⁴+q⁴-2p³q³+4p²q²
= p⁴+q⁴-+2p²q²
=(p²+q²)³
 \(x=\frac { -b\pm \sqrt { b^{ 2 }-4ac } }{ 2a } \)
\(\therefore \quad x=\quad \frac { q^{ 2 }-p^{ 2 }\pm \sqrt { p^{ 2 }+q^{ 2 } } }{ 2p^{ 2 } } \)
\(\therefore x_{ 1 }=\frac { q^{ 2 }-p^{ 2 }\pm p^{ 2 }+q^{ 2 } }{ 2p^{ 2 } } =\frac { q^{ 2 } }{ p^{ 2 } } \)
and \(x_{ 2 }=\frac { q^{ 2 }-p^{ 2 }-p^{ 2 }-q^{ 2 } }{ 2p^{ 2 } } =-1\)
The roots are \(\frac { q^{ 2 } }{ p^{ 2 } } -1\)

Total number of balls = 19
(i) Prime numbers from 1 to 19 are 2, 3, 5, 7, 9, 11, 13, 17, 19 = Total 8 prime numbers
\(\therefore\)  Probability of drawing a prime number = \(\frac{8}{19}\)
(ii) Numbers divisible by 3 or 5 are 3, 6, 9, 15, 18, 10, 5, 12 = Total 8 numbers
\(\therefore\) Probability of drawing a number divisible by 3 or 5 = \(\frac{8}{19}\)
(iii) Number divisible by 5 and 10 are 5, 10, 15 = Total 3
\(\therefore\)  Numbers which are neither divisible by 5 nor 10 are 19 - 3 = 16
 \(\therefore\) Probability of drawing a number which is neither divisible by 5 nor by 10 = \(\frac{16}{19}\)
(iv) Even numbers from 1- 19 are 2, 4, 6, 8, 10, 12, 14, 16, 18 [Total 9 even numbers]
\(\therefore\)  Probability of drawing an even number = \(\frac{9} {19}\)

Considers \(d_1=d_2+4\) or \(d_2=d_1+4\), where \(d_1\) and \(d_2\) are the common differences of the two arithmetic progressions.
Also, \(a_{124}=b_{42} \Rightarrow a_1+123 d_1=b_1+41 d_2\) where \(a_1, a_{124}, b_1\) and \(b_{42}\) are the \(1 \mathrm{st}, 124 \mathrm{th}, 1\) st and 42 nd terms of the two arithmetic progressions, respectively.
\(\Rightarrow \quad a_1+123 d_1=a_1+41 d_2 \quad\left[\because a_1=b_1\right]\)
Solve the above two equations and the value of \(d_1=-2\) and \(d_2=-6\) or \(d_1=2\) and \(d_2=6\).
One set of possible value of common difference is as follows
\(a_{124}  =b_{42} \)
\(\Rightarrow a_1+123\left(4+d_2\right)  =b_1+41 d_2\)
\(\Rightarrow \quad 492+123 d_2  =41 d_2 \Rightarrow d_2=-6\)

(i) Let number of white balls be n.
Total number of balls = 6 + 4 + n = 10 + n
Let E be the event of drawing a white ball.
Given, \(P(E)=\frac{1}{3}\)
\(\begin{aligned} \Rightarrow \frac{\text { Number of outcomes favourable to } E}{\text { Total number of outcomes }}=\frac{1}{3} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{n}{10+n}=\frac{1}{3} \Rightarrow n=5 \end{aligned}\)
(ii) Now, total number of balls = 10 + n = 10 + 5 = 15
Let x red balls are removed from the bag.
\(\therefore\)  Total number of balls in the bag = 15 - x
Now, \(\begin{gathered} P(E)=\frac{1}{2} \end{gathered}\)
\(\begin{gathered} \Rightarrow \frac{5}{15-x}=\frac{1}{2} \end{gathered}\)
\(\Rightarrow\)    x = 5

Total distance to be travelled by aircraft  = 2800 km
Let the original average speed be x km/h.
Original duration of the flight = \(\frac{2800}{x} \mathrm{~h}\)
It is given that the average speed is reduced by 100 km/h.
So, the reduced average speed = (x - 100) km/h
Time taken to cover the distance at the reduced
\(\text { speed }=\frac{2800}{x-100} h\)
According to the given condition,
\(\frac{2800}{x-100}-\frac{2800}{x}=\frac{30}{60}\)
\(\Rightarrow \frac{2800 x-2800 x+280000}{x(x-100)}=\frac{1}{2}\)
\(\Rightarrow\)            x(x-100) = 560000
\(\Rightarrow\)           x² - 100x - 560000 = 0
\(\Rightarrow\) x² - 800x + 700x - 560000 = 0
\(\Rightarrow\) x(x - 800) + 700(x - 800) = 0
\(\Rightarrow\)      (x - 800) (x + 700) = 0
\(\Rightarrow\)        x = 800,  x \(\neq\)- 700      [speed cannot be negative]
\(\Rightarrow\)       x = 800
\(\therefore\)     Original duration of flight = \(\frac{2800}{800} h=\frac{7}{2} h=3 \frac{1}{2} h\)

Here, we observe that 5 students have scored marks below 10, i.e it lies between class interval 0-10 and 9 students have scored marks below 20.
So, (9-5)=4 students lie in the class interval 10-20. Continuing in the same manner, we get the following frequency distribution table for given data.

Here, assumed mean(a)=45 and class width (h)=10
By step deviation method,
Mean \(\left( \overline { x } \right) =a+\left\{ \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right\} \times h\)
\(=45+\left\{ \frac { 29 }{ 85 } \right\} \times 10=45+\frac { 58 }{ 17 } \\ =45+3.41=48.41\)

In Figure (i), corresponding sides are not in proportion.
In Figure (ii), In \(\triangle LMN\)
\(\angle LMN+\angle MNL+\angle MLN\) = 180°
[by angle sum property of a triangle]
45° + \(\angle MNL\) + 57° = 180°
\(\angle MNL\) = 78°
\(\angle QPR=\angle LMN\) = 45°
\(\angle PQR = \angle MNL\) = 78°
So, \(\angle PQR\sim \angle MNL\)   [by AA similarity criterion]

Coordinates of midpoint of line joining two points (16, 4) and (36, 6) are:

(i) (c) Given, number of students in section A = 32 and number of students in section B = 36 The minimum number of books acquire for the class library
= LCM of (32,36) = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 = 2⁵ \(\times\) 3²= 32 \(\times\) 9 = 288
(ii) (b) Given, product of the two numbers
=LCM \(\times\) HCF
\(\therefore\) 32 \(\times\) 36 = LCM (32, 36) \(\times\) HCF (32, 36)
\(\Rightarrow\) 32 \(\times\) 36 = 288 \(\times\)HCF(32, 36)
\(\Rightarrow \operatorname{HCF}(32,36)=\frac{32 \times 36}{288}=4\)
(iii) (a) The prime factors of 36 are
36 = 2 \(\times\) 2 \(\times\)3 \(\times\) 3 = 2²\(\times\) 3²
(iv) (b) 7 \(\times\) 11  \(\times\) 13 \(\times\) 15  \(\times\)15 = 15015 + 15 = 15030
Hence, it is a composite number.
(v) (b) Given, p = ab² and q = a²b
LCM (p, q) = Product of the greatest power of each prime factor involved in the numbers, with highest power
= a² \(\times\)b²

(i) (d)
(ii) (b)
(iii) (a): x(x + 3) + 7 = 5x - 11
\(\Rightarrow x^{2}+3 x+7=5 x-11\)
\(\Rightarrow x^{2}-2 x+18=0 \) is a quadratic equation.
\((b) (x-1)^{2}-9=(x-4)(x+3)\)
\(\Rightarrow x^{2}-2 x-8=x^{2}-x-12\)
\(\Rightarrow x-4=0\) is not a quadratic equation.
\((c) x^{2}(2 x+1)-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}+x^{2}-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}-4 x^{2}+6=0\) is not a quadratic equation.
\((d) x(x-1)(x+7)=x(6 x-9)\)
\(\Rightarrow x^{3}+6 x^{2}-7 x=6 x^{2}-9 x\)
\(\Rightarrow x^{3}+2 x=0\)   is not a quadratic equation.
(iv) (d)
(v) (d)

(i) (a): For student Anu: Fixed charge + cost of food for 25 days = Rs 4500
i.e., x + 25y = 4500 For student Bindu:
Fixed charges + cost of food for 30 days = Rs 5200
i.e., x + 30y = 5200
(ii) (b): From above, we have a₁ = 1, b₁, = 25
\(c_{1}=-4500 \text { and } a_{2}=1, b_{2}=30, c_{2}=-5200 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{25}{30}=\frac{5}{6}, \frac{c_{1}}{c_{2}}=\frac{-4500}{-5200}=\frac{45}{52} \)
\(\Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution
(iii) (c) : We have x + 25y = 4500 .......(i)
and x + 30y = 5200                      ........(ii)
Subtracting (i) from (ii), we get
5y=700 \(\Rightarrow\) y=140
\(\therefore\) Cost of food per day is Rs 140
(iv) (c): We have, x + 25y = 4500
\(\Rightarrow\) x = 4500 - 25 x 140
\(\Rightarrow\) x = 4500 - 3500 = 1000
\(\therefore\) Fixed charges per month for the hostel is Rs 1000
(v) (d): We have, x = 1000, Y = 140 and Bindu takes food for 20 days.
\(\therefore\) Amount that Bindu has to pay = Rs (1000 + 20 x 140) = Rs 3800

(b) S = {HH, HT, TH, TT}
Favourable outcomes = {HH}
Total number of outcomes = 4
\(\therefore \text { Probability }=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{4}\)
Reason is true but not correct explanation of Assertion.

(d) Let the point C divides A (1, 2) and B(-1, 1) internally in the ratio 1 : 2.
Then, by section formula, we have
\(x=\frac{m_1 x_2+m_2 x_1}{m_1+m_2} \text { and } y=\frac{m_1 y_2+m_2 y_1}{m_1+m_2}\)
So,  \(x=\frac{2 \times 1+1 \times(-1)}{2+1} \text { and } y=\frac{2 \times 2+1 \times 1}{2+1}\)
\(\Rightarrow x=\frac{2-1}{3} \text { and } y=\frac{4+1}{3} \Rightarrow x=\frac{1}{3} \text { and } y=\frac{5}{3}\)
\(\therefore \quad x=\frac{1}{3} \text { and } y=\frac{5}{3}\)
\(\therefore\) Assertion (A) is false but Reason (R) is true.