Given pair of linear equations is
2x+3y=11   ..(i)
and 2x-4y=-24   ...(ii)
From Eq. (i) we have
4y=2x+24
\(\Rightarrow \quad y=\frac { 2x+24 }{ 4 } \)
\(\Rightarrow \quad y=\frac { x+12 }{ 2 } \quad \quad \quad ..(iii)\)
On substituting the value of y from Eq. (ii) in Eq. (i), we get
\(2x+3\left( \frac { x+12 }{ 2 } \right) =11\)
\(\Rightarrow\) 4x+3(x+12)=11 x 2
\(\Rightarrow\) 4x+3x+36=22 \(\Rightarrow\) 7x=22-36
\(\Rightarrow\) 7x=-14 \(\Rightarrow\) x=-2
On substituting x=-2 in Eq. (iii), we get
\(\Rightarrow \quad y=\frac { -2+12 }{ 2 } \Rightarrow \frac { 10 }{ 2 } \Rightarrow y=5\)
On substituting x=-2 and y=5 in the equation y=mx+3, we get
5=m x (-2)+3
\(\Rightarrow\)5=-2m+3
\(\Rightarrow\) 2m=3-5=-2
\(\Rightarrow\) m=-1

1902m

Let ABC is the right-angled triangle.
Let BC=x units and AB=y units.
Also AC Hypotenuse=29 units

Now Perimeter=70 units
\(\Rightarrow\) x+y+29=70
\(\Rightarrow\) x+y=41
\(\Rightarrow\) y=41-x    ....(i)
Also, AC²=AB²+BC²
\(\Rightarrow\) (29)²=(y)²+x²
\(\Rightarrow\) 841=(41-x)²+x² [Using (i)]
\(\Rightarrow\) 841=1681+x²-82x+x²
\(\Rightarrow\) 2x²-82x+840=0
\(\Rightarrow\) x²-41x+420=0
\(\Rightarrow\) (x-20)(x-21)=0
\(\Rightarrow\)  x=20 or x=21
Wnen x=20, y=41-x=21 and when x=21, y=41-21=20
\(\therefore\)  Length of other sides are 20 units and 21 units.

Let the speed of the passenger train be x km/hr. Then,
Speed of the express train = (x + 11)km/hr
Time taken by the passenger train to cover 132 km between Mysore to Bangalore \(\frac{132}{x} \mathrm{hr}\)
Time taken by the express train to cover 132 km between Mysore to Bangalore = \(=\frac{132}{x+11} \mathrm{hr}\)
Therefore,
\( \frac{132}{x}-\frac{132}{x+11}=1 \)
\( \frac{132(x+11)-132 x}{x(x+11)}=1 \)
\(\frac{132 x+1452-132 x}{x^{2}+11}=1 \)
\(\frac{1452}{x^{2}+11}=1 \)
1452 = x² + 11
x² + 11 - 1452 = 0
x2 - 33x + 44x - 1452 = 0
x(x - 33) + 44(x - 33) = 0
(x - 33)(x + 44) = 0
So, either 
x - 33 = 0
x = 33
Or
x + 44 = 0
x = -44
But, the speed of the train can never be negative.
Thus, when x = 33 then speed of express train
= x + 11
= 33 + 11
= 44
e speed of the passenger train is x = 33 km/hr
and the speed of the express train is x = 44 km/hr respectively.

According to the question, sum of zeroes = 3
Product of zeroes = \(-\frac{2}{5}\)
The required quadratic polynomial
= x² - x (Sum of zeroes) + Product of zeroes
= x² - x(3) - \(\frac{2}{5}\)
= x² - 3x - \(\frac{2}{5}\)
= \(\frac{1}{5}\left( 5{ x }^{ 2 }-15x-2 \right) \)
\(\therefore\)  The quadratic polynomial is \(\left( 5{ x }^{ 2 }-15x-2 \right) \frac{1}{5}\)

No

Let coordinates of R be (x,y)

\(AR=\frac { 3 }{ 4 } AB\)
But \(AR+RB=AB\quad \Rightarrow \quad \frac { 3 }{ 4 } AB+RB=AB\)
\(\Rightarrow \ RB=AB-\frac { 3 }{ 4 } AB=\frac { 4AB-3AB }{ 4 } =\frac { AB }{ 4 } \)
\(\frac { AR }{ RB } =\frac { \frac { 3 }{ 4 } AB }{ \frac { 1 }{ 4 } AB } =\frac { 3 }{ 4 } :\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \times \frac { 4 }{ 1 } =3:1\)
Thus, R divides AB in the ratio 3:1.
\(x=\frac { 3\times 0+1\times (-4) }{ 3+1 } =\frac { 0-4 }{ 4 } =\frac { -4 }{ 4 } =-1\) and \( y=\frac { 3\times 6+1\times 0 }{ 3+1 } =\frac { 18+0 }{ 4 } =\frac { 18 }{ 4 } =\frac { 9 }{ 2 } \)
Thus , coordinates of R are \(\left( -1,\frac { 9 }{ 2 } \right) \).

Let, Radius of the circle = r
\(\therefore\)  Circumference of the circle = \(2\pi r\)
A.T.Q \(2\pi r\) = 22 cm
\(\Rightarrow\) \(2\times {22\over 7}\times r=22\ \ \ \Rightarrow\ \ \ \ r={{22\times 7\over 2\times 22}}={7\over 2}cm\)
Area of quadrant of the circle \(={\pi r^2\theta\over 360^o}={22\over 7}\times {7\times 7\over 2\times 2}\times {90^o\over 360^o}={22\times 7\over 2\times 2\times 4}={77\over 8}cm^2\) 

Given,     AP is 3, 8, 13,........., 253
Here,       a = 3, d = 8 - 3 = 5
First term from the last = 253 then d = -5
a₂₀ = a + 19d = 253 + 19 (-5) = 253 - 95 = 158

Length of wiper blade = 30 cm = r
\(\theta\) = 105°
\(\therefore\) Area of cleaned by two blade
= 2 \(\times\) Area of sector formed by blade
\(\begin{aligned} & =2 \times \frac{\theta}{360^{\circ}} \times \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{2 \times 105^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(30)^2 \end{aligned}\)
= 825 \(\times\) 2 cm²
= 1650 cm²

(i) P (single digit number) = \(\frac{9}{100}\)
(ii) P (perfect square) = \(\frac{1}{10}\)
(iii) P (a number which is divisible by 7) = \(\frac{7}{50}\)

We have, \(\frac { \cos { { 60 }^{ 0 } } +\sin { { 45 }^{ 0 } } -\cot { { 30 }^{ 0 } } }{ \tan { { 60 }^{ 0 } } +\sec { { 45 }^{ 0 } } -cosec{ 30 }^{ 0 } } \)
\(=\frac { \frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 2 } } -\sqrt { 3 } }{ \sqrt { 3 } +\sqrt { 2 } -2 } =\frac { \frac { \sqrt { 2 } +2-2\sqrt { 2 } \times \sqrt { 3 } }{ 2\sqrt { 2 } } }{ \sqrt { 3 } +\sqrt { 2 } -2 } \)
\([ \because \cos { { 60 }^{ 0 } } =1/2,\sin { { 45 }^{ 0 } } =1/\sqrt { 2 } ,\cot { { 30 }^{ 0 } } =\sqrt { 3 }\)
\( \tan { { 60 }^{ 0 } } =\sqrt { 3 } ,\sec { { 45 }^{ 0 } } =\sqrt { 2 } \quad and\quad cosec{ 30 }^{ 0 }=2 ] \)
\(=\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 2 } (\sqrt { 3 } +\sqrt { 2 } -2) } \)
\(=\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 6 } +2\sqrt { 2 } \times \sqrt { 2 } -2\sqrt { 2 } \times 2 } =\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 6 } +4-4\sqrt { 2 } } \) 

Given, a and β are the zeroes of the polynomial
p(x)=ax²+bx+c.
Sum of zeroes, a+β=\(-b\over a\) and product of zeroes, aβ=\(c\over a\)
Now, a²β+aβ²=aβ(a+β)
\(=\frac { c }{ a } \times \frac { (b) }{ a } =\frac { -bc }{ a^{ 2 } } \)

Let us assume to the contrary that \(3+2\sqrt { 5 } \)  is a rational number. Then, it can be expressed in the form \(\frac{a}{b}\), where a, b are coprime integers and \(b\neq 0\)
Now,  \(3+2\sqrt { 5 } \) = a/b, where a,b are integers and \(b\neq 0\)
On rearranging, we get
\(2\sqrt { 5 } =\frac { a }{ b } -3\quad or\quad \sqrt { 5 } =\frac { a }{ 2b } -\frac { 3 }{ 2 } \)
Since, a, b are integers and \(b\neq 0\) , therefore \(\frac{a}{2b}\) is rational number and so \(\frac{a}{2b}\) - \(\frac{3}{2}\) is a rational number.
[since, difference of two rational numbers is also a rational number]
\(\Rightarrow \sqrt { 5 } \) is a rational number. But \(\sqrt { 5 } \) is an irrational number.
This shows that our assumption is incorrect.
So, \(3+2\sqrt { 5 } \) is irrational.

AB²=(-2-7)²+(5-10)² = (-9)²+(-5)² =81+25 =106
BC²=(3-(-2))²+(-4-5)²=(5)²+(-9)²=25+81=106
AC²=(3-7)²+(-4-10)²=(4)^2₊(14)²=16+196=212
Since AB²+BC²=AC²
∴ ABC is a right triangle.
AB = \(\sqrt { 106 } \) and BC = \(\sqrt { 106 } \)
∵ AB = BC
∴ ABC is an isosceles right triangle. 

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

(d) If Assertion is incorrect but Reasonis correct.

Total number of possible outcome, n(S) = 200
(i) (c): Let A be the event that number on selected card is divisible by 10.
A = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120,130, 140, 150,160, 170, 180, 190,200}
\(\Rightarrow\) n(A) = 20
\(\begin{equation} P(A)=\frac{n(A)}{n(S)}=\frac{20}{200}=\frac{1}{10} \end{equation}\)
(ii) (c) : Let B be the event that the number on the selected card is a prime number more than 100 but less than 150.
B = {101, 103, 107, 109, 113, 127, 131, 137, 139, 149}
\(\Rightarrow\)n(B) = 10
\(\begin{equation} P(B)=\frac{n(B)}{n(S)}=\frac{10}{200}=\frac{1}{20} \end{equation}\)
(iii) (c) : Let C be the event that the number on the selected card is a multiple number of 3.
C = {3, 6, 9,12, ..... ,192,195, 198}
\(\Rightarrow\) n(C) = 66
\(\begin{equation} P(C)=\frac{n(C)}{n(S)}=\frac{66}{200}=\frac{33}{100} \end{equation}\)
(iv) (a) : Let D be the event that the number on the selected card is a perfect square.
D = {I, 4, 9, 16,25,36,49,64,81, 100, 121, 144, 169, 196}
\(\Rightarrow\)n(D) = 14
\(\begin{equation} P(D)=\frac{n(D)}{n(S)}=\frac{14}{200}=\frac{7}{100} \end{equation}\)
(v) (a): Let E be the event that the number on the selected card is a perfect cube.
E = {I, 8, 27, 64, 125}
\(\Rightarrow\)n(E) = 5
\(\begin{equation} P(E)=\frac{5}{200}=\frac{1}{40} \end{equation}\)

(i) (c): \(\text { In } \Delta O P Q\), we have
\(\tan 60^{\circ}=\frac{P Q}{P O} \)
\(\Rightarrow \sqrt{3}=\frac{20}{P O} \)
\(\Rightarrow P O=\frac{20}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): In \(\Delta\)ORS, we have
\(\tan 30^{\circ}=\frac{R S}{O R} \Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{O R} \Rightarrow O R=20 \sqrt{3} \mathrm{~m}\)
(iii) (d): Clearly, width of the road = PR
\(\begin{array}{l} =P O+O R=\left(\frac{20}{\sqrt{3}}+20 \sqrt{3}\right) \mathrm{m} \\ =20\left(\frac{4}{\sqrt{3}}\right) \mathrm{m}=\frac{80}{\sqrt{3}} \mathrm{~m}=46.24 \mathrm{~m} \end{array}\)
(iv) (a): \(\text { In } \Delta O P Q \text { , if } \angle P O Q=45^{\circ} \text { , then }\)
\(\tan 45^{\circ}=\frac{P Q}{P O} \Rightarrow 1=\frac{20}{P O} \Rightarrow P O=20 \mathrm{~m}\)
(v) (b)

Here the smallest 3-digit number divisible by 7 is 105. So, the number of bacteria taken into consideration is 105, 112, 119, .... ,994 So, first term (a) = 105, d = 7 and last term = 994
(i) (c): t₅ = a + 4d = 105 + 28 = 133
(ii) (b): Let n samples be taken under consideration.
\(\because\) Last term = 994
\(\Rightarrow\) a + (n - 1)d = 994 \(\Rightarrow\) 105 + (n - 1)7 = 994 \(\Rightarrow\) n = 128
(iii) (a): Total number of bacteria in first 10 samples
\(=S_{10}=\frac{10}{2}[2(105)+9(7)]=1365\)
(iv) (a): t₇ from end = (128 - 7 + 1)^th term from beginning = 122^th term = 105 + 121(7) = 952
(v) (c): t₅₀ = 105 + 49 x 7 = 448