The given equations can be rewritten as
2x - 3y - 8 = 0 and 4x - 6y - 9 = 0
On comparing with standard form of pair of linear equations, we get
a₁ = 2, b₁ = -3, c₁ = -8
and a₂ = 4, b₂ = -6, c₂ = -9
Now, \(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2},\frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2} and \frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}\)
Thus, \(\frac{1}{2}=\frac{1}{2}\neq \frac{8}{9}\) i.e.,\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, the pair of linear equations is inconsistent.

Given, HCF of 306 and 657 = 9
We know that
LCM \(\times\) HCF = Product of two numbers
\(\Rightarrow\) LCM \(\times\) 9 = 306 \(\times\) 657
\(\Rightarrow\)   LCM = \( \frac{306 \times 657}{9}\)
= 34 \(\times\) 657 = 22338
\(\therefore\) LCM of 306 and 657 = 22338

Sum of zeroes,
\(S={3-\sqrt3\over 5}+{3+\sqrt3\over 5}={3-\sqrt3+3+\sqrt3\over 5}={6\over 5}\)
Product of zeroes
\(P=\left(3-\sqrt3\over5\right)\times\left(3+\sqrt3\over 5\right)={3^2-(\sqrt{3})\over 5\times5}={6\over 5}\)
Polynomial
\(p(x) =x^2+ P = x^2 -{6\over 5}x+{6\over 25}\)
=k(25x² - 30x + 6)

Let  \(5\sqrt { 6 } \)  be a rational number, which can be put in a/b, where b\(\neq \)0; a and b are co-prime
\(5\sqrt { 6 } =\frac { a }{ b } \)
\(\sqrt { 6 } =\frac { a }{ 5b } \)
⇒  \(\sqrt { 6 } \) =rational
But, we know that  \(\sqrt { 6 } \)   is an irrational number.
Thus, our assumption is wrong.
Hence, \(\sqrt { 6 } \) is an irrational number

Given, pair of equations is
5x+\(\lambda \)y=4
and 15x+3y=12
Here, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 5 }{ 15 } =\frac { 1 }{ 3 } ,\quad \frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { \lambda }{ 3 } \)  
and  \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -4 }{ -12 } =\frac { 1 }{ 3 } \)
Condition for infinitely many solutions is
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \frac { 1 }{ 3 } =\frac { \lambda }{ 3 } =\frac { 1 }{ 3 } \Rightarrow \lambda =1\)

2x² - 6x + 3 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac
= (-6)² - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation
\( x=\frac{-b \pm b^{2}-4 a c}{2 a} \)
\(=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(2)(3)}}{2(2)} \)
\(=\frac{6 \pm \sqrt{12}}{4}=\frac{6 \pm 2 \sqrt{3}}{4} \)
\(=\frac{3 \pm \sqrt{3}}{2} \)
Therefore the root are \(=\frac{3 \pm \sqrt{3}}{2} \)

9x²-17x+8=0
9x²-9x-8x+8=0
(9x-8)(x-1)=0
x=\(8\over 9\) or x=1

7x – 15y = 2  ..(1)
x + 2y = 3...(2)
Step 1 :We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (2) :
x + 2y = 3
and write it as x = 3 – 2y  ...(3)
Step 2 : Substitute the value of x in Equation (1). We get
7(3 – 2y) – 15y = 2
i.e., 21 – 14y – 15y = 2
i.e., – 29y = –19
Therefore,\(y=\frac{19}{29}\)  
Step 3 : Substituting this value of y in Equation (3), we get
\(x=3-2\left(\frac{19}{29}\right)=\frac{49}{29}\) 
Therefore, the solution is \(x=\frac{49}{29}, y=\frac{19}{29}\) 
Verification : Substituting \(x=\frac{49}{29}, y=\frac{19}{29}\) , you can verify that both the Equations (1) and (2) are satisfied.
To understand the substitution method more clearly, let us consider it stepwise:
Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and 10 below, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent
Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

We have, \(\alpha+\beta=3\) and \(\alpha \beta=-2\) Now, required polynomial is given by
\(x^{2}-\left(\frac{2 \alpha}{\beta}+\frac{2 \beta}{\alpha}\right) x+\left(\frac{2 \alpha}{\beta}\right) \cdot\left(\frac{2 \beta}{\alpha}\right)\) 
\(=x^{2}-2\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) x+4\) 
\(=x^{2} - 2\left(\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}\right) x+4\) 
Ans.x²+ 13x+ 4

We have 117 = 65 x 1 + 52
65 = 52 x 1 + 13
and 52 = 13 x 4 + 0
Hence, HCF = 13
65m-117 =13
⇒ 65m = 117 + 13 = 130
∴  m=\(\\ m=\frac { 130 }{ 65 } =2\)
Now,    65=13 x 5
117 =3²x 13
LCM = 13 x 5 x 3²= 585

Let the present age of Sagar be x yr and the age of Tiru be y yr.
5 yr ago, Sagar's age = (x - 5) yr and Tiru's age = (y - 5) yr
According to the given condition,
(x-5)=2(y-5)
\(\Rightarrow\) x-5=2y-10
\(\Rightarrow\) x-2y+5=0
After 10 yr, Sagar's age = (x + 10) yr and Tiru's age = (y + 10) yr
According to the given condition,
x+10=(y+10)=10
\(\Rightarrow\)x+10=y+20
\(\Rightarrow\) x-y-10=0
Thus, we get the following pair of linear equations
x-2y+5=0  ..(i)
x-y-10=0   ..(ii)
Now, let us draw the graphs of Eqs.(i) and (ii), by finding at least two solutions for each of these equations: The solutions of the equations are given in tables.
Table for x-2y+5=0 or \(y=\frac { x+5 }{ 2 } \) is

Table for x-y-10=0 or y=x-10 is

Plot the points A (5, 5) and B (-5,0) and join them to get the line AB. Similarly, plot the points C (5,-5) and D (10, 0) and join-them to get the line CD.

It is clear from the graph that, lines AB and CD intersect each other at point E (25,15).
So, x= 25 and y=15 is the required solution.
Hence, Sagar's present age = 25 yr and Tiru's present age = 15 yr.

Let \(f(y)=7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 }\)
 \(=\frac { 21y^{ 2 }-11y-2 }{ 3 } \)
\(=\frac { 21y^{ 2 }-14y+3y-2 }{ 3 } \)  [by splitting the middle term]
\(=\frac { 7y(3y-2)+1(3y-2) }{ 3 } \)
\(=\frac { 1 }{ 3 } (3y-2)(7y+1)\)
So, the value of \(7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 } \) is zero whe 3y-2=0 or 7y+1=0, i.e. when \(y=\frac { 2 }{ 3 } \) or \(y=\frac { 1 }{ 7 } \).
Thus, the zeroes\(=\frac { 2 }{ 3 } -\frac { 1 }{ 7 } =\frac { 14-3 }{ 21 } =\frac { 11 }{ 21 } =-\left( \frac { -11 }{ 3\times 7 } \right) \)
\(=-(1).\left( \frac { Coefficient \ of \ y }{ Coefficient \ of \ y^{ 2 } } \right) \) and product of zeroes\(=\left( \frac { 2 }{ 3 } \right) \left( -\frac { 1 }{ 7 } \right) =\frac { -2 }{ 21 } =\frac { -2 }{ 3\times 7 } \)
\(=\left( \frac { Constant \ term }{ Coefficient \ of \ y^{ 2 } } \right) \)
Hence, the relations between the zeroes and the coefficients of the polynomials is verified.

Let a and 2a are the zeroes of the polynomial 2x²-5x-(2k+1).
Then, \(\alpha +2\alpha =\frac { 5 }{ 2 } \Rightarrow 3\alpha =\frac { 5 }{ 2 } \Rightarrow \alpha =\frac { 5 }{ 6 } \) 
\(2\alpha =2\times \frac { 5 }{ 6 } =\frac { 5 }{ 3 } \)
So, the zeroes of the polynomial are \(\frac { 5 }{ 6 } \) and \(\frac { 5 }{ 3 } \)
Now, as \(\alpha \times 2\alpha =\frac { -2k-1 }{ 2 } \)
\(\frac { 5 }{ 6 } \times \frac { 5 }{ 3 } =\frac { -2k-1 }{ 2 } \Rightarrow \frac { 25 }{ 18 } =\frac { -2k-1 }{ 2 } \)
\(\Rightarrow \quad 25-9(-2k-1)\) 
\( \Rightarrow \quad 25=-18k-9\) 
\( \Rightarrow \quad 18k=-9-25\)

Here, n is a natural number and let 6ⁿ ends with 0.
Hence, 6ⁿ is divisible by 5.
But the prime factors of 6 are 2 and 3, so 5 is not a factor.
\(\Rightarrow \)  6ⁿ = (2 x 3)ⁿ
In the prime factorisation of 6ⁿ , 5 is not a factor.
By using the fundamental theorem of arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order, in which the prime factors occur.
So, our assumption, 6ⁿ ends with 0, is wrong.
Thus, there does not exist any natural number n, for which 6ⁿ ends with zero.

\(\frac { 4x }{ x-2 } -\frac { 3x }{ x-1 } =7\frac { 1 }{ 2 } \)
\(\Rightarrow \frac { 4x(x-1)-3x(x-2) }{ (x-2)(x-1) } =\frac { 15 }{ 2 } \)
\(\Rightarrow \frac { 4x^{ 2 }-4x-3x^{ 2 }+6x }{ x^{ 2 }-3x+2 } =\frac { 15 }{ 2 } \)
\(\Rightarrow 2(x^{ 2 }+2x)=15(x^{ 2 }-3x+2)\)
\(\Rightarrow 2x^{ 2 }+4x=15x^{ 2 }-45x+30\)
\(\Rightarrow 13x^{ 2 }+49x+30=0\)
\(\Rightarrow 13x-39x-10x+30=0\)
\(\Rightarrow (x-3)(13x-10)=0\)
\(\Rightarrow x=3\quad x=\frac { 10 }{ 13 } \)

Let the time taken by smaller tap to fill tank completely = x hrs
So, volume of tank filled by smaller tap in 1 hr \(=\frac{1}{x}\)
Volume of tank filled by larger tap in 1 hr \(=\frac{1}{x-2}\)
Now, time taken by both taps to fill \(=1 \frac{7}{8}=\frac{15}{8} \mathrm{hrs}\)
Tank filled by smaller tap in \(\frac{15}{8} \mathrm{hrs}=\frac{1}{x} \times \frac{15}{8}=\frac{15}{8 x}\)
Tank filled by larger tap in \(\frac{15}{8} \mathrm{hrs}=\frac{1}{x-2} \times \frac{15}{8}=\frac{15}{8(x-2)}\)
Therefore, \(\frac{15}{8 x}+\frac{15}{8(x-2)}=1 \Rightarrow \frac{15}{8}\left[\frac{1}{x \mid y}+\frac{1}{x-2}\right]=1\)
\(\Rightarrow \quad \frac{2(x-1)}{x^{2}-2 x}=\frac{8}{15}\)
⇒ 15(x -1) = 4(x² - 2x)
⇒ 23x = 4x² + 15
⇒ 4x² - 23x + 15 = 0
\( x=\frac{-(-23) \pm \sqrt{(-23)^{2}-4 \cdot 4 \cdot 15}}{2 \cdot 4} \)
\(x=\frac{23 \pm 17}{8} \)
Taking positive sign,
\(x=\frac{23+17}{8}=5\)
Taling negative sign,
\(x=\frac{23-17}{8}=\frac{3}{4}\)
When, taking x = 5
Time taken by smaller tap = 5 hrs
Time taken by larger tap = x - 2 = 5 - 2 = 3 hrs
When, taking x = \(\frac{3}{4}\)
Time taken by smaller tap \(=\frac{3}{4} \mathrm{hr}\)
Time taken by larger tap = x - 2
\(=\frac{3}{4}-2=\frac{-5}{4}\), Its not solution.
Hence, time taken by smaller tap = 5 hr and time taken by larger tap = 3 hr.

Let the fraction be \(\frac { x }{ y } \)
\(\frac { x-2 }{ y+1 } =\frac { 1 }{ 2 } \)
\(\Rightarrow\)  2x - 4 = y + 1
\(\Rightarrow\) 2x - y = 5        ......(i)
Also,  \(\frac { x+4 }{ y-3 } =\frac { 3 }{ 2 } \)
\(\Rightarrow\)  2x + 8 = 3y - 9
\(\Rightarrow\) 2x - 3y = - 17  ......(ii)
Subtracting this value of y in eqn. (i),
2x - 11 = 5
\(\therefore\) x = 8
Hence,  x = 8, y = 11
\(\therefore \) Fraction \(=\frac { 8 }{ 11 }\)

p(x) = 2x² + 5x + k
Sum of zeroes \(=-\frac { Coeff. \ of \ x }{ Coeff. \ of \ { x }^{ 2 } } \)
\(=\alpha+\beta=\frac{-5}{2}\)
and product of zeroes \(=\frac { Costant \ term }{ Coeff. \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad =\alpha\beta=\frac{k}{2}\)
According to the question,
\({ \alpha }^{ 2 }+{ \beta }^{ 2 }+\alpha \beta =\frac { 21 }{ 4 } \)
\(\Rightarrow \quad \left( \alpha +\beta \right) ^{ 2 }-2\alpha \beta +\alpha \beta =\frac { 21 }{ 4 } \)
\(\left[ \because \left( \alpha +\beta \right) ^{ 2 }={ \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta \quad \right] \)
\(\Rightarrow \quad \left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { k }{ 2 } =\frac { 21 }{ 4 } \)
\(\Rightarrow \quad ( \frac { 25 }{ 4 })-\frac { 21 }{ 4 } = \frac { k }{ 2 }\)
\(\Rightarrow \quad 1=\frac{k}{2}\Rightarrow k=2\)
Hence, k = 2

x=-2, y=-1

\(\frac { \alpha }{ \beta } +\frac { \alpha }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta \)
\(=\frac { \alpha ^{ 2 }+\beta ^{ 2 } }{ \alpha \beta } +2\left( \frac { \alpha +\beta }{ \alpha \beta } \right) +3\alpha \beta =-\frac { 2 }{ 3 } \)

Hint LCM of 48. 72 and 108 = 2 x 2 x 2 x 2 x 3 × 3 x 3 = 432
Therefore, after 432 s, they will change simultaneously. We know that 60 s = 1min
⇒ 1s = 1/60 min
⇒ 432 s = 432/60 min =7min 12 s
Hence, the lights change simultaneously at 7 : 07 : 12 am.

(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

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

(i) (b): Roots of the quadratic equation are 2 and -3.
\(\therefore\) The required quadratic equation is 
\((x-2)(x+3)^{n}=0 \Rightarrow x^{2}+x-6=0\)
(ii) (a): We have, 2x² + kx + 1 = 0
Since, -1/2 is the root of the equation, so it will satisfy the given equation
\(\therefore \quad 2\left(-\frac{1}{2}\right)^{2}+k\left(-\frac{1}{2}\right)+1=0 \Rightarrow 1-k+2=0 \Rightarrow k=3\)
(iii) (d): If the roots of the quadratic equations are opposites to each other, then coefficient of x (sum of roots) is 0.
So, both (a) and (b) have the coefficient of x = 0.
(iv) (c): The given equation is (x - 2)² + 19 = 0
\(\Rightarrow x^{2}-4 x+4+19=0 \Rightarrow x^{2}-4 x+23=0\)
(v) (b): If one root of a quadratic equation is irrational, then its other root is also irrational and also its conjugate i.e., if one root is p +.\(\sqrt(q)\) then its other root is p -.\(\sqrt(q)\).

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850
(ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900
(iii) (c) : We have 2x + 3y = 850 .........(i)
and 3x + 2y = 900 .........(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get
5y = 750 \(\Rightarrow\) Y = 150
Thus, price of one Physics book is Rs 150.
(iv) (d): From equation (i) we have, 2x + 3 x 150 = 850
\(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200
Hence, cost of one Mathematics book = Rs 200
(v) (a): From above, we have
\(a_{1} =2, b_{1}=3, c_{1}=-850 \)
\(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\)
\(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus system of linear equations has unique solution.

(i) (c): Here, the simplest form of given options are
125/441 = 5³/(3² x 7²), 77/210 = 11/(2 x 3 x 5),
15/1600 = 3/(2⁶ x 5) Out of all the given options, the denominator of option (c) alone has only 2 and 5 as factors. So, it is a terminating decimal.
(ii) (b): 23/(2³ x 5²) = 23/200 = 0.115
(iii) (a): 441/(2² x 5⁷ x 7²) = 9/(2² x 5⁷), which is a terminating decimal.
(iv) (d): The fraction form of a non-terminating recurring decimal will have at least one prime number other than 2 and 5 as its factors in denominator. So, p can take either of 3, 7 or 15.
(v) (a): Here denominator has only two prime factors i.e., 2 and 5 and hence it is a terminating decimal.