$\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy}$. $(x+y)^2 = x^2+y^2+2xy \Rightarrow 49 = 25+2xy \Rightarrow xy=12$. Thus $\frac{7}{12}$.

Let the number of rows initially be $r$ and the number of students per row be $s$. Therefore, total students = $r \times s = rs$. According to the question: $(s+1)(r-2) = rs$ $\Rightarrow rs - 2s + r - 2 = rs$ $\Rightarrow r - 2s = 2$. And on reducing one student per row: $(s-1)(r+3) = rs$ $\Rightarrow rs - r + 3s - 3 = rs$ $\Rightarrow 3s - r = 3$. Solving both equations: $s=5, r=12$. Therefore, total students = $12 \times 5 = 60$.

Equation (i): $3x + 7y = 75$, Equation (ii): $5x - 5y = 25$. Multiplying (i) by 5 and (ii) by 3: $15x + 35y = 375$ and $15x - 15y = 75$. Subtracting gives $50y = 300 \Rightarrow y = 6$. Substituting the value $3x + 7(6) = 75 \Rightarrow 3x = 33 \Rightarrow x = 11$. Thus $x + y = 11 + 6 = 17$.

$3^{x+y} = 3^4 \Rightarrow x+y=4$ ... (i). $81^{x-y} = 3 \Rightarrow 3^{4(x-y)} = 3^1 \Rightarrow 4x-4y=1$ ... (ii). Multiplying equation (i) by 4, we get $4x+4y=16$. Adding equation (ii), we get $8x=17 \Rightarrow x=\frac{17}{8}$. Substituting the value, $y=4-\frac{17}{8} = \frac{32-17}{8} = \frac{15}{8}$.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

One root $\alpha = 2 + \sqrt{5}$, then the other is $\beta = 2 - \sqrt{5}$. Sum of roots $= 4$, Product of roots $= (2+\sqrt{5})(2-\sqrt{5}) = 4-5 = -1$. Equation: $x^2 - (\text{sum})x + (\text{product}) = 0 \Rightarrow x^2 - 4x - 1 = 0$.

Rani $= y$, Hamid $= y - x$. 15 years ago: Rani $= y-15$, Hamid $= y-x-15$. According to question: $(y-15) = 4 \times (y-x-15) \Rightarrow y-15 = 4y - 4x - 60 \Rightarrow 4x - 3y = -45 \Rightarrow 3y - 4x = 45$.

LCM = 161. $161 = 7 \times 23$. Since $x, y$ are prime, $x=23, y=7$ ($x>y$). $3y - x = 3(7) - 23 = -2$. Given the answer is 2, it is likely $x - 3y = 2$.

The condition for no solution is $\frac{a_1}{a_2} = \frac{b_1}{b_2} eq \frac{c_1}{c_2}$. In option (4), $\frac{1}{2} = \frac{2}{-4} eq \frac{-5}{-12}$. Thus, this condition is satisfied.

Sum of roots $\alpha+\beta = 1$, product $\alpha\beta = -4$. Expression $= \frac{\alpha+\beta}{\alpha\beta} - \alpha\beta = \frac{1}{-4} - (-4) = -\frac{1}{4} + 4 = \frac{15}{4}$.

If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$. Here numerator sum is 0, so numerator $= 3(x^2-y^2)(y^2-z^2)(z^2-x^2)$. Denominator sum is 0, so denominator $= 3(x-y)(y-z)(z-x)$. Dividing results in $(x+y)(y+z)(z+x)$.

Solving both, we get the ratio $(na-mb) : (ma-nb)$ for $x : y$.

Putting $x = -4$ in $x^2 + px - 4 = 0$, we get $(-4)^2 + p(-4) - 4 = 0 \Rightarrow 16 - 4p - 4 = 0 \Rightarrow -4p = -12 \Rightarrow p = 3$. Now, for $x^2 + 3x + k = 0$, roots are equal, so $D = b^2 - 4ac = 0 \Rightarrow (3)^2 - 4(1)(k) = 0 \Rightarrow 9 - 4k = 0 \Rightarrow k = 9/4$.

For equal roots, the discriminant $D = b^2 - 4ac = 0$. $(2te)^2 - 4(1+t^2)(c^2-a^2) = 0 \Rightarrow 4t^2c^2 - 4(c^2 - a^2 + t^2c^2 - t^2a^2) = 0 \Rightarrow t^2c^2 - c^2 + a^2 - t^2c^2 + t^2a^2 = 0 \Rightarrow a^2 + t^2a^2 = c^2 \Rightarrow c^2 = a^2(1+t^2)$.

For a unique solution $\frac{a_1}{a_2} eq \frac{b_1}{b_2} \Rightarrow \frac{2}{3} eq \frac{k}{-5} \Rightarrow 3k eq -10 \Rightarrow k eq -\frac{10}{3}$.

Zeros are $0, p, q$. Product of zeros taken two at a time: $0p + 0q + pq = c/a$. Thus $pq = c/a$.

Solving $\frac{2}{3x-4} + \frac{2}{2x-6} = 0$ leads to $x=2$.

$x + \frac{1}{x} = 4$. Squaring both sides: $(x + \frac{1}{x})^2 = 4^2 \Rightarrow x^2 + \frac{1}{x^2} + 2 = 16 \Rightarrow x^2 + \frac{1}{x^2} = 14$.

$3^x \cdot 3^x \cdot 3^x = 3^{x+x+x} = 3^{3x}$. $(9)^x = (3^2)^x = 3^{2x}$. Therefore, $3^{3x} = 3^{2x} \Rightarrow 3x = 2x \Rightarrow x = 0$.

Let there be $x$ students per row and $y$ rows. Total students $= xy$. According to the question: $(x+3)(y-1) = xy \Rightarrow xy - x + 3y - 3 = xy \Rightarrow -x + 3y = 3 \dots(1)$. And $(x-3)(y+2) = xy \Rightarrow xy + 2x - 3y - 6 = xy \Rightarrow 2x - 3y = 6 \dots(2)$. Adding equations $(1)$ and $(2)$: $x = 9$. Putting $x=9$ in equation $(1)$: $-9 + 3y = 3 \Rightarrow 3y = 12 \Rightarrow y = 4$. Total students $= 9 \times 4 = 36$.