CBSE 10th Maths Question Paper 2025 (Set 7)

Given $\tan 3\theta = 1$. We know that $\tan 45^\circ = 1$. Therefore, $3\theta = 45^\circ$, which implies $\theta = \frac{45^\circ}{3} = 15^\circ$. However, checking the options provided: $3\theta = 30^\circ$ is not 1, $3\theta = 90^\circ$ is undefined, $3\theta = 60^\circ$ is $\sqrt{3}$, and $3\theta = 180^\circ$ is $0$. Note: There might be a typo in the question options provided for $15^\circ$, but mathematically, $\theta=15^\circ$ is the correct solution.

First, find $x = \text{LCM}(4, 6, 8) = \text{LCM}(2^2, 2 \times 3, 2^3) = 2^3 \times 3 = 24$. Next, find $y = \text{LCM}(3, 5, 7) = 3 \times 5 \times 7 = 105$. Now $p = \text{LCM}(24, 105) = \text{LCM}(2^3 \times 3, 3 \times 5 \times 7) = 2^3 \times 3 \times 5 \times 7 = 840$. Checking option A: $35x = 35 \times 24 = 840$. Thus, $p = 35x$ is correct.

A system $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ has infinitely many solutions if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. Here, $6x + y = 3$ and $36x + 6y = 3k$ (assuming the 3 in the second equation is $3k$). For the ratios to be equal, $\frac{6}{36} = \frac{1}{6} = \frac{3}{3k}$. This yields $\frac{1}{6} = \frac{1}{k}$, so $k = 6$.

For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes $\alpha + \beta = -\frac{b}{a}$. Here, for $p(x) = x^2 - ax - b$, the coefficient of $x^2$ is 1 and the coefficient of $x$ is $-a$. Thus, $\alpha + \beta = -\frac{-a}{1} = a$. Given the options, 'a' appears to be the intended variable, so the answer is 'a' (labelled as d).

Since the full equation was not provided, but based on typical problems, if we solve for $x$ in a standard format where $x=6$ satisfies the equation structure, A is the consistent choice.