CBSE 10th Maths Question Paper 2022 (Set 3)

For a quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots, the discriminant $D = b^2 - 4ac$ must be equal to $0$. Here, $a = (m-1)$, $b = 2(m-1)$, and $c = 1$. Substituting in the discriminant formula: $D = [2(m-1)]^2 - 4(m-1)(1) = 0$ $4(m-1)^2 - 4(m-1) = 0$ Divide by 4: $(m-1)^2 - (m-1) = 0$ $(m-1)[(m-1) - 1] = 0$ $(m-1)(m-2) = 0$ Thus, $m = 1$ or $m = 2$. However, if $m=1$, the equation becomes $0x^2 + 0x + 1 = 0$, which is not a quadratic equation. Therefore, $m$ must be $2$.

Given equation: $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$ We split the middle term: $\sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0$ Factor by grouping: $x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0$ $(x + \sqrt{3})(\sqrt{3}x + 7) = 0$ Setting each factor to zero: 1) $x + \sqrt{3} = 0 \Rightarrow x = -\sqrt{3}$ 2) $\sqrt{3}x + 7 = 0 \Rightarrow x = -\frac{7}{\sqrt{3}} = -\frac{7\sqrt{3}}{3}$ Thus, the roots are $-\sqrt{3}$ and $-\frac{7\sqrt{3}}{3}$.

Let Rohan's present age be $x$. Age 5 years ago = $x-5$. Age 7 years from now = $x+7$. The problem states: $(x-5) + (x+7) = 2x + 1$ $2x + 2 = 2x + 1$ $2 = 1$, which is impossible. Wait, re-reading the prompt: "Rohan's age 5 years ago, AND 7 years from now, IS one more than twice his present age." This implies the sum of his past age and future age is $2x+1$. However, if the question meant individual statements, it would be contradictory. Solving for the sum: $2x + 2 = 2x + 1$ implies no solution exists for this configuration. Assuming the intended question was $(x-5)(x+7) = 2x+1$ or a different sum, but as stated, the equation $2x+2 = 2x+1$ has no solution.

The 2-digit multiples of 6 are: $12, 18, 24, \dots, 96$. This is an AP where $a = 12$, $d = 6$, and $n = 12$. The sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n-1)d]$. $S_{12} = \frac{12}{2} [2(12) + (12-1)6]$ $S_{12} = 6 [24 + 66]$ $S_{12} = 6 [90] = 540$. The sum is 540.

Volume of sphere = $\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (10.5)^3$ Volume of one cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3.5)^2 (3)$ Number of cones = $\frac{\text{Volume of sphere}}{\text{Volume of cone}}$ Number = $\frac{\frac{4}{3} \pi (10.5)^3}{\frac{1}{3} \pi (3.5)^2 (3)} = \frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3}$ $= 4 \times 3 \times 3 \times 10.5 / 3 = 4 \times 3 \times 10.5 = 126$. So, 126 cones are formed.