CBSE 10th Maths Question Paper 2022 (Set 2)

The mode is the value that appears most frequently in a dataset. By observation of the frequency distribution, the class interval with the highest frequency is the modal class. Once identified, we use the formula: $$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$$ where $l$ is the lower limit of the modal class, $f_1$ is the frequency of the modal class, $f_0$ is the frequency of the preceding class, $f_2$ is the frequency of the succeeding class, and $h$ is the class width.

Let $O$ be the center of the concentric circles. Let $AB$ be the chord of the larger circle (radius $R = 4$ cm) touching the smaller circle (radius $r = 3$ cm) at point $P$. Since the radius is perpendicular to the tangent, $\triangle OPA$ is a right-angled triangle with $\angle OPA = 90^\circ$. Using Pythagoras theorem in $\triangle OPA$: $OA^2 = OP^2 + AP^2$ $4^2 = 3^2 + AP^2$ $16 = 9 + AP^2 \implies AP^2 = 7 \implies AP = \sqrt{7}$ cm. The length of the chord $AB = 2 \times AP = 2\sqrt{7}$ cm.

Given AP: $30, 24, 18, ...$ First term $a = 30$, Common difference $d = 24 - 30 = -6$, Number of terms $n = 30$. The sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n-1)d]$. $S_{30} = \frac{30}{2} [2(30) + (30-1)(-6)]$ $S_{30} = 15 [60 + 29(-6)]$ $S_{30} = 15 [60 - 174] = 15(-114) = -1710$.

Given $S_n = n(4n + 1) = 4n^2 + n$. The $n^{th}$ term $a_n = S_n - S_{n-1}$. $a_n = (4n^2 + n) - [4(n-1)^2 + (n-1)]$ $a_n = 4n^2 + n - [4(n^2 - 2n + 1) + n - 1]$ $a_n = 4n^2 + n - [4n^2 - 8n + 4 + n - 1]$ $a_n = 4n^2 + n - 4n^2 + 7n - 3 = 8n - 3$. For $n=1, a_1 = 5$. For $n=2, a_2 = 13$. For $n=3, a_3 = 21$. The AP is $5, 13, 21, ...$

Volume of cone = $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 7^2 \times 28$. Volume of sphere = $\frac{4}{3}\pi R^3$. Since the clay is reshaped, volumes are equal: $\frac{4}{3}\pi R^3 = \frac{1}{3}\pi (7^2)(28)$ $4R^3 = 49 \times 28$ $R^3 = 49 \times 7 = 343$ $R = \sqrt[3]{343} = 7$ cm.