CBSE 10th Maths Question Paper 2022 (Set 7)

The given AP is $5, 11, 17, ..., 203$. Here, the first term $a = 5$ and common difference $d = 11 - 5 = 6$. The nth term is given by $a_n = a + (n-1)d$. Substituting the given values: $203 = 5 + (n - 1)6$. $203 - 5 = (n - 1)6$ $198 = (n - 1)6$ $n - 1 = \frac{198}{6} = 33$ $n = 33 + 1 = 34$. Thus, there are 34 terms in the AP.

The $n^{th}$ term is $a_n = 5 - 3n$. To find the first term $a_1$: $a_1 = 5 - 3(1) = 2$. To find the common difference $d$, find $a_2$: $a_2 = 5 - 3(2) = -1$. $d = a_2 - a_1 = -1 - 2 = -3$. The sum of first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$. For $n = 20$: $S_{20} = \frac{20}{2}[2(2) + (20-1)(-3)]$. $S_{20} = 10[4 + 19(-3)] = 10[4 - 57] = 10[-53] = -530$.

The equation is $9x^2 - 6\sqrt{2}x + 2 = 0$. We can write this as $(3x)^2 - 2(3x)(\sqrt{2}) + (\sqrt{2})^2 = 0$. This is of the form $a^2 - 2ab + b^2 = (a-b)^2$. So, $(3x - \sqrt{2})^2 = 0$. $3x - \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{3}$. Since it is a quadratic equation, both roots are equal: $x = \frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3}$.

Volume of cuboidal solid $= 18 \times 22 \times 6 = 2376 \text{ cm}^3$. Radius of a spherical shot $= \frac{3}{2} = 1.5 \text{ cm}$. Volume of one spherical shot $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 3.375 = \frac{99}{7} \text{ cm}^3$. Number of shots $= \frac{\text{Total Volume}}{\text{Volume of one shot}} = 2376 \div \frac{99}{7} = 2376 \times \frac{7}{99} = 24 \times 7 = 168$.

Given the distribution, the mode is 24, which lies in the class interval 20-30. Therefore, the modal class is 20-30. We are given: Sum of frequencies = $4 + x + 20 + y + 6 = 50$ $x + y + 30 = 50$ $x + y = 20$ --- (1) The formula for mode is: $Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$ Where: $l = 20$ (lower limit of modal class) $f_1 = 20$ (frequency of modal class) $f_0 = x$ (frequency of class preceding modal class) $f_2 = y$ (frequency of class succeeding modal class) $h = 10$ (class size) Substituting the values: $24 = 20 + \left( \frac{20 - x}{2(20) - x - y} \right) imes 10$ $4 = \left( \frac{20 - x}{40 - (x + y)} \right) imes 10$ Since $x + y = 20$: $4 = \left( \frac{20 - x}{40 - 20} \right) imes 10$ $4 = \left( \frac{20 - x}{20} \right) imes 10$ $4 = \frac{20 - x}{2}$ $8 = 20 - x$ $x = 12$ Substitute $x = 12$ into (1): $12 + y = 20$ $y = 8$ Thus, the missing frequencies are $x = 12$ and $y = 8$.